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

Question

我试图连续分组和计算相同的信息:

#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时会产生错误

我试过搜索同样的错误,但我没有找到一个有效的答案,以帮助我解决我的问题.

请帮忙

Answer 1

请查看有关Group By的文档

使用mapper(dict或key函数,将给定函数应用于组,将结果作为系列返回)或通过一系列列的组系列

以前是从这里开始的

这是一个简单的例子:

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

你明白了,你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