Tips for golfing in Python

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).

• Use Python 2 for golfing not 3 Aug 9 '17 at 15:04
• @Chris_Rands That simply does not universally hold, as there are cases in which Python 3 allows for shorter submissions. Jul 15 '18 at 14:36
• @JonathanFrech Especially the new := operator in 3.8 Mar 17 '19 at 17:31
• Feb 2 '21 at 11:54

Common helper functions

These are some golfed implementations of number theoretic functions that come up in challenges a lot. Many of these are due to xnor, especially the “Wilson’s theorem prime machines” of the form lambda n,i=1,p=1. The coprime/totient functions are Dennis’s (explanation here).

It is instructive to study what exactly these are doing, so that you can adapt them to your needs or roll them into another recursive function. That often ends up being shorter than pasting these directly into your solution as-is!

All of these assume n is a positive integer. The ones marked with an asterisk produce the wrong result if n = 1. Furthermore, these snippets assume Python 2. For Python 3, you might need to replace / by // here and there.

# Function                                                   Output of f(360)
#========================================================================================
f=lambda n,i=2:n/i*and[f(n,i+1),[i]+f(n/i)][n%i<1]      # [2, 2, 2, 3, 3, 5] (slow!)
f=lambda n,i=2:n/i*and f(n,i+1)if n%i else[i]+f(n/i)    # [2, 2, 2, 3, 3, 5]
f=lambda n,i=2:n/i*and(n%i and f(n,i+1)or[i]+f(n/i))    # [2, 2, 2, 3, 3, 5]
f=lambda n,i=2:n<2and{1}or n%i and f(n,i+1)or{i}|f(n/i)    # {1, 2, 3, 5}
f=lambda n,i=2:n<2and{i}or n%i and f(n,i+1)or{i}|f(n/i,i)  #*{2, 3, 5}
f=lambda n,i=2:n/i and[f(n,i+1),i+f(n/i)][n%i<1]           # 2+2+2+3+3+5 (slow!)
f=lambda n,i=2:n/i and f(n,i+1)if n%i else i+f(n/i)        # 2+2+2+3+3+5
f=lambda n,i=2:n/i and(n%i and f(n,i+1)or i+f(n/i))        # 2+2+2+3+3+5
f=lambda n,i=1,p=1:n*and p%i*[i]+f(n-p%i,i+1,p*i*i)     # first n primes
f=lambda n,i=1,p=1:n*and p%i*[i]+f(n-1,i+1,p*i*i)       # primes <= n
f=lambda n,i=1,p=1:n/i and p%i*i+f(n,i+1,p*i*i)            # sum of primes <= n
f=lambda n,i=1,p=1:n/i and p%i+f(n,i+1,p*i*i)              # count primes <= n
f=lambda n,i=1,p=1:n and-~f(n-p%i,i+1,p*i*i)               # nth prime
f=lambda n:all(n%m for m in range(2,n))                    #*is n prime? (not recursive)
f=lambda n:1>>n or n*f(n-1)                                # factorial
f=lambda n:sum(k/n*k%n>n-2for k in range(n*n))             # totient phi(n) (not recursive)
f=lambda n:[k/n for k in range(n*n)if k/n*k%n==1]          # coprimes up to n (not recursive)

Try it online!

Additions and byte saves are very welcome!

• #7 has a redundant space after ). (Also it has a variant similar to #2.) Jan 4 '18 at 0:16
• Um ... can f=lambda n:1>>n or n*f(n-1) be lambda n:n<2or n*f(n-1), or am I going crazy? Nov 26 '18 at 21:09
• That looks like an acceptable alternative whenever it's acceptable that f(0) == f(1) == True rather than 1.
– Lynn
Nov 26 '18 at 23:08
• Oh, totally forgot about that >_<. Nov 28 '18 at 13:01
• In Python 3.8, from math import perm as f is shorter than factorial definition. and that gives a shorter way to check primality using Wilson's theorem if math is already imported (first used in codegolf.stackexchange.com/a/218738/69850, ls mentioned in codegolf.stackexchange.com/a/194319/69850) Feb 8 '21 at 0:38

Printing a string without a trailing newline in Python 3

Suppose you have a string s, and need to print it without a trailing newline. The canonical way of doing this would be

print(s,end='')

However, if we look at the documentation for print, we can see that print takes in a variable number of objects as its first parameter, with "variable number" including zero. This means that we can do

print(end=s)

instead, for a saving of 3 bytes.

Note that this only works when s is a string, since otherwise the conversion to string would be too expensive:

print(1,end='')
print(end=str(1))

Thanks to @Reticality for this tip.

• Note for  commenters:  does not work in Python 3 (This is a spam prevention comment) Jun 25 '16 at 13:56

Arithmetic tricks

Here are some arithmetic tricks which are either shorter or are more useful due to precedence rules.

Assumptions                  Version 1        Version 2
-------------------------------------------------------------------
n >= 0 float                 n==0             0**n
n >= 0 integer               n==0             1>>n
n >  0 integer               n!=1             1%n
n >  0 integer, Python 2     n==1             1/n
n, m float                   n!=m             n-m
• 1//n could still be useful in Python 3 for n==1 since they have different precedence. Jun 16 '15 at 21:11
>>> for i in range(x):s+=input()

if value of i is useless:

>>> for i in*x:s+=input()

or

>>> exec's+=input();'*x
• You can make the second example into for i in*x:s+=input() to save another space. Also, you can remove the space between the exec and the first quotation mark to get exec's+=input();'*x Apr 19 '11 at 6:12
• shouldn't the second line be: for i in*x:s+=input() Aug 5 '15 at 7:47

Multiple statements can be put on one line separated by ;. This can save a lot of whitespace from indentation.

while foo(a):
print a;a*=2

Or even better:

while foo(a):print a;a*=2
• you can save one more by putting this one all on one line Feb 3 '11 at 13:16
• Not always possible if you have other compound statements like ifs and whiles inside the while. Oct 29 '11 at 20:23

Assignment expressions

Assignment expressions are a powerful language feature introduced in Python 3.8 (TIO). Use the "walrus operator" := to assign a variable inline as part of expression.

>>> (n:=2, n+1)
(2, 3)

You can save an expression to a variable inside a lambda, where assignments are not ordinarily allowed. Compare:

def f(s):t=s.strip();return t+t[::-1]
lambda s:s.strip()+s.strip()[::-1]
lambda s:(t:=s.strip())+t[::-1]

An assignment expression can be used in a comprehension to iteratively update a value, storing the result after each step in a list or other collection. This example computes a running sum by updating the running total t.

>>> t=0
>>> l=[1,2,3]
>>> print([t:=t+x for x in l])
[1, 3, 6]
>>> t
6

This can be done in a lambda with the initial value as an optional argument:

>>> f=lambda l,t=0:[t:=t+x for x in l]
>>> f([1,2,3])
[1, 3, 6]

This function is reusable: each call with start t back at 0.

• This changes everything!
– Lynn
Feb 23 '19 at 19:56
• What is Python infamous for? Oh... nevermind. Mar 1 '19 at 22:42

Translating chars in a string

I've seen this situation pop up a few times, so I thought a tip would be good.

Suppose you have a string s and you want to translate some chars of s to other chars (think ROT-13 like ciphers). For a more concrete example, suppose we want to swap just the as and bs in a string, e.g.

"abacus" -> "babcus"

The naïve way to do this would be:

lambda s:s.replace('a','T').replace('b','a').replace('T','b')

Note how we need to introduce a temporary 'T' to get the swapping right.

With eval, we can shorten this a bit:

lambda s:eval("s"+".replace('%s','%s')"*3%tuple("aTbaTb"))

For this particular example, iterating char-by-char gives a slightly better solution (feel free to try it!). But even so, the winner is str.translate, which takes a dictionary of from: to code points:

# Note: 97 is 'a' and 98 is 'b'
lambda s:s.translate({97:98,98:97})

In Python 2 this only works for Unicode strings, so unfortunately the code here is slightly longer:

lambda s:(u''+s).translate({97:98,98:97})

Some important points which make str.translate so useful are:

• It's easily extendable.
• Any char not specified is untouched by default, e.g. the "cus" in "abacus" above.
• The to part of the dictionary can actually be a (Unicode) string as well, e.g. {97:"XYZ"} (u"XYZ" in Python 2) would turn abacus -> XYZbXYZcus. It can also be None, but that doesn't save any bytes compared to "" or u"".
• "ab".translate({97:None}) is longer than "ab".translate({97:""}). Oct 8 '15 at 8:16
• @T.Verron Thanks for pointing that out - I'm not sure why I added that there tbh... Oct 8 '15 at 8:24
• The second example you gave can be shortened: replace "aTbaTb" with "aTb"*2. Also, you neglected to mention maketrans here, which could significantly shorten translations involving many characters, e.g.: from string import*;t="abcdefghijklmABCDEFGHIJKLMZYXWVUTSRQPONzyxwvutsrqpon";lambda s:s.translate(maketrans(t,t[::-1])) Aug 18 '17 at 4:05
• tuple("aTbaTb") -> (*"aTb"*2,) Mar 31 '21 at 19:02
• Worth noting is that, while .translate accepts a mapping of unicode values to replacement characters or values, a string can be indexed the same as a dict with integer keys. Thus, {97:98,98:97} may be replaced with 'ab'*99 in the call to .translate, as b and a will be at each others' indices in the resulting string. Aug 20 '21 at 2:28

If you're doing somewhat more complex golfing that require something from the standard library to be used a lot, import x as y can save some space:

import itertools as i
i.groupby(...) # same as itertools.groupby
• You can also do from itertools import *. This uses up 2 more characters, but you hit equal immediately by typing groupby instead of i.groupby. If you use groupby twice, you just saved 2 characters! Jan 28 '11 at 7:52
• from blah import* (without the last whitespace) is even shorter. Jan 28 '11 at 8:42
• import* also gets you all the rest of the itertools. I wish it were shorter to use product/combinations/permutations though Feb 3 '11 at 13:14
• Yeah, the names in itertools are just way too descriptive, makes me sad. Aug 23 '11 at 21:50

Binomial coefficient

The binomial coefficient $\binom{n}{k} \ = \frac{n!}{k!(n-k)!}$ can be expressed arithmetically as

((2**n+1)**n>>n*k)%2**n

This works for $n,k \geq 0$, except for $n=k=0$ it gives $0$ rather than $1$. More generally, it works to use

(b+1)**n/b**k%b

(TIO), where $b$ is any value strictly greater than the result. The first expression uses $b=2^n$, which exceeds $\binom{n}{k}$ except for $n=k=0$.

Why does this work? Let's look at an example with b=1000. Then, for n=6, we have

(b+1)**n = 1001 ** 6 = 1006015020015006001

Note how triples of digits encode the binomial coefficients in the n=6 row of Pascal's triangle:

1   6  15  20  15   6   1
1 006 015 020 015 006 001

This works because the binomial coefficients are the coefficients of the polynomial

$$(b+1)^n = \sum_{k=0}^n\binom{n}{k}b^k$$

and so can be read off as digits in base b, as long no binomial coefficient exceeds b which would cause regrouping.

We can extract a given triple of digits, say for $\binom{6}{2}=15$, by floor-dividing by 1000000 to delete the last 6 digits leaving 1006015020015, then take %1000 to extract the last triplet 015. More generally, doing /b**k%b extracts the k-th digit from the end zero-indexed in base b, that is the digit with multiplier b**k.

• Why is this a tip for golfing in Python? Many languages have built-ins for exponentiation. Aug 10 '18 at 12:46

Safely get the first element

You can check if a possibly-empty list l starts with a value x by doing

l[:1]==[x]

This gives False on an empty list, while l==x gives an out-of-bounds error. Strings works similarly

s[:1]=='a'

In general, you can safely check the n'th element as

l[n:n+1]==[a]

or as l[n:][:1]==[a] when n is a long expression.

map can take multiple iterable arguments and apply the function in parallel.

a=[1,4,2,6,4]
b=[2,3,1,8,2]
map(lambda x,y:...,zip(a,b))

you can write

map(lambda x,y:...,a,b)
• I didn't know this and it's super useful! Feb 3 '15 at 14:59

Negating Booleans

So you have a Boolean... a real Boolean, not one represented as an integer. You have a condition where it needs to be negated, and you can't just go back and negate it where you got it (e.g. != instead of ==), maybe because you use it once straight and once negated.

Well, who says your Booleans aren't longing to be integers deep in their little hearts?

>>> False < 1
True
>>> True < 1
False

8 bytes, not counting the colon:

if not C:

6 bytes:

if C<1:

EDIT: 5 bytes, thanks to user202729 in the comments:

if~-C:

This works because:

>>> -False
0
>>> -True
-1
>>> ~-False
-1
>>> ~-True
0

You only need to indent nested control structures:

def baz(i):
if i==0:i=1;print i;bar()
while i:i+=foo(i-1)

You can use default arguments of a function to save some indentation, since

def f(a,l=[1,2,3]):
return sum(a==i for i in l)

is one byte shorter than

def f(a):
l=[1,2,3]
return sum(a==i for i in l)
• A small caution: if a list that is passed in as an optional argument is modified in the function (like with pop or l =1), that list will be changed in the outer scope too.
– xnor
Aug 22 '14 at 0:14
• Also, this is only needed if the function body contains nested block statements; otherwise, you can just put everything on one line to avoid indentation. Sep 26 '14 at 1:55

Replace a value in a list

To replace every entry of value a with b in a list L, use:

map({a:b}.get,L,L)

For example,

L=[1,2,3,1,2,3]
a=2
b=3
print map({a:b}.get,L,L)

[1, 3, 3, 1, 3, 3]  #Output

In Python 3, this returns a map object rather than a list. The list entries can be any hashable values (ints, floats, strings, tuples, etc).

Here's how this works. A dictionary's get method takes a key and default value, and returns the dictionary's entry for that key, using the default value is the key is not present. This method is mapped method over each entry in L both as the key and the default value, which results in

[{a:b}.get(x,x) for x in L]

If x is a, then the dictionary transforms it to b, and otherwise, it defaults to itself. You can perform multiple replacements at the same time using a larger dictionary.

Credit to twobit on Anarchy Golf for exposing me to this trick.

Use extended slicing to select one of two strings

>>> for x in-2,2:print"WoolrlledH"[::x]
...
Hello
World

vs

>>> for x in 0,1:print["Hello","World"][x]
...
Hello
World

Python doesn't have a built-in way to do vector (component-wise) addition except with libraries. Say a and b are two equal-length lists of numbers you want to add. Instead of the list comprehension

c=[a[i]+b[i]for i in range(len(a))]

you can use

c=map(sum,zip(a,b))

This produces an annoying map object in Python 3, but it's shorter even if you have to convert to a list.

– est
Aug 22 '17 at 1:44
• @est But this is code golf... Apr 17 '18 at 16:17

None arguments in Python builtins

map (Python 2 only)

Mapping with None in place of a function assumes the identity function instead. This allows it to be used as an alternative to itertools.izip_longest for zipping lists to the length of the longest list:

>>> L = [[1, 2], [3, 4, 5, 6], ]
>>> map(None,*L)
[(1, 3, 7), (2, 4, None), (None, 5, None), (None, 6, None)]

For visualisation (with . representing None):

1 2                1 3 7
3 4 5 6      ->    2 4 .
7                  . 5 .
. 6 .

filter

filter with None also assumes the identity function, thus removing falsy elements.

>>> L = ["", 1, 0, , [], None, (), (4, 2)]
>>> filter(None, L)
[1, , (4, 2)]

This is a bit better than a list comprehension:

filter(None,L)
[x for x in L if x]

However, as @KSab notes, if all elements are of the same type then there may be shorter alternatives, e.g. filter(str,L) if all elements are strings.

• I had no idea you could do this! In the case of the filter, something similar I have done in the past is filter(str,L) if L is all strings or filter(int,L) if all ints which in some cases could be shorter.
– KSab
May 30 '15 at 9:18

In Python 3, the built-in function open underwent some changes. In particular, its first argument

file is either a string or bytes object giving the pathname (absolute or relative to the current working directory) of the file to be opened or an integer file descriptor of the file to be wrapped.

(source)

That means

suffices to read all input from STDIN.

Try it online on Ideone.

• ... except on Windows :/ (OSError: [WinError 6] The handle is invalid) Feb 17 '16 at 7:14

import os

Which is shorter than

import sys

Note that this has a limitation on input length, but it's so ridiculously large I'd say we're safe from the angry mob.

• raw_input() is shorter. If you need to read once, just spelling it out is shorter than the import + os.read; if more than once, assign it to a single-character value. Apr 30 '11 at 3:17
• @Wooble: why use raw_input for golfing? just use input. Sep 1 '11 at 13:54
• @Lie: good point; although it depends on the problem specification whether having input evaluated would work, you can just stipulate that you're using python 3 (although then your print functions require a bit more space...) Sep 1 '11 at 15:20
• This doesn't work in python2.7/3. The number of bytes to read must be an integer. Oct 19 '13 at 15:46
• 1e9 -> 9e9. Jun 25 '16 at 13:52

Just found out two new things. First, input() can parse tuples, like 1, 2, 3 is equivalent to the tuple (1, 2, 3).

And if you need to convert a value to float, just multiply by 1.. Yes, 1. is valid syntax (At least in 2.6).

• I think it's worth noting that in Python 2, input(x) is basically the same thing as eval(raw_input(x)). Unsafe to use in practice, but good for code golfing. Jan 24 '14 at 2:33
• Quoting the OP: Please post one tip per answer. Jun 23 '14 at 13:56
• To further explicate the point of this answer: in a challenge where input format is flexible, for instance if reading in a bunch of numerical arguments, you can write input() only once. E.g. a,b,c=input() will read in three comma-separated arguments and assign them to a, b, and c Feb 1 '16 at 4:17

If you represent boolean values as numbers you can save characters. This is especially true for using -1 as True.

Bitty conditionals work (Truth table):

a  b   &  |  ^
0  0   0  0  0
0  -1  0 -1 -1
-1 0   0 -1 -1
-1 -1 -1 -1  0

And ~ works as not:

a ~a
0 -1
-1  0

Even though the - for initializing -1 costs one character, this can easily save characters overall.

Compare:

while~a:

to:

while not a:

Use .center in ASCII art

In drawing a symmetrical ASCII art, you can center-justify each line in a fixed width of spaces. For example, "<|>".center(7) gives ' <|> '. This can be shorter than computing how many spaces are needed to center it.

You can also pad with a different character by doing "<|>".center(7,'_')

• This is actually really cool! Nov 5 '16 at 0:41
• I wish JS had this function... Nov 5 '16 at 0:57
• Would f"{'<|>':^7}" not be shorter in python3.6+ for non-variable widths? Even more so when providing the character to center by, f"{'<|>':-^7}" vs "<|>".center(7,"-") Sep 13 '19 at 10:48

If you want to know the type of a variable x:

x*0is 0  # -> integer
x*0==0   # -> float (if the previous check fails)
x*0==""  # -> string
x*0==[]  # -> array

The shortest infinite for comprehension

You may know the trick to easily create an infinite generator using the two-argument form of iter:

(... for _ in iter(lambda:0,1))

where 0 and 1 can be any two non-equal values.

As @ovs once pointed out to me, you can replace lambda:0 with int, because int returns 0 when called with no arguments.

(... for _ in iter(int,1))

However, we can do one byte better!:

(... for()in iter(set,1))

This uses the fact that () is a valid L-value (assignment target) in Python as long as the RHS is an empty sequence, and set returns an empty set when called with no arguments.

You can also use str instead of set for the same byte count.

• related numpy trick Nov 20 '21 at 6:08

Iterating over indices in a list

Sometimes, you need to iterate over the indices of a list l in order to do something for each element that depends on its index. The obvious way is a clunky expression:

# 38 chars
for i in range(len(l)):DoStuff(i,l[i])

The Pythonic solution is to use enumerate:

# 36 chars
for i,x in enumerate(l):DoStuff(i,x)

But that nine-letter method is just too long for golfing.

Instead, just manually track the index yourself while iterating over the list.

# 32 chars
i=0
for x in l:DoStuff(i,x);i+=1

Here's some alternatives that are longer but might be situationally better

# 36 chars
# Consumes list
i=0
while l:DoStuff(i,l.pop(0));i+=1

# 36 chars
i=0
while l[i:]:DoStuff(i,l[i]);i+=1

Split into chunks

You can split a list into chunks of a given size using zip and iter, as explained in this SO question.

>>> l=range(12)
>>> zip(*[iter(l)]*4)
[(0, 1, 2, 3), (4, 5, 6, 7), (8, 9, 10, 11)]

Of course, substituting in l as zip(*[iter(range(12))]*4) gives the same result.

The 4 is the number of elements per chunk. If the length isn't a multiple of this, any elements in the remainder are not included. For example, l=range(13) would give the same result.

The result is a list of tuples. If your input is a string and you want to produce a list of strings, you can do

>>> l="Code_golf"
>>> map(''.join,zip(*[iter(l)]*3))
['Cod', 'e_g', 'olf'] # Python 3 would give a map object

When the list l is defined by a list comprehension, instead of converting to an iterable as iter(l), you can instead write it as a generator comprehension with (...) instead of [...].

>>> l=(n for n in range(18)if n%3!=1)
>>> zip(*[l]*4)
[(0, 2, 3, 5), (6, 8, 9, 11), (12, 14, 15, 17)]

This consumes the generator, so l will appear empty afterwards. Note as before that we can inline l as zip(*[(n for n in range(18)if n%3!=1)]*4).

use os.urandom() as a random source instead of random.randint()

• Doesn't this require use of ord() to get a number instead of character? len("ord(os.urandom(1))") -> 18 and len("random.randint()") -> 16
– jscs
May 1 '11 at 3:10
• @Josh, don't forget you need to import random vs import os. randint() needs 3 parameters anyway. If you need a list of random numbers, you can use map(ord,os.urandom(N)) Also, sometimes, you actually need a random char instead of a number May 1 '11 at 7:17
• Late to the party, but if you only need a few random numbers, try using id(id), substituting the inner id with any 3-letter builtin if you need more than one. 'abc'[id(id)%3] is 11 characters shorter than 'abc'[random.randrange(3)], not even counting the import statement. Apr 20 '13 at 2:04
• @Fraxtil id can be applied on mostly everything, such as 1 or []. Jul 5 '18 at 9:41

Avoid startswith

The string method startswith is too long. There are shorter ways to check if a string s starts with a prefix t of unknown length.

t<=s<t+'~'     #Requires a char bigger than any in s,t
s.find(t)==0
s[:len(t)]==t
s.startswith(t)

The second one is well-suited for the truth/falsity of the negation.

if s.find(t):
• s[:len(t)]==t is shorter if t has a constant length less than 100000. Jun 15 '17 at 18:48

Leak variables to save on assignment

Combining with this tip, suppose you have a situation like

for _ in*x:doSomething()
a="blah"

for a in["blah"]*x:doSomething()

to skip out on a variable assignment. However, be aware that

exec"doSomething();"*x;a="blah"

in Python 2 is just shorter, so this only really saves in cases like assigning a char (via "c"*x) or in Python 3.

However, where things get fun is with Python 2 list comprehensions, where this idea still works due to a quirk with list comprehension scope:

[doSomething()for a in["blah"]*x]

(Credits to @xnor for expanding the former, and @Lembik for teaching me about the latter)

• Does the last one not work in Python 3?
– xnor
Jun 26 '15 at 17:29
• @xnor in Python 3 exec requires parentheses, so it would be one character longer than the for... format. Jul 14 '17 at 7:10