Dear All, I am new to R and slowly learning how to use the system. The following code is an exercise I was trying. The intent is to generate 10 random samples of size 5 from a vector with integers 1:10 and 2 missing values. I then want to generate a matrix, for each sample which shows the frequency of missing values (NA) in each sample. My solution, using sapply is at the end. If anyone has the time and/or intrest to critique my method I'd be very grateful. I'm especially interested in knowing if there is a better way to accomplish this problem.> (x<-replicate(10,sample(c(1:10,rep(NA,2)),5)))[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] 3 NA 3 4 2 10 NA 4 5 4 [2,] 5 7 7 3 9 2 8 NA 7 9 [3,] NA 8 1 5 NA 7 10 2 NA 6 [4,] 2 NA 6 10 8 4 4 7 4 7 [5,] 7 9 10 8 3 6 1 NA 9 NA> # Since table will return only a single item of vaule FALSE > # if there are no missing values (NA) in a sample, sapply > # will return a list and not a matrix. > # So to get a matrix, the factor function needs to be used > # to identify possible results (FALSE, TRUE) for the table > # function. > sapply(1:10,function(i) table(factor(is.na(x[,i]),c(FALSE,TRUE))))[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] FALSE 4 3 5 5 4 5 4 3 4 4 TRUE 1 2 0 0 1 0 1 2 1 1>Thanks for your thoughts. Murray M Cooper, Ph.D. Richland Statistics 9800 N 24th St Richland, MI, USA 49083 Mail: richstat at earthlink.net
Dear Murray,
Here is one way: create a function that takes k sample()s from any vector
(e.g., x) and then calculates the number of NA values in it. Then replicate
the procedure as many times as you want.
# The function
foo <- function(x, k = 5){
xsample <- sample(x, k)
sum( is.na(xsample) )
}
# Vector of data
y <- c(1:10, NA, NA)
# The replication (10 replicates are used)
# replicate(10, foo(y) )
# [1] 1 2 1 1 0 0 1 0 1 1
HTH,
Jorge
On Wed, Jul 1, 2009 at 10:15 PM, Murray Cooper
<myrmail@earthlink.net>wrote:
> Dear All,
>
> I am new to R and slowly learning how to use the system.
>
> The following code is an exercise I was trying.
> The intent is to generate 10 random samples of size 5 from
> a vector with integers 1:10 and 2 missing values. I then want
> to generate a matrix, for each sample which shows the frequency
> of missing values (NA) in each sample. My solution, using sapply
> is at the end.
>
> If anyone has the time and/or intrest to critique my method I'd
> be very grateful. I'm especially interested in knowing if there is
> a better way to accomplish this problem.
>
> (x<-replicate(10,sample(c(1:10,rep(NA,2)),5)))
>>
> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
> [1,] 3 NA 3 4 2 10 NA 4 5 4
> [2,] 5 7 7 3 9 2 8 NA 7 9
> [3,] NA 8 1 5 NA 7 10 2 NA 6
> [4,] 2 NA 6 10 8 4 4 7 4 7
> [5,] 7 9 10 8 3 6 1 NA 9 NA
>
>> # Since table will return only a single item of vaule FALSE
>> # if there are no missing values (NA) in a sample, sapply
>> # will return a list and not a matrix.
>> # So to get a matrix, the factor function needs to be used
>> # to identify possible results (FALSE, TRUE) for the table
>> # function.
>> sapply(1:10,function(i) table(factor(is.na(x[,i]),c(FALSE,TRUE))))
>>
> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
> FALSE 4 3 5 5 4 5 4 3 4 4
> TRUE 1 2 0 0 1 0 1 2 1 1
>
>>
>>
> Thanks for your thoughts.
>
> Murray M Cooper, Ph.D.
> Richland Statistics
> 9800 N 24th St
> Richland, MI, USA 49083
> Mail: richstat@earthlink.net
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
[[alternative HTML version deleted]]
On Jul 1, 2009, at 9:15 PM, Murray Cooper wrote:> Dear All, > > I am new to R and slowly learning how to use the system. > > The following code is an exercise I was trying. > The intent is to generate 10 random samples of size 5 from > a vector with integers 1:10 and 2 missing values. I then want > to generate a matrix, for each sample which shows the frequency > of missing values (NA) in each sample. My solution, using sapply > is at the end. > > If anyone has the time and/or intrest to critique my method I'd > be very grateful. I'm especially interested in knowing if there is > a better way to accomplish this problem. > >> (x<-replicate(10,sample(c(1:10,rep(NA,2)),5))) > [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] > [1,] 3 NA 3 4 2 10 NA 4 5 4 > [2,] 5 7 7 3 9 2 8 NA 7 9 > [3,] NA 8 1 5 NA 7 10 2 NA 6 > [4,] 2 NA 6 10 8 4 4 7 4 7 > [5,] 7 9 10 8 3 6 1 NA 9 NA >> # Since table will return only a single item of vaule FALSE >> # if there are no missing values (NA) in a sample, sapply >> # will return a list and not a matrix. >> # So to get a matrix, the factor function needs to be used >> # to identify possible results (FALSE, TRUE) for the table >> # function. >> sapply(1:10,function(i) table(factor(is.na(x[,i]),c(FALSE,TRUE)))) > [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] > FALSE 4 3 5 5 4 5 4 3 4 4 > TRUE 1 2 0 0 1 0 1 2 1 1 > > Thanks for your thoughts.Murray, if I correctly understand what you want as an end result, then: > x [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] 3 NA 3 4 2 10 NA 4 5 4 [2,] 5 7 7 3 9 2 8 NA 7 9 [3,] NA 8 1 5 NA 7 10 2 NA 6 [4,] 2 NA 6 10 8 4 4 7 4 7 [5,] 7 9 10 8 3 6 1 NA 9 NA > colSums(is.na(x)) [1] 1 2 0 0 1 0 1 2 1 1 To take that in stages: > is.na(x) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] FALSE TRUE FALSE FALSE FALSE FALSE TRUE FALSE FALSE FALSE [2,] FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE FALSE FALSE [3,] TRUE FALSE FALSE FALSE TRUE FALSE FALSE FALSE TRUE FALSE [4,] FALSE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE [5,] FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE FALSE TRUE The above gives you either TRUE or FALSE at each position in the matrix. TRUE if the value is NA. The colSums() function is optimized for speed using C code, to calculate the sum of the values in each column. Since a TRUE is equal to 1 and a FALSE is equal to 0, using colSums() on the above intermediate step, gives you a column by column count of the NA values in each. > as.numeric(TRUE) [1] 1 > as.numeric(FALSE) [1] 0 See ?colSums for more information and the sister function rowSums(). HTH, Marc Schwartz