[R] Merging two data frames, but keeping NAs

arun smartpink111 at yahoo.com
Thu Dec 5 16:37:30 CET 2013


Hi,
Try ?join() 

library(plyr)
y$ref <- y$id
> join(x,y,by="ref")
   ref id val
1   NA NA  NA
2   NA NA  NA
3   NA NA  NA
4   10 NA  NA
5    9  9 106
6    8  8 107
7    7  7 105
8    6  6 104
9    5 NA  NA
10  NA NA  NA
11   1 NA  NA
12   2  2 101
13   3  3 102
14   4  4 103
15   5 NA  NA


A.K.


On Thursday, December 5, 2013 9:58 AM, Rainer M Krug <Rainer at krugs.de> wrote:
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Hi

My brain is giving up on this...

I have the following two data.frames:

  x <-  data.frame(ref=c(NA, NA, NA, 10:5, NA, 1:5))
  y <-  data.frame(id = c(2, 3, 4, 6, 7, 9, 8), val = 101:107)

Which look as follow:

> x
   ref
1   NA
2   NA
3   NA
4   10
5    9
6    8
7    7
8    6
9    5
10  NA
11   1
12   2
13   3
14   4
15   5
> y
  id val
1  2 101
2  3 102
3  4 103
4  6 104
5  7 105
6  9 106
7  8 107
> 

Now I want to merge y into x, but that

a) the sort order of x stays the same (sort=FALSE in merge()) and
b) the NAs stay

The result should look as follow (column id only here for clarity):

> result
   ref  id  val
1   NA  NA  NA
2   NA  NA  NA
3   NA  NA  NA
4   10  NA  NA
5    9   9   106
6    8   8   107
7    7   7   105
8    6   6   104
9    5  NA  NA
10  NA  NA  NA
11   1  NA  NA
12   2   2  101
13   3   3  102
14   4   4  103
15   5  NA  NA

merge(x, y, by.x="ref", by.y="id", sort=FALSE) leaves out the NA, but
otherwise it works:

> merge(x, y, by.x=1, by.y="id", sort=FALSE)
  ref val
1   9 106
2   8 107
3   7 105
4   6 104
5   2 101
6   3 102
7   4 103

Is there any way that I can tell merge() to keep the NA, or how can I
achieve what I want?

Thanks,

Rainer

- -- 
Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation
Biology, UCT), Dipl. Phys. (Germany)

Centre of Excellence for Invasion Biology
Stellenbosch University
South Africa

Tel :       +33 - (0)9 53 10 27 44
Cell:       +33 - (0)6 85 62 59 98
Fax :       +33 - (0)9 58 10 27 44

Fax (D):    +49 - (0)3 21 21 25 22 44

email:      Rainer at krugs.de

Skype:      RMkrug
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