Python Pandas ＜pandas.core.groupby.DataFrameGroupBy对象在...＞。

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问题描述

我正在尝试分组并计数相同的信息:

#Functions

def postal_saude ():
    global df, lista_solic

    #List of solicitantes in Postal Saude
    list_sol = [lista_solic["name1"], lista_solic["name2"]]

    #filter Postal Saude Solicitantes
    df = df[(df['Cliente']==lista_clientes["6"]) 
        & (df['Nome do solicitante'].isin(list_sol))]

    #Alphabetical order
    df = df.sort_index(by=['Nome do solicitante', 'nomeCorrespondente'])

    #Grouping data of column
    grouping = df.groupby('Tipo do serviços');

    print (grouping)


postal_saude()

当它到达 df.groupby 时会引起错误

我尝试搜索相同的错误，但是我没有找到有效的答案来帮助我解决问题.

推荐答案

查看有关组成的文档

使用映射器的组系列(dict或键功能，应用给定功能要组为串联)或一系列列
返回结果

上一堂取自

这是一个快速示例:

df = pd.DataFrame({'a':[1,1,1,2,2,2,3,3,3,3],'b':np.random.randn(10)})

df
   a         b
0  1  1.048099
1  1 -0.830804
2  1  1.007282
3  2 -0.470914
4  2  1.948448
5  2 -0.144317
6  3 -0.645503
7  3 -1.694219
8  3  0.375280
9  3 -0.065624

groups = df.groupby('a')

groups # Tells you what "df.groupby('a')" is, not an error
<pandas.core.groupby.DataFrameGroupBy object at 0x00000000097EEB38>

groups.count() # count the number of 1 present in the 'a' column
   b
a   
1  3
2  3
3  4

groups.sum() # sums the 'b' column values based on 'a' grouping

          b
a          
1  1.224577
2  1.333217
3 -2.030066

您明白了，可以使用我提供的第一个链接从这里构建.

df_count = groups.count()

df_count
   b
a   
1  3
2  3
3  4

type(df_count) # assigning the `.count()` output to a variable create a new df
pandas.core.frame.DataFrame

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问题描述

I am trying to group and count the same info in a row:

#Functions

def postal_saude ():
    global df, lista_solic

    #List of solicitantes in Postal Saude
    list_sol = [lista_solic["name1"], lista_solic["name2"]]

    #filter Postal Saude Solicitantes
    df = df[(df['Cliente']==lista_clientes["6"]) 
        & (df['Nome do solicitante'].isin(list_sol))]

    #Alphabetical order
    df = df.sort_index(by=['Nome do solicitante', 'nomeCorrespondente'])

    #Grouping data of column
    grouping = df.groupby('Tipo do serviços');

    print (grouping)


postal_saude()

When it gets to the df.groupby it raises an error

I have tried searching this same error but I have not found a valid answer to help me fix my problem.

推荐答案

Take a look at this documentation about Group By

Group series using mapper (dict or key function, apply given function to group, return result as series) or by a series of columns

The previous is taken from here

Here's a quick example:

df = pd.DataFrame({'a':[1,1,1,2,2,2,3,3,3,3],'b':np.random.randn(10)})

df
   a         b
0  1  1.048099
1  1 -0.830804
2  1  1.007282
3  2 -0.470914
4  2  1.948448
5  2 -0.144317
6  3 -0.645503
7  3 -1.694219
8  3  0.375280
9  3 -0.065624

groups = df.groupby('a')

groups # Tells you what "df.groupby('a')" is, not an error
<pandas.core.groupby.DataFrameGroupBy object at 0x00000000097EEB38>

groups.count() # count the number of 1 present in the 'a' column
   b
a   
1  3
2  3
3  4

groups.sum() # sums the 'b' column values based on 'a' grouping

          b
a          
1  1.224577
2  1.333217
3 -2.030066

You get the idea, you can build from here using the first link I provided.

df_count = groups.count()

df_count
   b
a   
1  3
2  3
3  4

type(df_count) # assigning the `.count()` output to a variable create a new df
pandas.core.frame.DataFrame

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