熊猫DataFrame Groupby两列并获取计数

问题:熊猫DataFrame Groupby两列并获取计数

我有以下格式的熊猫数据框:

df = pd.DataFrame([[1.1, 1.1, 1.1, 2.6, 2.5, 3.4,2.6,2.6,3.4,3.4,2.6,1.1,1.1,3.3], list('AAABBBBABCBDDD'), [1.1, 1.7, 2.5, 2.6, 3.3, 3.8,4.0,4.2,4.3,4.5,4.6,4.7,4.7,4.8], ['x/y/z','x/y','x/y/z/n','x/u','x','x/u/v','x/y/z','x','x/u/v/b','-','x/y','x/y/z','x','x/u/v/w'],['1','3','3','2','4','2','5','3','6','3','5','1','1','1']]).T
df.columns = ['col1','col2','col3','col4','col5']

df:

   col1 col2 col3     col4 col5
0   1.1    A  1.1    x/y/z    1
1   1.1    A  1.7      x/y    3
2   1.1    A  2.5  x/y/z/n    3
3   2.6    B  2.6      x/u    2
4   2.5    B  3.3        x    4
5   3.4    B  3.8    x/u/v    2
6   2.6    B    4    x/y/z    5
7   2.6    A  4.2        x    3
8   3.4    B  4.3  x/u/v/b    6
9   3.4    C  4.5        -    3
10  2.6    B  4.6      x/y    5
11  1.1    D  4.7    x/y/z    1
12  1.1    D  4.7        x    1
13  3.3    D  4.8  x/u/v/w    1

现在,我想按两列将其分组,如下所示:

df.groupby(['col5','col2']).reset_index()

输出:

             index col1 col2 col3     col4 col5
col5 col2                                      
1    A    0      0  1.1    A  1.1    x/y/z    1
     D    0     11  1.1    D  4.7    x/y/z    1
          1     12  1.1    D  4.7        x    1
          2     13  3.3    D  4.8  x/u/v/w    1
2    B    0      3  2.6    B  2.6      x/u    2
          1      5  3.4    B  3.8    x/u/v    2
3    A    0      1  1.1    A  1.7      x/y    3
          1      2  1.1    A  2.5  x/y/z/n    3
          2      7  2.6    A  4.2        x    3
     C    0      9  3.4    C  4.5        -    3
4    B    0      4  2.5    B  3.3        x    4
5    B    0      6  2.6    B    4    x/y/z    5
          1     10  2.6    B  4.6      x/y    5
6    B    0      8  3.4    B  4.3  x/u/v/b    6

我想按如下方式获取每一行的计数。预期Yield:

col5 col2 count
1    A      1
     D      3
2    B      2
etc...

如何获得我的预期输出?我想为每个“ col2”值找到最大的计数吗?

I have a pandas dataframe in the following format:

df = pd.DataFrame([[1.1, 1.1, 1.1, 2.6, 2.5, 3.4,2.6,2.6,3.4,3.4,2.6,1.1,1.1,3.3], list('AAABBBBABCBDDD'), [1.1, 1.7, 2.5, 2.6, 3.3, 3.8,4.0,4.2,4.3,4.5,4.6,4.7,4.7,4.8], ['x/y/z','x/y','x/y/z/n','x/u','x','x/u/v','x/y/z','x','x/u/v/b','-','x/y','x/y/z','x','x/u/v/w'],['1','3','3','2','4','2','5','3','6','3','5','1','1','1']]).T
df.columns = ['col1','col2','col3','col4','col5']

df:

   col1 col2 col3     col4 col5
0   1.1    A  1.1    x/y/z    1
1   1.1    A  1.7      x/y    3
2   1.1    A  2.5  x/y/z/n    3
3   2.6    B  2.6      x/u    2
4   2.5    B  3.3        x    4
5   3.4    B  3.8    x/u/v    2
6   2.6    B    4    x/y/z    5
7   2.6    A  4.2        x    3
8   3.4    B  4.3  x/u/v/b    6
9   3.4    C  4.5        -    3
10  2.6    B  4.6      x/y    5
11  1.1    D  4.7    x/y/z    1
12  1.1    D  4.7        x    1
13  3.3    D  4.8  x/u/v/w    1

Now I want to group this by two columns like following:

df.groupby(['col5','col2']).reset_index()

OutPut:

             index col1 col2 col3     col4 col5
col5 col2                                      
1    A    0      0  1.1    A  1.1    x/y/z    1
     D    0     11  1.1    D  4.7    x/y/z    1
          1     12  1.1    D  4.7        x    1
          2     13  3.3    D  4.8  x/u/v/w    1
2    B    0      3  2.6    B  2.6      x/u    2
          1      5  3.4    B  3.8    x/u/v    2
3    A    0      1  1.1    A  1.7      x/y    3
          1      2  1.1    A  2.5  x/y/z/n    3
          2      7  2.6    A  4.2        x    3
     C    0      9  3.4    C  4.5        -    3
4    B    0      4  2.5    B  3.3        x    4
5    B    0      6  2.6    B    4    x/y/z    5
          1     10  2.6    B  4.6      x/y    5
6    B    0      8  3.4    B  4.3  x/u/v/b    6

I want to get the count by each row like following. Expected Output:

col5 col2 count
1    A      1
     D      3
2    B      2
etc...

How to get my expected output? And I want to find largest count for each ‘col2’ value?


回答 0

紧跟@Andy的答案,您可以执行以下操作来解决第二个问题:

In [56]: df.groupby(['col5','col2']).size().reset_index().groupby('col2')[[0]].max()
Out[56]: 
      0
col2   
A     3
B     2
C     1
D     3

Followed by @Andy’s answer, you can do following to solve your second question:

In [56]: df.groupby(['col5','col2']).size().reset_index().groupby('col2')[[0]].max()
Out[56]: 
      0
col2   
A     3
B     2
C     1
D     3

回答 1

您正在寻找size

In [11]: df.groupby(['col5', 'col2']).size()
Out[11]:
col5  col2
1     A       1
      D       3
2     B       2
3     A       3
      C       1
4     B       1
5     B       2
6     B       1
dtype: int64

要获得与waitingkuo相同的答案(“第二个问题”),但要简洁一些,可以按级别分组:

In [12]: df.groupby(['col5', 'col2']).size().groupby(level=1).max()
Out[12]:
col2
A       3
B       2
C       1
D       3
dtype: int64

You are looking for size:

In [11]: df.groupby(['col5', 'col2']).size()
Out[11]:
col5  col2
1     A       1
      D       3
2     B       2
3     A       3
      C       1
4     B       1
5     B       2
6     B       1
dtype: int64

To get the same answer as waitingkuo (the “second question”), but slightly cleaner, is to groupby the level:

In [12]: df.groupby(['col5', 'col2']).size().groupby(level=1).max()
Out[12]:
col2
A       3
B       2
C       1
D       3
dtype: int64

回答 2

数据插入pandas数据框并提供列名

import pandas as pd
df = pd.DataFrame([['A','C','A','B','C','A','B','B','A','A'], ['ONE','TWO','ONE','ONE','ONE','TWO','ONE','TWO','ONE','THREE']]).T
df.columns = [['Alphabet','Words']]
print(df)   #printing dataframe.

这是我们的打印数据:

为了在pandas和counter中创建一组数据框
您需要再提供一个对分组进行计数的列,我们将该列称为dataframe中的“ COUNTER”

像这样:

df['COUNTER'] =1       #initially, set that counter to 1.
group_data = df.groupby(['Alphabet','Words'])['COUNTER'].sum() #sum function
print(group_data)

输出:

Inserting data into a pandas dataframe and providing column name.

import pandas as pd
df = pd.DataFrame([['A','C','A','B','C','A','B','B','A','A'], ['ONE','TWO','ONE','ONE','ONE','TWO','ONE','TWO','ONE','THREE']]).T
df.columns = [['Alphabet','Words']]
print(df)   #printing dataframe.

This is our printed data:

For making a group of dataframe in pandas and counter,
You need to provide one more column which counts the grouping, let’s call that column as, “COUNTER” in dataframe.

Like this:

df['COUNTER'] =1       #initially, set that counter to 1.
group_data = df.groupby(['Alphabet','Words'])['COUNTER'].sum() #sum function
print(group_data)

OUTPUT:


回答 3

仅使用单个groupby的惯用解决方案

(df.groupby(['col5', 'col2']).size() 
   .sort_values(ascending=False) 
   .reset_index(name='count') 
   .drop_duplicates(subset='col2'))

  col5 col2  count
0    3    A      3
1    1    D      3
2    5    B      2
6    3    C      1

说明

groupby size方法的结果是带有col5col2在索引中的Series 。从这里,您可以使用另一种groupby方法来查找其中的每个值的最大值, col2但是没有必要这样做。您可以简单地对所有值进行降序排序,然后只保留第一次col2使用drop_duplicates方法出现的行。

Idiomatic solution that uses only a single groupby

(df.groupby(['col5', 'col2']).size() 
   .sort_values(ascending=False) 
   .reset_index(name='count') 
   .drop_duplicates(subset='col2'))

  col5 col2  count
0    3    A      3
1    1    D      3
2    5    B      2
6    3    C      1

Explanation

The result of the groupby size method is a Series with col5 and col2 in the index. From here, you can use another groupby method to find the maximum value of each value in col2 but it is not necessary to do. You can simply sort all the values descendingly and then keep only the rows with the first occurrence of col2 with the drop_duplicates method.


回答 4

您是否要在数据帧中添加一个包含组计数的新列(例如’count_column’):

df.count_column=df.groupby(['col5','col2']).col5.transform('count')

(我选择了“ col5”,因为它不包含nan)

Should you want to add a new column (say ‘count_column’) containing the groups’ counts into the dataframe:

df.count_column=df.groupby(['col5','col2']).col5.transform('count')

(I picked ‘col5’ as it contains no nan)


回答 5

您可以只使用内置函数计数,然后使用groupby函数

df.groupby(['col5','col2']).count()

You can just use the built-in function count follow by the groupby function

df.groupby(['col5','col2']).count()