R 将difftime的乘积矢量化_R_Date

R 将difftime的乘积矢量化

r date

R 将difftime的乘积矢量化,r,date,R,Date,所以我有一个数据框，看起来像这样： ID Initialdate Finaldate 1405 2003-12-03 2010-12-07 7044 2004-12-08 2011-10-13 7219 2008-05-16 2009-06-04 18618 2004-06-17 2012-02-13 19900 2005-06-01 2008-06-11 20138 2010-01-20 2010-01-20 29067 2003-04-30 200

所以我有一个数据框，看起来像这样：

    ID Initialdate  Finaldate
  1405  2003-12-03 2010-12-07
  7044  2004-12-08 2011-10-13
  7219  2008-05-16 2009-06-04
 18618  2004-06-17 2012-02-13
 19900  2005-06-01 2008-06-11
 20138  2010-01-20 2010-01-20
 29067  2003-04-30 2004-09-10
 33546  2003-11-25 2008-10-10
 37321  2003-06-07 2006-03-20
 43028  2004-09-23 2008-07-25
 43591  2005-04-06 2005-11-15
 46749  2005-02-28 2005-05-16
 48846  2005-08-02 2005-08-02
114353  2002-05-17 2006-10-26
128180  2004-06-17 2010-06-21
128648  2003-05-07 2009-07-23
133337  2004-05-26 2012-07-26
149181  2002-10-19 2008-07-27
214079  2003-09-26 2007-05-20
215060  2006-04-17 2011-08-17
229816  2007-04-25 2011-09-24
238123  2007-11-26 2012-01-31
253776  2006-03-02 2012-04-19
258660  2010-03-25 2012-04-09
265356  2002-04-22 2002-04-22

unqualified<-(df[df$Duration <= '179.000',])

unqualified<-(df[df$Duration < 180,])

我用以下代码制作了第四列，其中包含最终日期和初始日期之间的差异，并将其清理如下：

df$Duration<-(difftime(df$Finaldate, df$Initialdate, units = "days"))
df$Duration<-as.numeric(df$Duration, units = "days")

我的计划是将持续时间数据矢量化，特别是那些少于180天的数据，然后使用新的数据帧从初始数据帧中删除这些ID，代码如下：

df_final这样如何：
    ID Initialdate  Finaldate
  1405  2003-12-03 2010-12-07
  7044  2004-12-08 2011-10-13
  7219  2008-05-16 2009-06-04
 18618  2004-06-17 2012-02-13
 19900  2005-06-01 2008-06-11
 20138  2010-01-20 2010-01-20
 29067  2003-04-30 2004-09-10
 33546  2003-11-25 2008-10-10
 37321  2003-06-07 2006-03-20
 43028  2004-09-23 2008-07-25
 43591  2005-04-06 2005-11-15
 46749  2005-02-28 2005-05-16
 48846  2005-08-02 2005-08-02
114353  2002-05-17 2006-10-26
128180  2004-06-17 2010-06-21
128648  2003-05-07 2009-07-23
133337  2004-05-26 2012-07-26
149181  2002-10-19 2008-07-27
214079  2003-09-26 2007-05-20
215060  2006-04-17 2011-08-17
229816  2007-04-25 2011-09-24
238123  2007-11-26 2012-01-31
253776  2006-03-02 2012-04-19
258660  2010-03-25 2012-04-09
265356  2002-04-22 2002-04-22

unqualified<-(df[df$Duration <= '179.000',])

unqualified<-(df[df$Duration < 180,])

unqualifiedremove引号在您的比较中，即，df$Duration好吧，我不觉得自己愚蠢，那很容易。谢谢你回答noob的问题！