If else statement for a list with matrices - r

Question

I am learning how to properly set up loops but still struggle with the correct indexing and syntax.
Below I have a list of two matrices, 2 columns x 3 rows each.
I want to look at the second column in each matrix, and create ideally a new column with values 1 if x>0.50, else = 0. Help will be much appreciated for both ways to do it, it should help me learn. Also, if you know any good reference of structuring loops in higher dimensional lists it would be much appreciated. Thanks so much.
a <- c(0.1,0.2,0.3)
b <- c(0.8,0.2,0.5)
c <- c(0.4,0.9,1.0)
d <- c(0.7,0.9,0.2)
ab <- cbind(a,b)
cd <- cbind(c,d)
abcd <- list(ab,cd)
presabs <- vector("list",ncol(y))
# Trying lapply but indexing something wrong
res <- lapply(abcd, function(x) if (x[,2]>0.5) {1} else {0})
# Other method also not working:
for (i in 1:length(abcd))
for (j in 1:length(a)){
{
if(abc[[i]][j]>0.50){
presabs[j] <- 1
} else {
presabs[j] <- 0
}
}
}

Answer 1

You can either use ifelse or in this case it is also not needed. The OP's question is creating a binary variable as the third column. This can be done using many variations,
lapply(abcd, function(x) cbind(x,new= +(x[,2]>0.5)))
or
lapply(abcd, function(x) cbind(x,new= (x[,2]>0.5)+0L))
Or
lapply(abcd, function(x) cbind(x,new= (x[,2]>0.5)*1))
Or
lapply(abcd, function(x) cbind(x,new= as.integer(x[,2]>0.5)))
If the values to be changed are different, for example
a <- 3
b <- 2
lapply(abcd, function(x) cbind(x, new= c(a, b)[(x[,2] > 0.5)+1L]))
Benchmarks
set.seed(25)
abcd1 <- lapply(1:60, function(i) matrix(rnorm(1e5*2), ncol=2))
viaChris <- function() lapply(abcd1, function(x) f(x, a=1, b=0, thresh =.5))
akrun <- function() lapply(abcd1, function(x) cbind(x, lab= +(x[,2] >0.5)))
system.time(viaChris())
# user system elapsed
#1.683 0.000 1.444
system.time(akrun())
# user system elapsed
# 0.481 0.000 0.322
library(microbenchmark)
microbenchmark(akrun(), viaChris(), unit='relative', times=20L)
#Unit: relative
# expr min lq mean median uq max neval cld
# akrun() 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 20 a
#viaChris() 3.726728 3.459581 3.475673 3.488114 3.400262 3.445557 20 b

Answer 2

The above answer works because logical values are coerced into 1 (for TRUE) and 0 (for FALSE). A more general solution might look something like:
lapply(abcd, function(x) cbind(x, ifelse(x[, 2] > .5, a, b)))
where a and b are numeric values you can specify. We can even be more general. For example:
## Define a general function that adds a new column of values
## based on whether or not the values in the i'th column of the
## matrix exceeds a threshold.
f = function(x, a, b, thresh, i = 2)
cbind(x, lab = ifelse(x[, i] > thresh, a, b))
## Apply the function above to each matrix in the list 'abcd', with
## a = 1, b = 0, and thresh = .5.
lapply(abcd, function(x) f(x, a = 1, b = 0, thresh = .5))