工作解决方案

Question

我有两个用于组装第三个的列表。我将list_one与list_two进行比较，如果list_one中的字段值位于list_two中，则list_two中的值都将复制到list_final中。如果list_two中缺少字段的值，那么我希望看到一个空值（None）放入list_final。 list_final将具有与list_one相同的项目数和顺序：

psutil

list_final的值应为：

list_one = ['one', 'two', 'three', 'four', 'five', 'six', 'seven']
list_two = [['seven','7'], ['five','5'], ['four','4'], ['three','3'], ['one','1']]
list_final = []

我最接近的是：

[['one','1'], [None,None], ['three','3'], ['four','4'], ['five','5'], [None,None], ['seven','7']]

但是只用list_final = [x if [x,0] in list_two else [None,None] for x in list_one]填充list_final。我看了一些教程，但我似乎无法围绕这个概念包围我的脑子。任何帮助将不胜感激。

Answer 1

您的代码中发生了什么：

list_final = [x if [x,0] in list_two else [None,None] for x in list_one]

取list_one并用
x（又名保持原样）[x,0]中存在list_two（但，见下文）
ELSE用[None, None]替换当前元素。

由于list_two不包含与[x,0]匹配的任何元素（在给定示例中以x为准），所有值都将替换为[None, None]。

工作解决方案

list_one = ['one', 'two', 'three', 'four', 'five', 'six', 'seven']
list_two = [['seven','7'], ['five','5'], ['four','4'], ['three','3'], ['one','1']]

# Turns list_two into a nice and convenient dict much easier to work with
# (Could be inline, but best do it once and for all)
list_two = dict(list_two)  # {'one': '1', 'three': '3', etc}

list_final = [[k, list_two[k]] if k in list_two else [None, None] for k in list_one]

我的另一方面：

获取您想要的内容，即[k, dict(list_two)[k]]
但只是尝试这样做k in list_two
ELSE将此条目替换为[None, None]。

Answer 2

你可以试试这个：

list_one = ['one', 'two', 'three', 'four', 'five', 'six', 'seven']
list_two = [['seven','7'], ['five','5'], ['four','4'], ['three','3'], ['one','1']]
final_list = [[None, None] if not any(i in b for b in list_two) else [c for c in list_two if i in c][0] for i in list_one]
print(final_list)

输出：

[['one', '1'], [None, None], ['three', '3'], ['four', '4'], ['five', '5'], [None, None], ['seven', '7']]

Python 3二维列表理解

2 个答案:

工作解决方案