Obtain vector of non-zero elements in sparse matrix, keeping both column and row names

user_17125683 0
Viewed 45

Suppose I have the following matrix:

mat <- matrix(data = c(1, 2, 3, 0, 0, 0, 0, 0, 0, 
                       0, 0, 0, 2, 3, 4, 0, 0, 0,
                       0, 0, 0, 0, 0, 0, 5, 6, 7),
              nrow = 9, 
              dimnames = list(c(paste0("x", 1:3),
                                paste0("y", 1:3),
                                paste0("z", 1:3)),
                              c("a", "b", "c")))


   a b c
x1 1 0 0
x2 2 0 0
x3 3 0 0
y1 0 2 0
y2 0 3 0
y3 0 4 0
z1 0 0 5
z2 0 0 6
z3 0 0 7

Instead of a matrix, I want a vector only keeping the non-zero elements.

red <- apply(mat, 1, function(x) x[x != 0])

x1 x2 x3 y1 y2 y3 z1 z2 z3 
 1  2  3  2  3  4  5  6  7

Is there a way to have the reduced vector keep the column names as well? Preferably the pattern "colname + character + rowname". See desired output below. Note that we do not know in advance how many columns/rows there will be, nor how they are named.

a=x1 a=x2 a=x3 b=y1 b=y2 b=y3 c=z1 c=z2 c=z3 
 1    2    3    2    3    4    5    6    7

Thank you in advance!

apply r
4 Answers

An idea can be,

library(dplyr)
library(tidyr)

data.frame(mat) %>% 
 tibble::rownames_to_column('id') %>% 
 pivot_longer(-1) %>% 
 filter(value != 0) %>% 
 unite(key, name, id, sep = '=')

# A tibble: 9 x 2
  key   value
  <chr> <dbl>
1 a=x1      1
2 a=x2      2
3 a=x3      3
4 b=y1      2
5 b=y2      3
6 b=y3      4
7 c=z1      5
8 c=z2      6
9 c=z3      7
user_56355800

A base solution with which(..., arr.ind = TRUE):

ind <- which(mat != 0, arr.ind = TRUE)
setNames(
  mat[ind],
  paste(colnames(mat)[ind[, 'col']], rownames(mat)[ind[, 'row']], sep = '=')
)

# a=x1 a=x2 a=x3 b=y1 b=y2 b=y3 c=z1 c=z2 c=z3 
#    1    2    3    2    3    4    5    6    7
user_100689850

Here is a tidyverse solution with using deframe():

library(tidyverse)

data.frame(mat) %>% 
  rownames_to_column() %>% 
  pivot_longer(-rowname) %>% 
  mutate(names = paste(name, rowname, sep="="), .keep="unused", .before=1) %>% 
  filter(value !=0) %>% 
  deframe()

a=x1 a=x2 a=x3 b=y1 b=y2 b=y3 c=z1 c=z2 c=z3 
   1    2    3    2    3    4    5    6    7
user_133216470

We could also use asplit and unlist it (but get .'s instead of ='s):

red <- unlist(asplit(mat, 2))
red <- red[red != 0]

Output:

a.x1 a.x2 a.x3 b.y1 b.y2 b.y3 c.z1 c.z2 c.z3 
   1    2    3    2    3    4    5    6    7

Or you could of course - at the cost of some elegance - use some regex, as suggested by @user321797 to get the desired output exactly:

names(red) <- gsub(names(red), pattern = "\\.", replacement = "=")

Output:

a=x1 a=x2 a=x3 b=y1 b=y2 b=y3 c=z1 c=z2 c=z3 
   1    2    3    2    3    4    5    6    7
user_47864660
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