Question

我有一个数据结构L = [（int，dict {key：values}），（int，dict {key：values}）...]。

给定输入列表[0,1]我想找到任何字典键，其中输入列表[0,1]的两个/所有值都存在。< / p>

目前，我的错误是如果我使用input_list = [0,1]，该函数将返回一个匹配，其中字典值只是[0]，其中值是[0,1]。只有这第二个结果是可取的。我觉得这是一个微小的变化，但我无法理解。为实现这一目标，我该做些什么改变？

代码

#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))

输出

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): []

谢谢：）

Answer 1

如果value以错误的方式包含input_list中的所有项目，您需要检查代码。 all(v in input_list for v in value)检查value中是否可以找到input_list中的所有项目。如果你改变它，反过来它将按预期工作：

all(v in value for v in input_list)

请注意，如果您要将input_list转换为set，则可以轻松检查input_list是value的子集。这将更容易理解和更有效：

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))

输出：

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来解决：

#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))

我建议你避免使用像[key for key, value in dict_a.items() if all(v in input_list for v in value)]这样的结构，因为它会让你的代码难以理解。

Answer 3

请阅读：

Check if two unordered lists are equal 和How can I compare two ordered lists in python?

根据您在检查相等性时要使用的语义，您可以使用一种解决方案或另一种解决方案。如果您要进行简单的有序列表比较，我会选择：

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))

我有一个包含字典的元组列表。如何在这些词典中编辑我的代码以在单独的列表中查找值？

3 个答案: