Programming FAQ

General questions

Is there a source code-level debugger with breakpoints and single-stepping?

Yes.

Several debuggers for Python are described below, and the built-in function breakpoint() allows you to drop into any of them.

The pdb module is a simple but adequate console-mode debugger for Python. It is part of the standard Python library, and is documented in the Library Reference Manual. You can also write your own debugger by using the code for pdb as an example.

The IDLE interactive development environment, which is part of the standard Python distribution (normally available as idlelib), includes a graphical debugger.

PythonWin is a Python IDE that includes a GUI debugger based on pdb. The PythonWin debugger colors breakpoints and has quite a few cool features such as debugging non-PythonWin programs. PythonWin is available as part of pywin32 project and as a part of the ActivePython distribution.

Eric is an IDE built on PyQt and the Scintilla editing component.

trepan3k is a gdb-like debugger.

Visual Studio Code is an IDE with debugging tools that integrates with version-control software.

There are a number of commercial Python IDEs that include graphical debuggers. They include:

• Wing IDE

• PyCharm

Are there tools to help find bugs or perform static analysis?

Yes.

Ruff, Pylint and Pyflakes do basic checking that will help you catch bugs sooner.

Static type checkers such as mypy, ty, Pyrefly, and pytype can check type hints in Python source code.

How can I create a stand-alone binary from a Python script?

You don’t need the ability to compile Python to C code if all you want is a stand-alone program that users can download and run without having to install the Python distribution first. There are a number of tools that determine the set of modules required by a program and bind these modules together with a Python binary to produce a single executable.

One is to use the freeze tool, which is included in the Python source tree as Tools/freeze. It converts Python byte code to C arrays; with a C compiler you can embed all your modules into a new program, which is then linked with the standard Python modules.

It works by scanning your source recursively for import statements (in both forms) and looking for the modules in the standard Python path as well as in the source directory (for built-in modules). It then turns the bytecode for modules written in Python into C code (array initializers that can be turned into code objects using the marshal module) and creates a custom-made config file that only contains those built-in modules which are actually used in the program. It then compiles the generated C code and links it with the rest of the Python interpreter to form a self-contained binary which acts exactly like your script.

The following packages can help with the creation of console and GUI executables:

• Nuitka (Cross-platform)

• PyInstaller (Cross-platform)

• PyOxidizer (Cross-platform)

• cx_Freeze (Cross-platform)

• py2app (macOS only)

• py2exe (Windows only)

Are there coding standards or a style guide for Python programs?

Yes. The coding style required for standard library modules is documented as PEP 8.

Core language

Why am I getting an UnboundLocalError when the variable has a value?

It can be a surprise to get the UnboundLocalError in previously working code when it is modified by adding an assignment statement somewhere in the body of a function.

This code:

>>> x = 10
>>> def bar():
...     print(x)
...
>>> bar()
10

works, but this code:

>>> x = 10
>>> def foo():
...     print(x)
...     x += 1

results in an UnboundLocalError:

>>> foo()
Traceback (most recent call last):
  ...
UnboundLocalError: cannot access local variable 'x' where it is not associated with a value

This is because when you make an assignment to a variable in a scope, that variable becomes local to that scope and shadows any similarly named variable in the outer scope. Since the last statement in foo assigns a new value to x, the compiler recognizes it as a local variable. Consequently when the earlier print(x) attempts to print the uninitialized local variable and an error results.

In the example above you can access the outer scope variable by declaring it global:

>>> x = 10
>>> def foobar():
...     global x
...     print(x)
...     x += 1
...
>>> foobar()
10

This explicit declaration is required in order to remind you that (unlike the superficially analogous situation with class and instance variables) you are actually modifying the value of the variable in the outer scope:

>>> print(x)
11

You can do a similar thing in a nested scope using the nonlocal keyword:

>>> def foo():
...    x = 10
...    def bar():
...        nonlocal x
...        print(x)
...        x += 1
...    bar()
...    print(x)
...
>>> foo()
10
11

What are the rules for local and global variables in Python?

In Python, variables that are only referenced inside a function are implicitly global. If a variable is assigned a value anywhere within the function’s body, it’s assumed to be a local unless explicitly declared as global.

Though a bit surprising at first, a moment’s consideration explains this. On one hand, requiring global for assigned variables provides a bar against unintended side-effects. On the other hand, if global was required for all global references, you’d be using global all the time. You’d have to declare as global every reference to a built-in function or to a component of an imported module. This clutter would defeat the usefulness of the global declaration for identifying side-effects.

Why do lambdas defined in a loop with different values all return the same result?

Assume you use a for loop to define a few different lambdas (or even plain functions), for example:

>>> squares = []
>>> for x in range(5):
...     squares.append(lambda: x**2)

This gives you a list that contains 5 lambdas that calculate x**2. You might expect that, when called, they would return, respectively, 0, 1, 4, 9, and 16. However, when you actually try you will see that they all return 16:

>>> squares[2]()
16
>>> squares[4]()
16

This happens because x is not local to the lambdas, but is defined in the outer scope, and it is accessed when the lambda is called — not when it is defined. At the end of the loop, the value of x is 4, so all the functions now return 4**2, that is 16. You can also verify this by changing the value of x and see how the results of the lambdas change:

>>> x = 8
>>> squares[2]()
64

In order to avoid this, you need to save the values in variables local to the lambdas, so that they don’t rely on the value of the global x:

>>> squares = []
>>> for x in range(5):
...     squares.append(lambda n=x: n**2)

Here, n=x creates a new variable n local to the lambda and computed when the lambda is defined so that it has the same value that x had at that point in the loop. This means that the value of n will be 0 in the first lambda, 1 in the second, 2 in the third, and so on. Therefore each lambda will now return the correct result:

>>> squares[2]()
4
>>> squares[4]()
16

Note that this behaviour is not peculiar to lambdas, but applies to regular functions too.

How do I share global variables across modules?

The canonical way to share information across modules within a single program is to create a special module (often called config or cfg). Just import the config module in all modules of your application; the module then becomes available as a global name. Because there is only one instance of each module, any changes made to the module object get reflected everywhere. For example:

config.py:

x = 0   # Default value of the 'x' configuration setting

mod.py:

import config
config.x = 1

main.py:

import config
import mod
print(config.x)

Note that using a module is also the basis for implementing the singleton design pattern, for the same reason.

What are the “best practices” for using import in a module?

In general, don’t use from modulename import *. Doing so clutters the importer’s namespace, and makes it much harder for linters to detect undefined names.

Import modules at the top of a file. Doing so makes it clear what other modules your code requires and avoids questions of whether the module name is in scope. Using one import per line makes it easy to add and delete module imports, but using multiple imports per line uses less screen space.

It’s good practice if you import modules in the following order:

1. standard library modules – such as sys, os, argparse, re

2. third-party library modules (anything installed in Python’s site-packages directory) – such as dateutil, requests, tzdata

3. locally developed modules

It is sometimes necessary to move imports to a function or class to avoid problems with circular imports. Gordon McMillan says:

Circular imports are fine where both modules use the “import <module>” form of import. They fail when the 2nd module wants to grab a name out of the first (“from module import name”) and the import is at the top level. That’s because names in the 1st are not yet available, because the first module is busy importing the 2nd.

In this case, if the second module is only used in one function, then the import can easily be moved into that function. By the time the import is called, the first module will have finished initializing, and the second module can do its import.

It may also be necessary to move imports out of the top level of code if some of the modules are platform-specific. In that case, it may not even be possible to import all of the modules at the top of the file. In this case, importing the correct modules in the corresponding platform-specific code is a good option.

Only move imports into a local scope, such as inside a function definition, if it’s necessary to solve a problem such as avoiding a circular import or are trying to reduce the initialization time of a module. This technique is especially helpful if many of the imports are unnecessary depending on how the program executes. You may also want to move imports into a function if the modules are only ever used in that function. Note that loading a module the first time may be expensive because of the one time initialization of the module, but loading a module multiple times is virtually free, costing only a couple of dictionary lookups. Even if the module name has gone out of scope, the module is probably available in sys.modules.

Why are default values shared between objects?

This type of bug commonly bites neophyte programmers. Consider this function:

def foo(mydict={}):  # Danger: shared reference to one dict for all calls
    ... compute something ...
    mydict[key] = value
    return mydict

The first time you call this function, mydict contains a single item. The second time, mydict contains two items because when foo() begins executing, mydict starts out with an item already in it.

It is often expected that a function call creates new objects for default values. This is not what happens. Default values are created exactly once, when the function is defined. If that object is changed, like the dictionary in this example, subsequent calls to the function will refer to this changed object.

By definition, immutable objects such as numbers, strings, tuples, and None, are safe from change. Changes to mutable objects such as dictionaries, lists, and class instances can lead to confusion.

Because of this feature, it is good programming practice to not use mutable objects as default values. Instead, use None as the default value and inside the function, check if the parameter is None and create a new list/dictionary/whatever if it is. For example, don’t write:

def foo(mydict={}):
    ...

but:

def foo(mydict=None):
    if mydict is None:
        mydict = {}  # create a new dict for local namespace

This feature can be useful. When you have a function that’s time-consuming to compute, a common technique is to cache the parameters and the resulting value of each call to the function, and return the cached value if the same value is requested again. This is called “memoizing”, and can be implemented like this:

# Callers can only provide two parameters and optionally pass _cache by keyword
def expensive(arg1, arg2, *, _cache={}):
    if (arg1, arg2) in _cache:
        return _cache[(arg1, arg2)]

    # Calculate the value
    result = ... expensive computation ...
    _cache[(arg1, arg2)] = result           # Store result in the cache
    return result

You could use a global variable containing a dictionary instead of the default value; it’s a matter of taste.

How can I pass optional or keyword parameters from one function to another?

Collect the arguments using the * and ** specifiers in the function’s parameter list; this gives you the positional arguments as a tuple and the keyword arguments as a dictionary. You can then pass these arguments when calling another function by using * and **:

def f(x, *args, **kwargs):
    ...
    kwargs['width'] = '14.3c'
    ...
    g(x, *args, **kwargs)

What is the difference between arguments and parameters?

Parameters are defined by the names that appear in a function definition, whereas arguments are the values actually passed to a function when calling it. Parameters define what kind of arguments a function can accept. For example, given the function definition:

def func(foo, bar=None, **kwargs):
    pass

foo, bar and kwargs are parameters of func. However, when calling func, for example:

func(42, bar=314, extra=somevar)

the values 42, 314, and somevar are arguments.

Why did changing list ‘y’ also change list ‘x’?

If you wrote code like:

>>> x = []
>>> y = x
>>> y.append(10)
>>> y
[10]
>>> x
[10]

you might be wondering why appending an element to y changed x too.

There are two factors that produce this result:

1. Variables are simply names that refer to objects. Doing y = x doesn’t create a copy of the list – it creates a new variable y that refers to the same object x refers to. This means that there is only one object (the list), and both x and y refer to it.

2. Lists are mutable, which means that you can change their content.

After the call to append(), the content of the mutable object has changed from [] to [10]. Since both the variables refer to the same object, using either name accesses the modified value [10].

If we instead assign an immutable object to x:

>>> x = 5  # ints are immutable
>>> y = x
>>> x = x + 1  # 5 can't be mutated, we are creating a new object here
>>> x
6
>>> y
5

we can see that in this case x and y are not equal anymore. This is because integers are immutable, and when we do x = x + 1 we are not mutating the int 5 by incrementing its value; instead, we are creating a new object (the int 6) and assigning it to x (that is, changing which object x refers to). After this assignment we have two objects (the ints 6 and 5) and two variables that refer to them (x now refers to 6 but y still refers to 5).

Some operations (for example y.append(10) and y.sort()) mutate the object, whereas superficially similar operations (for example y = y + [10] and sorted(y)) create a new object. In general in Python (and in all cases in the standard library) a method that mutates an object will return None to help avoid getting the two types of operations confused. So if you mistakenly write y.sort() thinking it will give you a sorted copy of y, you’ll instead end up with None, which will likely cause your program to generate an easily diagnosed error.

However, there is one class of operations where the same operation sometimes has different behaviors with different types: the augmented assignment operators. For example, += mutates lists but not tuples or ints (a_list += [1, 2, 3] is equivalent to a_list.extend([1, 2, 3]) and mutates a_list, whereas some_tuple += (1, 2, 3) and some_int += 1 create new objects).

In other words:

• If we have a mutable object (such as list, dict, set), we can use some specific operations to mutate it and all the variables that refer to it will see the change.

• If we have an immutable object (such as str, int, tuple), all the variables that refer to it will always see the same value, but operations that transform that value into a new value always return a new object.

If you want to know if two variables refer to the same object or not, you can use the is operator, or the built-in function id().

How do I write a function with output parameters (call by reference)?

Remember that arguments are passed by assignment in Python. Since assignment just creates references to objects, there’s no alias between an argument name in the caller and callee, and consequently no call-by-reference. You can achieve the desired effect in a number of ways.

1. By returning a tuple of the results:

   >>> def func1(a, b):
   ...     a = 'new-value'        # a and b are local names
   ...     b = b + 1              # assigned to new objects
   ...     return a, b            # return new values
   ...
   >>> x, y = 'old-value', 99
   >>> func1(x, y)
   ('new-value', 100)
   

   This is almost always the clearest solution.

2. By using global variables. This isn’t thread-safe, and is not recommended.

3. By passing a mutable (changeable in-place) object:

   >>> def func2(a):
   ...     a[0] = 'new-value'     # 'a' references a mutable list
   ...     a[1] = a[1] + 1        # changes a shared object
   ...
   >>> args = ['old-value', 99]
   >>> func2(args)
   >>> args
   ['new-value', 100]
   

4. By passing in a dictionary that gets mutated:

   >>> def func3(args):
   ...     args['a'] = 'new-value'     # args is a mutable dictionary
   ...     args['b'] = args['b'] + 1   # change it in-place
   ...
   >>> args = {'a': 'old-value', 'b': 99}
   >>> func3(args)
   >>> args
   {'a': 'new-value', 'b': 100}
   

5. Or bundle up values in a class instance:

   >>> class Namespace:
   ...     def __init__(self, /, **args):
   ...         for key, value in args.items():
   ...             setattr(self, key, value)
   ...
   >>> def func4(args):
   ...     args.a = 'new-value'        # args is a mutable Namespace
   ...     args.b = args.b + 1         # change object in-place
   ...
   >>> args = Namespace(a='old-value', b=99)
   >>> func4(args)
   >>> vars(args)
   {'a': 'new-value', 'b': 100}
   

   There’s almost never a good reason to get this complicated.

Your best choice is to return a tuple containing the multiple results.

How do you make a higher order function in Python?

You have two choices: you can use nested scopes or you can use callable objects. For example, suppose you wanted to define linear(a,b) which returns a function f(x) that computes the value a*x+b. Using nested scopes:

def linear(a, b):
    def result(x):
        return a * x + b
    return result

Or using a callable object:

class linear:

    def __init__(self, a, b):
        self.a, self.b = a, b

    def __call__(self, x):
        return self.a * x + self.b

In both cases,

taxes = linear(0.3, 2)

gives a callable object where taxes(10e6) == 0.3 * 10e6 + 2.

The callable object approach has the disadvantage that it is a bit slower and results in slightly longer code. However, note that a collection of callables can share their signature via inheritance:

class exponential(linear):
    # __init__ inherited
    def __call__(self, x):
        return self.a * (x ** self.b)

Object can encapsulate state for several methods:

class counter:

    value = 0

    def set(self, x):
        self.value = x

    def up(self):
        self.value = self.value + 1

    def down(self):
        self.value = self.value - 1

count = counter()
inc, dec, reset = count.up, count.down, count.set

Here inc(), dec() and reset() act like functions which share the same counting variable.

How do I copy an object in Python?

In general, try copy.copy() or copy.deepcopy() for the general case. Not all objects can be copied, but most can.

Some objects can be copied more easily. Dictionaries have a copy() method:

newdict = olddict.copy()

Sequences can be copied by slicing:

new_l = l[:]

How can I find the methods or attributes of an object?

For an instance x of a user-defined class, dir(x) returns an alphabetized list of the names containing the instance attributes and methods and attributes defined by its class.

How can my code discover the name of an object?

Generally speaking, it can’t, because objects don’t really have names. Essentially, assignment always binds a name to a value; the same is true of def and class statements, but in that case the value is a callable. Consider the following code:

>>> class A:
...     pass
...
>>> B = A
>>> a = B()
>>> b = a
>>> print(b)
<__main__.A object at 0x16D07CC>
>>> print(a)
<__main__.A object at 0x16D07CC>

Arguably the class has a name: even though it is bound to two names and invoked through the name B the created instance is still reported as an instance of class A. However, it is impossible to say whether the instance’s name is a or b, since both names are bound to the same value.

Generally speaking it should not be necessary for your code to “know the names” of particular values. Unless you are deliberately writing introspective programs, this is usually an indication that a change of approach might be beneficial.

In comp.lang.python, Fredrik Lundh once gave an excellent analogy in answer to this question:

The same way as you get the name of that cat you found on your porch: the cat (object) itself cannot tell you its name, and it doesn’t really care – so the only way to find out what it’s called is to ask all your neighbours (namespaces) if it’s their cat (object)…

….and don’t be surprised if you’ll find that it’s known by many names, or no name at all!

What’s up with the comma operator’s precedence?

Comma is not an operator in Python. Consider this session:

>>> "a" in "b", "a"
(False, 'a')

Since the comma is not an operator, but a separator between expressions the above is evaluated as if you had entered:

("a" in "b"), "a"

not:

"a" in ("b", "a")

The same is true of the various assignment operators (=, +=, and so on). They are not truly operators but syntactic delimiters in assignment statements.

Is there an equivalent of C’s “?:” ternary operator?

Yes, there is. The syntax is as follows:

[on_true] if [expression] else [on_false]

x, y = 50, 25
small = x if x < y else y

Before this syntax was introduced in Python 2.5, a common idiom was to use logical operators:

[expression] and [on_true] or [on_false]

However, this idiom is unsafe, as it can give wrong results when on_true has a false boolean value. Therefore, it is always better to use the ... if ... else ... form.

Is it possible to write obfuscated one-liners in Python?

Yes. Usually this is done by nesting lambda within lambda. See the following three examples, slightly adapted from Ulf Bartelt:

from functools import reduce

# Primes < 1000
print(list(filter(None,map(lambda y:y*reduce(lambda x,y:x*y!=0,
map(lambda x,y=y:y%x,range(2,int(pow(y,0.5)+1))),1),range(2,1000)))))

# First 10 Fibonacci numbers
print(list(map(lambda x,f=lambda x,f:(f(x-1,f)+f(x-2,f)) if x>1 else 1:
f(x,f), range(10))))

# Mandelbrot set
print((lambda Ru,Ro,Iu,Io,IM,Sx,Sy:reduce(lambda x,y:x+'\n'+y,map(lambda y,
Iu=Iu,Io=Io,Ru=Ru,Ro=Ro,Sy=Sy,L=lambda yc,Iu=Iu,Io=Io,Ru=Ru,Ro=Ro,i=IM,
Sx=Sx,Sy=Sy:reduce(lambda x,y:x+y,map(lambda x,xc=Ru,yc=yc,Ru=Ru,Ro=Ro,
i=i,Sx=Sx,F=lambda xc,yc,x,y,k,f=lambda xc,yc,x,y,k,f:(k<=0)or (x*x+y*y
>=4.0) or 1+f(xc,yc,x*x-y*y+xc,2.0*x*y+yc,k-1,f):f(xc,yc,x,y,k,f):chr(
64+F(Ru+x*(Ro-Ru)/Sx,yc,0,0,i)),range(Sx))):L(Iu+y*(Io-Iu)/Sy),range(Sy
))))(-2.1, 0.7, -1.2, 1.2, 30, 80, 24))
#    \___ ___/  \___ ___/  |   |   |__ lines on screen
#        V          V      |   |______ columns on screen
#        |          |      |__________ maximum of "iterations"
#        |          |_________________ range on y axis
#        |____________________________ range on x axis

Don’t try this at home, kids!

What does the slash(/) in the parameter list of a function mean?

A slash in the argument list of a function denotes that the parameters prior to it are positional-only. Positional-only parameters are the ones without an externally usable name. Upon calling a function that accepts positional-only parameters, arguments are mapped to parameters based solely on their position. For example, divmod() is a function that accepts positional-only parameters. Its documentation looks like this:

>>> help(divmod)
Help on built-in function divmod in module builtins:

divmod(x, y, /)
    Return the tuple (x//y, x%y).  Invariant: div*y + mod == x.

The slash at the end of the parameter list means that both parameters are positional-only. Thus, calling divmod() with keyword arguments would lead to an error:

>>> divmod(x=3, y=4)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: divmod() takes no keyword arguments

Numbers and strings

How do I specify hexadecimal and octal integers?

To specify an octal digit, precede the octal value with a zero, and then a lower or uppercase “o”. For example, to set the variable “a” to the octal value “10” (8 in decimal), type:

>>> a = 0o10
>>> a
8

Hexadecimal is just as easy. Simply precede the hexadecimal number with a zero, and then a lower or uppercase “x”. Hexadecimal digits can be specified in lower or uppercase. For example, in the Python interpreter:

>>> a = 0xa5
>>> a
165
>>> b = 0XB2
>>> b
178

Why does -22 // 10 return -3?

It’s primarily driven by the desire that i % j have the same sign as j. If you want that, and also want:

i == (i // j) * j + (i % j)

then integer division has to return the floor. C also requires that identity to hold, and then compilers that truncate i // j need to make i % j have the same sign as i.

There are few real use cases for i % j when j is negative. When j is positive, there are many, and in virtually all of them it’s more useful for i % j to be >= 0. If the clock says 10 now, what did it say 200 hours ago? -190 % 12 == 2 is useful; -190 % 12 == -10 is a bug waiting to bite.

How do I get int literal attribute instead of SyntaxError?

Trying to lookup an int literal attribute in the normal manner gives a SyntaxError because the period is seen as a decimal point:

>>> 1.__class__
  File "<stdin>", line 1
  1.__class__
   ^
SyntaxError: invalid decimal literal

The solution is to separate the literal from the period with either a space or parentheses.

>>> 1 .__class__
<class 'int'>
>>> (1).__class__
<class 'int'>

How do I convert a string to a number?

For integers, use the built-in int() type constructor, for example, int('144') == 144. Similarly, float() converts to a floating-point number, for example, float('144') == 144.0.

By default, these interpret the number as decimal, so that int('0144') == 144 holds true, and int('0x144') raises ValueError. int(string, base) takes the base to convert from as a second optional argument, so int( '0x144', 16) == 324. If the base is specified as 0, the number is interpreted using Python’s rules: a leading ‘0o’ indicates octal, and ‘0x’ indicates a hex number.

Do not use the built-in function eval() if all you need is to convert strings to numbers. eval() will be significantly slower and it presents a security risk: someone could pass you a Python expression that might have unwanted side effects. For example, someone could pass __import__('os').system("rm -rf $HOME") which would erase your home directory.

eval() also has the effect of interpreting numbers as Python expressions, so that, for example, eval('09') gives a syntax error because Python does not allow leading ‘0’ in a decimal number (except ‘0’).

How do I convert a number to a string?

For example, to convert the number 144 to the string '144', use the built-in type constructor str(). If you want a hexadecimal or octal representation, use the built-in functions hex() or oct(). For fancy formatting, see the f-strings and Format string syntax sections. For example, "{:04d}".format(144) yields '0144' and "{:.3f}".format(1.0/3.0) yields '0.333'.

How do I modify a string in place?

You can’t, because strings are immutable. In most situations, you should simply construct a new string from the various parts you want to assemble it from. However, if you need an object with the ability to modify in-place Unicode data, try using an io.StringIO object or the array module:

>>> import io
>>> s = "Hello, world"
>>> sio = io.StringIO(s)
>>> sio.getvalue()
'Hello, world'
>>> sio.seek(7)
7
>>> sio.write("there!")
6
>>> sio.getvalue()
'Hello, there!'

>>> import array
>>> a = array.array('w', s)
>>> print(a)
array('w', 'Hello, world')
>>> a[0] = 'y'
>>> print(a)
array('w', 'yello, world')
>>> a.tounicode()
'yello, world'

How do I use strings to call functions/methods?

There are various techniques.

• The best is to use a dictionary that maps strings to functions. The primary advantage of this technique is that the strings do not need to match the names of the functions. This is also the primary technique used to emulate a case construct:

  def a():
      pass
  
  def b():
      pass
  
  dispatch = {'go': a, 'stop': b}  # Note lack of parens for funcs
  
  dispatch[get_input()]()  # Note trailing parens to call function
  

• Use the built-in function getattr():

  import foo
  getattr(foo, 'bar')()
  

  Note that getattr() works on any object, including classes, class instances, modules, and so on.

  This is used in several places in the standard library, like this:

  class Foo:
      def do_foo(self):
          ...
  
      def do_bar(self):
          ...
  
  f = getattr(foo_instance, 'do_' + opname)
  f()
  

• Use locals() to resolve the function name:

  def myFunc():
      print("hello")
  
  fname = "myFunc"
  
  f = locals()[fname]
  f()
  

Is there an equivalent to Perl’s chomp() for removing trailing newlines from strings?

You can use S.rstrip("\r\n") to remove all occurrences of any line terminator from the end of the string S without removing other trailing whitespace. If the string S represents more than one line, with several empty lines at the end, the line terminators for all the blank lines will be removed:

>>> lines = ("line 1 \r\n"
...          "\r\n"
...          "\r\n")
>>> lines.rstrip("\n\r")
'line 1 '

Since this is typically only desired when reading text one line at a time, using S.rstrip() this way works well.

Is there a scanf() or sscanf() equivalent?

Not as such.

For simple input parsing, the easiest approach is usually to split the line into whitespace-delimited words using the split() method of string objects and then convert decimal strings to numeric values using int() or float(). split() supports an optional “sep” parameter which is useful if the line uses something other than whitespace as a separator.

For more complicated input parsing, regular expressions are more powerful than C’s sscanf and better suited for the task.

What does UnicodeDecodeError or UnicodeEncodeError error mean?

See the Unicode HOWTO.

Can I end a raw string with an odd number of backslashes?

A raw string ending with an odd number of backslashes will escape the string’s quote:

>>> r'C:\this\will\not\work\'
  File "<stdin>", line 1
    r'C:\this\will\not\work\'
    ^
SyntaxError: unterminated string literal (detected at line 1)

There are several workarounds for this. One is to use regular strings and double the backslashes:

>>> 'C:\\this\\will\\work\\'
'C:\\this\\will\\work\\'

Another is to concatenate a regular string containing an escaped backslash to the raw string:

>>> r'C:\this\will\work' '\\'
'C:\\this\\will\\work\\'

It is also possible to use os.path.join() to append a backslash on Windows:

>>> os.path.join(r'C:\this\will\work', '')
'C:\\this\\will\\work\\'

Note that while a backslash will “escape” a quote for the purposes of determining where the raw string ends, no escaping occurs when interpreting the value of the raw string. That is, the backslash remains present in the value of the raw string:

>>> r'backslash\'preserved'
"backslash\\'preserved"

Also see the specification in the language reference.

Performance

My program is too slow. How do I speed it up?

That’s a tough one, in general. First, here is a list of things to remember before diving further:

• Performance characteristics vary across Python implementations. This FAQ focuses on CPython.

• Behaviour can vary across operating systems, especially when talking about I/O or multi-threading.

• You should always find the hot spots in your program before attempting to optimize any code (see the profile module).

• Writing benchmark scripts will allow you to iterate quickly when searching for improvements (see the timeit module).

• It is highly recommended to have good code coverage (through unit testing or any other technique) before potentially introducing regressions hidden in sophisticated optimizations.

That being said, there are many tricks to speed up Python code. Here are some general principles which go a long way towards reaching acceptable performance levels:

• Making your algorithms faster (or changing to faster ones) can yield much larger benefits than trying to sprinkle micro-optimization tricks all over your code.

• Use the right data structures. Study documentation for the Built-in Types and the collections module.

• When the standard library provides a primitive for doing something, it is likely (although not guaranteed) to be faster than any alternative you may come up with. This is doubly true for primitives written in C, such as builtins and some extension types. For example, be sure to use either the list.sort() built-in method or the related sorted() function to do sorting (and see the Sorting Techniques for examples of moderately advanced usage).

• Abstractions tend to create indirections and force the interpreter to work more. If the levels of indirection outweigh the amount of useful work done, your program will be slower. You should avoid excessive abstraction, especially under the form of tiny functions or methods (which are also often detrimental to readability).

If you have reached the limit of what pure Python can allow, there are tools to take you further away. For example, Cython can compile a slightly modified version of Python code into a C extension, and can be used on many different platforms. Cython can take advantage of compilation (and optional type annotations) to make your code significantly faster than when interpreted. If you are confident in your C programming skills, you can also write a C extension module yourself.

See also

The wiki page devoted to performance tips.

What is the most efficient way to concatenate many strings together?

str and bytes objects are immutable, therefore concatenating many strings together is inefficient as each concatenation creates a new object. In the general case, the total runtime cost is quadratic in the total string length.

To accumulate many str objects, the recommended idiom is to place them into a list and call str.join() at the end:

chunks = []
for s in my_strings:
    chunks.append(s)
result = ''.join(chunks)

(Another reasonably efficient idiom is to use io.StringIO.)

To accumulate many bytes objects, the recommended idiom is to extend a bytearray object using in-place concatenation (the += operator):

result = bytearray()
for b in my_bytes_objects:
    result += b

Sequences (tuples/lists)

How do I convert between tuples and lists?

The type constructor tuple(seq) converts any sequence (actually, any iterable) into a tuple with the same items in the same order.

For example, tuple([1, 2, 3]) yields (1, 2, 3) and tuple('abc') yields ('a', 'b', 'c'). If the argument is a tuple, it does not make a copy but returns the same object, so it is cheap to call tuple() when you aren’t sure that an object is already a tuple.

The type constructor list(seq) converts any sequence or iterable into a list with the same items in the same order. For example, list((1, 2, 3)) yields [1, 2, 3] and list('abc') yields ['a', 'b', 'c']. If the argument is a list, it makes a copy just like seq[:] would.

What’s a negative index?

Python sequences are indexed with positive numbers and negative numbers. For positive numbers 0 is the first index 1 is the second index and so forth. For negative indices -1 is the last index and -2 is the penultimate (next to last) index and so forth. Think of seq[-n] as the same as seq[len(seq)-n].

Using negative indices can be very convenient. For example S[:-1] is all of the string except for its last character, which is useful for removing the trailing newline from a string.

How do I iterate over a sequence in reverse order?

Use the reversed() built-in function:

for x in reversed(sequence):
    ...  # do something with x ...

This won’t touch your original sequence, but build a new copy with reversed order to iterate over.

How do you remove duplicates from a list?

See the Python Cookbook for a long discussion of many ways to do this:

https://code.activestate.com/recipes/52560/

If you don’t mind reordering the list, sort it and then scan from the end of the list, deleting duplicates as you go:

if mylist:
    mylist.sort()
    last = mylist[-1]
    for i in range(len(mylist)-2, -1, -1):
        if last == mylist[i]:
            del mylist[i]
        else:
            last = mylist[i]

If all elements of the list may be used as set keys (that is, they are all hashable) this is often faster:

mylist = list(set(mylist))

This converts the list into a set, thereby removing duplicates, and then back into a list.

How do you remove multiple items from a list?

As with removing duplicates, explicitly iterating in reverse with a delete condition is one possibility. However, it is easier and faster to use slice replacement with an implicit or explicit forward iteration. Here are three variations:

mylist[:] = filter(keep_function, mylist)
mylist[:] = (x for x in mylist if keep_condition)
mylist[:] = [x for x in mylist if keep_condition]

The list comprehension may be fastest.

How do you make an array in Python?

Use a list:

["this", 1, "is", "an", "array"]

Lists are equivalent to C or Pascal arrays in their time complexity; the primary difference is that a Python list can contain objects of many different types.

The array module also provides methods for creating arrays of fixed types with compact representations, but they are slower to index than lists. Also note that NumPy and other third-party packages define array-like structures with various characteristics as well.

To get Lisp-style linked lists, you can emulate cons cells using tuples:

lisp_list = ("like",  ("this",  ("example", None) ) )

If mutability is desired, you could use lists instead of tuples. Here the analogue of a Lisp car is lisp_list[0] and the analogue of cdr is lisp_list[1]. Only do this if you’re sure you really need to, because it’s usually a lot slower than using Python lists.

How do I create a multidimensional list?

You probably tried to make a multidimensional array like this:

>>> A = [[None] * 2] * 3

This looks correct if you print it:

>>> A
[[None, None], [None, None], [None, None]]

But when you assign a value, it shows up in multiple places:

>>> A[0][0] = 5
>>> A
[[5, None], [5, None], [5, None]]

The reason is that replicating a list with * doesn’t create copies, it only creates references to the existing objects. The *3 creates a list containing 3 references to the same list of length two. Changes to one row will show in all rows, which is almost certainly not what you want.

The suggested approach is to create a list of the desired length first and then fill in each element with a newly created list:

A = [None] * 3
for i in range(3):
    A[i] = [None] * 2

This generates a list containing 3 different lists of length two. You can also use a list comprehension:

w, h = 2, 3
A = [[None] * w for i in range(h)]

Or, you can use an extension that provides a matrix datatype; NumPy is the best known.

How do I apply a method or function to a sequence of objects?

To call a method or function and accumulate the return values in a list, a list comprehension is an elegant solution:

result = [obj.method() for obj in mylist]

result = [function(obj) for obj in mylist]

To just run the method or function without saving the return values, a plain for loop will suffice:

for obj in mylist:
    obj.method()

for obj in mylist:
    function(obj)

Why does a_tuple[i] += [‘item’] raise an exception when the addition works?

This is because of a combination of the fact that augmented assignment operators are assignment operators, and the difference between mutable and immutable objects in Python.

This discussion applies in general when augmented assignment operators are applied to elements of a tuple that point to mutable objects, but we’ll use a list and += as our exemplar.

If you wrote:

>>> a_tuple = (1, 2)
>>> a_tuple[0] += 1
Traceback (most recent call last):
   ...
TypeError: 'tuple' object does not support item assignment

The reason for the exception should be immediately clear: 1 is added to the object a_tuple[0] points to (1), producing the result object, 2, but when we attempt to assign the result of the computation, 2, to element 0 of the tuple, we get an error because we can’t change what an element of a tuple points to.

Under the covers, what this augmented assignment statement is doing is approximately this:

>>> result = a_tuple[0] + 1
>>> a_tuple[0] = result
Traceback (most recent call last):
  ...
TypeError: 'tuple' object does not support item assignment

It is the assignment part of the operation that produces the error, since a tuple is immutable.

When you write something like:

>>> a_tuple = (['foo'], 'bar')
>>> a_tuple[0] += ['item']
Traceback (most recent call last):
  ...
TypeError: 'tuple' object does not support item assignment

The exception is a bit more surprising, and even more surprising is the fact that even though there was an error, the append worked:

>>> a_tuple[0]
['foo', 'item']

To see why this happens, you need to know that (a) if an object implements an __iadd__() magic method, it gets called when the += augmented assignment is executed, and its return value is what gets used in the assignment statement; and (b) for lists, __iadd__() is equivalent to calling extend() on the list and returning the list. That’s why we say that for lists, += is a “shorthand” for list.extend():

>>> a_list = []
>>> a_list += [1]
>>> a_list
[1]

This is equivalent to:

>>> result = a_list.__iadd__([1])
>>> a_list = result

The object pointed to by a_list has been mutated, and the pointer to the mutated object is assigned back to a_list. The end result of the assignment is a no-op, since it is a pointer to the same object that a_list was previously pointing to, but the assignment still happens.

Thus, in our tuple example what is happening is equivalent to: