更改pandas groupby使用的函数值

您可以重写您的函数-创建由布尔值掩码填充的新列,然后聚合 mean 最后一个倍数 100 具有 Series.mul :

n = 3

full_data['new'] = full_data['Volume'] <= n
full_data = full_data.groupby(['Id', 'Week_id'])['new'].mean().mul(100).reset_index()

带功能的解决方案:

def per(df, n):
    df['new'] = df['Volume'] <= n
    return df.groupby(['Id', 'Week_id'])['new'].mean().mul(100).reset_index()

编辑:解决方案来源 github :

full_data = pd.DataFrame({
        'Id':list('XXYYZZXYZX'),
         'Volume':[2,4,8,1,2,5,8,2,6,4],
         'Week_id':list('aaabbbabac')
})

print (full_data)

val = 5
def per(c):
    def f1(x):
        return x[(x<=c)].count() / x.count() * 100
    return f1

full_data2 = full_data.groupby(['Id', 'Week_id']).agg({'Volume': per(val)}).reset_index()
print (full_data2)
  Id Week_id      Volume
0  X       a   66.666667
1  X       c  100.000000
2  Y       a    0.000000
3  Y       b  100.000000
4  Z       a    0.000000
5  Z       b  100.000000

def percentage(x):
    return x[(x<=val)].count() / x.count() * 100

full_data1 = full_data.groupby(['Id', 'Week_id'], as_index=False).agg({'Volume': percentage})

print (full_data1)
  Id Week_id      Volume
0  X       a   66.666667
1  X       c  100.000000
2  Y       a    0.000000
3  Y       b  100.000000
4  Z       a    0.000000
5  Z       b  100.000000

我想出了一个最简洁的方法来解决我的问题:

def percentage(x):
    global c
    return x[(x<=c)].count() / x.count() * 100

c=5
full_data_5 = full_data.groupby(['Id', 'Week_id'], as_index=False).agg({'Volume': percentage})

c=7
full_data_7 = full_data.groupby(['Id', 'Week_id'], as_index=False).agg({'Volume': percentage})

c=9
full_data_9 = full_data.groupby(['Id', 'Week_id'], as_index=False).agg({'Volume': percentage})

# etc

但是,我使用的是全局变量,这是一个有争议的实践。