pandas 如何在熊猫中将日期列拆分为单独的日、月、年列
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How to split a date column into separate day , month ,year column in pandas
提问by Ratnesh
I have dataset, df.head(4):
我有数据集，df.head(4)：
Dewptm Fog Humidity Pressurem Tempm Wspdm Rainfall
datetime_utc
1996-11-01 11.666667 0.0 52.916667 -2659.666667 22.333333 2.466667 0
1996-11-02 10.458333 0.0 48.625000 1009.833333 22.916667 8.028571 0
1996-11-03 12.041667 0.0 55.958333 1010.500000 21.791667 4.804545 0
1996-11-04 10.222222 0.0 48.055556 1011.333333 22.722222 1.964706 0
Here is df.columns:
这是df.columns：
Index(['Dewptm', 'Fog', 'Humidity', 'Pressurem', 'Rain', 'Tempm', 'Wspdm',
'Rainfall'],
dtype='object')
How could I split datetime_utccolumn into the year, month and day column?
如何将datetime_utc列拆分为年、月和日列？
I tried:
我试过：
df["day"] = df['datetime_utc'].map(lambda x: x.day)
df["month"] = df['datetime_utc'].map(lambda x: x.month)
df["year"] = df['datetime_utc'].map(lambda x: x.year)
Error:
错误：
KeyError: 'datetime_utc'
KeyError: 'datetime_utc'
Also
还
pd.concat([df.drop('datetime_utc', axis = 1),
(df.datetime_utc.str.split("-).str[:3].apply(pd.Series)
.rename(columns={0:'year', 1:'month', 2:'day'}))], axis = 1)
I am getting error:
我收到错误：
KeyError: "['datetime_utc'] not found in axis" The problem I am facing is the column
datetime_utcis the default index column in my dataset, Please suggest me an approach.
KeyError: "['datetime_utc'] not found in axis" 我面临的问题是该列
datetime_utc是我数据集中的默认索引列，请建议我一种方法。
回答by Erfan
The problem is that datetime_utcis in your index instead a column, so you have to access your index to be able to make your new columns:
问题是datetime_utc在您的索引中而不是一列，因此您必须访问索引才能创建新列：
df['day'] = df.index.day
df['month'] = df.index.month
df['year'] = df.index.year
print(df)
Dewptm Fog Humidity Pressurem Tempm Wspdm \
datetime_utc
1996-11-01 11.666667 0.0 52.916667 -2659.666667 22.333333 2.466667
1996-11-02 10.458333 0.0 48.625000 1009.833333 22.916667 8.028571
1996-11-03 12.041667 0.0 55.958333 1010.500000 21.791667 4.804545
1996-11-04 10.222222 0.0 48.055556 1011.333333 22.722222 1.964706
Rainfall day month year
datetime_utc
1996-11-01 0 1 11 1996
1996-11-02 0 2 11 1996
1996-11-03 0 3 11 1996
1996-11-04 0 4 11 1996
If you want datetime_utcas a column you have to reset your index and then you can access the datetime methods with dt.month, dt.yearand dt.daylike following:
如果你想datetime_utc为一列，你必须重置索引，然后您可以访问的日期时间的方法dt.month，dt.year以及dt.day类似以下内容：
# Reset our index so datetime_utc becomes a column
df.reset_index(inplace=True)
# Create new columns
df['day'] = df['datetime_utc'].dt.day
df['month'] = df['datetime_utc'].dt.month
df['year'] = df['datetime_utc'].dt.year
print(df)
datetime_utc Dewptm Fog Humidity Pressurem Tempm Wspdm \
0 1996-11-01 11.666667 0.0 52.916667 -2659.666667 22.333333 2.466667
1 1996-11-02 10.458333 0.0 48.625000 1009.833333 22.916667 8.028571
2 1996-11-03 12.041667 0.0 55.958333 1010.500000 21.791667 4.804545
3 1996-11-04 10.222222 0.0 48.055556 1011.333333 22.722222 1.964706
Rainfall day month year
0 0 1 11 1996
1 0 2 11 1996
2 0 3 11 1996
3 0 4 11 1996
Noteif you index is not in datetimetype yet, use the following before you try to extract year, month and day:
请注意，如果您的索引datetime尚未键入，请在尝试提取年、月和日之前使用以下内容：
df.index = pd.to_datetime(df.index)
