python - 更改 pandas groupby 使用的函数中的值

Question

我正在执行以下操作:

def percentage(x):
    return x[(x<=5)].count() / x.count() * 100

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

但我想这样做groupby连续具有多个值，例如 x<=7 , x<=9 , x<=11等percentage功能。

除了编写多个函数并调用它们之外，最简单的方法是什么？

所以基本上我想避免做这样的事情:

def percentage_1(x):
    return x[(x<=5)].count() / x.count() * 100

full_data_1 = full_data.groupby(['Id', 'Week_id'], as_index=False).agg({'Volume': percentage_1})

def percentage_2(x):
    return x[(x<=7)].count() / x.count() * 100

full_data_2 = full_data.groupby(['Id', 'Week_id'], as_index=False).agg({'Volume': percentage_2})

# etc.

Answer 1

您可以重写您的函数 - 创建由 bool 掩码填充的新列，然后使用 Series.mul 聚合 mean 和最后一个倍数 100 :

n = 3

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

具有功能的解决方案:

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

编辑:来自github的解决方案:

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