How can I stop a for statement if one of them is true? Python

Question

I'm trying to check if the list have something in common. If they have one print "x" but if they don't print "y". But only one of them, not both. Code:

lista_1 = [1, 2, 4, 5, 6, 7, 8]
lista_2 = [10, 12, 16, 5, 3, 2]
for i in lista_1:
    if i in lista_2:
        print ('Tienen un elemento en común')
        break
        exit()
    if i not in lista_2:
        print ('No tienen ningún elemento en común')

With this way it prints x once and y once, but I want to print only one of them

I think this is just the wrong approach, with `set` operations or `any()`/`all()` you can do this much more efficiently — Chris_Rands, Apr 12 '17 at 08:14
Possible duplicate of [Test if lists share any items in python](http://stackoverflow.com/questions/3170055/test-if-lists-share-any-items-in-python) — Chris_Rands, Apr 12 '17 at 08:15
I've already read that question, but I don't want the same result as he does — Alberto Martínez, Apr 12 '17 at 08:18
To show only once the result, the checking process I've already did it, with this I mean to not show the 2 prints, to first check and then print — Alberto Martínez, Apr 12 '17 at 08:21

Neil · Accepted Answer · 2017-04-12T08:36:12.693

6

You need to iterate through the whole list before you can say that they have no elements in common, so I added a for-else statement. Additionally, I made lista_2 a set to increase efficiency.

lista_1 = [1, 2, 4, 5, 6, 7, 8]
set_2 = {10, 12, 16, 5, 3, 2}
for i in lista_1:
    if i in set_2:
        print('Tienen un elemento en común')
        break
else:
    print('No tienen ningún elemento en común')

edited Apr 12 '17 at 08:36

answered Apr 12 '17 at 08:09

Neil

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If you want this approach, then just use a `for-else` construct, no need for a flag, however, other approaches are better (see my comments on the question) – Chris_Rands Apr 12 '17 at 08:17
But even if they have something in common or not it prints no, why? – Alberto Martínez Apr 12 '17 at 08:17
@almartinez I just tested it and that's not the case. What input are you using? – Neil Apr 12 '17 at 08:17
@nfn neil I'm not using any input, the list are already defined – Alberto Martínez Apr 12 '17 at 08:19
@Chris_Rands Thanks for the input, I updated my code. Never knew that existed, I'll make sure to start using that. – Neil Apr 12 '17 at 08:20
@almartinez code updated, run it through the new code. It works. I"m assuming that Tienen en elmento en comun means Has an element in common. – Neil Apr 12 '17 at 08:21
thats what i wanted. Thanks you very much, the answer is just to put the else out of the for. – Alberto Martínez Apr 12 '17 at 08:23
Also you should make `lista_2` a `set`; this makes the membership testing O(l) rather than O(n), vastly improving speed on large datasets – Chris_Rands Apr 12 '17 at 08:28
@Chris_Rands Updated. – Neil Apr 12 '17 at 08:30
Good, (last comment I promise), probably better to call it `seta_2` now – Chris_Rands Apr 12 '17 at 08:35
@Chris_Rands Updated. – Neil Apr 12 '17 at 08:36
2

@Chris_Rands: I don't think that every spanish word is an english word with a trailing `a` :) – Eric Duminil Apr 12 '17 at 08:47

score 0 · Answer 2 · answered Apr 12 '17 at 08:14

0

The easiest way to check if 2 lists have common elements is to check intersection between the two:

common = list(set(lista_1).intersection(lista_2))
if(now common):
    print("no common elements")

answered Apr 12 '17 at 08:14

Gal Dreiman

3,969
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score 0 · Answer 3 · answered Apr 12 '17 at 08:19

An easier approach would be to use:

(list(set(lista_1) & set(lista_2)))

This will take both lists, convert them to unordered sets that can then be compared between one another with the & (and) operator. The result of this will be a set containing the items that match in both sets, which the outermost list() method will convert back into a list.

And if it's not empty, you've got matches. ;)

score 0 · Answer 4 · answered Apr 12 '17 at 08:22

0

You could do by looking at the intersection of the two corresponding sets:

 if set(lista_1) & set(lista_2):
    print('common elemens')
 else:
    print("no common elements")

answered Apr 12 '17 at 08:22

greole

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score 0 · Answer 5 · answered Apr 12 '17 at 08:45

0

To make the process faster and more concise, you could use a set (conjunto in Spanish, I believe) for fast lookup, and use any() to stop at the first found common element :

#encoding: utf-8

lista_1 = [1, 2, 4, 5, 6, 7, 8]
lista_2 = [10, 12, 16, 5, 3, 2]

conjunto_2 = set(lista_2)

comun = any(elemento_1 in conjunto_2 for elemento_1 in lista_1)

if comun:
    print('Tienen un elemento en común')
else:
    print ('No tienen ningún elemento en común')

answered Apr 12 '17 at 08:45

Eric Duminil

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@almartinez: Yes, it's been solved. My answer is more concise and possibly faster, though. – Eric Duminil Apr 12 '17 at 08:57

score -1 · Answer 6 · edited Apr 12 '17 at 13:43

-1

If you code 2 ifs, it will show both. You must use 1 if and 1 elif. Change the second if to elif or else.

edited Apr 12 '17 at 13:43

khelwood

55,782
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answered Apr 12 '17 at 10:22

Amir

1

How can I stop a for statement if one of them is true? Python

6 Answers6