Question

我有一个字典列表如下

list1 = [{'3': ['0'], '10': ['2'], '9': ['8'], '6': ['8']},
         {'3': ['5'], '9': ['0'], '2': ['3']},
         {'2':['10'],'10':['8'],'4' :['9']}]

我还有另一个列表

list2 = [0,1,2,3,5,6,7,8,9,10]

我想检查list2的每个值是否是list1的任何字典的键，然后如果它与键匹配，那么我希望所有值与这些键相关联并添加这些值。我如何以非常优化的方式在python中实现这一点，因为我的字典列表非常大。

Answer 1

from collections import defaultdict
dct = defaultdict(list)
for x in list1:
    for y,z in x.items():
        dct[int(y)].append(int(z[0]))
for x in list2:
    if x in dct:
        print sum(dct[x])

Answer 2

因为您的词典列表非常大，所以预处理是必要且有效的。

SumKey = {}
for item in list1:
    for key in item.keys():
        if key not in SumKey:
            SumKey[key] = 0
        SumKey[key] += item[key]

SumList = []
for item in list2:
    if item in SumKey.keys():
        SumList.append(SumKey[item])
    else:
        SumList.append(0)

Answer 3

以上答案非常pythonic，可能有点迟钝。

这是一个更具可读性（但不是最高性能！）：

list1=[{'3': ['0'], '10': ['2'], '9': ['8'], '6': ['8']},\
       {'3': ['5'], '9': ['0'], '2': ['3']},\
       {'2':['10'],'10':['8'],'4' :['9']}]

list2 = [0,1,2,3,5,6,7,8,9,10]
strlist2 = [str(x) for x in list2]

total = 0
for subdict in list1:
    filtered = [v for k, v in subdict.items() if k in strlist2]

    print(filtered)

    for sublist in filtered:
        print(sublist)

        for val in sublist:
            total += int(val)

print(total)

Answer 4

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