Python/Pandas 分箱数据 Timedelta答案

【问题标题】：Python/Pandas Binning Data TimedeltaPython/Pandas 分箱数据 Timedelta
【发布时间】：2018-04-06 10:01:32
【问题描述】：

我有一个包含两列的 DataFrame

    userID     duration
0   DSm7ysk    03:08:49
1   no51CdJ    00:35:50
2   ...

'duration' 类型为 timedelta。我试过使用

bins = [dt.timedelta(minutes = 0), dt.timedelta(minutes = 
        5),dt.timedelta(minutes = 10),dt.timedelta(minutes = 
        20),dt.timedelta(minutes = 30), dt.timedelta(hours = 4)]

labels = ['0-5min','5-10min','10-20min','20-30min','30min+']

df['bins'] = pd.cut(df['duration'], bins, labels = labels)

但是，分箱数据不使用指定的分箱，而是为帧中的每个持续时间创建的。

将 timedelta 对象分箱到不规则箱中的最简单方法是什么？还是我只是在这里遗漏了一些明显的东西？

【问题讨论】：

标签： python pandas datetime timedelta binning

【解决方案1】：

您可以在分箱前标准化到秒。这减少了对整数进行分箱的问题。

df = pd.DataFrame({'userID': ['A', 'B'],
                   'duration': pd.to_timedelta(['00:08:49', '00:35:50'])})

L = ['00:00:00', '00:05:00', '00:10:00', '00:20:00', '00:30:00', '04:00:00']

bins = pd.to_timedelta(L).total_seconds()
cats = ['0-5min', '5-10min', '10-20min', '20-30min', '30min+']

df['bins'] = pd.cut(df['duration'].dt.total_seconds(), bins, labels=cats)

print(df)

#    duration userID     bins
# 0  00:08:49      A  5-10min
# 1  00:35:50      B   30min+

【讨论】：

【解决方案2】：

它适用于我的 pandas 0.23.4

import pandas as pd
import numpy as np

df = pd.DataFrame({
    'userID': ['DSm7ysk', 'no51CdJ', 'foo', 'bar'],
    'duration': [pd.Timedelta('3 hours 8 minutes 49 seconds'), pd.Timedelta('35 minutes 50 seconds'), pd.Timedelta('1 minutes 13 seconds'), pd.Timedelta('6 minutes 43 seconds')]
})

bins = [
    pd.Timedelta(minutes = 0),
    pd.Timedelta(minutes = 5),
    pd.Timedelta(minutes = 10),
    pd.Timedelta(minutes = 20),
    pd.Timedelta(minutes = 30),
    pd.Timedelta(hours = 4)
]

labels = ['0-5min', '5-10min', '10-20min', '20-30min', '30min+']

df['bins'] = pd.cut(df['duration'], bins, labels = labels)

结果：

【讨论】：