Question

res[0]={'k3':org_dict[k3][0],'k2':org_dict[k2][0],'k1':org_dict[k1][0]} 
#res[0] value pos is in filter_data:[0,0,0]

res[1]={'k3':org_dict[k3][0],'k2':org_dict[k2][1],'k1':org_dict[k1][1]}
#res[1] value pos is in filter_data:[0,1,1]

res[2]={'k3':org_dict[k3][1],'k2':org_dict[k2][0],'k1':org_dict[k1][1]}
#res[2] value pos is in filter_data:[1,0,1]

期待结果：

res=[{'k3': 5, 'k2': 3, 'k1': 1},{'k3': 5, 'k2': 2, 'k1': 2},...]

...

例如：

dataType:'JSON'

非常感谢!!!

Answer 1

使用有序词典：

import collections
d = collections.OrderedDict([('k3',[5, 6]), ('k2', [3, 2]), ('k1',[1, 2])])
f_data = [[0, 0, 0], [0, 1, 1], [1, 0, 1], [1, 1, 0]]

迭代过滤器;在循环的顶部创建一个新的空字典;使用zip将过滤器中的索引与目标字典中的项相关联并迭代;将带有过滤值的item键添加到新词典中。

for f in f_data:
    q = collections.OrderedDict()
    for index, (k, v) in zip(f, d.items()):
        q[k] = v[index]
    print(q)

>>>
OrderedDict([('k3', 5), ('k2', 3), ('k1', 1)])
OrderedDict([('k3', 5), ('k2', 2), ('k1', 2)])
OrderedDict([('k3', 6), ('k2', 3), ('k1', 2)])
OrderedDict([('k3', 6), ('k2', 2), ('k1', 1)])
>>>

使用普通字典，您需要一个序列来将键与过滤器匹配：

f_data = [[ 0 ,  0 ,  0 ], [ 0 ,  1 ,  1 ], [ 1 ,  0 ,  1 ], [ 1 ,  1 ,  0 ]]
            |    |    |      |    |    |      |    |    |      |    |    |
          'k3'  'k2' 'k1'  'k3'  'k2' 'k1'  'k3'  'k2' 'k1'  'k3'  'k2' 'k1'

d = dict([('k3',[5, 6]), ('k2', [3, 2]), ('k1',[1, 2])])
key_order = ['k3', 'k2', 'k1']
for f in f_data:
    q = dict()
    for index, key in zip(f, key_order):
        q[key] = d[key][index]
    print(q)

>>>
{'k3': 5, 'k2': 3, 'k1': 1}
{'k3': 5, 'k2': 2, 'k1': 2}
{'k3': 6, 'k2': 3, 'k1': 2}
{'k3': 6, 'k2': 2, 'k1': 1}
>>>

如何使用列表数据过滤dict？

1 个答案: