使用集合中的计数器动态创建的字典的总和 [英] Sum values of dynamically created dictionaries using Counter from collections

问题描述

我有这个df:

import pandas as pd
a = [1,1,1,2,2,3,3,3,3,4,4,5,5,5]
b = ["pi","pi","k","J","pi","pi","k","k","J","pi","k","pi","k","pi"]
bin0 = [0,0,0,1,0,0,1,0,0,0,1,1,0,0]
bin1 = [1,1,1,0,1,0,0,1,1,0,0,0,1,0]
bin2 = [0,0,0,0,0,1,0,0,0,1,0,0,0,1]

df_test = pd.DataFrame({"a": a, "b": b,"bin0": bin0,"bin1": bin1,"bin2": 
bin2})

喜欢这个:

    a   b  bin0  bin1  bin2
0   1  pi     0     1     0
1   1  pi     0     1     0
2   1   k     0     1     0
3   2   J     1     0     0
...
12  5   k     0     1     0
13  5  pi     0     0     1

然后，我要根据此df创建字典，并对具有相同密钥的那些字典求和:

Then I want to create dictionaries from this df and sum over those dictionaries if they have the same key:

from collections import Counter

thismodule = sys.modules[__name__]

df1 = df_test.groupby(['a', 'b']).agg({'b':'size', 'bin0':'sum', 
'bin1':'sum', 'bin2':'sum'}).rename(columns={'b': 'cant', 'bin0': 'b0', 
'bin1': 'b1', 'bin2': 'b2'}).reset_index(drop = False)


for evt in df1.a.unique():
    name1 = 'dict_'+str(evt)
    name2 = 'col_'+str(evt)
    df_ = df1
    df_ = df_[df_.a==evt].drop('a', 1).set_index('b').to_dict('index')
    setattr(thismodule, name1, df_)
    setattr(thismodule, name2, col_)  

例如，获取:

df_1 = {'k': {'cant': 1, 'b0': 0, 'b1': 1, 'b2': 0}, 'pi': {'cant': 2, 
'b0': 0, 'b1': 2, 'b2': 0}}

col_1 = Counter({'k': {'cant': 1, 'b0': 0, 'b1': 1, 'b2': 0}, 'pi': 
{'cant': 2, 'b0': 1, 'b1': 0, 'b2': 1}})

最后，当我想对具有相同键的字典的值求和时，会出现错误:

Finally, when I want to sum over the values of the dictionaries which have the same key I get an error:

col_1 = eval("col_1")
col_2 = eval("col_2")

sumdict = col_1 +col_2
print(sumdict)

错误是:

newcount = count + other[elem]

TypeError: unsupported operand type(s) for +: 'dict' and 'dict'

推荐答案

这不是您要实现的目标:

Isn't it this what you want to achieve:

df_test.groupby(['a', 'b']).sum().reset_index().groupby('a').sum()

    bin0    bin1    bin2
a           
1   0   3   0
2   1   1   0
3   1   2   1
4   1   0   1
5   1   1   1

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