Python Pandas <pandas.core.groupby.DataFrameGroupBy 对象在 ...>
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Python Pandas <pandas.core.groupby.DataFrameGroupBy object at ...>
提问by Lucas Mascia
I am trying to group and count the same info in a row:
我正在尝试分组并连续计算相同的信息：
#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 servi?os');
print (grouping)
postal_saude()
When it gets to the df.groupbyit raises an error
当它到达df.groupby时会引发错误
I have tried searching this same error but I have not found a valid answer to help me fix my problem.
我试过搜索同样的错误，但我没有找到一个有效的答案来帮助我解决我的问题。
回答by Leb
Take a look at this documentation about Group By
看看这个关于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
使用映射器（字典或键函数，将给定函数应用于组，将结果作为系列返回）或按一系列列对系列进行分组
The previous is taken from here
上一篇摘自这里
Here's a quick example:
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
You get the idea, you can build from here using the first link I provided.
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
