How to make this code more efficient in R? - r

Question

I know this is a stupid question, but I'm kinda frustrated with my code because it takes so much time. Jere is one part of my code.
basically I have a matrix called "distance"...
a b c
1 2 5 7
2 6 8 4
3 9 2 3
and then lets say I have a column in a data frame, contains of {a,b,c}
c1 c2 c3
c ... ...
a
a just another column
b
c ... ...
so I want to do a match, I wanna make another matrix with ncol=nrow(distance), and nrow=nrow(c1). where replace the factor value with their distance value. Here's an example of the first column of matrix that I'm going to make
a will replaced by 2
b will replaced by 5
c will replaced by 7
and for the second column, i will take row number 2 from distance matrix, and so on... so the result will be like this
m1 m2 m3
7 4 3
2 6 9
2 6 9
5 8 2
7 4 3
That is just an easy example, and I'm running this code, but when it deals with large iterations, it's kinda stressful for me.
for(l in 1:ncol(d.cat)){
get.unique = sort(unique(d.cat[, l]))
for(j in 1:nrow(d.cat)){
value = as.character(d.cat[j, l])
index = which(get.unique == value)
d2[j,l] = (d[[l]][i, index])
}
}
d.cat is categorical data. And d[[...]] is the list of matrix distance for every column in d.cat.

Answer 1

Try to store the indices and do the updating in one go. Lets say your distance matrix is dmat and data frame is df and you want to create a matrix named newmat
a.ind = which(df$c1=="a")
b.ind = which(df$c1=="b")
c.ind = which(df$c1=="c")
newmat = matrix(0,nrow=length(df$c1),ncol=3)
newmat[a.ind,] = dmat[,1]
newmat[b.ind,] = dmat[,2]
newmat[c.ind,] = dmat[,3]

Answer 2

Here's some data
set.seed(123)
d = matrix(1:9, 3, dimnames=list(NULL, letters[1:3]))
df = data.frame(c1 = sample(letters[1:3], 10, TRUE), stringsAsFactors=FALSE)
and a solution
t(d[, match(df$c1, colnames(d))])
For example
> d
a b c
[1,] 1 4 7
[2,] 2 5 8
[3,] 3 6 9
> df$c1
[1] "a" "c" "b" "c" "c" "a" "b" "c" "b" "b"
> t(d[,match(df$c1, colnames(d))])
[,1] [,2] [,3]
a 1 2 3
c 7 8 9
b 4 5 6
c 7 8 9
c 7 8 9
a 1 2 3
b 4 5 6
c 7 8 9
b 4 5 6
b 4 5 6

Answer 3

Your data
mat <- matrix(c(2,6,9,5,8,2,7,4,3), nrow=3)
rownames(mat) <- 1:3
colnames(mat) <- letters[1:3]
library(dplyr)
set.seed(1)
df <- as.data.frame(matrix(sample(letters[1:3], 12, replace=TRUE), nrow=4)) %>%
setNames(paste0("c", 1:3))
# c1 c2 c3
# 1 a a b
# 2 b c a
# 3 b c a
# 4 c b a
Using purrr::map2_df, iterate through columns of df and columns of tmat
library(purrr)
tmat <- t(mat)
map2_df(df, seq_len(ncol(tmat)), ~tmat[,.y][.x])
# # A tibble: 4 x 3
# c1 c2 c3
# <dbl> <dbl> <dbl>
# 1 2. 6. 2.
# 2 5. 4. 9.
# 3 5. 4. 9.
# 4 7. 8. 9.

Answer 4

Here is my attempt using the tidyverse :
library(tidyverse)
# Lets create some example
distance <- data_frame(a = sample(1:10, 1000, T), b = sample(1:10, 1000, T), c = sample(1:10, 1000, T))
c1 <- data_frame(c1 = sample(letters[1:3], 1000, T), c2 = sample(letters[1:3], 1000, T))
# First rearrange a little bit your data to make it more tidy
distance2 <- distance %>%
mutate(i = seq_len(n())) %>%
gather(col, value, -i)
c2 <- c1 %>%
mutate(i = seq_len(n()) %>%
gather(col, value, -i)
# Now just join the data and spread it again
c12 %>%
left_join(distance2, by = c("i", "value" = "col")) %>%
select(i, col, value.y) %>%
spread(col, value.y)