Python 获取 DataFrame 的日期时间列的工作日/星期几
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Get weekday/day-of-week for Datetime column of DataFrame
提问by EP1986
I have a DataFrame dflike the following (excerpt, 'Timestamp' are the index):
我有一个如下df所示的数据帧（摘录，“时间戳”是索引）：
Timestamp Value
2012-06-01 00:00:00 100
2012-06-01 00:15:00 150
2012-06-01 00:30:00 120
2012-06-01 01:00:00 220
2012-06-01 01:15:00 80
...and so on.
I need a new column df['weekday']with the respective weekday/day-of-week of the timestamps.
我需要一个新列df['weekday']，其中包含时间戳的相应工作日/星期几。
How can I get this?
我怎样才能得到这个？
采纳答案by EdChum
Use the new dt.dayofweekproperty:
使用新dt.dayofweek属性：
In [2]:
df['weekday'] = df['Timestamp'].dt.dayofweek
df
Out[2]:
Timestamp Value weekday
0 2012-06-01 00:00:00 100 4
1 2012-06-01 00:15:00 150 4
2 2012-06-01 00:30:00 120 4
3 2012-06-01 01:00:00 220 4
4 2012-06-01 01:15:00 80 4
In the situation where the Timestampis your index you need to reset the index and then call the dt.dayofweekproperty:
在Timestamp是您的索引的情况下，您需要重置索引，然后调用该dt.dayofweek属性：
In [14]:
df = df.reset_index()
df['weekday'] = df['Timestamp'].dt.dayofweek
df
Out[14]:
Timestamp Value weekday
0 2012-06-01 00:00:00 100 4
1 2012-06-01 00:15:00 150 4
2 2012-06-01 00:30:00 120 4
3 2012-06-01 01:00:00 220 4
4 2012-06-01 01:15:00 80 4
Strangely if you try to create a series from the index in order to not reset the index you get NaNvalues as does using the result of reset_indexto call the dt.dayofweekproperty without assigning the result of reset_indexback to the original df:
奇怪的是，如果您尝试从索引创建一个系列以便不重置索引，您将获得NaN值，就像使用 的结果reset_index调用dt.dayofweek属性而不将结果分配reset_index回原始 df 一样：
In [16]:
df['weekday'] = pd.Series(df.index).dt.dayofweek
df
Out[16]:
Value weekday
Timestamp
2012-06-01 00:00:00 100 NaN
2012-06-01 00:15:00 150 NaN
2012-06-01 00:30:00 120 NaN
2012-06-01 01:00:00 220 NaN
2012-06-01 01:15:00 80 NaN
In [17]:
df['weekday'] = df.reset_index()['Timestamp'].dt.dayofweek
df
Out[17]:
Value weekday
Timestamp
2012-06-01 00:00:00 100 NaN
2012-06-01 00:15:00 150 NaN
2012-06-01 00:30:00 120 NaN
2012-06-01 01:00:00 220 NaN
2012-06-01 01:15:00 80 NaN
EDIT
编辑
As pointed out to me by user @joris you can just access the weekdayattribute of the index so the following will work and is more compact:
正如用户@joris 向我指出的那样，您可以访问weekday索引的属性，因此以下内容将起作用并且更加紧凑：
df['Weekday'] = df.index.weekday
df['Weekday'] = df.index.weekday
回答by QM.py
If the Timestampif a datatime, then you can just use:df['weekday'] = df['Timestamp'].apply(lambda x: x.weekday())
如果Timestampif a datatime，那么您可以使用：df['weekday'] = df['Timestamp'].apply(lambda x: x.weekday())
or
或者
df['weekday'] = pd.to_datetime(df['Timestamp']).apply(lambda x: x.weekday())
df['weekday'] = pd.to_datetime(df['Timestamp']).apply(lambda x: x.weekday())
回答by TiTo
In case somebody else has the same issue with a multiindexed dataframe, here is what solved it for me, based on @joris solution:
如果其他人对多索引数据帧有同样的问题，以下是基于@joris 解决方案为我解决的问题：
df['Weekday'] = df.index.get_level_values(1).weekday
for me date was the get_level_values(1)instead of get_level_values(0), which would work for the outer index.
对我来说 date 是get_level_values(1)代替get_level_values(0)，它适用于外部索引。
