Question

我有这样的字典，

data = {'04-01-2012': [{1: 0.93}, {2: 0.9195000000000001}, {3: 0.9090000000000001}, {4: 0.8985000000000002},
                       {5: 0.8880000000000002}, {6: 0.8775000000000003}, {7: 0.8670000000000003},
                       {8: 0.8565000000000004}, {9: 0.8460000000000004}],
        '12-01-2012': [{1: 0.96}],
        '07-01-2012': [{1: 0.96}, {2: 0.95}, {3: 0.94}, {4: 0.9299999999999999}, {5: 0.9199999999999999},
                       {6: 0.9099999999999999}],
        '06-01-2012': [{1: 0.945}, {2: 0.9365}, {3: 0.928}, {4: 0.9195000000000001}, {5: 0.9110000000000001},
                       {6: 0.9025000000000002}, {7: 0.8940000000000002}],
        '10-01-2012': [{1: 0.93}, {2: 0.9244}, {3: 0.9188}],
        '05-01-2012': [{1: 0.935}, {2: 0.926}, {3: 0.917}, {4: 0.908}, {5: 0.899}, {6: 0.89}, {7: 0.881}, {8: 0.872}],
        '11-01-2012': [{1: 0.945}, {2: 0.9325}],
        '02-01-2012': [{1: 0.94}, {2: 0.9299999999999999}, {3: 0.9199999999999999}, {4: 0.9099999999999999},
                       {5: 0.8999999999999999}, {6: 0.8899999999999999}, {7: 0.8799999999999999},
                       {8: 0.8699999999999999}, {9: 0.8599999999999999}, {10: 0.8499999999999999},
                       {11: 0.8399999999999999}],
        '03-01-2012': [{1: 0.955}, {2: 0.9455}, {3: 0.936}, {4: 0.9265000000000001}, {5: 0.9170000000000001},
                       {6: 0.9075000000000002}, {7: 0.8980000000000002}, {8: 0.8885000000000003},
                       {9: 0.8790000000000003}, {10: 0.8695000000000004}],
        '08-01-2012': [{1: 0.94}, {2: 0.9295}, {3: 0.919}, {4: 0.9085000000000001}, {5: 0.8980000000000001}],
        '01-01-2012': [{1: 0.95}, {2: 0.94}, {3: 0.9299999999999999}, {4: 0.9199999999999999}, {5: 0.9099999999999999},
                       {6: 0.8999999999999999}, {7: 0.8899999999999999}, {8: 0.8799999999999999},
                       {9: 0.8699999999999999}, {10: 0.8599999999999999}, {11: 0.8499999999999999},
                       {12: 0.8399999999999999}],
        '09-01-2012': [{1: 0.92}, {2: 0.91}, {3: 0.9}, {4: 0.89}]}

我需要迭代字典值并将所有 1＆＃39; ， 2＆＃39> 等组合在一起。

这是我目前的代码

from collections import defaultdict

final = defaultdict(list)

for k, v in data.items():
    new_data = next(iter(v))
    for m, n in new_data.items():
        final[m].append(n)

print(final)

# defaultdict(<class 'list'>, {1: [0.935, 0.92, 0.955, 0.96, 0.94, 0.93, 0.95, 0.96, 0.945, 0.94, 0.945, 0.93]})

它仅对所有1个进行分组，而不是对2个进行分组，依此类推。我做错了什么？

Answer 1

你忘了迭代许多小词典：

from collections import defaultdict

final = defaultdict(list)

for k, v in data.items():
    for d in v:                 # <-- this was missing
        for m, n in d.items():
            final[m].append(n)

print(final)

（你只调用next(...)，它只产生第一项。）

输出：

defaultdict（，{1：[0.96,0.935,0.93,0.945,0.96,0.95,9.93,0.94,0.945,0.955,0.94,0.92]，2：[0.926,0.9244,0.9365,0.95,0.94,0.9195000000000001， 0.9299999999999999,0.9325,0.9455,0.9295,0.91]，3：[0.917,0.9188,0.928,0.94,0.9299999999999999,0.9090000000000001,0.919999999999999999,0.936,0.919,0.9]，4：[0.908,0.9195000000000001,0.929999999999999999,0.9199999999999999,0.8985000000000002,0.9099999999999999， 0.9265000000000001，0.9085000000000001，0.89]，5：0.899，0.9110000000000001，0.9199999999999999，0.9099999999999999，0.8880000000000002，0.8999999999999999，0.9170000000000001，0.8980000000000001]，6：[0.89，0.9025000000000002，0.9099999999999999，0.8999999999999999，0.8775000000000003，0.8899999999999999，0.9075000000000002]，7：[0.881， 0.8940000000000002,0.8899999999999999,0.8670000000000003,0.8799999999999999,0.8980000000000002]，8：[0.872,0.8799999999999999,0.8565000000000004,0.8699999999999999,0.8885000000000003]，9：[0.8699999999999999,0.84600000 00000004,0.8599999999999999,0.8790000000000003]，10：[0.859999999999999999,0.8499999999999999,0.8695000000000004]，11：[0.849999999999999999,0.8399999999999999]，12：[0.8399999999999999]]）

Answer 2

    new_data = next(iter(v))`

这是出错的地方。这只返回你的iterable中的下一个值，在这种情况下是绑定到key的键值对：1。

你可以通过添加一个print语句来看到这个：

from collections import defaultdict

final = defaultdict(list)
print final
for k, v in data.items():

new_data = next(iter(v))
for m, n in new_data.items():
    print new_data

    final[m].append(n)

#{1: 0.96}
#{1: 0.935}
#{1: 0.93}
#{1: 0.945}
#{1: 0.96}
#{1: 0.95}
#{1: 0.93}
#{1: 0.94}
#{1: 0.945}
#{1: 0.955}
#{1: 0.94}
#{1: 0.92}

然后解决方案是实际抓取所有项目，而不仅仅是第一项，这可以通过很多方式完成，其中最简单，最丑陋的是嵌套另一个循环。

from collections import defaultdict

final = defaultdict(list)
for k, v in data.items():

    iterable = iter(v)
    for i in range(len(v)):
        new_data = iterable.next()
        for m, n in new_data.items():

            final[m].append(n)

print(final)

在嵌套字典Python中对数据进行分组

2 个答案: