From the documentation: List comprehensions provide a concise way to create lists. Common applications are to make new lists where each element is the result of some operations applied to each member of another sequence or iterable, or to create a subsequence of those elements that satisfy a certain condition. About your question, the list … Read more
You can use reduce (functools.reduce in Python3) if the integer is non-zero >>> x = [False, False, False, 10, False, False] >>> reduce(lambda i,j:i or j, x) 10 You can use a generator expression also here >>> x = [False, False, False, 10, False, False] >>> (i for i in x if i!=False).next() 10 Also … Read more
Avoiding the (explicit) nested loops: sum([[x]*5 for x in range(1,n+1)],[]) (Per @StevenRumbalski’s suggestion, a better approach would be itertools.chain(*[[x]*5 for x in range(1,n+1)]), which yields an iterable of the values you want.) [x for x in range(1,6)]*n
all() Returns true if all of the items in list are True >>> print(all([True, True, True, True])) True >>> print(all([False, True, True, False])) False In above problem we need to check if all vowels are present in a string (ex: businesswoman) using all as follows: >>> all(t in “businesswoman” for t in ‘aeiouu’) True Similarly … Read more
your syntax is slightly off coordinates=[[x,y,z] for x in range(4) for y in range(4) for z in range(4)]
It works like a foreach loop, so foreach (x in [1..3]) { foreach (y in [x .. x * 2]) { yield y; } } First x is 1, so y in [1 .. 2] Then x is 2, so y in [2 .. 4] Then x is 3, so y in [3 .. 6] … Read more
I would suggest to use a set. This contains only unique values: unique=list(set(letters)) print(unique) If you really would like to use a comprehension you can do this: s = set() [i for i in letters if not (i in s or s.add(i))] This works, because s.add() returns None