CODEWITHSUNDEEP
rmatrixvector
J. Doe
J. Doe

Reputation: 619

Merge vectors with different length into a matrix one after the other

Let's assume that we have the following three vectors.

L = c("1", "3")
K = c("2", "9", "2:9")
S = c("7")

Is there any way to combine them into a matrix that will look like the above ?

L    K    S
1    0    0
3    0    0
0    2    0
0    9    0
0    2:9  0
0    0    7

Thank you.

Upvotes: 4

Views: 2083

Answers (3)

Sotos
Sotos

Reputation: 51582

Here is an idea. First we create a list with all the vectors, and then a matrix with number of rows equal to the sum of all elements of the vectors, and number of columns equal to the number of vectors. We then use mapply to match the first vector's elements (of our list) with the first column of the matrix. Same with second and the third vectors and columns respectively. We then use that ( a logical matrix) to convert all the remaining unmatched elements of the matrix to 0.

l1 <- list(L, K, S)
m1 <- matrix(unlist(l1), nrow = sum(lengths(l1)), ncol = length(l1))
m1[!mapply(`%in%`, as.data.frame(m1), l1)] <- 0

m1
#     [,1] [,2]  [,3]
#[1,] "1"  "0"   "0" 
#[2,] "3"  "0"   "0" 
#[3,] "0"  "2"   "0" 
#[4,] "0"  "9"   "0" 
#[5,] "0"  "2:9" "0" 
#[6,] "0"  "0"   "7"

To address your comment and upgrade this to also work with same values appearing in multiple vectors, we follow the same logic but we do it on the indices of each column of the matrix based on the cumulative sequence of each vector. Since this got a bit more complicated, we can put it all in a function that accepts a list of the vectors as input i.e.

create_mat <- function(list){
  m1 <- matrix(unlist(list), nrow = sum(lengths(list)), ncol = length(list))
  m2 <- matrix(seq(nrow(m1)), ncol = ncol(m1), nrow = nrow(m1))
  l2 <- lapply(lengths(list), seq)
  v2 <- c(0, head(cumsum(lengths(list)), -1))
  l2 <-  Map(`+`, l2, v2)
  m1[!mapply(`%in%`, as.data.frame(m2), l2)] <- 0
  return(m1)
}

# Test with some values being same for multiple vectors,
M = c("1", "3")
N = c("1", "9", "2:9")
P = c("3")

create_mat(list(M, N, P))

#     [,1] [,2]  [,3]
#[1,] "1"  "0"   "0" 
#[2,] "3"  "0"   "0" 
#[3,] "0"  "1"   "0" 
#[4,] "0"  "9"   "0" 
#[5,] "0"  "2:9" "0" 
#[6,] "0"  "0"   "3"

Upvotes: 2

Lamia
Lamia

Reputation: 3875

Here's another way to do it: you first create a matrix of 0s, and then using two vectors of indexes k and 1:l, you input your values into it at the right locations.

l = length(c(L,K,S))
k = rep(1:3,times=c(length(L),length(K),length(S)))
m = matrix(0,ncol=3,nrow=l)
m[cbind(1:l,k)] = c(L,K,S)
     [,1] [,2]  [,3]
[1,] "1"  "0"   "0" 
[2,] "3"  "0"   "0" 
[3,] "0"  "2"   "0" 
[4,] "0"  "9"   "0" 
[5,] "0"  "2:9" "0" 
[6,] "0"  "0"   "7"

Edit: For a version that generalizes better to a larger number of input vectors, as per @DavidArenburg's comment, you could do:

l = list(L,K,S)
len = length(unlist(l))
k = rep(seq_along(l), lengths(l))
m = matrix(0, nrow=len, ncol=length(l))
m[cbind(1:len, k)] = unlist(l)

Upvotes: 5

JAD
JAD

Reputation: 2170

library(dplyr)

L = c("1", "3")
K = c("2", "9", "2:9")
S = c("7")

Ltable <- tibble(L=L)
Ktable <- tibble(K=K)
Stable <- tibble(S=S)
JoinedMatrix <- Ltable %>% bind_rows(Ktable) %>% bind_rows(Stable) %>% as.matrix() 
JoinedMatrix[which(is.na(JoinedMatrix))] <- "0"

bind_rows from dplyr allows you to bind dataframes (or tibbles) together. Since the three tibbles have differently named columns, they are kept as different columns after joining, filling the missing fields with NA. After that we simply replace all NAs with 0 and we're done.

> JoinedMatrix
     L   K     S  
[1,] "1" "0"   "0"
[2,] "3" "0"   "0"
[3,] "0" "2"   "0"
[4,] "0" "9"   "0"
[5,] "0" "2:9" "0"
[6,] "0" "0"   "7"

Upvotes: 1

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