问题：将DataFrame列类型从字符串转换为日期时间，格式为dd / mm / yyyy

如何将字符串的DataFrame列（以dd / mm / yyyy格式）转换为日期时间？

点击查看英文原文

How can I convert a DataFrame column of strings (in dd/mm/yyyy format) to datetimes?

回答 0

最简单的方法是使用to_datetime：

df['col'] = pd.to_datetime(df['col'])

它还dayfirst为欧洲时代提供了依据（但请注意，这并不严格）。

它在起作用：

In [11]: pd.to_datetime(pd.Series(['05/23/2005']))
Out[11]:
0   2005-05-23 00:00:00
dtype: datetime64[ns]

您可以传递特定格式：

In [12]: pd.to_datetime(pd.Series(['05/23/2005']), format="%m/%d/%Y")
Out[12]:
0   2005-05-23
dtype: datetime64[ns]
点击查看英文原文

The easiest way is to use to_datetime:

df['col'] = pd.to_datetime(df['col'])

It also offers a dayfirst argument for European times (but beware this isn’t strict).

Here it is in action:

In [11]: pd.to_datetime(pd.Series(['05/23/2005']))
Out[11]:
0   2005-05-23 00:00:00
dtype: datetime64[ns]

You can pass a specific format:

In [12]: pd.to_datetime(pd.Series(['05/23/2005']), format="%m/%d/%Y")
Out[12]:
0   2005-05-23
dtype: datetime64[ns]

回答 1

如果您的日期列是格式为’2017-01-01’的字符串，则可以使用pandas astype将其转换为日期时间。

df['date'] = df['date'].astype('datetime64[ns]')

或使用datetime64 [D]（如果您想要“天”精度而不是纳秒）

print(type(df_launath['date'].iloc[0]))

Yield

<class 'pandas._libs.tslib.Timestamp'> 与使用pandas.to_datetime时相同

您可以尝试使用其他格式，然后是’％Y-％m-％d’，但至少可以使用。

点击查看英文原文

If your date column is a string of the format ‘2017-01-01’ you can use pandas astype to convert it to datetime.

df['date'] = df['date'].astype('datetime64[ns]')

or use datetime64[D] if you want Day precision and not nanoseconds

print(type(df_launath['date'].iloc[0]))

yields

<class 'pandas._libs.tslib.Timestamp'> the same as when you use pandas.to_datetime

You can try it with other formats then ‘%Y-%m-%d’ but at least this works.

回答 2

如果要指定复杂的格式，可以使用以下内容：

df['date_col'] =  pd.to_datetime(df['date_col'], format='%d/%m/%Y')

format此处的更多详细信息：

点击查看英文原文

You can use the following if you want to specify tricky formats:

df['date_col'] =  pd.to_datetime(df['date_col'], format='%d/%m/%Y')

More details on format here:

回答 3

如果您的约会中有多种格式，别忘了设定infer_datetime_format=True以简化生活

df['date'] = pd.to_datetime(df['date'], infer_datetime_format=True)

资料来源：pd.to_datetime

或者，如果您想要定制的方法：

def autoconvert_datetime(value):
    formats = ['%m/%d/%Y', '%m-%d-%y']  # formats to try
    result_format = '%d-%m-%Y'  # output format
    for dt_format in formats:
        try:
            dt_obj = datetime.strptime(value, dt_format)
            return dt_obj.strftime(result_format)
        except Exception as e:  # throws exception when format doesn't match
            pass
    return value  # let it be if it doesn't match

df['date'] = df['date'].apply(autoconvert_datetime)
点击查看英文原文

If you have a mixture of formats in your date, don’t forget to set infer_datetime_format=True to make life easier

df['date'] = pd.to_datetime(df['date'], infer_datetime_format=True)

Source: pd.to_datetime

or if you want a customized approach:

def autoconvert_datetime(value):
    formats = ['%m/%d/%Y', '%m-%d-%y']  # formats to try
    result_format = '%d-%m-%Y'  # output format
    for dt_format in formats:
        try:
            dt_obj = datetime.strptime(value, dt_format)
            return dt_obj.strftime(result_format)
        except Exception as e:  # throws exception when format doesn't match
            pass
    return value  # let it be if it doesn't match

df['date'] = df['date'].apply(autoconvert_datetime)
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