277
\$\begingroup\$

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.

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  • 30
    \$\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
  • 5
    \$\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\$ – moinudin Jan 28 '11 at 12:28
  • 54
    \$\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
  • 4
    \$\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

147 Answers 147

4
\$\begingroup\$

Take multi-line input

Use list(iter(input,eof)) to take multi-line input. eof can be any string that you want to stop taking input on if you see it. An example would be eof = ''. The python 2 version is list(iter(raw_input,eof)), however you may want to use sys.stdin.readlines() instead if you have already imported sys.

| improve this answer | |
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  • \$\begingroup\$ list(iter(input,eof)) is shorter by one, you don't need that space. Also worth noting that this is only for Python 3 - Python 2 is 4 bytes longer because raw_input. \$\endgroup\$ – Mego Nov 30 '15 at 3:15
  • \$\begingroup\$ Is shorter than what? \$\endgroup\$ – J Atkin Nov 30 '15 at 3:33
  • \$\begingroup\$ Than what you have posted, because it has an extraneous space. \$\endgroup\$ – Mego Nov 30 '15 at 3:34
  • \$\begingroup\$ Oh, I see. I assume the reader would remove the space though. I will edit it out now. \$\endgroup\$ – J Atkin Nov 30 '15 at 3:38
4
\$\begingroup\$

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.)

| improve this answer | |
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4
\$\begingroup\$

Difference of two parallel expressions

Say you have a challenge to find the difference of some characteristic on two inputs. Your solution has the form lambda a,b:e(b)-e(a), where e is some long expression you've written. Repeating e twice is wasteful, so what do you do?

Here are templates sorted by length. Assume that e stands for a long expression, not one that's already defined as a function. Also assume inputs can be taken in either order.

31 bytes*

lambda*l:eval('e(%s)-'*2%l+'0')

*Requires that e only mentions its variable once. Assumes -e(x) negates the whole expression, otherwise requires parens like -(e(x)) for two more bytes.

34 bytes

f=lambda a,*b:e(a)-(b>()and f(*b))

36 bytes

lambda a,b:d(b)-d(a)
d=lambda x:e(x)

36 bytes

a,b=[e(x)for x in input()]
print b-a

37 bytes

r=0
for x in input():r=e(x)-r
print r

39 bytes

lambda*l:int.__sub__(*[e(x)for x in l])
| improve this answer | |
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4
\$\begingroup\$

Use Splat (*) to pass a bunch of single character strings into a function

For example:

a.replace("a","b")
a.replace(*"ab")    -2 bytes

some_function("a","b","c")
some_function(*"abc")       -5 bytes

In fact, if you have n single-character strings, you will save 3(n - 1) - 1 or 3n - 4 bytes by doing this (because each time, you remove the "," for each one and add a constant *).

| improve this answer | |
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4
\$\begingroup\$

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
| improve this answer | |
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4
\$\begingroup\$

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.


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.


The Python 3.8 "walrus" assignment expressions allow a short expression to give pairs, though with an extra initial element.

>>> p=''
>>> [(p,p:=c)for c in"golf"]
[('', 'g'), ('g', 'o'), ('o', 'l'), ('l', 'f')]
| improve this answer | |
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3
\$\begingroup\$

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

| improve this answer | |
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  • \$\begingroup\$ What kind of restrictions are you thinking about? I don't think Python in itself has any kind of restriction on exec, so you must be referring to problem statements? \$\endgroup\$ – hallvabo Jan 28 '11 at 8:43
  • \$\begingroup\$ @hallvabo, you're right it's eval() I was most likely thinking of. You can't do eval("print 1") because print 1 is a statement. I'll update the post. \$\endgroup\$ – JPvdMerwe Jan 28 '11 at 10:23
  • \$\begingroup\$ As a general rule, this is worth trying when you need to do something n times, but don't care about the index. As soon as you need to initialise a loop variable and increment it, it ends up slightly longer than a for loop \$\endgroup\$ – gnibbler Feb 3 '11 at 20:35
  • 2
    \$\begingroup\$ In Python 3 you can do eval("print(1)") since print() is now a function. \$\endgroup\$ – trichoplax Apr 20 '14 at 23:02
  • 2
    \$\begingroup\$ In Python 3 you can also do exec("print(1)") since exec() is now a function. \$\endgroup\$ – CalculatorFeline May 28 '17 at 5:37
3
\$\begingroup\$

Was somewhat mentioned but I want to expand:

[a,b],[c,d]=[[1,2],[3,4]]

works as well as simple a,b=[1,2]. Another great thing is to use ternary operator (similiar to C-like ?:)

x if x<3 else y

and no one mentioned map. Map will call first function given as first argument on each item from second argument. For example assume that a is a list of strings of integers (from user input for example):

sum(map(int,a)) 

will make sum of all integers.

| improve this answer | |
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  • 5
    \$\begingroup\$ Quoting the OP: Please post one tip per answer. \$\endgroup\$ – nyuszika7h Jun 23 '14 at 13:53
  • \$\begingroup\$ x if cond else y == cond and x or y. \$\endgroup\$ – user202729 Jul 5 '18 at 9:34
3
\$\begingroup\$

If you rely on data (mostly for kolmogorov-complexity problems), use the built-in zip encoding/decoding and store the data in a file (add +1 for the filename):

open('f','rb').read().decode('zip')

If you have to store the data in the source code, then you need to encode the zip with base64 and do:

"base64literal".decode('base64').decode('zip')

These don't necessarily save characters in all instances, though.

| improve this answer | |
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3
\$\begingroup\$

When your program needs to return a value, you might be able to use a yield, saving one character:

def a(b):yield b

However, to print it you'd need to do something like

for i in a(b):print i
| improve this answer | |
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  • 1
    \$\begingroup\$ If only a single value is yielded, print next(i()) will work too. \$\endgroup\$ – nyuszika7h Jun 23 '14 at 13:35
  • \$\begingroup\$ Or just do print[*a(b)] \$\endgroup\$ – Esolanging Fruit Jun 5 '17 at 0:59
3
\$\begingroup\$

Abusing try/except blocks can sometimes save characters, especially for exiting out of nested loops or list comprehensions. This:

for c in s:
 for i in l:
  q=ord(c)==i
  if q:print i,c;break
 if q:break

... can become this, saving 3 characters:

try:
 for c in s:
  for i in l:
   if ord(c)==i:print i,c;1/0
except:0

... which in this particular instance can be compressed even further using list comprehensions:

try:[1/(ord(c)-i)for c in s for i in l]
except:print i,c

For an example, see e.g. https://codegolf.stackexchange.com/a/36492/16766.

| improve this answer | |
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3
\$\begingroup\$

When using Python 3, for your final print statement, use exit to save one char (note: this prints to STDERR, so you might not be able to use this):

print('x')
exit('x')

exit even adds a trailing newline. There is one caveat, however: exit(some_integer) will not print.

| improve this answer | |
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  • \$\begingroup\$ Does this print to STDOUT or STDERR? This sounds like it would be the latter, but in most challenges only the former is allowed. \$\endgroup\$ – Sp3000 Apr 19 '15 at 4:24
  • \$\begingroup\$ @Sp3000 I don't know... how can I tell? \$\endgroup\$ – Justin Apr 19 '15 at 5:18
  • \$\begingroup\$ Seems to be stderr \$\endgroup\$ – Sp3000 Apr 19 '15 at 5:32
  • \$\begingroup\$ exit(some_integer) is valid because exit code is a default output form. \$\endgroup\$ – pppery Jun 23 at 15:20
3
\$\begingroup\$

Shorter isinstance

Instead of

isinstance(x,C) # 15 bytes

there are several alternatives:

x.__class__==C  # 14 bytes
'a'in dir(x)    # 12 bytes, if the class has a distinguishing attribute 'a'
type(x)==C      # 10 bytes, doesn't work with old-style classes
'K'in`x`        # 8 bytes, only in python 2, if no other classes contain 'K'
                # watch out for false positives from the hex address

Some of them may save extra bytes depending on the context, because you can eliminate a space before or after the expression.

Thanks Sp3000 for contributing a couple of tips.

| improve this answer | |
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3
\$\begingroup\$

Avoid the repeat argument of itertools.product

As @T.Verron points out, in most cases (e.g. ranges and lists), you can instead do

product(*[x]*n)

However, even if you have a generator which you can only use once, like a Python 3 map, the repeat argument is still unnecessary. In such a case you can use itertools.tee:

product(x,repeat=n)
product(*tee(x,n))

For n = 2 you don't even need to include n, since 2 is the default argument to tee.

| improve this answer | |
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  • \$\begingroup\$ product(*[L]*n) is even shorter. \$\endgroup\$ – T. Verron Oct 8 '15 at 11:59
  • \$\begingroup\$ @T.Verron The initial use case I had in mind was something like a Python 3 map, for which that wouldn't work, but I'll add it as a note thanks :) \$\endgroup\$ – Sp3000 Oct 8 '15 at 12:05
  • 1
    \$\begingroup\$ Doesn't feersum have an open 500-point bounty for using itertools at all in a sufficiently old question? \$\endgroup\$ – lirtosiast Oct 8 '15 at 13:17
3
\$\begingroup\$

Optional empty sequence argument

Suppose we want to write a recursive function that prepends to a sequence (e.g. list, tuple) each time. For example, the Python 3 program

def f(n,k,L=[]):n and[f(n-1,k,[b]+L)for b in range(k)]or print(L)

works like itertools.product, taking n,k and printing all length n lists of numbers taken from range(k). (Example thanks to @xnor)

If we don't need L to be a list specifically, we can save on the optional empty list argument by making use of unpacking, like so:

def f(n,k,*T):n and[f(n-1,k,b,*T)for b in range(k)]or print(T)

where T is now a tuple instead. In the general case, this saves 3 bytes!

In Python 3.5+, this also works if we're appending to the end of a sequence, i.e. we can change f(n-1,k,L+[b]) to f(n-1,k,*T,b). The latter is a syntax error in earlier versions of Python though.

| improve this answer | |
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3
\$\begingroup\$

Use slicing + assignment instead of mutator methods

l.insert(x,y) # before
l[x:x]=y,     # after

l.reverse()   # before
l[::-1]=l     # after

l.append(x)   # before
l[L:]=x,      # after (where L is any integer >= len(l))

l[:]=x        # set the contents of l to the contents of x

EDIT: thanks to @quintopia for pointing this out, these are statements, not expressions. The mutator methods are void functions, so they are expressions which evaluate to None. This means that things like [l.reverse() for x in L] and condition or l.reverse() are valid, whereas [l[::-1]=l for x in L] and condition or l[::-1]=l are not.

| improve this answer | |
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  • \$\begingroup\$ Note that this cannot always be done. For instance, when using or or and as conditionals, assignment is not allowed, but append and the like are. \$\endgroup\$ – quintopia Jan 13 '16 at 5:34
  • \$\begingroup\$ l[:]=x is a bit unpythonic - same number of bytes, but I reckon l=x[:] is better practice. Also, [::-1] is listed as a tip here, L[:x]+=y, is better for the general case where x might be an expression (but same byte count if it's just x), and l+=x, is listed here for append. \$\endgroup\$ – Sp3000 Feb 17 '16 at 7:10
  • \$\begingroup\$ @Sp3000 l[:]=x and l=x[:] do different things. The former mutates the list itself, i.e. all references to that list, whereas the latter sets the variable l to a copy of x \$\endgroup\$ – Cyoce Feb 17 '16 at 7:47
  • \$\begingroup\$ @Cyoce Oh, nevermind then - my eyes are going to have to get used to seeing that :P \$\endgroup\$ – Sp3000 Feb 17 '16 at 8:01
3
\$\begingroup\$

Easiest way to swap two values

>>> a=5
>>> b=4
>>> a,b=b,a
>>> a
4
>>> b
5
| improve this answer | |
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3
\$\begingroup\$

Access list, while building it inside comprehension

In python version 2.4 (and <2.3 with some tweaks) it is possible to access list, from list comprehension. Source #1, Source #2 (Safari, Python Cookbook, 2nd edition)

Python creates secret name _[1] for list, while it is created and store it in locals. Also names _[2], _[3]... are used for nested lists.

So to access list, you may use locals()['_[1]'].

In earlier versions this is not enough. You'll need to use locals()['_[1]'].__self__

I couldn't find evidence, that somethins like that is possible in versions >2.4

Don't think, that it might be usefull often, but who knows! At least it helps with building one-liners.

Example:

# Remove duplicates from a list:
>>> L = [1,2,2,3,3,3]
>>> [x for x in L if x not in locals()['_[1]']]
[1,2,3]
| improve this answer | |
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  • 5
    \$\begingroup\$ This still exists in later Pythons. In Python 3, it's called '.0', and it's a generator. \$\endgroup\$ – isaacg Aug 18 '17 at 6:42
3
\$\begingroup\$

Shorter way to copy/clone a list

(not deep clone. For deep clone see this answer)

(credit to this answer)

a=x[:]
b=[*x]
c=x*1

Try it online!

| improve this answer | |
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3
\$\begingroup\$

Slicing tricks for hard-wired lists of strings

Warning: Python is the language which worships readability above all else; so coding this way is a Mortal Sin.

This sort of thing comes up a lot; such as here where for a given digit in 0<=d<=9, we can get the 7-bit segment b value as a hex string from the list

b=['7e','30','6d','79','33','5b','5f','70','7f','7b'][d]

If the length of such a list is more than just a few elements, you're usually better off at least using split because you can replace a bunch of "','"s with a single character " " as delimiter. E.g.:

b='7e 30 6d 79 33 5b 5f 70 7f 7b'.split()[d]

This can be used for almost any list of strings (possibly at a small additional cost using a delimiter such as ",").

But if in addition, the strings we are selecting for all have the same length k (k==2 in our example), then with the magic of Python slicing, we can write the above as:

b='7e306d79335b5f707f7b'[2*d:][:2]

which saves a lot of bytes because we don't need character delimiters at all. But in that case, usually even shorter would be:

b='7367355777e0d93bf0fb'[d::10]
| improve this answer | |
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3
\$\begingroup\$

Unpacking in Python3

If you only need the first few values in the array

>>> a, b, *c = [1, 2, 3, 4, 5]
>>> a
1
>>> b
2
>>> c
[3, 4, 5]

Same applies to when you need last few values

>>> *a, b, c = [1, 2, 3, 4, 5]
>>> a
[1, 2, 3]
>>> b
4
>>> c
5

Or even with the first few and last few

>>> a, *b, c = [1, 2, 3, 4, 5]
>>> a
1
>>> b
[2, 3, 4]
>>> c
5
| improve this answer | |
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  • 3
    \$\begingroup\$ I was experimenting with this, and I found out you can also initialise an empty array if the number of values is one less than the number of variables, such as a,*b,c=1,2 or a,*b,c="ab" \$\endgroup\$ – Jo King Oct 22 '18 at 11:25
3
\$\begingroup\$

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 -> +
| improve this answer | |
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3
\$\begingroup\$

Using exec to remove repeated print

This is not quite often applicable, but can save some bytes, especially in ASCII art. Take the following code, which prints the flag where n=4.

# 43 bytes                  |  ***
n=input()                   |  **
while~-n:n-=1;print'*'*n    |  *
print'|'                    |  |

Notice that we repeat print twice. We can remove this using exec in the following code, saving 3 bytes.

# 40 bytes
n=input()
exec"'*'*n;n-=1;print"*n+"'|'"
| improve this answer | |
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2
\$\begingroup\$

Run your code through an space-remover, like this one:

#Pygolfer
a=raw_input()
for i in [i for i in range(len(a)) if a[i]==" "]:
    try:b=a[:i]+a[i+1:];eval(b);a=b;print a
    except:pass

(This just tries to remove the spaces one by one, and try if the code still works. Please still do check your code manually.)

Manual things to do: print'string' works.

[str(i)for i in(1,2,3,4)] works.

etc.

| improve this answer | |
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2
\$\begingroup\$

You can generate pseudo random numbers using hash.

hash('V~')%10000

Will print 2014.

| improve this answer | |
\$\endgroup\$
  • 1
    \$\begingroup\$ Prints 9454 for me. \$\endgroup\$ – nyuszika7h Jun 23 '14 at 13:07
  • 2
    \$\begingroup\$ Python 2.7.2 always returns 2014, but Python 3.4.0 returns more a more random number per session, like 6321, 3744, and 5566. \$\endgroup\$ – Cees Timmerman Aug 13 '14 at 8:34
  • \$\begingroup\$ Pretty similar to this. \$\endgroup\$ – user202729 Jul 5 '18 at 9:42
2
\$\begingroup\$

Try a lambda expression

By default, submissions may be functions and functions may be anonymous. A lambda expression is often the shortest framework for input/output. Compare:

lambda s:s+s[::-1]
def f(s):return s+s[::-1]
s=input();print s+s[::-1]

(These concatenate a string with its reverse.)

The big limitation is that the body of a lambda must be a single expression, and so cannot contain assignments. For built-ins, you can do assignments like e=enumerate outside the function body or as an optional argument.

This doesn't work for expressions in terms of the inputs. But, note that using a lambda might still be worth repeating a long expression.

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

The lambda is shorter even though we save a char in the named function by having it print rather than return. The break-even point for two uses is length 12.

However, if you have many assignments or complex structures like loops (that are hard to make recursive calls), you're probably be better off taking the hit and write a named function or program.

| improve this answer | |
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2
\$\begingroup\$

cmp in Python 2

Say you want to output P if x>0, N if x<0, and Z if x==0.

"ZPN"[cmp(x,0)]

Try it online

This function was removed in Python 3.0.1, although it remained in Python 3.0 by mistake.

| improve this answer | |
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  • 1
    \$\begingroup\$ Congratulations on the 100th answer to this question! \$\endgroup\$ – NinjaBearMonkey Mar 2 '16 at 20:51
2
\$\begingroup\$

Abuse of or in lambdas

I'm surprised this isn't in here yet, but if you need a multi-statement lambda, or evaluates both of its operands, as opposed to and which doesn't evaluate the second one if the first one is not True. For instance, a contrived example, to print the characters in a string one by one with an interval:

list(
    map(
        (lambda i: 
            sleep(.06) or print(i) or print(ord(i)) # all of these get executed
        ), 
        "compiling... "
    )
)
            

In this case it isn't shorter, but I've found it to be, sometimes.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Related \$\endgroup\$ – Sp3000 Mar 4 '16 at 3:21
  • \$\begingroup\$ @Sp3000 *sighs* I tried searching this question for or, lambda, evaluation etc but didn't see that \$\endgroup\$ – cat Mar 4 '16 at 3:24
  • \$\begingroup\$ lambda i:[sleep(.06),print(i),print(ord(i))] \$\endgroup\$ – user202729 Jul 5 '18 at 9:37
2
\$\begingroup\$

Omit needless spaces

Python tokens only need to separated by a space for

  • A letter followed by a letter
  • A letter followed by a digit

In all other cases, the space can be omitted (with a few exceptions). Here's a table.

  L D S
 +-----
L|s s n
D|n - n
S|n n n    

First token is row, second token is column
L: Letter
D: Digit
S: Symbol

s: space
n: no space
-: never happens (except multidigit numbers)

Examples

Letter followed by letter: Space

not b
for x in l:
lambda x:
def f(s):
x in b"abc"

Letter followed by digit: Space

x or 3
while 2<x:

Letter followed by symbol: No space

c<d
if~x:
x and-y
lambda(a,b):
print"yes"
return[x,y,z]

Digit followed by letter: No space

x+1if x>=0else 2
0in l

(Some versions of Python 2 will fail on a digit followed by else or or.)

Digit followed by digit: Never occurs

Consecutive digits make a multidigit number. I am not aware of any situation where two digits would be separated by a space.

Digit followed by symbol: No space

3<x
12+n
l=0,1,2

A space is needed for 1 .__add__ and other built-ins of integers, since otherwise the 1. is parsed as a float.

Symbol followed by letter: No space

~m
2876<<x&1
"()"in s

Symbol followed by digit: No space

-1
x!=2

Symbol followed by symbol: No space

x*(a+b)%~-y
t**=.5
{1:2,3:4}.get()
"% 10s"%"|"
| improve this answer | |
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  • \$\begingroup\$ In general, you can't have a letter after a digit if it confuses the lexer. A digit followed by e is expected to be a float literal, so something like 1else wouldn't work for versions of python that support exponents in the literal. Similarly, as 0o is the prefix of an octal literal, o can follow any digit but 0. For the complete lexical rules, refer to docs.python.org/2/reference/lexical_analysis.html \$\endgroup\$ – xsot Aug 1 '16 at 5:20
  • \$\begingroup\$ @xsot Do you know what versions of Python will parse this way? I remember a comment thread on this, but I can't find it. \$\endgroup\$ – xnor Aug 1 '16 at 5:51
  • \$\begingroup\$ I've always assumed these rules so I'm not sure which versions of python deviate from them. Possibly the older ones, if any. \$\endgroup\$ – xsot Aug 1 '16 at 6:12
  • \$\begingroup\$ @xsot It works on 2.7.10 but fails on Anachy's 2.7.2. \$\endgroup\$ – xnor Aug 1 '16 at 6:28
2
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Use IDLE 3.3 to take multiline input

In IDLE versions 3.1 to 3.3, the command input() reads an entire multiline string like "line1\nline2", rather than a single line at a time as per the spec. This was fixed in version 3.4.

Calling input() only once is very convenient for golfing. Whether one can take advantage of this is debatable, but I think it is an acceptable interpreter- or environment-specific behavior.

| improve this answer | |
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  • \$\begingroup\$ Wait... so how do you... sort of, enter the string if Enter adds a newline rather than EOF? Does CTRL+D work on Windows? \$\endgroup\$ – cat Mar 4 '16 at 3:00
  • 1
    \$\begingroup\$ @tac I copy-paste it into IDLE. \$\endgroup\$ – xnor Mar 4 '16 at 3:30

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