Question

I have a datafile:

https://dl.dropbox.com/u/22681355/example.csv

Read file:

example<-read.csv("example.csv")
example<-example[,-1]

example[,1] contains a list of numbers increasing in numerical order.
example[,2] contains another set of numbers

First I would like to identify the numbers in example[,2] that are no listed in example[,1]

diff<-setdiff(example[,2],example[,1])

Now that I know these values I would like to insert them into example[,1] leaving existing values in example[,1] and example[,2] intact.

A short example would be:

Example[,1]   Example[,2]
1             1000
1             50
1             3
1             90
1             25
3             4
5             2
5             7
etc           etc

After I run setdiff() I get the numbers not in the first column but in the second.

Now I would like to place them in example[,1] to produce the following output:

Example[,1]   Example[,2]
1             1000
1             50
1             3
1             90
1             25
2             NA
3             4
4             NA
5             2
5             7
etc           etc

So basically placing them in numerical order but leaving everything else intact.

Part 1 excellently solved by Joris Meys!

I have two further questions:

/////////////////////////////////////////////
////////////////////////////////////////////

1:

Can the same be done if there is an additional third column but I don’t want to do anything with it?

e.g.:

ORIGINAl

 Example[,1]   Example[,2] Example[,3]
 1             1000        37
 1             50          18
 1             3           54
 1             90          72
 1             25          23
 3             4           15
 5             2           20
 5             7           9
 etc           etc

Desired OUTPUT:

Example[,1]   Example[,2]  Example[,3]
1             1000         37
1             50           18
1             3            54
1             90           72
1             25           23
2             NA           NA
3             4            15
4             NA           NA
5             2            20
5             7            19
etc           etc

2:

Instead of adding NA in example[,2] to cases where example[,1] doesnt have the value from example[,2] for example example[,1] doesn’t have number ’30’ then I would like to search for whether example[,2] has number’30’and see what value example[,1] has in that row then add it to example[,2] instead of the NA’s.

for example:

Example[,1]   Example[,2]  Example[,3]
1             1000         37
1             50           18
1             3            54
1             90           72
1             25           23
2             NA           NA
3             4            15
4             NA           NA
5             2            20
5             7            19
etc           etc

Instead of NA’s have:

Example[,1]   Example[,2]  Example[,3]
1             1000         37
1             50           18
1             3            54
1             90           72
1             25           23
2            5            20
3             4            15
4            3           15
5             2            20
5             7            19
etc           etc

Answer 1

The following approch also works if your matrix has more than two columns. It’s an extension of Joris Meys’ solution.

Example <- matrix(c(1,1,1,1,1,3,5,5,
                    1000,50,3,90,25,4,2,7,37,18,54,72,23,15,20,9),ncol=3)


diffs <- setdiff(Example[,2], Example[,1])
new_mat <- rbind(Example,
                 matrix(c(diffs,
                          rep(NA, length(diffs) * (ncol(Example) - 1))), 
                        ncol = ncol(Example)))
solution <- new_mat[order(new_mat[,1]),]

The result:

      [,1] [,2] [,3]
 [1,]    1 1000   37
 [2,]    1   50   18
 [3,]    1    3   54
 [4,]    1   90   72
 [5,]    1   25   23
 [6,]    2   NA   NA
 [7,]    3    4   15
 [8,]    4   NA   NA
 [9,]    5    2   20
[10,]    5    7    9
[11,]    7   NA   NA
[12,]   25   NA   NA
[13,]   50   NA   NA
[14,]   90   NA   NA
[15,] 1000   NA   NA

Once you have created this matrix, it’s easy to generate a new one without NAs:

solution2 <- solution
solution2[is.na(solution2)] <- Example[match(sort(diffs), Example[,2]), -2]

The result:

      [,1] [,2] [,3]
 [1,]    1 1000   37
 [2,]    1   50   18
 [3,]    1    3   54
 [4,]    1   90   72
 [5,]    1   25   23
 [6,]    2    5   20
 [7,]    3    4   15
 [8,]    4    3   15
 [9,]    5    2   20
[10,]    5    7    9
[11,]    7    5    9
[12,]   25    1   23
[13,]   50    1   18
[14,]   90    1   72
[15,] 1000    1   37