Argthtjtr

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.

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

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]

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.

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)