Question

这可能很简单，但我无法理解它。

我有一个列表，例如：

list_a = [['Peter', '2016'],['David', '2020'],['Ulrik', '2018'],['Lars', '2017'],['Dave', '2019'],['Ann', '2015']]

然后用户可以选择一些项目，比如 ['Peter', '2016'] 和 ['Lars', '2017']。然后将选择存储为索引，如：

user_selection = [0,3]

现在，我希望对 list_a 进行排序：

list_a.sort(key=itemgetter(1), reverse=False)

该列表现在按日期排序，如下所示：

list_a = [['Ann', '2015'], ['Peter', '2016'], ['Lars', '2017'], ['Ulrik', '2018'], ['Dave', '2019'],['David', '2020']]

但是，正如您从我的标题中猜到的，user_selection 现在是“错误的”，指的是用户没有选择的项目。我如何“更新”选择（或将其与列表一起排序）使其变为： user_selection = [1,2] ?

Answer 1

重做我的答案：

这里有一个完整的例子：

list_a = [['Peter', '2016'],['David', '2020'],['Ulrik', '2018'],['Lars', '2017'],['Dave', '2019'],['Ann', '2015']]

这是您的清单。

如果 user_selection 现在关注：

user_selection = [0,3]

您可以执行以下操作：

user_selected_lists = [] # a list where we safe the inputs from list_a which the user selected

for selected in user_selection:
   user_selected_lists.append(list_a[selected]) # this takes the item at the index the user selected and safes it into the `user_selected_lists`

# Now you have the both lists the user selected in the `user_selected_lists`

new_indexes = [] # the new indexes of the selected lists
for selected_list in user_selected_lists:
    new_indexes.append(list_a.index(selected))  # this gets the new index from the new list of each list we safed from the first list.

然后就可以了。没试过，但逻辑应该可行。

或者简单的方法。

list_a_duplicate = list_a    # make a duplicate from list_a

list_a.sort()   # sort the list as you wish

# If you want now to know what index 0 and 3 are NOW after sorting you can just do following.

old_indexes_now = []   # the new indexes
for index in user_selection:
   list_item = list_a_duplicate[index] # get each item of the old list
   new_index = list_a.index(list_item) # get the index of the item from the new list
   old_indexes_now.append(new_index)

我希望你现在可以理解它。

排序列表后如何保留选择/获取新索引？

1 个答案: