Compare list in python to detect an equality

Question

First, I am a novice at python programming and attempted much research within other questions but none that I could find that relate to something like this (all others were a bit more advanced) --- That said moving on. The solution needed: Go through two two integer lists and compare for equality. Ideally I want it to continue going through the lists over and over till there is an equality (more on this after showing code). The number will be generated in `list2` over and over till there is an equality. Explanation to code: I have two lists that are generated via a random number generation. The lists are not equal in size. So `list1` has 500 entries and `list2` will have different amounts varying from 1 to 100. ``` #current attempt to figure out the comparison. if (list1 = list2): print(equalNumber) ``` Maybe I do not know much about looping, but I want it to loop through the list, I really do not know where to start from. Maybe I'm not using a loop like a for loop or while? This is m…

Accepted Answer

There are several possible interpretations of your question.

1) Loop over the lists pairwise, stopping when a pair is equal:

```
>>> s = [10, 14, 18, 20, 25]
>>> t = [55, 42, 18, 12, 4]
>>> for x, y in zip(s, t):
        if x == y:
            print 'Equal element found:', x
            break

Equal element found: 18
```

2) Loop over a list, stopping when any element is equal to any other element in the first list. This is a case where sets are useful (they do fast membership testing):

```
>>> s = {18, 20, 25, 14, 10}
>>> for x in t:
        if x in s:
            print 'Equal element found', x
            break

Equal element found 18
```

3) Loop over both like element-wise and compare their values:

```
>>> s = [10, 14, 18, 20, 25]
>>> t = [55, 42, 18, 12, 4]
>>> [x==y for x, y in zip(s, t)]
[False, False, True, False, False]
```

Compare list in python to detect an equality

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