如何在 Python 中的 list/s 中查找特定的 item/s答案

【问题标题】：How to find specific item/s in list/s in Python如何在 Python 中的 list/s 中查找特定的 item/s
【发布时间】：2022-08-13 21:23:29
【问题描述】：

我是新手，我想从 Python 的列表中的列表中找到特定的项目索引。它仅适用于一个列表中的多个列表，我想知道您是否知道很简单的搜索方法，以查找所有现有和新添加的列表以查找项目的特定索引，而不是在两个列表中。我会很乐意提供帮助。

main_list = [[0, 1, 1, 3], [1, \'b\', \'c\', \'d\'], [0, \'b\', 1, \'d\']]

print(f\'MAIN LIST: {main_list}\')

find_in_list = False # put number of list if specific search is needed ### find_in_list = 1
find_item = \'b\' # item that you want to find in one or multiple lists

count_number_of_lists = -1

add_item_index = []
store_items_index = []
item_index_of_store_items_index = 0
list_index_and_item_index = []

for sub_list_index, sub_list_others in enumerate(main_list):

    count_number_of_lists += 1

    if find_in_list == False:
        sub_list = main_list[sub_list_index] # no specific list

        for sub_list_index_2, sub_list_others_2 in enumerate(sub_list):
            if sub_list_others_2 == find_item:  # find exactly same item as needed to be found
                add_item_index.append(sub_list_index_2)
                print(add_item_index)

    else:
        sub_list = main_list[find_in_list] # specific list

        for sub_list_index_2, sub_list_others_2 in enumerate(sub_list):
            if sub_list_others_2 == find_item: # find exactly same item as needed to be found
                add_item_index.append(sub_list_index_2)

    store_items_index.append(add_item_index)
    add_item_index = []
    for sub_index, sub_other in enumerate(store_items_index):
        item_index_of_store_items_index = store_items_index[sub_index]

    list_index_and_item_index.append([count_number_of_lists, item_index_of_store_items_index])
    if find_in_list == False:
        print(f\'LIST INDEX: {count_number_of_lists}, ITEM INDEX: {item_index_of_store_items_index}\')

if find_in_list == False:
    print(f\'List of [LIST INDEX and, [ITEM INDEX]]: {list_index_and_item_index}\') # all LIST INDEX and ITEM INDEX in one list
    print(f\'ALL ITEMS INDEX: {store_items_index}\') # all index of items in one list

else:
    print(f\'LIST INDEX: {count_number_of_lists}, ITEM INDEX: {item_index_of_store_items_index}\')

标签： python list

【解决方案1】：

main_list = [[0, 1, 1, 3], [1, 'b', 'c', 'd'], [0, 'b', 1, 'd']]

find_in_list = False
find_item = 'b'

if not find_in_list:
    list_indices = list(range(len(main_list)))
else:
    list_indices = [find_in_list]

res = []
for i in list_indices:
    occurrences = []
    for j, e in enumerate(main_list[i]):
        if e == find_item:
            occurrences.append(j)
    res.append([i, occurrences])
print(res)

印刷

[[0, []], [1, [1]], [2, [1]]]

【讨论】：

非常感谢，它非常有帮助。我对此仍然很陌生，看看我还需要学习多少才能使我的代码变得更好。谢谢！

【解决方案2】：

如果在子列表中找不到“要查找的元素”，我宁愿显示零。也可以通过列表理解来解决此任务：

lst = [[0, 1, 1, 3], [1, 'b', 'c', 'd'], [0, 'b', 1, 'd']]
to_find = "b"
print([[ind, [sub_lst.count(to_find)]] for ind, sub_lst in enumerate(lst)])

输出：

[[0, [0]], [1, [1]], [2, [1]]]

【讨论】：