从两个二维列表中删除重复项 [英] removing duplicates from two 2 dimensional list

问题描述

我正在寻找一种解决方案，以从python中的两个2​​d列表中删除重复项，所以找不到，所以这是我的问题:
我有两个列表，例如

i searched for a solution to remove deplicates from two 2d list in python i couldn't find so here my question:
i have two lists, for example

[[1,2],[3,5],[4,4],[5,7]]

[[1,3],[4,4],[3,5],[3,5],[5,6]]

预期结果:

[[1,2],[1,3],[5,7],[5,6]]

我要删除与其他列表的值完全匹配的列表中的列表.

I want to remove list inside on the lists that match EXACTLY the values of the other list.

我的脚本:

def filter2dim(firstarray, secondarray):
    unique = []
    for i in range(len(firstarray)):
       temp=firstarray[i]
       for j in range(len(secondarray)):
           if(temp == secondarray[j]):
              break
           elif(j==(len(secondarray)-1)):
               unique.append(temp)
    for i in range(len(secondarray)):
       temp=secondarray[i]
       for j in range(len(firstarray)):
           if(temp == firstarray[j]):
              break
           elif(j==(len(firstarray)-1)):
               unique.append(secondarray[i])
    return 

如果您修复它并解释您的操作，将非常有用. 谢谢，问候

Please if you fix it and explain what you did it will be greateful. Thank you, Best Regards

推荐答案

用元组替换您的2项列表，并且您可以使用set操作(因为元组是不可变的，列表不是不可变的，并且set项必须是不可变的):

Replace your 2-item lists with tuples and you can use set operations (because tuples are immutable and lists not, and set items must be immutable):

a = {(1,2),(3,5),(4,4),(5,7)}
b = {(1,3),(4,4),(3,5),(3,5),(5,6)}
print(a.symmetric_difference(b)) # {(1, 2), (1, 3), (5, 6), (5, 7)}

请注意，这也会删除每个列表中的重复项，因为它们是已设置的，并且顺序将被忽略.

Note this also removes duplicates within each list because they are sets, and order is ignored.

如果您需要以编程方式将列表转换为元组，则列表理解就可以了:

If you need to programatically convert your lists into tuples, a list comprehension works just fine:

list_a = [[1,2],[3,5],[4,4],[5,7]]
set_a = {(i, j) for i, j in a}
print(set_a) # {(1, 2), (4, 4), (5, 7), (3, 5)}   

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