Question

我正在尝试按时间戳分组。首先，我必须将获得的时间（字符串）转换为日期时间。转换日期时间后，我注意到尽管给出了熊猫添加日期的特定格式，但我不需要日期。我正在努力删除此内容，仅保留时间对象，但没有成功。我为删除日期所做的任何操作都会将dtype返回给无法执行groupby的对象。

示例数据：

https://miratrix.co.uk/          00:01:55
https://miratrix.co.uk/          00:02:02
https://miratrix.co.uk/          00:02:45
https://miratrix.co.uk/          00:01:22
https://miratrix.co.uk/          00:02:02
https://miratrix.co.uk/app-marketing-agency/          00:02:23
https://miratrix.co.uk/get-in-touch/          00:02:26
https://miratrix.co.uk/get-in-touch/          00:00:18
https://miratrix.co.uk/get-in-touch/          00:02:37
https://miratrix.co.uk/          00:00:31
https://miratrix.co.uk/          00:02:00
https://miratrix.co.uk/app-store-optimization-...          00:02:25
https://miratrix.co.uk/          00:03:36
https://miratrix.co.uk/app-marketing-agency/          00:02:09
https://miratrix.co.uk/get-in-touch/          00:02:14
https://?page_id=16198/          00:00:15
https://videos/channel/UCAQfRNzXGD4BQICkO1KQZUA/          00:09:07
https://miratrix.co.uk/get-in-touch/          00:01:39
https://miratrix.co.uk/app-marketing-agency/          00:01:07

到目前为止我尝试过的事情

*Returned Object*
ga_organic['NEW Avg. Time on Page'] = pd.to_datetime(ga_organic['Avg. Time on Page'], format="%H:%M:%S").dt.time

*Returned Datetime but when trying to sample only time it returned an object*
ga_organic['NEW Avg. Time on Page'] = ga_organic['Avg. Time on Page'].astype('datetime64[ns]')

ga_organic['NEW Avg. Time on Page'].dt.time

我有一种关于日期时间的感觉，我不知道，这就是为什么我有这个问题。欢迎任何帮助或指示。

#### Update ####

感谢ALollz为时间戳提供解决方案。

ga_organic['NEW Avg. Time on Page'] = pd.to_timedelta(ga_organic['Avg. Time on Page'])

但是使用GroupBy通过此方法时，我仍然遇到相同的错误：

avg_time = ga_organic.groupby(ga_organic.index)['NEW Avg. Time on Page'].mean()

错误：“数据错误：没有要聚合的数字类型”

是否有用于处理分组日期时间的特定功能？

Answer 1

似乎groupby无法将timedelta64识别为数字类型。有几种解决方法，可以使用numeric_only=False或使用total_seconds。

import pandas as pd

#df = pd.read_clipboard(header=None)
#df[1] = pd.to_timedelta(df[1])

df.groupby(df.index//2)[1].mean()
#DataError: No numeric types to aggregate

# To fix pass `numeric_only=False`
df.groupby(df.index//2)[1].mean(numeric_only=False)
#0   00:01:58.500000
#1   00:02:03.500000
#2   00:02:12.500000
#3          00:01:22
#4          00:01:34
#5   00:02:12.500000
#6   00:02:52.500000
#7   00:01:14.500000
#8          00:05:23
#9          00:01:07
#Name: 1, dtype: timedelta64[ns]

在float中使用简单的.total_seconds值：

df[1] = df[1].dt.total_seconds()

df.groupby(df.index//2)[1].mean()
#0    118.5
#1    123.5
#2    132.5
#3     82.0
#4     94.0
#5    132.5
#6    172.5
#7     74.5
#8    323.0
#9     67.0
#Name: 1, dtype: float64

可以使用pd.to_timedelta指定unit='s'来将其转换回

Datetime dtype是Object而不是Datetime

1 个答案: