Python 熊猫计数（不同）等价物
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/15411158/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Pandas count(distinct) equivalent
提问by Adriano Almeida
I am using pandas as a db substitute as I have multiple databases (oracle, mssql, etc) and I am unable to make a sequence of commands to a SQL equivalent.
我使用 Pandas 作为数据库替代品，因为我有多个数据库（oracle、mssql 等），而且我无法为 SQL 等价物制作一系列命令。
I have a table loaded in a DataFrame with some columns:
我在 DataFrame 中加载了一个表，其中包含一些列：
YEARMONTH, CLIENTCODE, SIZE, .... etc etc
In SQL, to count the amount of different clients per year would be:
在 SQL 中，要计算每年不同客户端的数量是：
SELECT count(distinct CLIENTCODE) FROM table GROUP BY YEARMONTH;
And the result would be
结果是
201301 5000
201302 13245
How can I do that in pandas?
我怎么能在熊猫中做到这一点？
采纳答案by Dan Allan
I believe this is what you want:
我相信这就是你想要的：
table.groupby('YEARMONTH').CLIENTCODE.nunique()
Example:
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
回答by Roman Kh
Interestingly enough, very often len(unique())is a few times (3x-15x) faster than nunique().
有趣的是，通常len(unique())比nunique().
回答by Enthusiast
Here is another method, much simple, lets say your dataframe name is daatand column name is YEARMONTH
这是另一种方法，非常简单，假设您的数据框名称是daat，列名称是YEARMONTH
daat.YEARMONTH.value_counts()
回答by YOBEN_S
Using crosstab, this will return more information than groupbynunique
使用crosstab，这将返回比groupbynunique
pd.crosstab(df.YEARMONTH,df.CLIENTCODE)
Out[196]:
CLIENTCODE 1 2 3
YEARMONTH
201301 2 1 0
201302 1 2 1
After a little bit modify ,yield the result
稍加修改后，产生结果
pd.crosstab(df.YEARMONTH,df.CLIENTCODE).ne(0).sum(1)
Out[197]:
YEARMONTH
201301 2
201302 3
dtype: int64
回答by Gangaraju
I am also using nuniquebut it will be very helpful if you have to use an aggregate function like 'min', 'max', 'count' or 'mean'etc.
我也在使用，nunique但如果您必须使用诸如'min', 'max', 'count' or 'mean'等的聚合函数，它将非常有帮助。
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
回答by Vivek Ananthan
With new pandas version, it is easy to get as dataframe
使用新的 Pandas 版本，很容易获得数据框
unique_count = pd.groupby(['YEARMONTH'], as_index=False).agg(uniq_CLIENTCODE =('CLIENTCODE',pd.Series.count))
回答by Ramon
Here an approach to have count distinct over multiple columns. Let's have some data:
这是一种在多列上计数不同的方法。让我们有一些数据：
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
Now, list the columns of interest and use groupby in a slightly modified syntax:
现在，列出感兴趣的列并以稍微修改的语法使用 groupby：
columns = ['YEAR_MONTH', 'PRODUCT_CODE']
table[columns].groupby(table['CLIENT_CODE']).nunique()
We obtain:
我们获得：
YEAR_MONTH PRODUCT_CODE CLIENT_CODE
1 2 3
2 2 3
3 1 1
回答by Vivek Payasi
Distinct of column along with aggregations on other columns
不同列以及其他列上的聚合
To get the distinct number of values for any column (CLIENTCODEin your case), we can use nunique. We can pass the input as a dictionary in aggfunction, along with aggregations on other columns:
要获得任何列的不同数量的值（CLIENTCODE在您的情况下），我们可以使用nunique. 我们可以在agg函数中将输入作为字典传递，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)
