将Pandas GroupBy对象转换为DataFrame

Question

我从这样的输入数据开始

df1 = pandas.DataFrame( { 
    "Name" : ["Alice", "Bob", "Mallory", "Mallory", "Bob" , "Mallory"] , 
    "City" : ["Seattle", "Seattle", "Portland", "Seattle", "Seattle", "Portland"] } )

打印时显示如下:

   City     Name
0   Seattle    Alice
1   Seattle      Bob
2  Portland  Mallory
3   Seattle  Mallory
4   Seattle      Bob
5  Portland  Mallory

分组很简单:

g1 = df1.groupby( [ "Name", "City"] ).count()

和打印产生一个GroupBy对象:

                  City  Name
Name    City
Alice   Seattle      1     1
Bob     Seattle      2     2
Mallory Portland     2     2
        Seattle      1     1

但我最终想要的是另一个包含GroupBy对象中所有行的DataFrame对象.换句话说,我希望得到以下结果:

                  City  Name
Name    City
Alice   Seattle      1     1
Bob     Seattle      2     2
Mallory Portland     2     2
Mallory Seattle      1     1

我无法在pandas文档中看到如何实现这一点.任何提示都会受到欢迎.

Answer 1

g1这是一个DataFrame.它有一个分层索引,但是:

In [19]: type(g1)
Out[19]: pandas.core.frame.DataFrame

In [20]: g1.index
Out[20]: 
MultiIndex([('Alice', 'Seattle'), ('Bob', 'Seattle'), ('Mallory', 'Portland'),
       ('Mallory', 'Seattle')], dtype=object)

也许你想要这样的东西？

In [21]: g1.add_suffix('_Count').reset_index()
Out[21]: 
      Name      City  City_Count  Name_Count
0    Alice   Seattle           1           1
1      Bob   Seattle           2           2
2  Mallory  Portland           2           2
3  Mallory   Seattle           1           1

或类似的东西:

In [36]: DataFrame({'count' : df1.groupby( [ "Name", "City"] ).size()}).reset_index()
Out[36]: 
      Name      City  count
0    Alice   Seattle      1
1      Bob   Seattle      2
2  Mallory  Portland      2
3  Mallory   Seattle      1

Answer 2

我想略微改变Wes给出的答案,因为版本0.16.2要求as_index=False.如果不设置它,则会得到一个空数据帧.

来源:

如果它们是命名列,则聚合函数不会返回聚合的组,as_index=True默认情况下为when .分组列将是返回对象的索引.

as_index=False如果它们是命名列,则传递将返回您聚合的组.

聚合函数是那些减少返回的对象的尺寸,例如:mean,sum,size,count,std,var,sem,describe,first,last,nth,min,max.这就是当你这样做DataFrame.sum()并回来时会发生的事情Series.

nth可以作为减速器或过滤器,请参见此处.

import pandas as pd

df1 = pd.DataFrame({"Name":["Alice", "Bob", "Mallory", "Mallory", "Bob" , "Mallory"],
                    "City":["Seattle","Seattle","Portland","Seattle","Seattle","Portland"]})
print df1
#
#       City     Name
#0   Seattle    Alice
#1   Seattle      Bob
#2  Portland  Mallory
#3   Seattle  Mallory
#4   Seattle      Bob
#5  Portland  Mallory
#
g1 = df1.groupby(["Name", "City"], as_index=False).count()
print g1
#
#                  City  Name
#Name    City
#Alice   Seattle      1     1
#Bob     Seattle      2     2
#Mallory Portland     2     2
#        Seattle      1     1
#

编辑:

在版本0.17.1及更高版本,您可以使用subset的count和reset_index与参数name在size:

print df1.groupby(["Name", "City"], as_index=False ).count()
#IndexError: list index out of range

print df1.groupby(["Name", "City"]).count()
#Empty DataFrame
#Columns: []
#Index: [(Alice, Seattle), (Bob, Seattle), (Mallory, Portland), (Mallory, Seattle)]

print df1.groupby(["Name", "City"])[['Name','City']].count()
#                  Name  City
#Name    City                
#Alice   Seattle      1     1
#Bob     Seattle      2     2
#Mallory Portland     2     2
#        Seattle      1     1

print df1.groupby(["Name", "City"]).size().reset_index(name='count')
#      Name      City  count
#0    Alice   Seattle      1
#1      Bob   Seattle      2
#2  Mallory  Portland      2
#3  Mallory   Seattle      1

count和之间的差异size是size计算NaN值而count不是.

Answer 3

简单地说,这应该完成任务:

import pandas as pd

grouped_df = df1.groupby( [ "Name", "City"] )

pd.DataFrame(grouped_df.size().reset_index(name = "Group_Count"))

这里,grouped_df.size()提取唯一的groupby计数,reset_index()方法重置你想要的列的名称.最后,调用pandas Dataframe()函数来创建DataFrame对象.

Answer 4

也许我误解了这个问题,但是如果你想将groupby转换回数据帧,你可以使用.to_frame().当我这样做时我想重置索引,所以我也包括了那个部分.

示例代码与问题无关

df = df['TIME'].groupby(df['Name']).min()
df = df.to_frame()
df = df.reset_index(level=['Name',"TIME"])

Answer 5

下面的解决方案可能更简单：

df1.reset_index().groupby( [ "Name", "City"],as_index=False ).count()

Answer 6

这会以与普通方法相同的顺序返回序数级别/索引groupby()。它与 @NehalJWani 在他的评论中发布的答案基本相同，但存储在一个变量中，并reset_index()调用它的方法。

fare_class = df.groupby(['Satisfaction Rating','Fare Class']).size().to_frame(name = 'Count')
fare_class.reset_index()

此版本不仅返回相同的百分比数据（这对于统计数据很有用），而且还包含 lambda 函数。

fare_class_percent = df.groupby(['Satisfaction Rating', 'Fare Class']).size().to_frame(name = 'Percentage')
fare_class_percent.transform(lambda x: 100 * x/x.sum()).reset_index()

      Satisfaction Rating      Fare Class  Percentage
0            Dissatisfied        Business   14.624269
1            Dissatisfied         Economy   36.469048
2               Satisfied        Business    5.460425
3               Satisfied         Economy   33.235294

例子：

Answer 7

我发现这对我有用.

import numpy as np
import pandas as pd

df1 = pd.DataFrame({ 
    "Name" : ["Alice", "Bob", "Mallory", "Mallory", "Bob" , "Mallory"] , 
    "City" : ["Seattle", "Seattle", "Portland", "Seattle", "Seattle", "Portland"]})

df1['City_count'] = 1
df1['Name_count'] = 1

df1.groupby(['Name', 'City'], as_index=False).count()

Answer 8

我已经聚合了大量明智的数据并存储到数据框

almo_grp_data = pd.DataFrame({'Qty_cnt' :
almo_slt_models_data.groupby( ['orderDate','Item','State Abv']
          )['Qty'].sum()}).reset_index()

Answer 9

关键是使用reset_index（）方法。

使用：

import pandas

df1 = pandas.DataFrame( { 
    "Name" : ["Alice", "Bob", "Mallory", "Mallory", "Bob" , "Mallory"] , 
    "City" : ["Seattle", "Seattle", "Portland", "Seattle", "Seattle", "Portland"] } )

g1 = df1.groupby( [ "Name", "City"] ).count().reset_index()

现在，您在g1中有了新的数据框：