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

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 2

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)

Answer 3

Use powers of the imaginary unit to calculate sines and cosines.

For example, given an angle d in degrees, you can calculate the sine and cosine as follows:

p=1j**(d/90.)
s=p.real
c=p.imag

This can also be used for related functions such as the side length of a unit n-gon:

l=abs(1-1j**(4./n))

Answer 4

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 5

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 6

Abuse the fact that in case of an expression yielding True boolean operators return the first value that decides about the outcome of the expression instead of a boolean:

>>> False or 5
5

is pretty straightforward. For a more complex example:

>>> i = i or j and "a" or ""

i's value remains unchanged if it already had a value set, becomes "a" if j has a value or in any other case becomes an empty string (which can usually be omitted, as i most likely already was an empty string).

Answer 7

Lets play with some list tricks

a=[5,5,5,5,5,5,5]

can be written as:

a=[5]*7

It can be expanded in this way. Lets, say we need to do something like

for i in range(0,100,3):a[i]=5

Now using the slicing trick we can simply do:

a[0:100:3]=[5]*(1+99//3)

Answer 8

Logical short-circuiting in recursive functions

A detailed guide

I had worked with short-circuiting and/or's for a while without really grasping how they work, just using b and x or y just as a template. I hope this detailed explanation will help you understand them and use them more flexibly.

---

Recursive named lambda functions are often shorter than programs that loop. For evaluation to terminate, there must be control flow to prevent a recursive call for the base case. Python has a ternary condition operator that fits the bill.

f=lambda x:base_value if is_base_case else recursive_value

Note that list selection won't work because Python evaluates both options. Also, regular if _: isn't an option because we're in a lambda.

---

Python has another option to short-circuit, the logical operator keywords and and or. The idea is that

True or b == True
False and b == False

so Python can skip evaluate b in these cases because the result is known. Think of the evaluation of a or b as "Evaluate a. If it's True, output a. Otherwise, evaluate and output b." So, it's equivalent to write

a or b
a if a else b

It's the same for a or b except we stop if a is False.

a and b
a if (not a) else b

You might wonder why we didn't just write False if (not a) else b. The reason is that this works for non-Boolean a. Such values are first converted to a Boolean. The number 0, None, and the empty list/tuple/set become False, and are so called "Falsey". The rest are "Truthy".

So, a or b and a and b always manages to produce either a or b, while forming a correct Boolean equation.

(0 or 0) == 0
(0 or 3) == 3
(2 or 0) == 2
(2 or 3) == 2
(0 and 0) == 0
(0 and 3) == 0
(2 and 0) == 0
(2 and 3) == 3
('' or 3) == 3
([] and [1]) == []
([0] or [1]) == [0]

---

Now that we understand Boolean short-circuiting, let's use it in recursive functions.

f=lambda x:base_value if is_base_case else recursive_value

The simplest and most common situation is when the base is something like f("") = "", sending a Falsey value to itself. Here, it suffices to do x and with the argument.

For example, this function doubles each character in a string, f("abc") == "aabbcc".

f=lambda s:s and s[0]*2+f(s[1:])

Or, this recursively sums the cubes of numbers 1 through n, so f(3)==36.

f=lambda n:n and n**3+f(n-1)

Another common situation is for your function to take non-negative numbers to lists, with a base case of 0 giving the empty list. We need to transform the number to a list while preserving Truthiness. One way is n*[5], where the list can be anything nonempty. This seems silly, but it works.

So, the following returns the list [1..n].

f=lambda n:n*[5]and f(n+1)+[n]  

Note that negative n will also give the empty list, which works here, but not always. For strings, it's similar with any non-empty string. If you've previously defined such a value, you can save chars by using it.

More generally, when your base value is an empty list, you can use the arithmetic values True == 1 and False == 0 to do:

[5]*(is_not_base_case)and ...

TODO: Truthy base value

---

TODO: and/or

Answer 9

One trick I have encountered concerns returning or printing Yes/No answers:

 print 'YNeos'[x::2]

x is the condition and can take value 0 or 1.

I found this rather brilliant.

Answer 10

List comprehension.

shortList = []
for x in range(10):
    shortList += [x * 2]

can be shortened into

shortList = [x*2 for x in range(10)]

Or even shorter:

shortList = range(0,20,2)

Answer 11

If you are doing something small in a for loop whose only purpose is to invoke a side effect (pop, print in Python 3, append), it might be possible to translate it to a list-comprehension. For example, from Keith Randall's answer here, in the middle of a function, hence the indent:

  if d>list('XXXXXXXXX'):
   for z in D:d.pop()
   c=['X']

Can be converted to:

  if d>list('XXXXXXXXX'):
   [d.pop()for z in D]
   c=['X']

Which then allows this golf:

  if d>list('XXXXXXXXX'):[d.pop()for z in D];c=['X']

An if within a for works just as well:

for i in range(10):
 if is_prime(i):d.pop()

can be written as

[d.pop()for i in range(10)if is_prime(i)]

Answer 12

Use eval to iterate

Say you want to apply f composed k times to the number 1, then print the result.

This can be done via an exec loop,

n=1
exec("n=f(n);"*k)
print(n)

which runs code like n=1;n=f(n);n=f(n);n=f(n);n=f(n);n=f(n);print(n).

But, it's one character shorter to use eval

print(eval("f("*k+'1'+")"*k))

which evaluates code like f(f(f(f(f(1))))) and prints the result.

This does not save chars in Python 2 though, where exec doesn't need parens but eval still does. It does still help though when f(n) is an expression in which n appears only once as the first or last character, letting you use only one string multiplication.

Answer 13

Use map for side effects

Usually you use map to transform a collection

>> map(ord,"abc")
[97, 98, 99]

But you can also use it to repeatedly act on object by a built-in method that modifies it.

>> L=[1,2,3,4,5]
>> map(L.remove,[4,2])
[None, None]
>> L
[1, 3, 5]

Be aware that the calls are done in order, so earlier ones might mess up later ones.

>> L=[1,2,3,4,5]
>> map(L.pop,[0,1])
[1, 3]
>> L
[2, 4, 5]

Here, we intended to extract the first two elements of L, but after extracting the first, the next second element is the original third one. We could sort the indices in descending order to avoid this.

An advantage of the evaluation-as-action is that it can be done inside of a lambda. Be careful in Python 3 though, where map objects are not evaluated immediately. You might need an expression like 0in map(...) to force evaluation.

Answer 14

Object method as answer

Your submission can be a method of an object

range(123,790,111).count

This defines an anonymous function much shorter than

lambda n:range(123,790,111).count(n)

The object method is a valid function that meets our definition. For example, it could be bound and called

f=range(123,790,111).count
print f(99)

Because it avoids a costly lambda, this saves characters even rewriting from

lambda n:n in range(123,790,111)

Consider using an object method when your solution is a simple two-input function of your input and some concrete object. You can use dir() to get a list of methods of an object. Note in particular methods like .__add__ that are called for an operator like +. Most infix operators correspond to a method.

Other examples:

"prefix{}suffix".format
lambda s:"prefix"+s+"suffix"

2 .__rpow__    #Space for lexer
lambda n:n*n

[0,0].__le__
lambda l:[0,0]<=l

["North","East","South","West"].pop
lambda n:["North","East","South","West"][n]

You can even something save bytes with two input by currying. For example, compare

lambda l:expression_in_l.count
lambda n,l:n in expression_in_l

where expression_in_l produces a list with no duplicates and has favorable spacing and precedence.

Answer 15

When mapping a function on a list in Python 3, instead of doing [f(x)for x in l] or list(map(f,l)), do [*map(f,l)].

It works for all other functions returning generators too (like filter).

The best solution is still switching to Python 2 though

Answer 16

A condition like

s = ''
if c:
    s = 'a'

can be written as

s = c*'a'

and there is possibly a need for parenthesis for condition.

This can also be combined with other conditions as (multiple ifs)

s = c1*'a' + c2*'b'

or (multiple elifs)

s = c1*'a' or c2*'b'

For example FizzBuzz problem's solution will be

for i in range(n):
    print((i%3<1)*"Fizz"+(i%5<1)*"Buzz" or i)

Answer 17

Format() works on dates:

"We are in year: %s" % (date.strftime('%y'))

Becomes:

"We are in year: {:%y}".format(date)

Or even better with f-strings:

f"We are in year: {date:%y}"

Answer 18

Build a string instead of joining

To concatenate strings or characters, it can be shorter to repeatedly append to the empty string than to join.

23 chars

s=""
for x in l:s+=f(x)

25 chars

s="".join(f(x)for x in l)

Assume here that f(x) stands for some expression in x, so you can't just map.

But, the join may be shorter if the result doesn't need saving to a variable or if the for takes newlines or indentation.

Answer 19

String keys to dicts

For a dictionary with string keys which also happen to be valid Python variable names, you can get a saving if there's at least three items by using dict's keyword arguments:

{'a':1,'e':4,'i':9}
dict(a=1,e=4,i=9)

The more string keys you have, the more quote characters you'll save, so this is particularly beneficial for large dictionaries (e.g. for a kolmogorov challenge).

Answer 20

When squaring single letter variables, it is shorter to times it by itself

>>> x=30
>>> x*x
900

Is one byte shorter than

>>> x=30
>>> x**2
900

Answer 21

If transforming from list to a tuple or set to a set, list or tuple is needed, as of Python 3.5 you can use the splat operator:

tuple(iterable) -> (*iterable,)  (-3 bytes)
set(iterable)   -> {*iterable}   (-2 bytes)
list(iterable)  -> [*iterable]   (-3 bytes)

If you're doing this as well as appending/prepending, you can do the following for an extra bonus:

iterable+[1]       -> *iterable,1          (-2 bytes, 3 for tuples)
iterable+iter2     -> *iterable,*iterable2 (+1 byte, 0 for tuples, though can combine types)
[1]+iterable+[1]   -> 1,*iterable,1        (-3 bytes, -4 for tuples)
iterable+[1]+iter2 -> *iterable,1,*iter2   (0 bytes, -1 for tuples)

Basically, ,* instead of , gives a +1 byte penalty and , instead of ,[] gives -2 bytes.

This shows [1,*iterable,1] is a golfier way of doing [1]+iterable+[1] by one byte, even when we're not doing any type conversion.

And just for fun, {*{}} is the same length as set() for challenges without letters.

Answer 22

Check if a number is a power of 2

Check whether a positive integer n is a perfect power of 2, that is one of 1, 2, 4, 8, 16, ..., with any of these expression:

n&~-n<1
n&-n==n
2**n%n<1

This is shorter than converting to binary or using a loop. The last one also works for checking, say, powers of 3.

Answer 23

Be aware of all, any and map:

if isdigit(a) and isdigit(b) and isdigit(c)
if all(map(isdigit,[a,b,c]))

Answer 24

Sometimes you need convert boolean expression into integer (0/1) Simple use this Boolean (in examples below c > 0) in ariphmetic

a=b+(c>0)
a+=c>0
a=sum(c>0 for c in b) # space in "0 for" may be omitted

And sometimes you need simple convert boolean to int (for example for printing or convert to binary string). In programm you may use some variants

1 if c>0 else 0
c>0and 1or 0
(0,1)[c>0]
int(c>0)

but shortest way is

+(c>0)

Answer 25

If you need to import a lot of modules you can reassign __import__ to something shorter, this also has the advantage of being able to name imports anything you want.

i=__import__;s=i('string');x=i('itertools');

Answer 26

Cut out newlines wherever you can.

At the top-level, it doesn't matter.

a=1
b=9

Takes the same amount of characters as:

a=1;b=9

In the first case you have a newline instead of a ;. But in function bodies, you save however deep the nesting level is:

def f():
 a=1;b=9

Actually in this case, you can have them all on one line:

def f():a=1;b=9

If you have an if or a for, you can likewise have everything on one line:

if h:a=1;b=9
for g in(1,2):a=1;b=9

But if you have a nesting of control structures (e.g. if in a for, or if in a def), then you need the newline:

if h:for g in(1,2):a=1;b=9 #ERROR

if h:
 for g in(1,2):a=1;b=9 # SAUL GOODMAN

Answer 27

Find the n'th number meeting a condition

Many sequence challenges ask you to find the n'th number in a sequence of increasing positive integers. When you have a expression p(i) that checks membership, you can do this with the recursive function:

f=lambda n,i=1:n and-~f(n-p(i),i+1)

Note that expression p(i) must give 0 or 1, not just Falsey or Truthy. The outputs are one-indexed, so say for the sequence of primes, it would give

f(1) = 2
f(2) = 3
...

For 0-indexed outputs, shift the base case

f=lambda n,i=1:n+1and-~f(n-p(i),i+1)

The recursive function f=lambda n,i=1:n and-~f(n-p(i),i+1) works by decrementing the required count n each time it gets a hit, and incrementing the output value each time for each value it checks. It might seem weird to redundantly track i, but it's longer to do:

f=lambda n,i=1:n and f(n-p(i),i+1)or~-i

Also compare the natural list strategy (zero-indexed here)

lambda n:[i for i in range(n*n)if p(i)][n]

(You might need a larger bound than n*n.)

Answer 28

Iterate over adjacent pairs

It's common to want to iterate over adjacent pairs of items in a list or string, i.e.

"golf" -> [('g','o'), ('o','l'), ('l','f')]

There's a few methods, and which is shortest depends on specifics.

Shift and zip

## 47 bytes
l=input()
for x,y in zip(l,l[1:]):do_stuff(x,y)

Create a list of adjacent pairs, by removing the first element and zipping the original with the result. This is most useful in a list comprehension like

sum(abs(x-y)for x,y in zip(l,l[1:]))

You can also use map with a two-input function, though note that the original list is no longer truncated.

## Python 2
map(cmp,l[:-1],l[1:])

Keep the previous

## 41 bytes, Python 3
x,*l=input()
for y in l:do_stuff(x,y);x=y

Iterate over the elements of the list, remembering the element from a previous loop. This works best with Python 3's ability to unpack to input into the initial and remaining elements.

If there's an initial value of x that serves as a null operation in do_stuff(x,y), you can iterate over the whole list.

## 39 bytes
x=''
for y in input():do_stuff(x,y);x=y

Truncate from the front

## 46 bytes
l=input()
while l[1:]:do_stuff(*l[:2]);l=l[1:]

Keep shortening the list and act on the first two elements. This works best when your operation is better-expressed on a length-two list or string than on two values.

Python 3 can do shorter with unpacking. You can shorten to while l: if your operation is harmless on singletons.

---

I've written these all as loops, but they also lend to a recursive functions. You can also adjust to get cyclic pairs by putting the first element at the end of the list, or as the initial previous-value.

Answer 29

Formatting a Matrix

I've seen this code to get a character matrix (2D array) as a string.

'\n'.join(''.join(i)for i in M)

It's shorter to use a map instead:

'\n'.join(map(''.join,M))

If you're printing the result, it's shortest to use a for loop:

print('\n'.join(map(''.join,M)))
for i in M:print(*i,sep='')      # -5 bytes

If you're using Python 2, you can't use the print trick, but you can still use the for loop:

for i in M:print(''.join(i))     # -3 bytes

Answer 30

Sometimes you can use Python's exec-statement combined with string repetition, to shorten loops. Unfortunately, you can't often use this, but when you can you can get rid of a lot of long loop constructs. Additionally, because exec is a statement you can't use it in lambdas, but eval() might work (but eval() is quite restricted in what you can do with it) although it's 2 characters longer.

Here is an example of this technique in use: GCJ 2010 1B-A Codegolf Python Solution