Pandas DataFrame:删除所有leap年中的特定日期 [英] Pandas DataFrame: Delete specific date in all leap years

问题描述

以下序列是我获得的pandas DataFrame的摘录:

The following sequence is an extract of the pandas DataFrame that I've got:

>>> df_t
              value
2011-01-31    -5.575000
2011-03-31     7.700000
2011-05-31    15.966667
2011-07-31    10.683333
2011-08-31    10.454167
2011-10-31     9.320833
2011-12-31    -0.358333
2012-01-31   -11.550000
2012-03-31     1.700000
2012-05-31    12.333333
2012-07-31    12.816667
2012-08-31    11.837500
2012-10-31     2.733333
2012-12-31     4.075000
2013-01-31     2.450000
2013-03-31    -4.262500
2013-05-31    11.491667
2013-07-31    14.812500
2013-08-31    13.920833
2013-10-31     4.125000
2013-12-31     0.075000 

如何在每个leap年删除3月31日? 我尝试过类似的事情:

How can I delete March 31st in every leap year? I tried something like:

def isleap(year):
return year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)

if isleap(df_t.index.year):
        df_t=df_t[df_t.index.dayofyear!=91]

...但是很明显，这在我脑海中太简单了.唯一的解决方案是遍历整个数据框并在每一步进行检查，如果年份是a年，日期是一年的第91天，还是有任何更简单的解决方案可用?

...but obviously, this was too straightforward in my head. Is the only solution to loop through the whole dataframe and check at every step if the year is a leap year and the date is 91st day of year or is there any easier solution available?

问题不在于如何确定年份是否为a年，而是(如果是的话)删除上述数据框中的3月31日.

The issue is not how to determine whether a year is a leap year, but, if so, to delete March 31st in the above dataframe.

推荐答案

下面是一个以矢量化方式执行此操作的示例.您应注意，and和or不适合布尔向量，请使用&和|.

Here is an example to do that in a vectorized way. You shall note that and and or are not appropriate for a vector of booleans, use & and | instead.

import pandas as pd
import numpy as np

s = pd.Series(np.random.randn(600), index=pd.date_range('1990-01-01', periods=600, freq='M'))

Out[76]: 
1990-01-31   -0.7594
1990-02-28   -0.1311
1990-03-31    1.2031
1990-04-30    1.1999
1990-05-31   -2.4399
               ...  
2039-08-31   -0.3554
2039-09-30   -0.3265
2039-10-31   -0.3832
2039-11-30   -1.4139
2039-12-31   -0.3086
Freq: M, dtype: float64


def is_leap_and_MarchEnd(s):
    return (s.index.year % 4 == 0) & ((s.index.year % 100 != 0) | (s.index.year % 400 == 0)) & (s.index.month == 3) & (s.index.day == 31)

mask = is_leap_and_MarchEnd(s)
s[mask]
Out[77]: 
1992-03-31    0.7834
1996-03-31    0.3121
2000-03-31   -1.2050
2004-03-31    0.6017
2008-03-31    0.1045
               ...  
2020-03-31    1.1037
2024-03-31    0.5139
2028-03-31   -0.8116
2032-03-31   -0.6939
2036-03-31   -1.1999
dtype: float64

# do delete these row
s[~mask]

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