根据每个嵌套列表中的第二个元素从嵌套中对元素进行排序? [英] Sorting elements from a nested based on the second element in each nested list?

问题描述

因此，我有一个嵌套列表，其中包含单词和数字，如以下示例所示:

So I have a nested list that contains words and numbers like the following example:

nested_list = [['This', 1],['is' , 2],['a', 3],['list', 4]]

我还有一个数字列表:

number_list = [2,3]

我想根据天气生成两个嵌套列表，列表的第二个元素在数字列表中包含一个数字.

I want to generate a two nested lists based on weather the second element of the list contains a number in the list of numbers.

我想输出为:

list1 = [['is', 2],['a', 3]] #list one has values that matched the number_list
list2 = [['This', 1],['list', 4]] #list two has values that didn't match the number_list

我正在使用for循环遍历列表，但我希望有更好的方法.

I was using a for loop to iterate through the list but I was hoping that there was a better way.

推荐答案

使用两个列表理解:

>>> nested_list = [['This', 1],['is' , 2],['a', 3],['list', 4]]
>>> number_list = [2,3]
>>> list1 = [item for item in nested_list if item[1] in number_list]
>>> list2 = [item for item in nested_list if item[1] not in number_list]
>>> list1
[['is', 2], ['a', 3]]
>>> list2
[['This', 1], ['list', 4]]

使用字典(仅需要单次迭代):

Using a dict( only single iteration is required):

>>> dic = {'list1':[], 'list2':[]}
for item in nested_list:
    if item[1] in number_list:
        dic['list1'].append(item)
    else:
        dic['list2'].append(item)
...         
>>> dic['list1']
[['is', 2], ['a', 3]]
>>> dic['list2']
[['This', 1], ['list', 4]]

如果number_list很大，则先将其转换为set以提高效率.

If number_list is huge then convert it to a set first to improve efficiency.

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