On May 21, 2011, at 9:12 AM, Dimitri Liakhovitski wrote:
Hello!
I've tried for a while - but can't figure it out. I have data frame x:
y=c("a","b","c","d","e")
z=c("m","n","o","p","r")
a=c(0,0,1,0,0)
b=c(2,0,0,0,0)
c=c(0,0,0,4,0)
x<-data.frame(y,z,a,b,c,stringsAsFactors=F)
str(x)
Some of the values in columns a,b, and c are >0:
I need to write a loop through all the cells in columns a,b,c that are
0 (only through them).
For each of those cells, I need to know:
1. Name of the column it is in
apply(x[,3:5], 1, function(z) if(any(z >0) ){
names(x)[2+which(z >0)]
} else {
"none" })
[1] "b" "none" "a" "cc" "none"
2 The entry of column y that is in the same row
apply(x, 1, function(z) if(any(z[3:5] >0) ){ z[1] } else { "none" })
[1] "a" "none" "c" "d" "none"
there might be pitfalls about which I am unaware since z will be
coerced to a character vector. Generally the character comparisons
with ">" will be "as expected" when the values were originally numeric.
> ("-3" > 0)
[1] FALSE
> ("0.1" > 0)
[1] TRUE
3 The entry of column z that is in the same row
apply(x, 1, function(z) if(any(z[3:5] >0) ){ z[2] } else { "none" })
[1] "m" "none" "o" "p" "none"
If you want to use NA instead of "none" I don't foresee any problems.
--
David
It'd be good to save this info in a data frame somehow - so that I
could loop through rows of this data frame.
To explain what I need it for eventually: I have a different data
frame "large.df" that has the same columns (variables) - but with many
more entries than "x". Something like:
large.df<-expand.grid(y,z)
names(large.df)<-c("y","z")
set.seed(123)
large.df$a<-sample(0:5,75,replace=T)
set.seed(234)
large.df$b<-sample(0:5,75,replace=T)
set.seed(345)
large.df$c<-sample(0:5,75,replace=T)
large.df$y<-as.character(large.df$y)
large.df$z<-as.character(large.df$z)
large.df<-large.df[order(large.df$y,large.df$z),]
row.names(large.df)<-1:nrow(large.df)
(large.df);str(large.df)
1. Find the first cell in x that is > 0 (in this case - it's x[3,"a"].
2. Find all the corresponding cells in the large.df - in this case,
it's:
large.df[large.df$y %in% "c" & large.df$z %in% "o","a"]
and those 3 values can be found in rows 37:39 of large.df, in column
"a".
3. Take those 3 values and add to them the corresponding value in x
(in this case = 1) divided by their length (in this case = 3).
4. Do the same for the other cells in x that are >0.
The final result will be (sorry for lengthy code):
large.df[large.df$y %in% "c" & large.df$z %in%
"o","a"]<-large.df[large.df$y %in% "c" & large.df$z %in%
"o","a"]+x[3,"a"]/3
large.df[large.df$y %in% "a" & large.df$z %in%
"m","b"]<-large.df[large.df$y %in% "a" & large.df$z %in%
"m","b"]+x[1,"b"]/3
large.df[large.df$y %in% "d" & large.df$z %in%
"p","c"]<-large.df[large.df$y %in% "d" & large.df$z %in%
"p","c"]+x[4,"c"]/3
(large.df)
(It just happens that at the end I divide by 3 - it could be anything
that is length(large.df[large.df$y %in% "c" & large.df$z %in%
"o","a"]), etc.
Thanks a lot for your suggestions!
--
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com
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David Winsemius, MD
West Hartford, CT
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
