What general tips do you have for golfing in Python? I'm looking for ideas which can be applied to code-golf problems and which are also at least somewhat specific to Python (e.g. "remove comments" is not an answer).

Please post one tip per answer.

  • 27
    \$\begingroup\$ Oh, I can see a whole set of questions like this one coming for each language... \$\endgroup\$ – R. Martinho Fernandes Jan 28 '11 at 4:26
  • 4
    \$\begingroup\$ @Marthinho I agree. Just started a C++ equivalent. I don't think its a bad thing though, as long as we don't see the same answers re-posted across many of these question types. \$\endgroup\$ – marcog Jan 28 '11 at 12:28
  • 51
    \$\begingroup\$ Love the question but I have to keep telling myself "this is ONLY for fun NOT for production code" \$\endgroup\$ – Greg Guida Dec 21 '11 at 0:08
  • 2
    \$\begingroup\$ Shouldn't this question be a community wiki post? \$\endgroup\$ – user8397947 May 29 '16 at 15:35
  • 3
    \$\begingroup\$ @dorukayhan Nope; it's a valid code-golf tips question, asking for tips on shortening python code for CG'ing purposes. Such questions are perfectly valid for the site, and none of these tags explicitly says that the question should be CW'd, unlike SO, which required CG challenges to be CW'd. Also, writing a good answer, and finding such tips always deserves something, that is taken away if the question is community wiki (rep). \$\endgroup\$ – Erik the Outgolfer Sep 9 '16 at 14:48

145 Answers 145

1 2 3 4 5

Pathlib from shorter files manipulations:

# get current dir, go up one dir, go down one dir, list all "py" files
# and get the whole file bytes

import os.path as p
import glob

d = p.join(p.dirname(p.abspath(__file__))), 'foo', '*.py')
for x in glob.glob(d):
    with open(x, 'rb') as f:


import pathlib as p

d = p.Path(__file__).absolute().parent / 'foo'
for f in d.glob('*.py'):

List all substrings

You can generate the contiguous substrings of a string s with a recursive function (Python 3 for [*s]).

f=lambda s:[*s]and[s]+f(s[1:])+f(s[:-1])

This will repeat substrings multiple times, but can be made a set to avoid repeats.

f=lambda s:{*s}and{s}|f(s[1:])|f(s[:-1])

To assign to a tuple, don't use parentheses. For example, a=1,2,3 assigns a to the tuple (1, 2, 3). b=7, assigns b to the tuple (7,). This works in both Python 2 and Python 3.


Python 2 and 3 differences

Recent challange pushed me to search for differences in two major versions of python. More precisely - same code, that returns different results in different versions. This might be helpful in other polyglot challenges.

1) Strings and bytes comparisson

  • Python 2: '' == b''
  • Python 3: '' != b''

2) Rounding (Luis Mendo answer)

  • Python 2: round(1*0.5) = 1.0
  • Python 3: round(1*0.5) = 0

3) Division (Jonathan Allan answer)

  • Python 2: 10/11 = 0
  • Python 3: 10/11 = 0.9090909090909091

4) Suggestions?

  • 1
    \$\begingroup\$ comparisson, challange \$\endgroup\$ – Jonathan Frech Aug 28 '18 at 20:11

Replace not with 1-

In python, the negation operator not wastes bytes, so we have to find a shorter way. Negation can be implemented as subtraction from 1 (obtained from my Keg experiece), which saves 1 byte. (Also, True and False can be alternatively represented as 1 and 0 internally, so this will not matter much.)

Compare this program:

lambda s:not(s[0]+s[-1]).isdigit()

With this program:

lambda s:1-(s[0]+s[-1]).isdigit()

Some straightforward tricks that might help:

and -> *
or -> +

Math built-ins in 3.8

In addition to the famous walrus operator, Python 3.8 introduces useful new math features.

Modular inverse

The modular-power built-in pow(base, exp, mod) can now compute the modular inverse using exp=-1. This requires that base and mod are relatively prime integers.

>>> pow(38, -1, 97)
>>> 23 * 38 % 97 == 1

exp may be any negative integer, which lets you compute modular powers of the modular inverse.

The math library has useful new functions for combinatorics and distances. Access them with import math. Or, write from math import* to import them without the math., which is worth it if you write more than one call.


  • math.comb(n,k): The binomial coefficient n choose k, which equals n! / (k! * (n - k)!).
  • math.perm(n,k): The number of ordered choices of k elements from n, which is n! / (n - k)!. Calling just math.perm(n) gives n!, useful as a synonym for math.factorial(n).
  • math.prod(l): The product of the elements of a list or iterable l, analogous to sum.

Distances and square roots

  • math.dist(p,q): The Euclidean distance between two points p and q. The inputs p and q must be lists or iterables that are the same length.
  • math.hypot(*p): The Euclidean distance from a point p to zero. Now takes any number of arguments; before it took only two. For some reason, arguments still must be splatted like math.hypot(1,2,3) rather than math.hypot([1,2,3]).
  • math.isqrt(n): The integer square root, that is the floor of the square root of n. Requires than n is a non-negative integer. Usually n**.5//1 suffices instead, but this gives exact integer output rather than a float.

If you have multidimensional array of numbers and for instance need to count all numbers greater than n.

First flatten the array, then apply filter function to match condition:

flatten=lambda l: sum(map(flatten,l),[]) if isinstance(l,list) else [l]
len(filter(lambda x:x>n,flatten(l)))

To find the all the indexes of a certain element in a list l, use

filter(lambda x:l[x]==element,range(len(l)))

To find the next index after a certain index:


To find the nth index:

list(filter(lambda x:l[x]==element,range(len(l))))[n]

Convert modules to lists

This will work for CPython (probably both 2 and 3), and lets you maybe shave a few bytes if you need to use a lot of different functions and classes with long names from the same module, but you aren't using any of them often enough to rename individually. You'll have to do some research first to figure out which magic numbers give you which functions. Example (rot13):

print s[28](b[53](),l)

Translated back to plain python, this is the same as:

print raw_input().translate(l)

which is obviously much shorter, but it should be clear how this methodology would eventually save bytes on much longer, more complicated programs that use more modules.


Use a list if you have multiple choices based on int

Say for example you have some output that will be 1, 0, or -1 and you need a different output for each case. You could do something like this:

print('0'if x==0else('1'if x>0else'-1'))

However, the better way is to use x as an index to a list like so:


which is 16 bytes shorter.

  • \$\begingroup\$ for numbers that won't be used in concatenation: print([0,1,-1][x]) \$\endgroup\$ – Felipe Nardi Batista Jul 12 '17 at 14:00

You can assign to a list inside of a for loop.

For example:

L=[1, 2, 3, 4, 6]
queue = [None]*len(L)
for e, queue[e] in enumerate(L):
    print("adding", queue[e], "to processing queue")

This can also be helpful if you need to switch the object you're assigning to.

class Foo:
    def __init__(self):
        self.x = None
a = Foo()
b = Foo()
for q, (lambda x: a if x%2==0 else b)(q).x in enumerate(range(10)):
    print(a.x, b.x)

Checking the length of a list

Empty      : `a==[]` but just checking if it's non empty and swapping the if and the else can be shorter 
Non-Empty  : `a` (assuming it is in a situation where it will be interpreted as a boolean)
len(a)>i   : `a>a[:i]` if the list is non-empty
  • 1
    \$\begingroup\$ Additionally because [] is falsy, if want to check if a list is not empty, you can simply do if a:. \$\endgroup\$ – Backerupper Dec 22 '17 at 21:08
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    \$\begingroup\$ The second one doesn't seem shorter? Although it can save a following space. But I think 1==len(a) also works for that. \$\endgroup\$ – Ørjan Johansen Dec 22 '17 at 21:44
  • \$\begingroup\$ The second one gives an error on the empty list. Assuming nonempty, a<a[:2] is shorter. \$\endgroup\$ – xnor Dec 23 '17 at 5:25

Fuse list comprehensions

You can write

[f(x)for x in l]+[g(x)for x in l]


sum([[f(x),g(x)]for x in l],[])

It gets even better with more comprehensions or if you have to take out more values

If you need to expand a list you can even turn l+[f(x)for x in l]+[g(x)for x in l] into sum([[f(x),g(x)]for x in l],l)

  • \$\begingroup\$ I don't think the two snippets return the same result. For example, [i for i in range(10)]+[-i for i in range(10)] returns [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, -1, -2, -3, -4, -5, -6, -7, -8, -9], while sum([[i,-i]for i in range(10)],[]) returns [0, 0, 1, -1, 2, -2, 3, -3, 4, -4, 5, -5, 6, -6, 7, -7, 8, -8, 9, -9]. \$\endgroup\$ – Erik the Outgolfer Jun 29 '18 at 22:29

Large hard coded numbers can be represented in larger bases, but there is a trade off. Higher bases only become worthwhile after a certain cutoff.

The only three bases you're likely to need to worry about are 10, 16, and 36. These are the cutoffs:

1000000000000 (13 bytes)                            -> 0xe8d4a51000 (12 bytes)
0x10000000000000000000000000000000000000 (40 bytes) -> int("9gmd8o3gbbaz3m2ydgtgwn9qo6xog",36) (39 bytes)

In Python,

True == 1   # true
False == 0  # true



returns 1 if b is larger than a. If not, returns -1.

More golfing,


returns exactly same value as mentioned above.

Useful when modify iterator index by comparable values.

  1. from lib import func as F

  2. from lib import*;F=func

  3. import lib;F=lib.func

#2 is better than #1 except in rare cases where something in lib clobbers another name that's important to you.

#3 uses lib twice, winning with short library names.

  • 1
    \$\begingroup\$ from lib import*;F=func is shorter than import lib;F=lib.func for six letter names; it is always shorter than from lib import func as F. \$\endgroup\$ – Dennis Dec 13 '18 at 4:03
  • \$\begingroup\$ @Dennis the import* trick is covered in a couple of tips farther up the list from here \$\endgroup\$ – Sparr Dec 13 '18 at 21:16
  • \$\begingroup\$ That doesn't prevent you from comparing import lib;F=lib.func to from lib import*;F=func in this answer. \$\endgroup\$ – Dennis Dec 13 '18 at 21:44
  • \$\begingroup\$ @Dennis you have a zillion rep on this site; you could edit it if you thought it would make the answer better \$\endgroup\$ – Sparr Dec 14 '18 at 5:02

Shortening a%b==a if b has a constant sign

For two expressions a and b, where each one results in an int (or long in Python 2) or float, you can replace these:



with these, if b is positive:



or these, if b is negative:



I'm presenting two expressions for each case because sometimes you may want to use one over the other to eliminate a space to separate expression b from an adjacent token. They both have the same precedence, so you're not usually going to need to surround the second expression with () if you don't need to do so to the first one.

This is useful if expression a is more than 1 byte long or b is negative, because it removes one occurrence of a from the expression. If \$a,b\$ are the lengths of expressions a and b respectively, and \$l\$ is the length of the original expression, the resulting expression will be \$l-a+1\$ bytes long. Note that this method is always going to be shorter than assigning expression a to a separate variable.


For example,


can be replaced with


for a total saving of 4 bytes.


Let's define the operator \$x\mathbin\%y\$ for \$x,y\in\mathbb Q\$.

Every rational number \$a\$ can be represented as \$a=bq+r\$, where \$q\in\mathbb Z,0\le r<|b|\$. Therefore, we can define an operator \$a\mathbin\%b\$, where the result has the same sign as \$b\$:

$$a=bq+r,q\in\mathbb Z,0\le r<|b|\\a\mathbin\%b=\begin{cases}\begin{align}r\quad b>0\\-r\quad b<0\end{align}\end{cases}$$

This represents the % operator in Python, which calculates the remainder of the division of two numbers. a % b is the same as abs(a) % b, and the result has the same sign as the divisor, b. For the \$a\mathbin\%b\$ operator, this equality holds:

$$(a\pm b)\mathbin\%b=a\mathbin\%b$$


$$a=bq+r\leftrightarrow a\pm b=bq+r\pm b=(bq\pm b)+r=b(q\pm1)+r$$

Moreover, for \$b>0\$, we have:

$$a\mathbin\%b=a\leftrightarrow r=a\leftrightarrow0\le a<b$$

Proof for \$r=a\leftarrow0\le a<b\$:

$$0\le a<b\leftrightarrow0\le bq+r<b\leftrightarrow bq=0\leftrightarrow a=r$$

Similarly, for \$b<0\$, we have \$b<a\le0\$.

Therefore, \$a\mathbin\%b=a\leftrightarrow\begin{cases}\begin{align}0\le a<b\quad b>0\\b<a\le0\quad b<0\end{align}\end{cases}\$, or, equivalently, \$(0\le a<b)\lor(b<a\le0)\$.


I discovered a clever trick used here.

Instead of using the for loop to repeat multiple times, repeat exec multiple times.


Compare this with

print"\n".join(">"+"+"*(i/16)+"[<"+"+"*16+">-]<"+"+"*(i%16)if i>31 else"+"*i for i in range(256))
  • 1
    \$\begingroup\$ There's already an answer for this trick. Admittedly it has no examples. \$\endgroup\$ – Ørjan Johansen Aug 12 '19 at 3:01

Recursive functions that print

Functions are allowed to print as programs do. A recursive function that prints can be shorter than both a pure function and a pure program.

Compare these Python 2 submissions to make a list of iteratively floor-halving a number while it's positive, like 10 -> [10, 5, 2, 1].

# 30 bytes: Program 
while n:print n;n/=2

# 29 bytes: Function
f=lambda n:n*[0]and[n]+f(n/2)

# 27 bytes: Function that prints
def g(n):1/n;print n;g(n/2)

Try it online!

The function that prints uses 1/n to terminate with error on hitting n=0 after having printing the desired numbers. This saves characters over the program's while and the pure function's base case, giving it the edge in byte count. Often, the termination can be shorter as part of the expression to print or the recursive call. It might even happen on its own for free, like terminating on an empty string when the first character is read.

The key property of our function here is that we're repeatedly applying an operation and listing the results at each step, in order. Additional variables can still be used this way by having them as optional inputs to the function that are passed in the recursive call. Moreover, because we're def'ing a function rather than writing a lambda, we can put statements such as variable assignments in its body.


Note: The below makes sense only in the program is scored as characters, not as bytes.

I haven't seen this, though somebody may have posted it somewhere.

I needed to have some long literal ASCII strings in the code so somehow shortening them (as characters, not bytes) would be beneficial. After some experiments I came up with what I call the "Chinese reencoding". I call it that way because ASCII characters mostly seem to be squashed in unicode code points that represent Chinese characters. You take an ASCII string S, encode it in bytes as ASCII, and then decode it in UTF16-BE, like that:


The resulting string is half the length. It has to be big endian, as the reverse reencoding may not work - and on most systems the shorter 'utf16' is little endian. You also may need to add a character like space if the original string has odd length, but many times this is OK. Also, for non ASCII characters this does not save length, because they result in too big unicode code points that are represented in the liong form ("\uXXXX")

In you code, use the following:


in order to get the original longer string, where [E] is the literal shortened string. This costs 29 additional characters, so the original string has to be longer than 58, obviously.

One example - below is my 12 days of Christmas (it can be shortened additionally, but let's use that as an example):

for i in range(12):print('On the %s day of Christmas\nMy true love sent to me\n%s'%('First Second Third Fourth Fifth Sixth Seventh Eighth Ninth Tenth Eleventh Twelfth'.split()[i],'\n'.join('Twelve Drummers Drumming,+Eleven Pipers Piping,+Ten Lords-a-Leaping,+Nine Ladies Dancing,+Eight Maids-a-Milking,+Seven Swans-a-Swimming,+Six Geese-a-Laying,+Five Gold Rings,+Four Calling Birds,+Three French Hens,+Two Turtle Doves, and+A Partridge in a Pear Tree.\n'.split('+')[11-i:])))

It's 477 characters long. Let's apply the "Chinese" trick to the two longer string:

r=lambda s:s.encode('utf-16be').decode();for i in range(12):print('On the %s day of Christmas\nMy true love sent to me\n%s'%(r('䙩牳琠卥捯湤⁔桩牤⁆潵牴栠䙩晴栠卩硴栠卥癥湴栠䕩杨瑨⁎楮瑨⁔敮瑨⁅汥癥湴栠呷敬晴栠').split()[i],'\n'.join(r('呷敬癥⁄牵浭敲猠䑲畭浩湧Ⱛ䕬敶敮⁐楰敲猠偩灩湧Ⱛ呥渠䱯牤猭愭䱥慰楮本⭎楮攠䱡摩敳⁄慮捩湧Ⱛ䕩杨琠䵡楤猭愭䵩汫楮本⭓敶敮⁓睡湳ⵡⵓ睩浭楮本⭓楸⁇敥獥ⵡⵌ慹楮本⭆楶攠䝯汤⁒楮杳Ⱛ䙯畲⁃慬汩湧⁂楲摳Ⱛ周牥攠䙲敮捨⁈敮猬⭔睯⁔畲瑬攠䑯癥猬\u2061湤⭁⁐慲瑲楤来\u2069渠愠健慲⁔牥攮ਠ').split('+')[11-i:])))

That's 362, including the lambda (it happens to be worth it, as it is used twice).

Now, all code is mostly ASCII characters, so you may have already guessed that you can use that with exec. There is higher overhead - 43 chars for "exec(''.encode('utf-16be').decode())" (in addition to the whole compressed program) and you may need to double escape some escaped characters in your literal strings (like '\n' in mine has to become '\n'). As a bonus you can always easily add that one space. The compressed porogram looks like:


and it's 299 characters long. You can see some high code points can always appear. I have not found a way to eliminate them, as the added handling code is not worth the benefit.

This is a cheap trick, in fact, but it can always be applied on top of your solution when the program is longish and there are no or few non-ASCII characters. Often you can devise a custom encoding that can stuff more than two ASCII chars in an unicode one, but it is specific for the task.

  • \$\begingroup\$ Most code-golf questions are scored in bytes, not characters \$\endgroup\$ – pppery Sep 7 '19 at 22:00
  • \$\begingroup\$ Interesting, I don't do much and the ones that I have done were scored in characters - but it makes sense. \$\endgroup\$ – Petar Donchev Sep 7 '19 at 22:02

1 or 0 can act as boolean operators in Python:

func = lambda x:1 if x//2==x/2 else 0
while 1:
    if func(n):

Which is 10 characters shorter than:

func = lambda x:True if x//2==x/2 else False
while True:
    if func(n):
  • 1
    \$\begingroup\$ Isn't this pointless in func? You could just have func= lambda x:x//2-x/2 and reverse the consequences of the if (the subtraction will give 0 if they are the same, +/-1 if they are different). It might be better to point out that any python type that has a definition for __bool__ or __non_zero__ could be used instead of a boolean. \$\endgroup\$ – FryAmTheEggman Oct 23 '14 at 20:24
  • 2
    \$\begingroup\$ Here's a tip for golfing everywhere: don't return booleans with a conditional! return True if condition else False can always be simplified to return condition, or if the condition isn't a boolean and you need a boolean, use return bool(condition) or return condition!=0 if it's a number. \$\endgroup\$ – Cyoce Feb 17 '16 at 8:02
  • \$\begingroup\$ @Cyoce This deserves to have many votes. \$\endgroup\$ – Erik the Outgolfer Jun 25 '16 at 6:43
  • \$\begingroup\$ Also, the while True is unnecessary. It could be shortened to while func(n):print('Hello')\nexit() where \n is the new line character \$\endgroup\$ – Cyoce Jul 7 '16 at 2:38

Not read all the answers but you can instead of

if x==3:
    print "yes"
    print "no"


print "yes" if x==3 else "no"
  • 4
    \$\begingroup\$ We already have this in shorter form. \$\endgroup\$ – xnor Apr 23 '15 at 8:14
  • \$\begingroup\$ this is specific to print function as asked in the question I guess \$\endgroup\$ – user3340707 Apr 23 '15 at 8:16
  • 6
    \$\begingroup\$ print"yneos"[x!=3::2] based on this answer \$\endgroup\$ – Jakube Apr 23 '15 at 8:28
  • \$\begingroup\$ @xnor That isn't always applicable, for instance if using recursion. \$\endgroup\$ – Ogaday Feb 16 '16 at 12:25

Slicing can assign:

Converting BGR pixels array to RGB:

l = len(pixels)
for idx in range(0, l - 2, 3):
    pixels[idx + 2], pixels[idx] = pixels[idx], pixels[idx + 2]


l = len(pixels)
pixels[2:l:3], pixels[0:l:3] = pixels[0:l:3], pixels[2:l:3]

It's not just shorter. It's 5 times faster. Except on pypy, where the first version is 2x times faster :)

  • \$\begingroup\$ You don't really need to get the length, pixels[2::3] should work too \$\endgroup\$ – wastl Jun 2 '18 at 19:26

Tricks with dicts:

# create a dict from iterable
d = {k: None for k in iterable}
# merge 2 dicts
# access a key, if it doesn't exist set a default value and return it
if 'foo' not in d:
    res = d['foo'] = 'bar'
    res = 'bar
    res = d['foo']


d = {**dict.fromkeys(iterable), **d1, **d2}
res = d.setdefault('foo', 'bar')

Or if you need repeated access:

import collections as c
d = c.ChainMap(dict.fromkeys(iterable), d1, d2, {'foo': 'bar'})
res = d['foo']

dict.get as a first-class function

# all keys in the dict G with a truthy value
[k for k in G if G[k]]

# all keys in the dict G with a falsy value
[k for k in G if not G[k]]

{…, **d} to merge dicts in Python 3

# merge two dicts
merged: {**a,**b} # the order lets you decide which overrides which

# set a defaut value

Extract elements from a list using ::

# This is a very specific tip when you want to get both an element at
# the near beginning of a list and one somewhere near the end.
# For example, let's assume you want to take the elements at indices 6 and 37
# from a list L of length 40 (it's important it's < 6+37*2)

# ok
# equivalent
# 36 = 5 + 31

Set literals

set() is 5 chars just to create an empty set. If you can have an initial element e, you can save 2 chars with {e}.

Generators instead of comprehension lists in function calls

# assuming the function iterates on its argument
f([x**2 for x in range(4)])
f(x**2 for x in range(4))
1 2 3 4 5

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