Python: finding an element in a list [duplicate]

Question

This question already has answers here:

How can I find the index for a given item in a list? (47 answers)

Closed 12 years ago.

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?

• Caveman: Can unmark this one as duplicated, the questions and answers don't handle the case of class instances that match a criteria.
• mikemaccana: Also finding the index of an item in a list is not the same thing as finding the item in a list.

Answer 1

From Dive Into Python:

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

• Kedar.Aitawdekar: But this code gives error when element is not in the list.In current example context if I search for 'three' (i.e: li.index('three')) gives error.
• Peter G: You can catch the error to detect when something isn't in the list. try: li.index("three") except ValueError: found = false
• yogesh mhetre: if you really want to find the index then, first check element in array if True then only do => li.index("example")

Answer 2

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 3

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

Answer 4

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.

• berkus: el = [x for x in mylist if x.attr == "foo"] if el: do something with el[0] solves the "existence" problem
• Thiago Lages de Alencar: el = next((x for x in mylist if x.attr == "foo"), None) if you want to receive a value even when doesn't exist
• Sebastian: [*[label_set for label_set in prediction_label_sets if x == 3], None][0] If you want it to fallback to none, without using an iterator. Not sure if it's simpler or harder.

Answer 5

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

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

• ierdna: doesn't work for multidimensional arrays if you're looking for a specific row, e.g. arr = np.array([[0,0,1],[0,1,0],[1,0,0]]) if you do np.where(arr == [0,1,0]) it doesn't give the right result
• mLstudent33: @ierdna, don't you need to supply axis=0 or axis=1?

Answer 6

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

Answer 7

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 8

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')

Answer 9

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 10

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]

• recursive: Get pythonic: def global_index(lst, test): return (idx for idx, val in enumerate(lst) if test(val) )
• John Fouhy: filter(lambda x: x>3, [1,2,3,4,5,6])