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

转载 作者:行者123 更新时间:2023-12-04 22:58:39 25 4
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我正在尝试对行中的相同信息进行分组和计数:

#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()

当到达到 df.group时,将引发错误



我尝试搜索相同的错误,但找不到有效的答案来帮助解决问题。

最佳答案

看一下有关Group By的文档

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



前一个取自 here

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

您明白了,可以使用我提供的第一个链接从此处进行构建。
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

关于Python Pandas <...上的pandas.core.groupby.DataFrameGroupBy对象>,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33440640/

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