如何在 Pandas 中转换 datetime 列的时区?

2022-01-13 00:00:00 python pandas datetime datetime-format timestamp

问题描述

我有一列(非索引列)，其中包含日期时间.例如，前五个条目如下所示:

I have a column (non-index column) that has datetimes inside it. For example, the first five entries look something like this:

[Timestamp('2018-11-15 19:57:55'),
 Timestamp('2018-11-15 19:59:46'),
 Timestamp('2018-11-15 20:00:59'),
 Timestamp('2018-11-15 20:01:41'),
 Timestamp('2018-11-15 20:01:54')]

我想将条目从 UTC 转换为太平洋时区.假设该列被称为 times 我目前正在执行以下操作:

I want to convert the entries from UTC to the Pacific timezone. Assuming the column is called times I am currently doing the following:

times.dt.tz_localize('GMT').dt.tz_convert('America/Los_Angeles')

虽然这成功地将列从 UTC 转换为 PST，但输出包含我不想要的无关组件.如下所示:

While this successfully converts the column from UTC to PST, the output has extraneous components that I do not want. It looks like the following:

[Timestamp('2018-11-15 11:57:55-0800', tz='America/Los_Angeles'),
 Timestamp('2018-11-15 11:59:46-0800', tz='America/Los_Angeles'),
 Timestamp('2018-11-15 12:00:59-0800', tz='America/Los_Angeles'),
 Timestamp('2018-11-15 12:01:41-0800', tz='America/Los_Angeles'),
 Timestamp('2018-11-15 12:01:54-0800', tz='America/Los_Angeles')]

如何从时间戳中删除或忽略 -0800?谢谢！

How do I remove or ignore the -0800 from the timestamps? Thanks!

解决方案

只需添加最后一步.tz_localize(None):

import pandas as pd
d = pd.Series(['2018-11-15 19:57:55', '2018-11-15 19:59:46'])
d = pd.to_datetime(d)
d
0   2018-11-15 19:57:55
1   2018-11-15 19:59:46
dtype: datetime64[ns]

d_pacific_tz_aware = d.dt.tz_localize("GMT").dt.tz_convert('America/Los_Angeles')
d_pacific_tz_aware
0   2018-11-15 11:57:55-08:00
1   2018-11-15 11:59:46-08:00
dtype: datetime64[ns, America/Los_Angeles]


d_pacific_tz_naive = d.dt.tz_localize("GMT").dt.tz_convert('America/Los_Angeles').dt.tz_localize(None)
d_pacific_tz_naive
0   2018-11-15 11:57:55
1   2018-11-15 11:59:46
dtype: datetime64[ns]

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