Skip to content

CONFUSSING WITH select[!miss] <- 1:sum(!miss)

5 messages · greg holly, Rui Barradas, William Dunlap +1 more

Original
greg holly
# 
Dear All;

I am very new in R and try to understand the logic for a program has been
run sucessfully. Here select[!miss] <- 1:sum(!miss) par is confussing me. I
need to understandand the logic behind this commend line.

Thanks in advance for your help,

Greg


miss <- apply(is.na(ph[,c("M1","X1","X2","X3")]),1, any)
select <- integer(nrow(ph))
select[!miss] <- 1:sum(!miss)
Rui Barradas
# 
Hello,

The first command line produces a logical vector with TRUE if at least 
one row element of ph is NA and FALSE otherwise.
The second creates a vector of zeros with length equal to nrow(ph).
Now the third command line. ! negates miss, so TRUE becomes FALSE and 
vice-versa. sum(!miss) counts how many not misses are there and 
1:sum(!miss) creates a vector 1, 2, ..., sum(!miss). To see this print 
each of these components one by one:

print(miss)
print(!miss)
print(sum(!miss))
etc

And select[!miss] uses a logical index into 'select' to set only the 
values of 'select' where !miss is TRUE equal to 1, 2, 3, ...

I believe you should read R-intro.pdf that comes with your installation 
of R more carefully, specially sections 2.4 and 2.7.

Hope this helps,

Rui Barradas

Em 06-12-2016 18:18, greg holly escreveu:
William Dunlap
# 
R is interactive so you can print the intermediate results:
Y=c(11,12,13,14,NA), row.names=paste0("R",1:5))
M1 X1 X2 X3  Y
R1  1  1  1  1 11
R2 NA  2  2  2 12
R3  3  3 NA  3 13
R4  4  4  4  4 14
R5  5  5  5  5 NA
R1    R2    R3    R4    R5
FALSE  TRUE  TRUE FALSE FALSE
[1] 0 0 0 0 0
[1] 3
[1] 1 0 0 2 3

Then you can look in the Introduction to R document or ask about the steps
that confuse you.


Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Dec 6, 2016 at 10:18 AM, greg holly <mak.hholly at gmail.com> wrote:

            

      
      
      
    

        

      
      
    

        

  
  


  


      

    

    

      
      

        


  

        

      Jim Lemon
    

    

      
      #
    

  


  

      
Hi Greg,
What is happening is easy to see:

 ph<-matrix(sample(1:100,40),ncol=4)
 colnames(ph)<-c("M1","X1","X2","X3")
 ph[sample(1:10,3),1]<-NA
 ph
       M1 X1 X2 X3
 [1,]  34 98  3 35
 [2,]  13 66 74 68
 [3,]  NA 22 99 79
 [4,]  94  6 80 36
 [5,]  18  9 16 65
 [6,]  NA 29 56 90
 [7,]  41 23  7 55
 [8,] 100 93 71 70
 [9,]  NA 61  8 57
[10,]  25  4 47 60
# get a logical vector showing which rows contain NA
 miss <- apply(is.na(ph[,c("M1","X1","X2","X3")]),1, any)
 miss
 [1] FALSE FALSE  TRUE FALSE FALSE  TRUE FALSE FALSE  TRUE FALSE
# create a vector of zeros the length of the number of rows
 select <- integer(nrow(ph))
 select
 [1] 0 0 0 0 0 0 0 0 0 0
# get the indices for the rows that do _not_ have NAs
 select[!miss] <- 1:sum(!miss)
 select
 [1] 1 2 0 3 4 0 5 6 0 7

If this is to select the rows without NAs, it may be easier to do:

which(!miss)
[1]  1  2  4  5  7  8 10

Jim

        
On Wed, Dec 7, 2016 at 5:18 AM, greg holly <mak.hholly at gmail.com> wrote:

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      greg holly
    

    

      
      #
    

  


  

      
Hi Jim, Rui and William;

I do appreciate for your explanations and help. These are very helpful.
Regards,

Hayrettin

        
On Tue, Dec 6, 2016 at 4:06 PM, Jim Lemon <drjimlemon at gmail.com> wrote: