月相差YYYYMM pandas [英] Month difference YYYYMM Pandas

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本文介绍了月相差YYYYMM pandas 的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我在数据框中有两个日期列,它们是浮点型的,所以我将其转换为日期格式YYYYMM.现在我必须找出两个月之间的差额 他们.我尝试了以下操作,但是却出错了.

I had two date columns in the data frame, which was of float type, So I converted it in to date format YYYYMM. Now I have to find the difference of months between them. I tried the below, but I goves error.

df['Date_1'] = pd.to_datetime(df['Date_1'], format = '%Y%m%d').dt.strftime('%Y%m') #Convert float to YYYYMM Format
df['Date_2'] = pd.to_datetime(df['Date_2'], format='%Y%m.0').dt.strftime('%Y%m') #Convert float to YYYYMM Format
df['diff'] = df['Date_1'] - df['Date_2'] #Gives error

推荐答案

我认为需要减去由

I think need subtract periods created byto_period :

df = pd.DataFrame({'Date_1':[20150810, 20160804],
                   'Date_2':[201505.0, 201602.0]})

print (df)
     Date_1    Date_2
0  20150810  201505.0
1  20160804  201602.0

df['Date_1'] = pd.to_datetime(df['Date_1'], format = '%Y%m%d').dt.to_period('m') 
df['Date_2'] = pd.to_datetime(df['Date_2'], format='%Y%m.0').dt.to_period('m') 
df['diff'] = df['Date_1'] - df['Date_2'] 
print (df)
   Date_1  Date_2 diff
0 2015-08 2015-05    3
1 2016-08 2016-02    6

另一种解决方案是将Date_1转换为每月的第一天:

Another solution is convert Date_1 to first day of month:

df['Date_1'] = pd.to_datetime(df['Date_1'], format = '%Y%m%d') - pd.offsets.MonthBegin()
df['Date_2'] = pd.to_datetime(df['Date_2'], format='%Y%m.0')
df['diff'] = df['Date_1'] - df['Date_2'] 
print (df)
      Date_1     Date_2     diff
0 2015-08-01 2015-05-01  92 days
1 2016-08-01 2016-02-01 182 days

这篇关于月相差YYYYMM pandas 的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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