How to find the last occurrence of an item in a Python list

Question

Say I have this list:

li = ["a", "b", "a", "c", "x", "d", "a", "6"]

As far as help showed me, there is not a builtin function that returns the last occurrence of a string (like the reverse of index ). So basically, how can I find the last occurrence of "a" in the given list?

Answer 1

If you are actually using just single letters like shown in your example, then str.rindex would work handily. This raises a ValueError if there is no such item, the same error class as list.index would raise. Demo:

>>> li = ["a", "b", "a", "c", "x", "d", "a", "6"]
>>> ''.join(li).rindex('a')
6

For the more general case you could use list.index on the reversed list:

>>> len(li) - 1 - li[::-1].index('a')
6

The slicing here creates a copy of the entire list. That's fine for short lists, but for the case where li is very large, efficiency can be better with a lazy approach:

def list_rindex(li, x):
    for i in reversed(range(len(li))):
        if li[i] == x:
            return i
    raise ValueError("{} is not in list".format(x))

One-liner version:

next(i for i in reversed(range(len(li))) if li[i] == 'a')

Answer 2

A one-liner that's like Ignacio's except a little simpler/clearer would be

max(loc for loc, val in enumerate(li) if val == 'a')

It seems very clear and Pythonic to me: you're looking for the highest index that contains a matching value. No nexts, lambdas, reverseds or itertools required.

Answer 3

Many of the other solutions require iterating over the entire list. This does not.

def find_last(lst, elm):
  gen = (len(lst) - 1 - i for i, v in enumerate(reversed(lst)) if v == elm)
  return next(gen, None)

Edit: In hindsight this seems like unnecessary wizardry. I'd do something like this instead:

def find_last(lst, sought_elt):
    for r_idx, elt in enumerate(reversed(lst)):
        if elt == sought_elt:
            return len(lst) - 1 - r_idx

Answer 4

>>> (x for x in reversed(list(enumerate(li))) if x[1] == 'a').next()[0]
6

>>> len(li) - (x for x in enumerate(li[::-1]) if x[1] == 'a').next()[0] - 1
6

Answer 5

I like both wim's and Ignacio's answers. However, I think itertools provides a slightly more readable alternative, lambda notwithstanding. (For Python 3; for Python 2, use xrange instead of range ).

>>> from itertools import dropwhile
>>> l = list('apples')
>>> l.index('p')
1
>>> next(dropwhile(lambda x: l[x] != 'p', reversed(range(len(l)))))
2

This will raise a StopIteration exception if the item isn't found; you could catch that and raise a ValueError instead, to make this behave just like index .

Defined as a function, avoiding the lambda shortcut:

def rindex(lst, item):
    def index_ne(x):
        return lst[x] != item
    try:
        return next(dropwhile(index_ne, reversed(range(len(lst)))))
    except StopIteration:
        raise ValueError("rindex(lst, item): item not in list")

It works for non-chars too. Tested:

>>> rindex(['apples', 'oranges', 'bananas', 'apples'], 'apples')
3

Answer 6

With dict

You can use the fact that dictionary keys are unique and when building one with tuples only the last assignment of a value for a particular key will be used. As stated in other answers, this is fine for small lists but it creates a dictionary for all unique values and might not be efficient for large lists.

dict(map(reversed, enumerate(li)))["a"]

6

Answer 7

I came here hoping to find someone had already done the work of writing the most efficient version of list.rindex , which provided the full interface of list.index (including optional start and stop parameters). I didn't find that in the answers to this question, or here , or here , or here . So I put this together myself... making use of suggestions from other answers to this and the other questions.

def rindex(seq, value, start=None, stop=None):
  """L.rindex(value, [start, [stop]]) -> integer -- return last index of value.
  Raises ValueError if the value is not present."""
  start, stop, _ = slice(start, stop).indices(len(seq))
  if stop == 0:
    # start = 0
    raise ValueError('{!r} is not in list'.format(value))
  else:
    stop -= 1
    start = None if start == 0 else start - 1
  return stop - seq[stop:start:-1].index(value)

The technique using len(seq) - 1 - next(i for i,v in enumerate(reversed(seq)) if v == value) , suggested in several other answers, can be more space-efficient: it needn't create a reversed copy of the full list. But in my (offhand, casual) testing, it's about 50% slower.

Answer 8

last_occurence=len(yourlist)-yourlist[::-1].index(element)-1

就这么简单。无需导入或创建函数。

Answer 9

lastIndexOf = lambda array, item: len(array) - (array[::-1].index(item)) - 1

Answer 10

Love @alcalde's solution, but faced ValueError: max() arg is an empty sequence if none of the elements match the condition.

To avoid the error set default=None :

max((loc for loc, val in enumerate(li) if val == 'a'), default=None)

Answer 11

Use a simple loop:

def reversed_index(items, value):
    for pos, curr in enumerate(reversed(items)):
        if curr == value:
            return len(items) - pos - 1
    raise ValueError("{0!r} is not in list".format(value))

Answer 12

如果列表很小，您可以计算所有索引并返回最大的：

index = max(i for i, x in enumerate(elements) if x == 'foo')

Answer 13

Here is a function for finding the last occurrence of an element in a list. A list and an element are passed to the function.

li = ["a", "b", "a", "c", "x", "d", "a", "6"]
element = "a"

def last_occurrence(li,element):
    for i in range(len(li)-1,0,-1):
        if li[i] == element:
            return i

    return -1

last_occ = last_occurrence(li, element)
if (last_occ != -1):
    print("The last occurrence at index : ",last_occ)
else:
    print("Element not found")

Inside the last_occurrence function a for loop is used with range . which will iterate the list in reverse order. if the element of the current index match the searched element, the function will return the index . In case after comparing all elements in the list the searched element is not found function will return -1 .

Answer 14

def rindex(lst, val):
    try:
        return next(len(lst)-i for i, e in enumerate(reversed(lst), start=1) if e == val)
    except StopIteration:
        raise ValueError('{} is not in list'.format(val))

Answer 15

val = [1,2,2,2,2,2,4,5].

If you need to find last occurence of 2

last_occurence = (len(val) -1) - list(reversed(val)).index(2)

Answer 16

Here's a little one-liner for obtaining the last index, using enumerate and a list comprehension:

li = ["a", "b", "a", "c", "x", "d", "a", "6"]
[l[0] for l in enumerate(li) if l[1] == "a"][-1]

Answer 17

from array import array
fooa = array('i', [1,2,3])
fooa.reverse()  # [3,2,1]
fooa.index(1)
>>> 2