问我有一个包含字典的元组列表。如何编辑代码以在这些字典中的单独列表中找到值？
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Stack Overflow用户

提问于 2016-11-21 19:10:30

回答 3查看 99关注 0票数 3

我有一个数据结构L= (int，dict{key: value})，(int，dict{key: value}).

给定输入列表0,1，我希望找到任何字典键，其中输入列表0,1的两个/所有值都存在。

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，其中字典值正好，值为0,1。只有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。我该怎么做才能让这一切发生？

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): []

谢谢您：)

python

dictionary

tuples

回答 3

Stack Overflow用户

回答已采纳

发布于 2016-11-21 19:25:44

您可以使用代码检查value是否以错误的方式包含input_list中的所有项。all(v in input_list for v in value)检查value中的所有项是否可以从input_list中找到。如果你用相反的方式改变它，它就会像你预期的那样起作用：

all(v in value for v in input_list)

注意，如果要将input_list转换为set，可以很容易地检查input_list是否是value的子集。这将更容易理解和更有效率：

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): []

票数 2

Stack Overflow用户

发布于 2016-11-21 19:31:24

您可以使用set减法来解决：

#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))

我建议您避免使用像这个[key for key, value in dict_a.items() if all(v in input_list for v in value)]这样的构造，因为它会使您的代码难以理解。

票数 0

Stack Overflow用户

发布于 2016-11-21 19:24:17

请阅读：

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

根据您在检查是否相等时要使用的语义，您可以使用一种或另一种解决方案。如果您要进行简单的排序列表比较，我将使用以下方法：

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))