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I have been trying to get the year month as a new column to my following example data set. 

       time site
1  1/01/2000    A
2  1/02/2000    A
3  1/03/2000    A
4  1/04/2000    A
5  1/05/2000    A
6  1/06/2000    A
7  1/07/2000    A
8  1/08/2000    A
9  1/09/2000    A
10 1/10/2000    A
11 1/11/2000    A
12 1/12/2000    A
13 1/01/2001    A
14 1/02/2001    A
15 1/03/2001    A
16 1/04/2001    A
17 1/05/2001    A
18 1/06/2001    A
19 1/07/2001    A
20 1/08/2001    A
21 1/09/2001    A
22 1/10/2001    A

The data set can be accessed using the following dput. 

dta<-structure(list(time = structure(c(1L, 3L, 5L, 7L, 9L, 11L, 13L, 
15L, 17L, 19L, 21L, 22L, 2L, 4L, 6L, 8L, 10L, 12L, 14L, 16L, 
18L, 20L), .Label = c("1/01/2000", "1/01/2001", "1/02/2000", 
"1/02/2001", "1/03/2000", "1/03/2001", "1/04/2000", "1/04/2001", 
"1/05/2000", "1/05/2001", "1/06/2000", "1/06/2001", "1/07/2000", 
"1/07/2001", "1/08/2000", "1/08/2001", "1/09/2000", "1/09/2001", 
"1/10/2000", "1/10/2001", "1/11/2000", "1/12/2000"), class = "factor"), 
    site = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "A", class = "factor")), .Names = c("time", 
"site"), class = "data.frame", row.names = c(NA, -22L))

I tried with the following codes and returns the same month for all 12. 

library(zoo)
dta$y_m = as.yearmon(as.Date(dta$time,"%m/%d/%Y"))

1] "Jan 2000" "Jan 2000" "Jan 2000" "Jan 2000" "Jan 2000" "Jan 2000" "Jan 2000"
 [8] "Jan 2000" "Jan 2000" "Jan 2000" "Jan 2000" "Jan 2000" "Jan 2001" "Jan 2001"
[15] "Jan 2001" "Jan 2001" "Jan 2001" "Jan 2001" "Jan 2001" "Jan 2001" "Jan 2001"
[22] "Jan 2001"

Can anybody help me please to get it right. Thank you

sriya
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1 Answers1

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Try this:

require(zoo)
as.yearmon(dta$time, format="%d/%m/%Y")

Output:

 [1] "Jan 2000" "Feb 2000" "Mar 2000" "Apr 2000" "May 2000" "Jun 2000"
 [7] "Jul 2000" "Aug 2000" "Sep 2000" "Oct 2000" "Nov 2000" "Dec 2000"
[13] "Jan 2001" "Feb 2001" "Mar 2001" "Apr 2001" "May 2001" "Jun 2001"
[19] "Jul 2001" "Aug 2001" "Sep 2001" "Oct 2001"
www
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  • It works with the example data set, but not with my actual data set. What might be the problem please. I copied several rows from the actual data set for this question. – sriya Sep 13 '17 at 01:08
  • The output is the same as Jan for all. – sriya Sep 13 '17 at 01:50
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    @sriya - I'd like to help, but I don't know what the error might be without seeing the data or the error. If the data you're using is different than the sample data, please supply the data that's causing the error. It can just be the date column. – www Sep 13 '17 at 02:06
  • Thanks, I can provide the data with the time column. It has got several thousands rows. How can I send it please? – sriya Sep 13 '17 at 02:11
  • I just tried the code you provided with the actual data set but only with date column. It works, that's really weird, so any guess for this problem please? – sriya Sep 13 '17 at 02:20
  • @sriya - The main thing is that it works. That's good. No, I'm not sure what the problem might have been without seeing it. Now all you have to do is store the results into a column like you did above: dta$y_m = as.yearmon(dta$time, format="%d/%m/%Y"). After that, please remember to mark this as a solution. – www Sep 13 '17 at 02:27