如何在保持順序的同時根據其中一個串列中的重復專案從2個串列中洗掉專案？

我有 2 個串列：

list_1 = [[1,1], [1,3], [1,1], [1,4]]
list_2 = ["string_1", "string_2", "string_3", "string_4"]

我的目標是洗掉重復的子串列list_1以及其中的字串與洗掉的子串列list_2具有相同的索引，同時保持子串列的順序。

我發現我可以洗掉重復的子串列，同時保持這個 SO中子串列的順序：

from itertools import * 
def unique_everseen(iterable, key=None):
    "List unique elements, preserving order. Remember all elements ever seen."
    seen = set()
    seen_add = seen.add
    if key is None:
        for element in filterfalse(seen.__contains__, iterable):
            seen_add(element)
            yield element
    else:
        for element in iterable:
            k = key(element)
            if k not in seen:
                seen_add(k)
                yield element

list(unique_everseen(list_1, key=frozenset))

但目前尚不清楚如何匹配洗掉的子串列索引list_2

最佳輸出：

new_list_1 = [[1,1], [1,3], [1,4]]
new_list_2 = ["string_1", "string_2", "string_4"]

uj5u.com熱心網友回復：

您可以使用zip：

list_1 = [[1,1], [1,3], [1,1], [1,4]]
list_2 = ["string_1", "string_2", "string_3", "string_4"]

output_1, output_2 = [], []

seen = set()

for sublst, s in zip(list_1, list_2):
    if (tup := tuple(sublst)) in seen:
        continue

    seen.add(tup)

    output_1.append(sublst)
    output_2.append(s)

print(output_1) # [[1, 1], [1, 3], [1, 4]]
print(output_2) # ['string_1', 'string_2', 'string_4']

請注意，您需要將tuple(sublst)（而不是sublst自身）存盤在 setseen中，因為sublstlike[1,1]是不可散列的。

uj5u.com熱心網友回復：

res1 = []
res2 = []

for i in range(len(list_1)):
    if list_1[i] not in list_1[:i]:
        res1.append(list_1[i])
        res2.append(list_2[i])

或者

res1, res2 = map(list, zip(*[(list_1[i], list_2[i]) for i in range(len(list_1)) if list_1[i] not in list_1[:i]]))

uj5u.com熱心網友回復：

您只需要保留一個額外的串列來跟蹤 list_1 中的重復項，然后使用理解串列來縮短結果串列。

list_1 = [[1,1], [1,3], [1,1], [1,4]]
list_2 = ["string_1", "string_2", "string_3", "string_4"]

selector = [True] * len(list_1)
for i in range(len(list_1)):
    for j in range(i):
        if list_1[j] == list_1[i]:
            selector[i] = False
            break

new_list_1 = [element for i, element in enumerate(list_1) if selector[i]]
new_list_2 = [element for i, element in enumerate(list_2) if selector[i]]

uj5u.com熱心網友回復：

解決方案set()和zip

def deduplicate(items):
     seen = set()
     for a, b in items:
         a = tuple(a)
         if not a in seen:
             seen.add(a)
             yield list(a), b

list_1 = [[1,1], [1,3], [1,1], [1,4]]
list_2 = ["string_1", "string_2", "string_3", "string_4"]        
tmp = list(deduplicate(zip(list_1, list_2)))
print([x[0] for x in tmp])
print([x[1] for x in tmp])


# [[1, 1], [1, 3], [1, 4]]
# ['string_1', 'string_2', 'string_4']

uj5u.com熱心網友回復：

另一種基于您的代碼的實作，只需修改您的函式以回傳唯一元素的索引而不是元素本身：

from itertools import *

list_1 = [[1,1], [1,3], [1,1], [1,4]]
list_2 = ["string_1", "string_2", "string_3", "string_4"]
 
def unique_everseen(iterable):
#   Lists indexes of unique elements in iterable, preserving order.
    seen = set()
    seen_add = seen.add
    for element in iterable:
        k = frozenset(element)
        if k not in seen:
            seen_add(k)
            yield iterable.index(element)

unique_indexes = [i for i in unique_everseen(list_1)]
print(unique_indexes)
filtered_list_1 = [list_1[i] for i in unique_indexes]
filtered_list_2 = [list_2[i] for i in unique_indexes]
print(filtered_list_1)
print(filtered_list_2)

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