On Sep 2, 2011, at 11:18 PM, Maya Joshi wrote:
Dear R experts.
I might be missing something obvious. I have been trying to fix this
problem
for some weeks. Please help.
#data
ped <- c(rep(1, 4), rep(2, 3), rep(3, 3))
y <- rnorm(10, 8, 2)
# variable set 1
M1a <- sample (c(1, 2,3), 10, replace= T)
M1b <- sample (c(1, 2,3), 10, replace= T)
M1aP1 <- sample (c(1, 2,3), 10, replace= T)
M1bP2 <- sample (c(1, 2,3), 10, replace= T)
# variable set 2
M2a <- sample (c(1, 2,3), 10, replace= T)
M2b <- sample (c(1, 2,3), 10, replace= T)
M2aP1 <- sample (c(1, 2,3), 10, replace= T)
M2bP2 <- sample (c(1, 2,3), 10, replace= T)
# variable set 3
M3a <- sample (c(1, 2,3), 10, replace= T)
M3b <- sample (c(1, 2,3), 10, replace= T)
M3aP1 <- sample (c(1, 2,3), 10, replace= T)
M3bP2 <- sample (c(1, 2,3), 10, replace= T)
mydf <- data.frame (ped, M1a,M1b,M1aP1,M1bP2, M2a,M2b,M2aP1,M2bP2,
M3a,M3b,M3aP1,M3bP2, y)
# functions and further calculations
mmat <- matrix
(c("M1a","M2a","M3a","M1b","M2b","M3b","M1aP1","M2aP1","M3aP1",
"M1bP2","M2bP2","M3bP2"), ncol = 4)
# first function
myfun <- function(x) {
x<- as.vector(x)
ot1 <- ifelse(mydf[x[1]] == mydf[x[3]], 1, -1)
You really ought to explain what you are trying to do. This code will
compare two lists. The list from mydf[x2]] will be the column
mydf["M2b"] which will have as its first element the four element
vector assigned above. I am guessing that was not what you wanted.
Notice this simple case using the "[" function as you are attempting
throws an error:
> list(M2b = c(1,2,3)) == list(M2a = c(1,2,3))
Error in list(M2b = c(1, 2, 3)) == list(M2a = c(1, 2, 3)) :
comparison of these types is not implemented'
So ... now that your code has failed to explain what you wanted why
don't your try some natural language explanations.
Notice that the "[[" function is generally what people want when
extracting from lists:
> list(M2b = c(1,2,3))[["M2b"]] == list(M2a = c(1,2,3))[["M2a"]]
[1] TRUE TRUE TRUE
ot2 <- ifelse(mydf[x[2]] == mydf[x[4]], 1, -1)
qt <- ot1 + ot2
return(qt)
}
qt <- apply(mmat, 1, myfun)
ydv <- c((y - mean(y))^2)
qtd <- data.frame(ped, ydv, qt)
# second function
myfun2 <- function(dataframe) {
vydv <- sum(ydv)*0.25
sumD <- sum(ydv * qt)
Rt <- vydv / sumD
return(Rt)
}
# using plyr
require(plyr)
dfsumd1 <- ddply(mydf,.(mydf$ped),myfun2)
Here are 2 issues:
(1) The output just one, I need the output for all three set of
variables
(as listed above)
An incredibly vague description.
(2) all three values of dfsumd is returning to same for all level
of ped:
1,2, 3
I sympathize with those forced to adopt the English language, but that
is the standard this decade. So givne your apparent difficulties, you
need to exert more effort at making explicit what is _supposed_ to be
returned.
Means that the function is applied to whole dataset but only
replicated in
output !!!
I tried with plyr not being lazy but due to my limited R knowledge,
If you
have a different suggestion, you are welcome too.
Thank you in advance...
Maya
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and provide commented, minimal, self-contained, reproducible code.
