pandas 从熊猫数据框中删除闰年
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Remove leap year day from pandas dataframe
提问by user308827
I have the foll. dataframe:
我有一个愚蠢的。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
I would like to drop all rows corresponding to Feb 29th, resulting in foll. data frame:
我想删除对应于 2 月 29 日的所有行，从而导致 foll。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
2012-02-28 125.5010
2016-01-07 125.5010
2016-01-08 125.5010
Right now, I am just doing it manually:
现在，我只是手动完成：
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.
df.drop(df.index[['2012-02-29']]). 我怎样才能使它适用于所有年份，而不必手动指定行索引。
采纳答案by jezrael
IIUC you can mask it and remove by loc:
IIUC 你可以屏蔽它并通过loc以下方式删除：
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
回答by Fabio Lamanna
If your dataframe has already the datetimecolumn as index you can:
如果您的数据框已经将该datetime列作为索引，您可以：
df = df[~((df.index.month == 2) & (df.index.day == 29))]
this should remove the rows containing the day February 29th for all years.
这应该删除所有年份中包含 2 月 29 日这一天的行。
回答by Markus Weninger
You can see the date as stringand see if it ends with 02-29:
您可以将日期视为string并查看它是否以以下结尾02-29：
df = df[~df.index.str.endswith('02-29')]
Using this method, you can use any string-comparism method like contains, etc.
使用此方法，您可以使用任何字符串比较方法，例如contains等。
