Как найти последнее вхождение в список python

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.

There are many ways to find out the first index of element in the list as Python in its language provides index() function that returns the index of first occurrence of element in list. But if one desires to get the last occurrence of element in list, usually a longer method has to be applied. Let’s discuss certain shorthands to achieve this particular task. 

Method #1 : Using join() + rfind() This is usually the hack that we can employ to achieve this task. Joining the entire list and then employing string function rfind() to get the first element from right i.e last index of element in list.

The index of last element occurrence: 10

Time Complexity: O(n), where n is the length of the input list. This is because we’re using join() + rfind() which has a time complexity of O(n) in the worst case.
Auxiliary Space: O(n), as we’re using additional space res other than the input list itself with the same size of input list.

Method #2: Using List Slice + index() Using list slicing we reverse the list and use the conventional index method to get the index of first occurrence of element. Due to the reversed list, the last occurrence is returned rather than the first index of list. 

The index of last element occurrence: 10

Method #3 : Using max() + enumerate() We use enumerate function to get the list of all the elements having the particular element and then max() is employed to get max i.e last index of the list. 

The index of last element occurrence: 10

Time complexity: O(n)

Auxiliary space: O(n), where n is length of list

Method #4 : Using replace() and index() methods

The index of last element occurrence: 10

Time complexity: O(n)

Auxiliary space: O(n), where n is length of list

Method #5 : Using try/except and index():

Here is another approach to finding the last occurrence of an element in a list, using try/except and index():

This approach uses a loop to repeatedly call the index() method on the list, starting from the index after the previous occurrence of val and going to the end of the list. When val is not found, the index() method raises a ValueError, which is caught by the try block and causes the function to return the last index at which val was found. If val is not found in the list, the function returns -1.

Time complexity: O(n)

Auxiliary space: O(1)

Last Updated :
17 Apr, 2023

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In this tutorial, you’ll learn how to use the Python list index method to find the index (or indices) of an item in a list. The method replicates the behavior of the indexOf() method in many other languages, such as JavaScript. Being able to work with Python lists is an important skill for a Pythonista of any skill level. We’ll cover how to find a single item, multiple items, and items meetings a single condition.

By the end of this tutorial, you’ll have learned:

• How the Python list.index() method works
• How to find a single item’s index in a list
• How to find the indices of all items in a list
• How to find the indices of items matching a condition
• How to use alternative methods like list comprehensions to find the index of an item in a list

Python List Index Method Explained

The Python list.index() method returns the index of the item specified in the list. The method will return only the first instance of that item. It will raise a ValueError is that item is not present in the list.

Let’s take a look at the syntax of the index() method:

# The list.index() Method Explained
list.index(
    element,    # The element to search for
    start,      # The index to start searching at
    end         # The index to end searching at
)

Let’s break these parameters down a little further:

• element= represents the element to be search for in the list
• start= is an optional parameter that indicates which index position to start searching from
• end= is an optional parameter that indicates which index position to search up to

The method returns the index of the given element if it exists. Keep in mind, it will only return the first index. Additionally, if an item doesn’t exist, a ValueError will be raised.

In the next section, you’ll learn how to use the .index() method.

Find the Index Position of an Item in a Python List

Let’s take a look at how the Python list.index() method works. In this example, we’ll search for the index position of an item we know is in the list.

Let’s imagine we have a list of the websites we open up in the morning and we want to know at which points we opened 'datagy'.

# Finding the index of an item in a list
a_list = ['datagy', 'twitter', 'facebook', 'twitter', 'tiktok', 'youtube']
print(a_list.index('datagy'))

# Returns: 0

We can see that the word 'datagy' was in the first index position. We can see that the word 'twitter' appears more than once in the list. In the next section, you’ll learn how to find every index position of an item.

Finding All Indices of an Item in a Python List

In the section above, you learned that the list.index() method only returns the first index of an item in a list. In many cases, however, you’ll want to know the index positions of all items in a list that match a condition.

Unfortunately, Python doesn’t provide an easy method to do this. However, we can make use of incredibly versatile enumerate() function and a for-loop to do this. The enumerate function iterates of an item and returns both the index position and the value.

Let’s see how we can find all the index positions of an item in a list using a for loop and the enumerate() function:

# Finding all indices of an item in a list
def find_indices(search_list, search_item):
    indices = []
    for (index, item) in enumerate(search_list):
        if item == search_item:
            indices.append(index)

    return indices

a_list = ['datagy', 'twitter', 'facebook', 'twitter', 'tiktok', 'youtube']
print(find_indices(a_list, 'twitter'))

# Returns: [1, 3]

Let’s break down what we did here:

1. We defined a function, find_indices(), that takes two arguments: the list to search and the item to find
2. The function instantiates an empty list to store any index position it finds
3. The function then loops over the index and item in the result of the enumerate() function
4. For each item, the function evaludates if the item is equal to the search term. If it is, the index is appended to the list
5. Finally, this list is returned

We can also shorten this list for a more compact version by using a Python list comprehension. Let’s see what this looks like:

# A shortened function to return all indices of an item in a list
def find_indices(search_list, search_item):
    return [index for (index, item) in enumerate(search_list) if item == search_item]

a_list = ['datagy', 'twitter', 'facebook', 'twitter', 'tiktok', 'youtube']
print(find_indices(a_list, 'twitter'))

# Returns: [1, 3]

One of the perks of both these functions is that when an item doesn’t exist in a list, the function will simply return an empty list, rather than raising an error.

Find the Last Index Position of an Item in a Python List

In this section, you’ll learn how to find the last index position of an item in a list. There are different ways to approach this. Depending on the size of your list, you may want to choose one approach over the other.

For smaller lists, let’s use this simpler approach:

# Finding the last index position of an item in a list
def find_last_index(search_list, search_item):
    return len(search_list) - 1 - search_list[::-1].index(search_item)

a_list = ['datagy', 'twitter', 'facebook', 'twitter', 'tiktok', 'youtube']

print(find_last_index(a_list, 'twitter'))

# Returns: 3

In this approach, the function subtracts the following values:

• len(search_list) returns the length of the list
• 1, since indices start at 0
• The .index() of the reversed list

There are two main problems with this approach:

1. If an item doesn’t exist, an ValueError will be raised
2. The function makes a copy of the list. This can be fine for smaller lists, but for larger lists this approach may be computationally expensive.

Let’s take a look at another approach that loops over the list in reverse order. This saves the trouble of duplicating the list:

# A less simple, but more memory efficient way of finding the last index of an item
def find_last_index(search_list, search_item):
    i = len(search_list) - 1
    while i >= 0:
        if search_list[i] == search_item:
            return i
        else:
            i -= 1
            
a_list = ['datagy', 'twitter', 'facebook', 'twitter', 'tiktok', 'youtube']

print(find_last_index(a_list, 'twitter'))

# Returns: 3

In the example above we loop over the list in reverse, by starting at the last index. We then evaluate if that item is equal to the search term. If it is we return the index position and the loop ends. Otherwise, we decrement the value by 1 using the augmented assignment operator.

Index of an Element Not Present in a Python List

By default, the Python list.index() method will raise a ValueError if an item is not present in a list. Let’s see what this looks like. We’ll search for the term 'pinterest' in our list:

# Searching for an item that doesn't exist
a_list = ['datagy', 'twitter', 'facebook', 'twitter', 'tiktok', 'youtube']

print(a_list.index('pinterest'))

# Raises: ValueError: 'pinterest' is not in list

When Python raises this error, the entire program stops. We can work around this by nesting it in a try-except block.

Let’s see how we can handle this error:

# Handling an error when an item doesn't exist
a_list = ['datagy', 'twitter', 'facebook', 'twitter', 'tiktok', 'youtube']

try:
    print(a_list.index('pinterest'))
except ValueError:
    print("Item doesn't exist!")

# Returns: Item doesn't exist!

Working with List Index Method Parameters

The Python list.index() method also provides two additional parameters, start= and stop=. These parameters, respectively, indicate the positions at which to start and stop searching.

Let’s say that we wanted to start searching at the second index and stop at the sixth, we could write:

# Using Start and Stop Parameters in list.index()
a_list = ['datagy', 'twitter', 'facebook', 'twitter', 'tiktok', 'youtube']

print(a_list.index('twitter', 2, 6))

# Returns: 3

By instructing the method to start at index 2, the method skips over the first instance of the string 'twitter'.

Finding All Indices of Items Matching a Condition

In this final section, we’ll explore how to find the index positions of all items that match a condition. Let’s say, for example, that we wanted to find all the index positions of items that contain the letter 'y'. We could use emulate the approach above where we find the index position of all items. However, we’ll add in an extra condition to our check:

# Finding Indices of Items Matching a Condition
def find_indices(search_list, search_item):
    return [index for (index, item) in enumerate(search_list) if search_item in item]


a_list = ['datagy', 'twitter', 'facebook', 'twitter', 'tiktok', 'youtube']
print(find_indices(a_list, 'y'))

# Returns:
# [0, 5]

The main difference in this function to the one shared above is that we evaluate on a more “fuzzy” condition.

Conclusion

In this tutorial, you learned how to use the index list method in Python. You learned how the method works and how to use it to find the index position of a search term. You also learned how to find the index positions of items that exist more than once, as well as finding the last index position of an item.

Finally, you learned how to handle errors when an item doesn’t exist as well as how to find the indices of items that match a condition.

Additional Resources

To learn more about related topics, check out the tutorials below:

• Python Lists: A Complete Overview
• Python Zip Lists – Zip Two or More Lists in Python
• Python IndexError: List Index Out of Range Error Explained
• Python List Index: Official Documentation

Списки полезны по-разному по сравнению с другими типами данных из-за их универсальности. В этой статье мы рассмотрим одну из самых распространенных операций со списками — поиск индекса элемента.

Мы рассмотрим различные сценарии поиска элемента, то есть нахождение первого, последнего и всех вхождений элемента. А также что происходит, когда искомого элемента не существует.

Использование Функции index()

Все операции, упомянутые в предыдущем абзаце, можно выполнить с помощью встроенной функции index(). Синтаксис этой функции:

index(element[, start[, end]])

Параметр element, естественно, представляет собой элемент который мы ищем. Параметры start и end являются необязательными и представляют диапазон индексов, в котором мы ищем element.

Значение по умолчанию для start — 0 (поиск с начала), а значение по умолчанию для end — это количество элементов в списке (поиск до конца списка).

Функция возвращает первую позицию element в списке, которую она могла найти, независимо от того, сколько равных элементов осталось после первого вхождения.

Нахождение первого появления элемента

Использование функции index() без установки каких-либо значений для start и end даст нам первое вхождение искомого element:

my_list = ['a', 'b', 'c', 'd', 'e', '1', '2', '3', 'b']

first_occurrence = my_list.index('b')
print("First occurrence of 'b' in the list: ", first_occurrence)

Что даст нам ожидаемый результат:

First occurrence of 'b' in the list: 1

Поиск всех вхождений элемента

Чтобы найти все вхождения элемента, мы можем использовать необязательный параметр start, чтобы мы выполняли поиск только в определенных сегментах списка.

Например, предположим, что первое вхождение элемента в index 3. Чтобы найти следующий, нам нужно будет продолжить поиск первого появления этого элемента после индекса 3. Мы будем повторять этот процесс, меняя место начала поиска, пока мы найдем новые вхождения элемента:

my_list = ['b', 'a', 2, 'n', False, 'a', 'n', 'a']

all_occurrences = []
last_found_index = -1
element_found = True

while element_found:
    try:
        last_found_index = my_list.index('a', last_found_index + 1)
        all_occurrences.append(last_found_index)
    except ValueError:
        element_found = False
    
if len(all_occurrences) == 0:
    print("The element wasn't found in the list")
else:
    print("The element was found at: " + str(all_occurrences))

Запуск этого кода даст нам:

The element was found at: [1, 5, 7]

Здесь нам пришлось использовать блок try, так как функция index() выдает ошибку, когда не может найти указанный element в заданном диапазоне. Это может быть необычно для разработчиков, которые больше привыкли к другим языкам, поскольку такие функции обычно возвращают -1 / null, когда элемент не может быть найден.

Однако в Python мы должны быть осторожны и использовать блок try при использовании этой функции.

Другой, более изящный способ сделать то же самое — использовать понимание списка и полностью игнорировать функцию index():

my_list = ['b', 'a', 2, 'n', False, 'a', 'n', 'a']

all_occurrences = [index for index, element in enumerate(my_list) if element == 'a']

print("The element was found at: " + str(all_occurrences))

Что даст нам тот же результат, что и раньше. У этого подхода есть дополнительное преимущество в том, что он не использует блок try.

Нахождение последнего появления элемента

Если вам нужно найти последнее вхождение элемента в списке, есть два подхода, которые вы можете использовать с функцией index():

1. Переверните список и найдите первое вхождение в перевернутом списке
2. Просмотрите все вхождения элемента и отслеживайте только последнее вхождение

Что касается первого подхода, если бы мы знали первое вхождение element в обратном списке, мы могли бы найти позицию последнего вхождения в исходном. В частности, мы можем сделать это, вычтя reversed_list_index - 1 из длины исходного списка:

my_list = ['b', 'a', 2, 'n', False, 'a', 'n', 'a']

reversed_list_index = my_list[::-1].index('n')
# or alteratively:
# reversed_list_index2 = list(reversed(my_list)).index('n')

original_list_index = len(my_list) - 1 - reversed_list_index

print(original_list_index)

Что даст нам желаемый результат:

Что касается второго подхода, мы могли бы настроить код, который мы использовали, чтобы найти все вхождения, и отслеживать только последнее обнаруженное вхождение:

my_list = ['b', 'a', 2, 'n', False, 'a', 'n', 'a']

last_occurrence = -1
element_found = True

while element_found:
    try:
        last_occurrence = my_list.index('n', last_occurrence + 1)
    except ValueError:
        element_found = False
    
if last_occurrence == -1:
    print("The element wasn't found in the list")
else:
    print("The last occurrence of the element is at: ", last_occurrence)

Что даст нам тот же результат:

Вывод

Мы рассмотрели некоторые из наиболее распространенных способов использования функции index() и способы избежать ошибки в некоторых случаях.

Помните о потенциально необычном поведении функции index(), когда она выдает ошибку вместо возврата -1 / None, когда элемент не найден в списке.

13 ответов

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

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

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

next(i for i in range(len(li)-1, -1, -1) if li[i] == 'a')

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

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

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

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

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")

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

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

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

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

6

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)

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

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))

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

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

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))