Python列表列表，刪除重復項

Question

嘗試刪除列表列表中的重復項並打印不重復的列表。

原始清單

a = [['country',['America_1','America_2','America_3','America_4','England_5','England_6'],['apple_1_more','orange_1_more']],['country',['Brazil_2','Brazil_3','Brazil_1','Brazil_4','Mexico_1','Mexico_3','Mexico_2'],['grapes_1_less','banana_1_more']]]

尋找輸出：

[['country', ['America', 'England'], ['orange_more', 'apple_more']], ['country', ['Mexico', 'Brazil'], ['grapes_less', 'banana_more']]]

但得到：

[['country', ['America', 'England'], ['orange_more', 'apple_more']], ['country', ['America', 'England', 'Mexico', 'Brazil'], ['orange_more', 'grapes_less', 'banana_more', 'apple_more']]]

碼：：

 a = [['country',['America_1','America_2','America_3','America_4','England_5','England_6'],['apple_1_more','orange_1_more']],['country',['Brazil_2','Brazil_3','Brazil_1','Brazil_4','Mexico_1','Mexico_3','Mexico_2'],['grapes_1_less','banana_1_more']]]
aa ={}
aaa=[]
aaaa=[]
aaaaa=[]
for i in a:
    for j in i[1]:
        j=j.split('_',1)[0]
        aaa.append(j)
    for k in i[2]:
        k=k.split('_',2)[0]+'_'+k.split('_',2)[2]
        aaaa.append(k)
    aa['country'] = [i[0],list(set(aaa)),list(set(aaaa))]
    aaaaa.append(aa['country'])
print (aaaaa)

Answer 1

使用列表推導，將每個子列表中的第二項與set() ：

a = [['country',['America','America','America','America','England','England']],['country',['Brazil','Brazil','Brazil','Brazil','Mexico','Mexico','Mexico']]]

a = [[i, list(set(j))] for i, j in a]
print(a)

輸出：

[['country', ['England', 'America']], ['country', ['Brazil', 'Mexico']]]

由於集合是無序的，因此這可能不會保留內部列表的順序，因此您可能需要考慮這一點。

Answer 2

您可以嘗試以下方法：

a = [['country',['America','America','America','America','England','England']],['country',['Brazil','Brazil','Brazil','Brazil','Mexico','Mexico','Mexico']]]



print(list(map(lambda x:[x[0],list(set(x[1:][0]))],a)))

輸出：

[['country', ['England', 'America']], ['country', ['Mexico', 'Brazil']]]

您的變量名稱非常混亂，仍然嘗試了新方法，您可以嘗試以下方法：

a = [['country',['America_1','America_2','America_3','America_4','England_5','England_6'],['apple_1_more','orange_1_more']],['country',['Brazil_2','Brazil_3','Brazil_1','Brazil_4','Mexico_1','Mexico_3','Mexico_2'],['grapes_1_less','banana_1_more']]]


final_data=[]
for i in a:
    sub_data=[]

    for j in i[1:]:
        d = {}

        for m in j:
            data=m.split('_')[0]
            d[data]=data

        sub_data.append(list(d.keys()))
    final_data.append(['country',*sub_data])
print(final_data)

輸出：

[['country', ['America', 'England'], ['orange', 'apple']], ['country', ['Brazil', 'Mexico'], ['banana', 'grapes']]]

如果您的數據格式總是這樣，則可以嘗試以下操作：

更新

a = [['country',['America_1','America_2','America_3','America_4','England_5','England_6'],['apple_1_more','orange_1_more']],['country',['Brazil_2','Brazil_3','Brazil_1','Brazil_4','Mexico_1','Mexico_3','Mexico_2'],['grapes_1_less','banana_1_more']]]


final_data=[]
for i in a:
    sub_data=[]
    sub_extra=[]

    for j in i[1:2]:
        sub_extra.append(i[2])
        d = {}

        for m in j:
            data=m.split('_')[0]
            d[data]=data

        sub_data.extend([list(d.keys()),*sub_extra])
    final_data.append(['country',*sub_data])
print(final_data)

輸出：

[['country', ['America', 'England'], ['apple_1_more', 'orange_1_more']], ['country', ['Mexico', 'Brazil'], ['grapes_1_less', 'banana_1_more']]]

Answer 3

使用此遞歸函數可刪除多級數組中的重復項：

def dup(input_):
    if isinstance(input_, list):
        try:
            input_ = list(set([i.split('_')[0] if not isinstance(i, list) else i for i in input_]))
        except TypeError:
            pass
        for child in input_:
            input_[input_.index(child)] = dup(child)

    return input_

Answer 4

這就是我要做的。

country_list1 = [a[0[0]]]
country_list2 = [a[1[0]]]
duplicates = [country for country in country_list1 in country_list2]
non_duplicates = [country for country in country_list1 not in country_list2]

這將為您提供重復的名稱和非重復的名稱這是考慮到兩個名稱中的名稱區分大小寫