我有一個包含字典的元組列表。 如何在這些詞典中編輯我的代碼以在單獨的列表中查找值？

Question

我有一個數據結構L = [（int，dict {key：values}），（int，dict {key：values}）...]。

給定輸入列表[0,1]我想找到任何字典鍵，其中存在輸入列表[0,1]的兩個/所有值。

目前，我的錯誤是，如果我使用input_list = [0,1]，該函數將返回一個匹配，其中字典值只是[0]，其中值是[0,1]。 只有這第二個結果是可取的。 我覺得這是一個微小的變化，但我無法理解。 為了實現這一目標，我該怎么做？

碼

#Python3
L = [(0, {0: [0], 1: [0, 1], 2: [0, 2], 3: [0, 3], 4: [0, 4]}), (1, {0: [1, 0], 1: [1], 2: [5, 1, 2,], 3: [1, 3], 4: [1, 4]}), (2, {0: [2, 0], 1: [2, 1], 2: [2], 3: [2, 3], 4: [2, 4]}), (3, {0: [3, 0], 1: [3, 1], 2: [3, 2], 3: [3], 4: [3, 4]}), (4, {0: [4, 0], 1: [4, 1], 2: [4, 2], 3: [4, 3], 4: [4]})]

#input_list = (eval(input('Enter your list: ')))
#input_list = ([0,1])
print('Input: ' + str(input_list))
for tupl in L:
    dict_a = (tupl[1])
    matching_key = ([key for key, value in dict_a.items() if all(v in input_list for v in value)])
    print('Node: ' + str(tupl[0]) + ' Match at key(s): ' + str(matching_key))

產量

L = [(0, {0: [0], 1: [0, 1], 2: [0, 2], 3: [0, 3], 4: [0, 4]}), (1, {0: [1, 0], 1: [1], 2: [5, 1, 2,], 3: [1, 3], 4: [1, 4]}), (2, {0: [2, 0], 1: [2, 1], 2: [2], 3: [2, 3], 4: [2, 4]}), (3, {0: [3, 0], 1: [3, 1], 2: [3, 2], 3: [3], 4: [3, 4]}), (4, {0: [4, 0], 1: [4, 1], 2: [4, 2], 3: [4, 3], 4: [4]})]

Enter your list: [0,1]
Input: [0, 1]
Node: 0 Match at key(s): [0, 1]
Node: 1 Match at key(s): [0, 1]
Node: 2 Match at key(s): []
Node: 3 Match at key(s): []
Node: 4 Match at key(s): []

Enter your list: [1,5,2]
Input: [1, 5, 2]
Node: 0 Match at key(s): []
Node: 1 Match at key(s): [1, 2]
Node: 2 Match at key(s): [1, 2]
Node: 3 Match at key(s): []
Node: 4 Match at key(s): []

謝謝：）

Answer 1

你有代碼檢查value包含input_list的所有項目。 all(v in input_list for v in value)檢查可以從input_list找到value中的所有項。 如果你改變它，反過來它將按預期工作：

all(v in value for v in input_list)

請注意，如果要將input_list轉換為set ，則可以輕松檢查input_list是否為value的子集。 這將更容易理解，更有效：

L = [(0, {0: [0], 1: [0, 1], 2: [0, 2], 3: [0, 3], 4: [0, 4]}), (1, {0: [1, 0], 1: [1], 2: [5, 1, 2,], 3: [1, 3], 4: [1, 4]}), (2, {0: [2, 0], 1: [2, 1], 2: [2], 3: [2, 3], 4: [2, 4]}), (3, {0: [3, 0], 1: [3, 1], 2: [3, 2], 3: [3], 4: [3, 4]}), (4, {0: [4, 0], 1: [4, 1], 2: [4, 2], 3: [4, 3], 4: [4]})]

input_list = set([0,1])

for tupl in L:
    dict_a = tupl[1]
    matching_key = [key for key, value in dict_a.items() if input_list <= set(value)]
    print('Node: ' + str(tupl[0]) + ' Match at key(s): ' + str(matching_key))

輸出：

Node: 0 Match at key(s): [1]
Node: 1 Match at key(s): [0]
Node: 2 Match at key(s): []
Node: 3 Match at key(s): []
Node: 4 Match at key(s): []

Answer 2

您可以使用set subtraction來解決：

#Python3
L = [(0, {0: [0], 1: [0, 1], 2: [0, 2], 3: [0, 3], 4: [0, 4]}), (1, {0: [1, 0], 1: [1], 2: [5, 1, 2,], 3: [1, 3], 4: [1, 4]}), (2, {0: [2, 0], 1: [2, 1], 2: [2], 3: [2, 3], 4: [2, 4]}), (3, {0: [3, 0], 1: [3, 1], 2: [3, 2], 3: [3], 4: [3, 4]}), (4, {0: [4, 0], 1: [4, 1], 2: [4, 2], 3: [4, 3], 4: [4]})]

input_list = (eval(input('Enter your list: ')))
#input_list = [1,5,2]
print('Input: ' + str(input_list))
for tupl in L:
    dict_a = (tupl[1])
    matching_key = []
    for key, lst in tupl[1].items():
        if not (set(lst) - set(input_list)):
            matching_key.append(key)
    print('Node: ' + str(tupl[0]) + ' Match at key(s): ' + str(matching_key))

我建議你避免像這樣的構造[key for key, value in dict_a.items() if all(v in input_list for v in value)]因為它混淆了，它使你的代碼難以理解。

Answer 3

請閱讀：

檢查兩個無序列表是否相等 ， 如何在python中比較兩個有序列表？

根據您在檢查相等性時要使用的語義，可以使用一種解決方案或另一種解決方案。 如果你想進行簡單的有序列表比較，我會選擇：

for tupl in L:
    dict_a = (tupl[1])
    if dict_a != input_list:
        continue
    print('Node: ' + str(tupl[0]) + ' Match at key(s): ' + str(matching_key))