Python: finding an element in a list [duplicate]

What is a good way to find the index of an element in a list in Python?
Note that the list may not be sorted.

Is there a way to specify what comparison operator to use?

2

10 Answers

From Dive Into Python:

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

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

0

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.

3

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

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

2

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")

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]

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