Tips for golfing in Python

Question

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.

Answer 1

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*[0]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*[0]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*[0]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*[0]and p%i*[i]+f(n-p%i,i+1,p*i*i)     # first n primes
f=lambda n,i=1,p=1:n*[0]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!

Answer 2

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.

Answer 3

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

Answer 4

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

if value of i is useless:

>>> for i in[0]*x:s+=input()

or

>>> exec's+=input();'*x

Answer 5

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

Answer 6

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.

Answer 7

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"".

Answer 8

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

Answer 9

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

Try it online!

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.

Answer 10

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[0]==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.

Answer 11

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

Instead of

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)

Answer 12

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

Answer 13

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.

Answer 14

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)

Answer 15

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)

Answer 16

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.

Answer 17

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

Answer 18

Adding vectors

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.

Answer 19

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], [7]]
>>> 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, [5], [], None, (), (4, 2)]
>>> filter(None, L)
[1, [5], (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.

Answer 20

Reading multi-line input

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

open(0).read()

suffices to read all input from STDIN.

Try it online on Ideone.

Answer 21

Use os.read to read all input:

import os
s=os.read(0,1e9)

Which is shorter than

import sys
s=sys.stdin.read()

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.

Answer 22

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).

Answer 23

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:

Answer 24

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,'_')

Answer 25

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

Answer 26

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).

Answer 27

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

Answer 28

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

Answer 29

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):

Answer 30

Leak variables to save on assignment

Combining with this tip, suppose you have a situation like

for _ in[0]*x:doSomething()
a="blah"

You can instead do:

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)