Python: finding an element in a list [duplicate]

Question

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

Here is another way using list comprehension (some people might find it debatable). It is very approachable for simple tests, e.g. comparisons on object attributes (which I need a lot):

el = [x for x in mylist if x.attr == "foo"][0]

Of course this assumes the existence (and, actually, uniqueness) of a suitable element in the list.

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

I use function for returning index for the matching element (Python 2.6):

def index(l, f):
     return next((i for i in xrange(len(l)) if f(l[i])), None)

Then use it via lambda function for retrieving needed element by any required equation e.g. by using element name.

element = mylist[index(mylist, lambda item: item["name"] == "my name")]

If i need to use it in several places in my code i just define specific find function e.g. for finding element by name:

def find_name(l, name):
     return l[index(l, lambda item: item["name"] == name)]

And then it is quite easy and readable:

element = find_name(mylist,"my name")