我使用熊猫作为数据库替代品,因为我有多个数据库(Oracle, SQL Server等),我无法使一个SQL等量命令序列。

我有一个表加载在一个DataFrame与一些列:

YEARMONTH, CLIENTCODE, SIZE, etc., etc.

在SQL中,计算每年不同客户端的数量将是:

SELECT count(distinct CLIENTCODE) FROM table GROUP BY YEARMONTH;

结果就是

201301    5000
201302    13245

我如何在熊猫中做到这一点?


当前回答

To obtain the number of different clients and sizes per year (i.e. number of unique values of multiple columns), use a list of columns: df.groupby('YEARMONTH')[['CLIENTCODE', 'SIZE']].nunique() Actually, the result from the above code can be obtained using SQL syntax on df using pandasql (a module built on pandas that lets you query pandas DataFrames using SQL syntax). #! pip install pandasql from pandasql import sqldf sqldf(""" SELECT COUNT(DISTINCT CLIENTCODE), COUNT(DISTINCT SIZE) FROM df GROUP BY YEARMONTH """) If you want to keep YEARMONTH as a column, i.e. the analog of the following SQL query SELECT YEARMONTH, COUNT(DISTINCT CLIENTCODE), COUNT(DISTINCT SIZE) FROM df GROUP BY YEARMONTH in pandas is the following (set as_index to False): df.groupby('YEARMONTH', as_index=False)[['CLIENTCODE', 'SIZE']].nunique() If you need to set custom names to the aggregated columns, i.e. the analog of the following SQL query: SELECT YEARMONTH, COUNT(DISTINCT CLIENTCODE) AS `No. clients`, COUNT(DISTINCT SIZE) AS `No. size` FROM df GROUP BY YEARMONTH use named aggregation in pandas: ( df.groupby('YEARMONTH', as_index=False) .agg(**{'No. clients':('CLIENTCODE', 'nunique'), 'No. size':('SIZE', 'nunique')}) )

其他回答

不同的列以及其他列上的聚合

要获得任何列(在您的例子中是CLIENTCODE)的不同值数量,我们可以使用nunique。我们可以在agg函数中传递输入作为字典,以及其他列上的聚合:

grp_df = df.groupby('YEARMONTH').agg({'CLIENTCODE': ['nunique'],
                                      'other_col_1': ['sum', 'count']})

# to flatten the multi-level columns
grp_df.columns = ["_".join(col).strip() for col in grp_df.columns.values]

# if you wish to reset the index
grp_df.reset_index(inplace=True)

这里有一种方法可以使多个列上的计数不同。让我们看一些数据:

data = {'CLIENT_CODE':[1,1,2,1,2,2,3],
        'YEAR_MONTH':[201301,201301,201301,201302,201302,201302,201302],
        'PRODUCT_CODE': [100,150,220,400,50,80,100]
       }
table = pd.DataFrame(data)
table

CLIENT_CODE YEAR_MONTH  PRODUCT_CODE
0   1       201301      100
1   1       201301      150
2   2       201301      220
3   1       201302      400
4   2       201302      50
5   2       201302      80
6   3       201302      100

现在,列出感兴趣的列,并在稍微修改的语法中使用groupby:

columns = ['YEAR_MONTH', 'PRODUCT_CODE']
table[columns].groupby(table['CLIENT_CODE']).nunique()

我们获得:

YEAR_MONTH  PRODUCT_CODE CLIENT_CODE
1           2            3
2           2            3
3           1            1

我相信这就是你想要的:

table.groupby('YEARMONTH').CLIENTCODE.nunique()

例子:

In [2]: table
Out[2]: 
   CLIENTCODE  YEARMONTH
0           1     201301
1           1     201301
2           2     201301
3           1     201302
4           2     201302
5           2     201302
6           3     201302

In [3]: table.groupby('YEARMONTH').CLIENTCODE.nunique()
Out[3]: 
YEARMONTH
201301       2
201302       3

有趣的是,len(unique())通常比nunique()快几倍(3 -15倍)。

创建一个数据透视表并使用非唯一级数函数:

ID = [ 123, 123, 123, 456, 456, 456, 456, 789, 789]
domain = ['vk.com', 'vk.com', 'twitter.com', 'vk.com', 'facebook.com',
          'vk.com', 'google.com', 'twitter.com', 'vk.com']
df = pd.DataFrame({'id':ID, 'domain':domain})
fp = pd.pivot_table(data=df, index='domain', aggfunc=pd.Series.nunique)
print(fp)

输出:

               id
domain
facebook.com   1
google.com     1
twitter.com    2
vk.com         3