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python - 在python中使用集合和列表创建一个新列表

问题描述

我有以下动态数据，但下面是它可能的示例：

# An incoming List of lists with only ints
# For exampe the incoming_list could be:
incoming_list = [[1,2]. [3]]

# The indexes are like so:
0: [1,2]
1: [3]

然后我有一个 check_list ，其中包含一些示例数据（将是动态的），如下所示：

# A List[int] for check_list
check_list= [3]

然后，如果任何传入列表数据在其索引中，我需要获取incoming_list 上的第一个int：

# If the following were input:
incoming_list = [[1,2]. [3]]
check_list= [3]

# Then a new_list would be:
new_list = [3] because incoming_list has a list with 3 in it, and the 
first element of that list is 3

##############################################################

# Another example ff the following were input:
incoming_list = [[1,2]. [3]]
check_list= [2,3]

# Then a new_list would be:
new_list = [1,3] because incoming_list has a 2 in the first index and 
its first value is 1 and because incoming list has a 3 in the second index 
and its first and only value is 3

我试图用一组列表组合来做到这一点，但我认为列表部分把它搞砸了：

new_list = list(set(v for k, v in incoming_lists if int(k) in check_list))

任何想法如何使这些干净和优雅？

标签： pythonpython-3.xpython-3.7

解决方案

O(1)在一般情况下，将列表预处理为可以在线性时间内完成的查找结构可能是有意义的：

lookup = {}
for lst in reversed(incoming_list):
    for x in lst:
        lookup[x] = lst[0]

然后你可以简单地使用这个：

result = [lookup[x] for x in check_list]

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