熊猫计算 groupby 结果中的特定值

Question

我刚刚学习了如何将 Pandas DataFrame 与日期时间索引按日期分组。如何计算 groupby 每天返回的特定值（“通过”或“失败”）的数量？

我的目标是计算每日产量：产量 = passed_count/(passed_count + failed_count)。

import pandas as pd
import random

data = [random.choice(['Passed','Failed']) for i in range(100)]
times = pd.date_range('20180101',freq='3H',periods=100)

df = pd.DataFrame({'A':data},index=times)
group = df.groupby(pd.Grouper(freq='1D'))

Answer 1

我认为你需要GroupBy.size 或SeriesGroupBy.value_counts，用unstack 重塑，最后除以sum：

np.random.seed(142)

data = np.random.choice(['Passed','Failed'], size=100)
times = pd.date_range('20180101',freq='3H',periods=100)

df = pd.DataFrame({'A':data},index=times)
#print (df)

---

df1 = df.groupby([pd.Grouper(freq='1D'), 'A']).size().unstack(fill_value=0)
#alternative 
#df1 = df.groupby(pd.Grouper(freq='1D'))['A'].value_counts().unstack(fill_value=0)
df1['new'] = df1['Passed'].div(df1.sum(axis=1))
#alternative
#df1['new'] = df1['Passed'].div(df1['Passed'].add(df1['Failed']))

print (df1)
A           Failed  Passed    new
2018-01-01       5       3  0.375
2018-01-02       5       3  0.375
2018-01-03       4       4  0.500
2018-01-04       4       4  0.500
2018-01-05       4       4  0.500
2018-01-06       3       5  0.625
2018-01-07       4       4  0.500
2018-01-08       5       3  0.375
2018-01-09       4       4  0.500
2018-01-10       4       4  0.500
2018-01-11       4       4  0.500
2018-01-12       7       1  0.125
2018-01-13       0       4  1.000