2015-07-21 98 views
2

如果我有這樣的熊貓差值

Date   value 
04 May 2015  1 
06 May 2015  1 
07 May 2015  1 
11 May 2015  1 
11 May 2015  1 

一個數據幀如何獲得日期索引的差異?即低於第三西:

Date   value Diff 
04 May 2015  1  NA 
06 May 2015  1  2 
07 May 2015  1  1 
11 May 2015  1  4 
11 May 2015  1  0 
+0

的可能的複製[計算大熊貓據幀指數之間的時間差(https://stackoverflow.com/questions/16777570/calculate熊貓數據框索引之間的時間差異) – jimijazz

回答

3

你的意思是這樣的:

df["Diff"] = df.index 
df["Diff"] = (df['Diff'] - df['Diff'].shift()) 

print(df) 
      value Diff 
Date      
2015-05-04  1 NaT 
2015-05-06  1 2 days 
2015-05-07  1 1 days 
2015-05-11  1 4 days 
2015-05-11  1 0 days 
+0

就是這樣。其實,我也意識到我需要在應用上面的代碼之前從字符串格式化日期(即用df [「Diff」] = pd.to_datetime(df.index,dayfirst = True)替換第一行) – dram