Pandas `.to_pydatetime()` 在 DataFrame 中不起作用 [英] Pandas `.to_pydatetime()` not working inside a DataFrame

问题描述

我有像 '03-21-2019' 这样的字符串，我想将其转换为原生 Python 日期时间对象:即 datetime.datetime 类型.通过 pandas 转换很容易:

I have strings like '03-21-2019' that I want to convert to the native Python datetime object: that is, of the datetime.datetime type. The conversion is easy enough through pandas:

import pandas as pd
import datetime as dt

date_str = '03-21-2019'
pd_Timestamp = pd.to_datetime(date_str)
py_datetime_object = pd_Timestamp.to_pydatetime()
print(type(py_datetime_object))

结果

<class 'datetime.datetime'>

这正是我想要的，因为我想通过从另一个中减去其中一个来计算 timedelta - 在本机 Python datetime.datetime 班级.但是，我的数据在 pd.DataFrame 中.当我尝试以下代码时:

This is precisely what I want, since I want to compute timedelta's by subtracting one of these from another - perfectly well-defined in the native Python datetime.datetime class. However, my data is in a pd.DataFrame. When I try the following code:

import pandas as pd
import datetime as dt

df = pd.DataFrame(columns=['Date'])
df.loc[0] = ['03-21-2019']
df['Date'] = df['Date'].apply(lambda x:
                              pd.to_datetime(x).to_pydatetime())
print(type(df['Date'].iloc[0]))

结果是

<class 'pandas._libs.tslibs.timestamps.Timestamp'>

这是 WRONG 类型，我终生无法弄清楚为什么只有部分 lambda 表达式是得到评估(即字符串到熊猫时间戳)，而不是最后一部分(即熊猫时间戳到日期时间.datetime).如果我显式定义函数，而不是使用 lambda 表达式，它也不起作用:

This is the WRONG type, and I can't for the life of me figure out why only part of the lambda expression is getting evaluated (that is, string-to-pandas-Timestamp), and not the last part (that is, pandas-Timestamp-to-datetime.datetime). It doesn't work if I define the function explicitly, either, instead of using a lambda expression:

import pandas as pd
import datetime as dt


def to_native_datetime(date_str: str) -> dt.datetime:
    return pd.to_datetime(date_str).to_pydatetime()


df = pd.DataFrame(columns=['Date'])
df.loc[0] = ['03-21-2019']
df['Date'] = df['Date'].apply(to_native_datetime)
print(type(df['Date'].iloc[0]))

结果和之前一样.它肯定在执行函数的一部分，因为结果不再是字符串.但我想要本机 Python datetime.datetime 对象，我看不到它.这看起来像是 pandas 中的一个错误，但我当然愿意将其视为我的用户错误.

The result is the same as before. It's definitely doing part of the function, as the result is not a string anymore. But I want the native Python datetime.datetime object, and I see no way of getting it. This looks like a bug in pandas, but I'm certainly willing to see it as user error on my part.

为什么我不能从 pandas.DataFrame 字符串列中获取本机 datetime.datetime 对象?

Why can't I get the native datetime.datetime object out of a pandas.DataFrame string column?

我看过这个话题和这个，但他们都没有回答我的问题.

I have looked at this thread and this one, but neither of them answer my question.

:还有更奇怪的事情:

import pandas as pd
import datetime as dt


def to_native_datetime(date_str: str) -> dt.datetime:
    return dt.datetime.strptime(date_str, '%m-%d-%Y')


df = pd.DataFrame(columns=['Date'])
df.loc[0] = ['03-21-2019']
df['Date'] = df['Date'].apply(to_native_datetime)
print(type(df['Date'].iloc[0]))

这里我什至没有使用 pandas 来转换字符串，我 STILL 得到一个

Here I'm not even using pandas to convert the string, and I STILL get a

<class 'pandas._libs.tslibs.timestamps.Timestamp'>

别说了！

非常感谢您的时间！

[进一步编辑]:显然，在这个线程，在 Nehal J Wani 的回答中，当您分配到 pd.DataFrame 时，pandas 会自动转换回其原生日期时间格式.这不是我想听到的，但显然，当我读出 pd.DataFrame 时，我将不得不即时转换.

[FURTHER EDIT]: Apparently, in this thread, in Nehal J Wani's answer, it comes out that pandas automatically converts back to its native datetime format when you assign into a pd.DataFrame. This is not what I wanted to hear, but apparently, I'm going to have to convert on-the-fly when I read out of the pd.DataFrame.

推荐答案

根据您的实际目标，您有几个没有直接提及的选项.

Depending on what your actual goal is, you've a couple options you didn't mention directly.

1) 如果您有一个静态日期时间对象或一列(pandas)时间戳，并且您愿意处理 Timedelta 的 Pandas 版本(pandas._libs.tslibs.timedeltas.Timedeltacode>)，可以直接在pandas中做减法:

1) If you have a static datetime object or a column of (pandas) Timestamps, and you're willing to deal with the Pandas version of a Timedelta (pandas._libs.tslibs.timedeltas.Timedelta), you can do the subtraction directly in pandas:

df = pd.DataFrame(columns=['Date'])
df.loc[0] = [pd.to_datetime('03-21-2019')]
df.loc[:, 'Offset'] = pd.Series([datetime.now()])
df.loc[:, 'Diff1'] = df['Offset'] - df['Date']
df.loc[:, 'Diff2'] = df['Date'] - datetime.now()

2) 如果您不关心 Dataframes，但愿意处理列表/numpy 数组，则可以通过对系列而不是单个元素进行操作，将日期时间转换为 Python 原生日期时间.下面，arr 是 datetime.datetime 对象的 numpy.ndarray.您可以使用 list(arr) 将其更改为日期时间的常规列表:

2) If you don't care about Dataframes, but are willing to deal with lists / numpy arrays, you can convert the datetimes to python-native datetimes by operating on the series rather than on individual elements. Below, arr is a numpy.ndarray of datetime.datetime objects. You can change it to a regular list of datetime with list(arr):

arr = df['Date'].dt.to_pydatetime()

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