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?
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2Can unmark this one as duplicated, the questions and answers don't handle the case of class instances that match a criteria.Caveman– Caveman2020-06-22 16:15:39 +00:00Commented Jun 22, 2020 at 16:15
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1Also finding the index of an item in a list is not the same thing as finding the item in a list.mikemaccana– mikemaccana2021-02-05 14:28:48 +00:00Commented Feb 5, 2021 at 14:28
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10 Answers
From Dive Into Python:
>>> li
['a', 'b', 'new', 'mpilgrim', 'z', 'example', 'new', 'two', 'elements']
>>> li.index("example")
5
3 Comments
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")
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
Comments
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
Comments
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 Comments
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.assuming you want to find a value in a numpy array, I guess something like this might work:
Numpy.where(arr=="value")[0]
2 Comments
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?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
Comments
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")
Comments
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')
Comments
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
Comments
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]
answered Mar 3, 2009 at 2:06
SingleNegationElimination
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2 Comments
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])