Python: finding an element in a list [duplicate]

Answer 1

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")

Answer 2

I found this by adapting some tutos. Thanks to google, and to all of you ;)

def findall(L, test):
    i=0
    indices = []
    while(True):
        try:
            # next value in list passing the test
            nextvalue = filter(test, L[i:])[0]

            # add index of this value in the index list,
            # by searching the value in L[i:] 
            indices.append(L.index(nextvalue, i))

            # iterate i, that is the next index from where to search
            i=indices[-1]+1
        #when there is no further "good value", filter returns [],
        # hence there is an out of range exeption
        except IndexError:
            return indices

A very simple use:

a = [0,0,2,1]
ind = findall(a, lambda x:x>0))

[2, 3]

P.S. scuse my english

Answer 3

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

Answer 4

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]