CODEWITHSUNDEEP
python
bnicholl
bnicholl

Reputation: 326

How to check if an integer at a given location is greater than every other integer within that list?

I am attempting to see if an integer at a given location is greater than every other integer within that list. For example:

values = [2,5,3,1,6]
if values[0] < all other integers
    print something

Keep in mind, I need to see if that specific index is less than all of the other indexes in the list, therefore, using something like min(values) will not work. A list such as

values = [2,5,3,1,6,1]

has no single minimum value; any given index should return False.

Upvotes: 0

Views: 68

Answers (3)

dawg
dawg

Reputation: 103844

You can use any to assert if any one item in the list meets a condition. Just skip the one entry in question like so:

def f(li, idx):
    return any(e>li[idx] for i, e in enumerate(li) if i!=idx)   

>>> f([2,5,3,1,6], 0)
True
>>> f([2,5,3,1,6], 4)
False

You can reverse < to > or whatever to fit your use. (Or add a not)

If you want to assert that the given index has a relationship with all other list elements, use all:

def f2(li, idx):
    return all(e>li[idx] for i, e in enumerate(li) if i!=idx)

>>> f2([2,5,3,1,6,1], 3)
False

Upvotes: 3

martineau
martineau

Reputation: 123463

You could use the built-in any() function like this:

values = [2,5,3,1,6]
loc = 4
if not any((values[i] > values[loc]) for i in range(len(values)) if i != loc):
    print('something')

Upvotes: 1

Prune
Prune

Reputation: 77847

Use the all operator to iterate over a sequence. In this case, you also have to eliminate checking against self. The boolean expression would be:

>>> values = [2,5,3,1,6]
>>> given_loc = 0

>>> all ([values[given_loc] < values[i] \
        for i in range(len(values)) \
            if i != given_loc])
False
>>> given_loc = 3
>>> all ([values[given_loc] < values[i] for i in range(len(values)) if i != given_loc])
True

Upvotes: 1

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