Question

我有这个数据框；

import pandas as pd
import numpy as np

df = pd.DataFrame({
    'Client':np.random.choice(['Customer_A', 'Customer_B'], 1000),
    'Product':np.random.choice( ['Guns', 'Ammo', 'Armour'], 1000),
    'Value':(np.random.randn(1000))
})
Categoricals = ['Client', 'Product']
df[Categoricals] = df[Categoricals].astype('category')
df = df.drop_duplicates()
df

我想要这个结果；

# Non-anonymous function for Anomaly limit
def Anomaly (x):
    Q3 = np.nanpercentile(x, q = 75)
    Q1 = np.nanpercentile(x, q = 25)
    IQR = (Q3 - Q1)
    return (Q3 + (IQR * 2.0))

# Non-anonymous function for CriticalAnomaly limit
def CriticalAnomaly (x):
    Q3 = np.nanpercentile(x, q = 75)
    Q1 = np.nanpercentile(x, q = 25)
    IQR = (Q3 - Q1)
    return (Q3 + (IQR * 3.0))


# Define metrics
Metrics = {'Value':['count', Anomaly, CriticalAnomaly]}

# Groupby has more than 1 grouping column, so agg can only accept non-anonymous functions
Limits = df.groupby(['Client', 'Product']).agg(Metrics)
Limits

但是在大型数据集上速度较慢，因为函数“ Anomaly”和“ CriticalAnomaly”必须重新计算Q1，Q3和IQR两次，而不是一次。通过将两个功能组合在一起，可以使其更快。但是结果输出到1列而不是2列。

# Combined anomaly functions
def CombinedAnom (x):
    Q3 = np.nanpercentile(x, q = 75)
    Q1 = np.nanpercentile(x, q = 25)
    IQR = (Q3 - Q1)
    Anomaly = (Q3 + (IQR * 2.0))
    CriticalAnomaly = (Q3 + (IQR * 3.0))
    return (Anomaly, CriticalAnomaly)

# Define metrics
Metrics = {'Value':['count', CombinedAnom]}

# Groupby has more than 1 grouping column, so agg can only accept non-anonymous functions
Limits = df.groupby(['Client', 'Product']).agg(Metrics)
Limits

如何创建组合函数，以便将结果分成两列？

Answer 1

如果您使用apply而不是agg，则可以返回一个Series，将其分解成列：

def f(g):
    return pd.Series({
        'c1': np.sum(g.b),
        'c2': np.prod(g.b)
    })

df = pd.DataFrame({'a': list('aabbcc'), 'b': [1,2,3,4,5,6]})
df.groupby('a').apply(f)

来源：

到

优化groupby agg函数以返回多个结果列

1 个答案: