从 pandas 数据框中删除leap年 [英] Remove leap year day from pandas dataframe

问题描述

我有傻瓜.数据框:

datetime
2012-01-01    125.5010
2012-01-02    125.5010
2012-01-03    125.5010
2012-02-04    125.5010
2012-02-05    125.5010
2012-02-29    125.5010
2012-02-28    125.5010
2016-01-07    125.5010
2016-01-08    125.5010
2016-02-29     81.6237

我想删除所有与2月29日相对应的行，结果是foll.数据框:

I would like to drop all rows corresponding to Feb 29th, resulting in foll. data frame:

datetime
2012-01-01    125.5010
2012-01-02    125.5010
2012-01-03    125.5010
2012-02-04    125.5010
2012-02-05    125.5010
2012-02-28    125.5010
2016-01-07    125.5010
2016-01-08    125.5010

现在，我只是手动进行操作

Right now, I am just doing it manually:

df.drop(df.index[['2012-02-29']]).我如何才能使其在所有年份都有效，而不必手动指定行索引.

df.drop(df.index[['2012-02-29']]). How can I make it so that it works for all years, without haveing to manually specify row index.

推荐答案

IIUC，您可以屏蔽它并通过

IIUC you can mask it and remove by loc:

def is_leap_and_29Feb(s):
    return (s.index.year % 4 == 0) & 
           ((s.index.year % 100 != 0) | (s.index.year % 400 == 0)) & 
           (s.index.month == 2) & (s.index.day == 29)

mask = is_leap_and_29Feb(df)
print mask
#[False False False False False  True False False False  True]

print df.loc[~mask]
#            datetime
#2012-01-01   125.501
#2012-01-02   125.501
#2012-01-03   125.501
#2012-02-04   125.501
#2012-02-05   125.501
#2012-02-28   125.501
#2016-01-07   125.501
#2016-01-08   125.501

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