Selection elements of a list based on another 'True'/'False' list

问题

I have two lists of the same length. The first one contains strings. The second one - strings that can be either 'True' or 'False'.

If the nth element of the second list is 'True', I want to append the nth element of the first list to another list.

So if I have:

List1:

('sth1','sth2','sth3','sth4') 

List2:

('True','False','True','False')

The outcome should be List3:

('sth1','sth3').

How can I intersect two list in that way?

回答1:

Use zip:

result = [x for x, y in zip(xs, ys) if y == 'True']

Example:

xs = ('sth1','sth2','sth3','sth4')
ys = ('True','False','True','False')
result = [x for x, y in zip(xs, ys) if y == 'True']
result
['sth1', 'sth3']

回答2:

Or use numpy:

import numpy as np
filtered = np.array(List1)[np.array(List2)]

Btw this only works if the elements inside List2 are True/False, not if they are "True"/"False".

回答3:

If you didn't know about zip :

l1 = ('sth1','sth2','sth3','sth4')
l2 = ('True','False','True','False')
l = [x for i,x in enumerate(l1) if l2[i]=='True']
print l
#=> ['sth1', 'sth3']

It would be shorter with a tuple of booleans :

l1 = ('sth1','sth2','sth3','sth4')
l2 = (True,False,True,False)

l = [x for i,x in enumerate(l1) if l2[i]]
print l

回答4:

Simplest way is to use itertools.compress method as follows.

import itertools

list1 = ('sth1','sth2','sth3','sth4')
list2 = ('True','False','True','False')
list2 = map(lambda x: x == 'True', list2)

result = list(itertools.compress(list1, list2))

compress method returns an iterator, so you that is why you need to wrap the iterator object in list

I hope it helps.

来源：https://stackoverflow.com/questions/41762592/selection-elements-of-a-list-based-on-another-true-false-list