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
我也使用nunique，但如果你必须使用'min'， 'max'， 'count'或'mean'等聚合函数，这将是非常有用的。
df.groupby('YEARMONTH')['CLIENTCODE'].transform('nunique') #count(distinct)
df.groupby('YEARMONTH')['CLIENTCODE'].transform('min') #min
df.groupby('YEARMONTH')['CLIENTCODE'].transform('max') #max
df.groupby('YEARMONTH')['CLIENTCODE'].transform('mean') #average
df.groupby('YEARMONTH')['CLIENTCODE'].transform('count') #count
其他回答
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')}) )
现在你也可以在Python中使用dplyr语法来做到这一点:
>>> from datar.all import f, tibble, group_by, summarise, n_distinct
>>>
>>> data = tibble(
... CLIENT_CODE=[1,1,2,1,2,2,3],
... YEAR_MONTH=[201301,201301,201301,201302,201302,201302,201302]
... )
>>>
>>> data >> group_by(f.YEAR_MONTH) >> summarise(n=n_distinct(f.CLIENT_CODE))
YEAR_MONTH n
<int64> <int64>
0 201301 2
1 201302 3
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
我也使用nunique，但如果你必须使用'min'， 'max'， 'count'或'mean'等聚合函数，这将是非常有用的。
df.groupby('YEARMONTH')['CLIENTCODE'].transform('nunique') #count(distinct)
df.groupby('YEARMONTH')['CLIENTCODE'].transform('min') #min
df.groupby('YEARMONTH')['CLIENTCODE'].transform('max') #max
df.groupby('YEARMONTH')['CLIENTCODE'].transform('mean') #average
df.groupby('YEARMONTH')['CLIENTCODE'].transform('count') #count
使用crosstab，这将返回比groupby nunique更多的信息:
pd.crosstab(df.YEARMONTH,df.CLIENTCODE)
Out[196]:
CLIENTCODE 1 2 3
YEARMONTH
201301 2 1 0
201302 1 2 1
稍加修改后，得到如下结果:
pd.crosstab(df.YEARMONTH,df.CLIENTCODE).ne(0).sum(1)
Out[197]:
YEARMONTH
201301 2
201302 3
dtype: int64
