问题:如何将熊猫数据框中的日期转换为“日期”数据类型?

我有一个熊猫数据框,其中一列包含格式为日期的字符串 YYYY-MM-DD

例如 '2013-10-28'

目前该dtype列的是object

如何将列值转换为Pandas日期格式?

I have a Pandas data frame, one of the column contains date strings in the format YYYY-MM-DD

For e.g. '2013-10-28'

At the moment the dtype of the column is object.

How do I convert the column values to Pandas date format?


回答 0

使用类型

In [31]: df
Out[31]: 
   a        time
0  1  2013-01-01
1  2  2013-01-02
2  3  2013-01-03

In [32]: df['time'] = df['time'].astype('datetime64[ns]')

In [33]: df
Out[33]: 
   a                time
0  1 2013-01-01 00:00:00
1  2 2013-01-02 00:00:00
2  3 2013-01-03 00:00:00

Use astype

In [31]: df
Out[31]: 
   a        time
0  1  2013-01-01
1  2  2013-01-02
2  3  2013-01-03

In [32]: df['time'] = df['time'].astype('datetime64[ns]')

In [33]: df
Out[33]: 
   a                time
0  1 2013-01-01 00:00:00
1  2 2013-01-02 00:00:00
2  3 2013-01-03 00:00:00

回答 1

基本上等同于@waitingkuo,但我将to_datetime在这里使用(它看起来更干净一些,并提供了一些其他功能,例如dayfirst):

In [11]: df
Out[11]:
   a        time
0  1  2013-01-01
1  2  2013-01-02
2  3  2013-01-03

In [12]: pd.to_datetime(df['time'])
Out[12]:
0   2013-01-01 00:00:00
1   2013-01-02 00:00:00
2   2013-01-03 00:00:00
Name: time, dtype: datetime64[ns]

In [13]: df['time'] = pd.to_datetime(df['time'])

In [14]: df
Out[14]:
   a                time
0  1 2013-01-01 00:00:00
1  2 2013-01-02 00:00:00
2  3 2013-01-03 00:00:00

处理ValueError小号
如果碰上的情况下做

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

抛出一个

ValueError: Unknown string format

这意味着您具有无效(不可强制)的值。如果可以将它们转换为pd.NaT,可以在其中添加一个errors='coerce'参数to_datetime

df['time'] = pd.to_datetime(df['time'], errors='coerce')

Essentially equivalent to @waitingkuo, but I would use to_datetime here (it seems a little cleaner, and offers some additional functionality e.g. dayfirst):

In [11]: df
Out[11]:
   a        time
0  1  2013-01-01
1  2  2013-01-02
2  3  2013-01-03

In [12]: pd.to_datetime(df['time'])
Out[12]:
0   2013-01-01 00:00:00
1   2013-01-02 00:00:00
2   2013-01-03 00:00:00
Name: time, dtype: datetime64[ns]

In [13]: df['time'] = pd.to_datetime(df['time'])

In [14]: df
Out[14]:
   a                time
0  1 2013-01-01 00:00:00
1  2 2013-01-02 00:00:00
2  3 2013-01-03 00:00:00

Handling ValueErrors
If you run into a situation where doing

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

Throws a

ValueError: Unknown string format

That means you have invalid (non-coercible) values. If you are okay with having them converted to pd.NaT, you can add an errors='coerce' argument to to_datetime:

df['time'] = pd.to_datetime(df['time'], errors='coerce')

回答 2

我想象大量数据从CSV文件输入到Pandas中,在这种情况下,您可以简单地在初始CSV读取期间转换日期:

dfcsv = pd.read_csv('xyz.csv', parse_dates=[0])其中0表示日期所在的列。如果希望日期成为索引,
也可以, index_col=0在其中添加。

参见https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.read_csv.html

I imagine a lot of data comes into Pandas from CSV files, in which case you can simply convert the date during the initial CSV read:

dfcsv = pd.read_csv('xyz.csv', parse_dates=[0]) where the 0 refers to the column the date is in.
You could also add , index_col=0 in there if you want the date to be your index.

See https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.read_csv.html


回答 3

现在你可以做 df['column'].dt.date

请注意,对于日期时间对象,如果您没有看到它们均为00:00:00的小时数,则说明它不是熊猫。那是iPython笔记本,试图使事情看起来更漂亮。

Now you can do df['column'].dt.date

Note that for datetime objects, if you don’t see the hour when they’re all 00:00:00, that’s not pandas. That’s iPython notebook trying to make things look pretty.


回答 4

执行此操作的另一种方法,如果您有多个要转换为日期时间的列,则此方法效果很好。

cols = ['date1','date2']
df[cols] = df[cols].apply(pd.to_datetime)

Another way to do this and this works well if you have multiple columns to convert to datetime.

cols = ['date1','date2']
df[cols] = df[cols].apply(pd.to_datetime)

回答 5

如果要获取DATE而不是DATETIME格式:

df["id_date"] = pd.to_datetime(df["id_date"]).dt.date

If you want to get the DATE and not DATETIME format:

df["id_date"] = pd.to_datetime(df["id_date"]).dt.date

回答 6

在某些情况下,可能需要将日期转换为其他频率。在这种情况下,我建议按日期设置索引。

#set an index by dates
df.set_index(['time'], drop=True, inplace=True)

此后,您可以更轻松地转换为最需要的日期格式类型。在下面,我依次转换为多种日期格式,最终以每个月初的一组每日日期结束。

#Convert to daily dates
df.index = pd.DatetimeIndex(data=df.index)

#Convert to monthly dates
df.index = df.index.to_period(freq='M')

#Convert to strings
df.index = df.index.strftime('%Y-%m')

#Convert to daily dates
df.index = pd.DatetimeIndex(data=df.index)

为了简洁起见,我没有显示在上面的每一行之后都运行以下代码:

print(df.index)
print(df.index.dtype)
print(type(df.index))

这给了我以下输出:

Index(['2013-01-01', '2013-01-02', '2013-01-03'], dtype='object', name='time')
object
<class 'pandas.core.indexes.base.Index'>

DatetimeIndex(['2013-01-01', '2013-01-02', '2013-01-03'], dtype='datetime64[ns]', name='time', freq=None)
datetime64[ns]
<class 'pandas.core.indexes.datetimes.DatetimeIndex'>

PeriodIndex(['2013-01', '2013-01', '2013-01'], dtype='period[M]', name='time', freq='M')
period[M]
<class 'pandas.core.indexes.period.PeriodIndex'>

Index(['2013-01', '2013-01', '2013-01'], dtype='object')
object
<class 'pandas.core.indexes.base.Index'>

DatetimeIndex(['2013-01-01', '2013-01-01', '2013-01-01'], dtype='datetime64[ns]', freq=None)
datetime64[ns]
<class 'pandas.core.indexes.datetimes.DatetimeIndex'>

It may be the case that dates need to be converted to a different frequency. In this case, I would suggest setting an index by dates.

#set an index by dates
df.set_index(['time'], drop=True, inplace=True)

After this, you can more easily convert to the type of date format you will need most. Below, I sequentially convert to a number of date formats, ultimately ending up with a set of daily dates at the beginning of the month.

#Convert to daily dates
df.index = pd.DatetimeIndex(data=df.index)

#Convert to monthly dates
df.index = df.index.to_period(freq='M')

#Convert to strings
df.index = df.index.strftime('%Y-%m')

#Convert to daily dates
df.index = pd.DatetimeIndex(data=df.index)

For brevity, I don’t show that I run the following code after each line above:

print(df.index)
print(df.index.dtype)
print(type(df.index))

This gives me the following output:

Index(['2013-01-01', '2013-01-02', '2013-01-03'], dtype='object', name='time')
object
<class 'pandas.core.indexes.base.Index'>

DatetimeIndex(['2013-01-01', '2013-01-02', '2013-01-03'], dtype='datetime64[ns]', name='time', freq=None)
datetime64[ns]
<class 'pandas.core.indexes.datetimes.DatetimeIndex'>

PeriodIndex(['2013-01', '2013-01', '2013-01'], dtype='period[M]', name='time', freq='M')
period[M]
<class 'pandas.core.indexes.period.PeriodIndex'>

Index(['2013-01', '2013-01', '2013-01'], dtype='object')
object
<class 'pandas.core.indexes.base.Index'>

DatetimeIndex(['2013-01-01', '2013-01-01', '2013-01-01'], dtype='datetime64[ns]', freq=None)
datetime64[ns]
<class 'pandas.core.indexes.datetimes.DatetimeIndex'>

回答 7

尝试使用pd.to_datetime函数将行之一转换为时间戳,然后使用.map将公式映射到整个列

Try to convert one of the rows into timestamp using the pd.to_datetime function and then use .map to map the formular to the entire column


回答 8

 #   Column          Non-Null Count   Dtype         
---  ------          --------------   -----         
 0   startDay        110526 non-null  object
 1   endDay          110526 non-null  object

import pandas as pd

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

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

 #   Column          Non-Null Count   Dtype         
---  ------          --------------   -----         
 0   startDay        110526 non-null  datetime64[ns]
 1   endDay          110526 non-null  datetime64[ns]
 #   Column          Non-Null Count   Dtype         
---  ------          --------------   -----         
 0   startDay        110526 non-null  object
 1   endDay          110526 non-null  object

import pandas as pd

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

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

 #   Column          Non-Null Count   Dtype         
---  ------          --------------   -----         
 0   startDay        110526 non-null  datetime64[ns]
 1   endDay          110526 non-null  datetime64[ns]

回答 9

为了完整起见,可能不是最直接的另一种选择,有点类似于@SSS提出的选择,但使用datetime库是:

import datetime
df["Date"] = df["Date"].apply(lambda x: datetime.datetime.strptime(x, '%Y-%d-%m').date())

For the sake of completeness, another option, which might not be the most straightforward one, a bit similar to the one proposed by @SSS, but using rather the datetime library is:

import datetime
df["Date"] = df["Date"].apply(lambda x: datetime.datetime.strptime(x, '%Y-%d-%m').date())

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