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Store lookup tables as magic numbers Say you want to hardcode a Boolean lookup table, like which of the first twelve English numbers contain an n. 0: False 1: True 2: False 3: False 4: False 5: False 6: False 7: True 8: False 9: True 10:True 11:True 12:False Then, you can implement this lookup table concisely as: 3714>>i&1 with the resulting 0 ...


Collapse two numerical loops into one Say you're iterating over the cells of an m*n grid. Instead of two nested for loops, one for the row and one of the columns, it's usually shorter to use a single loop to iterate over the m*n cells of the grid. You can extract the row and column of the cell inside the loop. Original code: for i in range(m): for j in ...


For integer n, you can write n+1 as -~n n-1 as ~-n because the bit flip ~x equals -1-x. This uses the same number of characters, but can indirectly cut spaces or parens for operator precedence. Compare: while n-1: #Same as while n!=1 while~-n: c/(n-1) c/~-n or f(n)+1 or-~f(n) (n-1)/10+(n-1)%10 ~-n/10+~-n%10 The operators ~ and unary - are ...


A nice way to convert an iterable to list on Python 3: imagine you have some iterable, like i = (1,2,3,4) i = range(4) i = (x**2 for x in range(5)) But you need a list: x=list(i) #the default way *x,=i #using starred assignment -> 4 char fewer It's very useful to make a list of chars out of a string s=['a','b','c','d','e'] s=list('abcde') *s,='...


This isn't the same solution as llhuii's, but it's also 42 bytes long. n=0;exec'print n;n^=(n^n+2)%3/2;n+=2;'*400 Try it online! Thanks to @JonathanFrech, we're now at 40 bytes. n=0;exec'print n;n=n+2^(n^n+2)/2%3;'*400 Try it online! There's another byte to be saved, for a total of 39. n=0;exec'print n;n=n+2^-(n^n+2)%3;'*400 Try it online!


Instead of range(x), you can use the * operator on a list of anything, if you don't actually need to use the value of i: for i in[1]*8:pass as opposed to for i in range(8):pass If you need to do this more than twice, you could assign any iterable to a variable, and multiply that variable by the range you want: r=1, for i in r*8:pass for i in r*1000:pass ...


You can use the good old alien smiley face to reverse sequences: [1, 2, 3, 4][::-1] # => [4, 3, 2, 1]


Extended iterable unpacking ("Starred assignment", Python 3 only) The best way to explain this is via an example: >>> a,*b,c=range(5) >>> a 0 >>> b [1, 2, 3] >>> c 4 We've already seen a use for this — turning an iterable into a list in Python 3: a=list(range(10)) *a,=range(10) Here are a few more uses. Getting the ...


For ages it bothered me that I couldn't think of a short way to get the entire alphabet. If you use range enough that R=range is worth having in your program, then [chr(i+97)for i in R(26)] is shorter than the naive 'abcdefghijklmnopqrstuvwxyz' , but otherwise it's longer by a single character. It haunted me that the clever one that required some ...


Python 2 - 12 characters print 8**999 The decimal representation of all even numbers from 0 to 100 can be found in the output: ...


43 bytes (4**n+1)**n%4**n**2/n&2**(2*n*n+n)/-~2**n<1 Try it online! The method is similar to Dennis' second (deleted) answer, but this answer is easier to be proved correct. Proof Short form The most significant digit of (4**n+1)**n%4**n**2 in base \$2^n\$ that is not divisible by \$n\$ will make the next (less significant) digit in (4**n+1)**n%...


Python 3 does not have cmp. For golfing, you can do 11 chars (a>b)-(a<b) which loses 3 chars over cmp(a,b). Amusingly, this is also an "official" workaround. The What's New in Python 3 page says "(If you really need the cmp() functionality, you could use the expression (a > b) - (a < b) as the equivalent for cmp(a, b).)"


Although python doesn't have switch statements, you can emulate them with dictionaries. For example, if you wanted a switch like this: switch (a): case 1: runThisCode() break case 2: runThisOtherCode() break case 3: runThisOtherOtherCode() break You could use if statements, or you could use this: ...


When you have two boolean values, a and b, if you want to find out if both a and b are true, use * instead of and: if a and b: #7 chars vs if a*b: #3 chars if either value is false, it will evaluate as 0 in that statement, and an integer value is only true if it is nonzero.


Python 3, 22 (Possibly not allowed) If the challenge is "to create a python script which prints the even numbers from 0 to 100" and not "to create a python script which prints the even numbers from 0 to 100, newline separated", then the shortest solution is: print(*range(0,101,2)) Remember, it's very important in code golf not to put any limitations on ...


No problemo! You can alias a method, but you have to know how to use it: >>> r=str.replace >>> a='hello' >>> r(r(r(r(a,'h','j'),'e','i'),'l','m'),'o','y') 'jimmy' The key is that you have to pass self explicitly, because the alias is a kind of function that takes an extra argument that takes self: >>> type(r) <type '...


Getting 39 bytes This is an explanation of how I got a 39-byte solution, which Dennis and JonathanFrech found separately as well. Or, rather, it explains how one could arrive at the answer in hindsight, in a way that's much nicer than my actual path to it, which was full of muddy reasoning and dead ends. n=0 exec"print n;n=n+2^-(n+2^n)%3;"*400 Writing ...


Exploit Python 2 string representations Python 2 lets you convert an object x to its string representation `x` at a cost of only 2 chars. Use this for tasks that are easier done on the object's string than the object itself. Join characters Given a list of characters l=['a','b','c'], one can produce ''.join(l) as `l`[2::5], which saves a byte. The ...


Use ~ to index from the back of a list If L is a list, use L[~i] to get the i'th element from the back. This is the i'th element of the reverse of L. The bit complement ~i equals -i-1, and so fixes the off-by-one error from L[-i].


Ceil and Floor If you ever want to get the rounded-up result for a division, much like you'd do with // for floor, you could use math.ceil(3/2) for 15 or the much shorter -(-3//2) for 8 bytes. math.floor(n) : 13 bytes+12 for import n//1 : 4 bytes math.ceil(n) : 12 bytes+12 for import -(-n//1) : 8 bytes


12 chars/assignment + 9 chars of overhead V=vars() # do once at the start of the program V["xy"[b]]=z Note that this only works at global scope, it does not work inside a function.


Choosing one of two numbers based on a condition You already know to use the list selection [x,y][b] with a Boolean b for the ternary expression y if b else x. The variables x, y, and b can also be expressions, though note that both x and y are evaluated even when not selected. Here's some potential optimizations when x and y are numbers. [0,y][b] -> y*...


Change import * to import* If you haven't heard, import* saves chars! from math import* is only 1 character longer than import math as m and you get to remove all instances of m. Even one time use is a saver!


Use += instead of append and extend A.append(B) can be shortened to: A+=B, B, here creates a one-element tuple which can be used to extend A just like [B] in A+=[B]. A.extend(B) can be shortened to: A+=B


PEP448 – Additional Unpacking Generalizations With the release of Python 3.5, manipulation of lists, tuples, sets and dicts just got golfier. Turning an iterable into a set/list Compare the pairs: set(T) {*T} list(T) [*T] tuple(T) (*T,) Much shorter! Note, however, that if you just want to convert something to a list and assign it to a variable, ...


Python2 26 i=0;exec"print i;i+=2;"*51 independent discovery of @bitpwner's solution


Total: 104 101 99 chars Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10


Length tradeoff reference I've think it would be useful to have a reference for the character count differences for some common alternative ways of doing things, so that I can know when to use which. I'll use _ to indicate an expression or piece of code. Assign to a variable: +4 x=_;x _ So, this breaks even if you Use _ a second time: _ has length 5 Use ...


Here's a whole bunch of micro-optimisations you can do: Use .split() to create a long list (-17 bytes): c=['yellow','blue','white','green','Black', 'purple', 'silver', 'cyan', 'magenta', 'red'] c='yellow blue white green Black purple silver cyan magenta red'.split() Remove extraneous brackets (-2 bytes): l+=y[0]*(random.randint(n[o],n[o+1])) l+=y[0]*...


54 52 50 49 48 45 39 bytes Removed 4 bytes thanks to Dennis. The latest version is inspired by the "some reason" in xnor's answer. class t(int):__add__=type a=b=t() c=d=0

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