问题:熊猫分组和

我正在使用此数据框:

Fruit   Date      Name  Number
Apples  10/6/2016 Bob    7
Apples  10/6/2016 Bob    8
Apples  10/6/2016 Mike   9
Apples  10/7/2016 Steve 10
Apples  10/7/2016 Bob    1
Oranges 10/7/2016 Bob    2
Oranges 10/6/2016 Tom   15
Oranges 10/6/2016 Mike  57
Oranges 10/6/2016 Bob   65
Oranges 10/7/2016 Tony   1
Grapes  10/7/2016 Bob    1
Grapes  10/7/2016 Tom   87
Grapes  10/7/2016 Bob   22
Grapes  10/7/2016 Bob   12
Grapes  10/7/2016 Tony  15

我想按名称然后按水果进行汇总,以获得每个名称的水果总数。

Bob,Apples,16 ( for example )

我尝试按名称和水果分组,但是如何获取水果总数。

I am using this data frame:

Fruit   Date      Name  Number
Apples  10/6/2016 Bob    7
Apples  10/6/2016 Bob    8
Apples  10/6/2016 Mike   9
Apples  10/7/2016 Steve 10
Apples  10/7/2016 Bob    1
Oranges 10/7/2016 Bob    2
Oranges 10/6/2016 Tom   15
Oranges 10/6/2016 Mike  57
Oranges 10/6/2016 Bob   65
Oranges 10/7/2016 Tony   1
Grapes  10/7/2016 Bob    1
Grapes  10/7/2016 Tom   87
Grapes  10/7/2016 Bob   22
Grapes  10/7/2016 Bob   12
Grapes  10/7/2016 Tony  15

I want to aggregate this by name and then by fruit to get a total number of fruit per name.

Bob,Apples,16 ( for example )

I tried grouping by Name and Fruit but how do I get the total number of fruit.


回答 0

用途GroupBy.sum

df.groupby(['Fruit','Name']).sum()

Out[31]: 
               Number
Fruit   Name         
Apples  Bob        16
        Mike        9
        Steve      10
Grapes  Bob        35
        Tom        87
        Tony       15
Oranges Bob        67
        Mike       57
        Tom        15
        Tony        1

Use GroupBy.sum:

df.groupby(['Fruit','Name']).sum()

Out[31]: 
               Number
Fruit   Name         
Apples  Bob        16
        Mike        9
        Steve      10
Grapes  Bob        35
        Tom        87
        Tony       15
Oranges Bob        67
        Mike       57
        Tom        15
        Tony        1

回答 1

你也可以使用agg函数

df.groupby(['Name', 'Fruit'])['Number'].agg('sum')

Also you can use agg function,

df.groupby(['Name', 'Fruit'])['Number'].agg('sum')

回答 2

如果要保留原始列FruitName,请使用reset_index()。否则,FruitName将成为指数的一部分。

df.groupby(['Fruit','Name'])['Number'].sum().reset_index()

Fruit   Name       Number
Apples  Bob        16
Apples  Mike        9
Apples  Steve      10
Grapes  Bob        35
Grapes  Tom        87
Grapes  Tony       15
Oranges Bob        67
Oranges Mike       57
Oranges Tom        15
Oranges Tony        1

如其他答案所示:

df.groupby(['Fruit','Name'])['Number'].sum()

               Number
Fruit   Name         
Apples  Bob        16
        Mike        9
        Steve      10
Grapes  Bob        35
        Tom        87
        Tony       15
Oranges Bob        67
        Mike       57
        Tom        15
        Tony        1

If you want to keep the original columns Fruit and Name, use reset_index(). Otherwise Fruit and Name will become part of the index.

df.groupby(['Fruit','Name'])['Number'].sum().reset_index()

Fruit   Name       Number
Apples  Bob        16
Apples  Mike        9
Apples  Steve      10
Grapes  Bob        35
Grapes  Tom        87
Grapes  Tony       15
Oranges Bob        67
Oranges Mike       57
Oranges Tom        15
Oranges Tony        1

As seen in the other answers:

df.groupby(['Fruit','Name'])['Number'].sum()

               Number
Fruit   Name         
Apples  Bob        16
        Mike        9
        Steve      10
Grapes  Bob        35
        Tom        87
        Tony       15
Oranges Bob        67
        Mike       57
        Tom        15
        Tony        1

回答 3

其他两个答案都能满足您的需求。

您可以使用该pivot功能将数据排列在一个漂亮的表中

df.groupby(['Fruit','Name'],as_index = False).sum().pivot('Fruit','Name').fillna(0)



Name    Bob     Mike    Steve   Tom    Tony
Fruit                   
Apples  16.0    9.0     10.0    0.0     0.0
Grapes  35.0    0.0     0.0     87.0    15.0
Oranges 67.0    57.0    0.0     15.0    1.0

Both the other answers accomplish what you want.

You can use the pivot functionality to arrange the data in a nice table

df.groupby(['Fruit','Name'],as_index = False).sum().pivot('Fruit','Name').fillna(0)



Name    Bob     Mike    Steve   Tom    Tony
Fruit                   
Apples  16.0    9.0     10.0    0.0     0.0
Grapes  35.0    0.0     0.0     87.0    15.0
Oranges 67.0    57.0    0.0     15.0    1.0

回答 4

df.groupby(['Fruit','Name'])['Number'].sum()

您可以选择不同的列来对数字求和。

df.groupby(['Fruit','Name'])['Number'].sum()

You can select different columns to sum numbers.


回答 5

您可以将设置groupbyindex ,然后使用sumlevel

df.set_index(['Fruit','Name']).sum(level=[0,1])
Out[175]: 
               Number
Fruit   Name         
Apples  Bob        16
        Mike        9
        Steve      10
Oranges Bob        67
        Tom        15
        Mike       57
        Tony        1
Grapes  Bob        35
        Tom        87
        Tony       15

You can set the groupby column to index then using sum with level

df.set_index(['Fruit','Name']).sum(level=[0,1])
Out[175]: 
               Number
Fruit   Name         
Apples  Bob        16
        Mike        9
        Steve      10
Oranges Bob        67
        Tom        15
        Mike       57
        Tony        1
Grapes  Bob        35
        Tom        87
        Tony       15

回答 6

.agg()函数的变体;提供以下功能:(1)持久化类型DataFrame,(2)应用平均值,计数,求和等,以及(3)在保持易读性的同时在多个列上启用groupby。

df.groupby(['att1', 'att2']).agg({'att1': "count", 'att3': "sum",'att4': 'mean'})

用你的价值观…

df.groupby(['Name', 'Fruit']).agg({'Number': "sum"})

A variation on the .agg() function; provides the ability to (1) persist type DataFrame, (2) apply averages, counts, summations, etc. and (3) enables groupby on multiple columns while maintaining legibility.

df.groupby(['att1', 'att2']).agg({'att1': "count", 'att3': "sum",'att4': 'mean'})

using your values…

df.groupby(['Name', 'Fruit']).agg({'Number': "sum"})

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