python中的組合filter（）無法按預期工作

Question

據我了解，以下代碼應輸出[['b']] 。 而是輸出[['a', 'exclude'], ['b']] 。

這是python中的錯誤，還是我誤會了某些東西？

lists_to_filter = [
    ['a', 'exclude'],
    ['b']
]
# notice that when 'exclude' is the last element, the code returns the expected result
for exclude_label in ['exclude', 'something']:
    lists_to_filter = (labels_list for labels_list in lists_to_filter if exclude_label not in labels_list)
    # notice that changing the line above to the commented line below (i.e. expanding the generator to a list) 
    # will make the code output the expected result, 
    # i.e. the issue is only when using filter on another filter, and not on a list
    # lists_to_filter = [labels_list for labels_list in lists_to_filter if exclude_label not in labels_list]
lists_to_filter = list(lists_to_filter)
print(lists_to_filter)

Answer 1

發生這種情況是因為lists_of_filter僅在循環外部進行迭代。 在循環之外，您具有exclude_label == 'something' ，這就是為什么您得到意外結果的原因。 要檢查它，您可以在一行中添加exclude_label = 'exclude' ：

lists_to_filter = [
    ['a', 'exclude'],
    ['b']
]

for exclude_label in ['exclude', 'something']:
    lists_to_filter = (labels_list for labels_list in lists_to_filter if exclude_label not in labels_list)

exclude_label = 'exclude'
lists_to_filter = list(lists_to_filter)
print(lists_to_filter)  # [['b']]

生成器表達式的文檔說：“ 后續的for子句和最左邊的for子句中的任何過濾條件都不能在封閉范圍內進行評估，因為它們可能取決於從最左邊的iterable獲得的值。 ” 在您的情況下if exclude_label ...的過濾條件取決於for exclude_label in ...循環中從for exclude_label in ...獲得的值。

Answer 2

因為您在循環中多次分配了lists_to_filter ，所以它只會返回最后的結果。 不包括'something' ， ['exclude', 'something']的最后一個元素

您可以使用all來實現您的目標：

lists_to_filter = [labels_list for labels_list in lists_to_filter if all(exclude_label not in labels_list for exclude_label in ['exclude', 'something'])]

或展開all邏輯：

    result = []

    for labels_list in lists_to_filter:
        exclude = False
        for exclude_label in ['exclude', 'something']:
            if exclude_label in labels_list:
                exclude = True
        if not exclude:
            result.append(labels_list)

    print(result)

或者您可以根據標題使用filter ：

lists_to_filter = list(filter(lambda x: all(exclude_label not in x for exclude_label in exclude_labels), lists_to_filter))

輸出：

[['b']]

希望對您有所幫助，如有其他問題，請發表評論。 :)