Python: finding an element in a list

From Dive Into Python:

>>> li
['a', 'b', 'new', 'mpilgrim', 'z', 'example', 'new', 'two', 'elements']
>>> li.index("example")
5

If you just want to find out if an element is contained in the list or not:

>>> li
['a', 'b', 'new', 'mpilgrim', 'z', 'example', 'new', 'two', 'elements']
>>> 'example' in li
True
>>> 'damn' in li
False

The best way is probably to use the list method .index.

For the objects in the list, you can do something like:

def __eq__(self, other):
    return self.Value == other.Value

with any special processing you need.

You can also use a for/in statement with enumerate(arr)

Example of finding the index of an item that has value > 100.

for index, item in enumerate(arr):
    if item > 100:
        return index, item

Source

assuming you want to find a value in a numpy array, I guess something like this might work:

Numpy.where(arr=="value")[0]

There is the index method, i = array.index(value), but I don't think you can specify a custom comparison operator. It wouldn't be hard to write your own function to do so, though:

def custom_index(array, compare_function):
    for i, v in enumerate(array):
        if compare_function(v):
            return i

The index method of a list will do this for you. If you want to guarantee order, sort the list first using sorted(). Sorted accepts a cmp or key parameter to dictate how the sorting will happen:

a = [5, 4, 3]
print sorted(a).index(5)

Or:

a = ['one', 'aardvark', 'a']
print sorted(a, key=len).index('a')

how's this one?

def global_index(lst, test):
    return ( pair[0] for pair in zip(range(len(lst)), lst) if test(pair[1]) )

Usage:

>>> global_index([1, 2, 3, 4, 5, 6], lambda x: x>3)
<generator object <genexpr> at ...>
>>> list(_)
[3, 4, 5]