Loop within string in r to output expressions with vectorized values

Question

Example data:

> DF
   A  B  C
1 11 22 88
2 11 22 47
3  2 30 21
4  3 30 21

> r
[1] "A==A[i] & B==B[i] "           "A==A[i] & C==C[i] "          
[3] "B==B[i] & C==C[i] "           "A==A[i] & B==B[i] & C==C[i] "

Executed code:

> output=list()
> for (j in r){
+   for (i in 1:nrow(DF)){
+    
+     output[[j]][i]=j  
+   }
+ }
> output
$`A==A[i] & B==B[i] `
[1] "A==A[i] & B==B[i] " "A==A[i] & B==B[i] " "A==A[i] & B==B[i] "
[4] "A==A[i] & B==B[i] "

$`A==A[i] & C==C[i] `
[1] "A==A[i] & C==C[i] " "A==A[i] & C==C[i] " "A==A[i] & C==C[i] "
[4] "A==A[i] & C==C[i] "

$`B==B[i] & C==C[i] `
[1] "B==B[i] & C==C[i] " "B==B[i] & C==C[i] " "B==B[i] & C==C[i] "
[4] "B==B[i] & C==C[i] "

$`A==A[i] & B==B[i] & C==C[i] `
[1] "A==A[i] & B==B[i] & C==C[i] " "A==A[i] & B==B[i] & C==C[i] "
[3] "A==A[i] & B==B[i] & C==C[i] " "A==A[i] & B==B[i] & C==C[i] "

> output=purrr::flatten_chr(output)
> output
 [1] "A==A[i] & B==B[i] "           "A==A[i] & B==B[i] "          
 [3] "A==A[i] & B==B[i] "           "A==A[i] & B==B[i] "          
 [5] "A==A[i] & C==C[i] "           "A==A[i] & C==C[i] "          
 [7] "A==A[i] & C==C[i] "           "A==A[i] & C==C[i] "          
 [9] "B==B[i] & C==C[i] "           "B==B[i] & C==C[i] "          
[11] "B==B[i] & C==C[i] "           "B==B[i] & C==C[i] "          
[13] "A==A[i] & B==B[i] & C==C[i] " "A==A[i] & B==B[i] & C==C[i] "
[15] "A==A[i] & B==B[i] & C==C[i] " "A==A[i] & B==B[i] & C==C[i] "

My target was to get the same output with the specific values A[i], B[i] and C[i] from DF, i.e., the final output would be the following:

> output
[1] "A==11 & B==22 "            "A==11 & B==22 "          
[3] "A==2 & B==30 "             "A==3 & B==30 "          
[5] "A==11 & C==88 "            "A==11 & C==47 "          
[7] "A==2 & C==21 "             "A==3 & C==21 "          
[9] "B==22 & C==88 "            "B==22 & C==47 "          
[11] "B==30 & C==21 "           "B==30 & C==21 "          
[13] "A==11 & B==22 & C==88 "   "A==11 & B==22 & C==47 "
[15] "A==2 & B==30 & C==21 "    "A==3 & B==30 & C==21 "

I would be grateful if someone could help me with this matter.

Romain Francois · Accepted Answer · 2017-11-17 11:38:23Z

2

The difficulty here comes from the way the problem is specified with the r vector, and it would be possible to modify Milan's answer to better match the expected result.

r <- c(
  "A==A[i] & B==B[i]", 
  "A==A[i] & C==C[i]",
  "B==B[i] & C==C[i]", 
  "A==A[i] & B==B[i] & C==C[i]"
)

Instead of that, I'll take another route, given that in each element of r, you only need to identify the names of the columns. For example, on the first case, you only need A and B.

library(purrr)
library(glue)
library(rlang)
names <- c("A", "B")
DF <- data.frame(
  A = c(11, 11, 2, 3),
  B = c(22, 22, 30, 30),
  C = c(88, 47, 21, 21)
)

map( names, ~{
  glue( "{name} == {values}", name = ., values = DF[[.]] )
})
#> [[1]]
#> A == 11
#> A == 11
#> A == 2
#> A == 3
#> 
#> [[2]]
#> B == 22
#> B == 22
#> B == 30
#> B == 30

you can then reduce it to only get one string per line of your data frame:

map( names, ~{
  glue( "{name} == {values}", name = ., values = DF[[.]] )
}) %>% 
  reduce( paste, sep = " & ")
#> [1] "A == 11 & B == 22" "A == 11 & B == 22" "A == 2 & B == 30" 
#> [4] "A == 3 & B == 30"

Wrapping it all up into a function that uses tidy eval to get the names to begin with:

tests <- function(data, ...){
  names <- map_chr(quos(...), f_name)

  map( names, ~{
    glue( "{name} == {values}", name = ., values = data[[.]] )
  }) %>% 
    reduce(paste, sep = " & " )

}

So that you can get the results by calling tests 4 times:

c(
  tests( DF, A, B), 
  tests( DF, A, C), 
  tests( DF, B, C), 
  tests( DF, A, B, C)
)
#>  [1] "A == 11 & B == 22"           "A == 11 & B == 22"          
#>  [3] "A == 2 & B == 30"            "A == 3 & B == 30"           
#>  [5] "A == 11 & C == 88"           "A == 11 & C == 47"          
#>  [7] "A == 2 & C == 21"            "A == 3 & C == 21"           
#>  [9] "B == 22 & C == 88"           "B == 22 & C == 47"          
#> [11] "B == 30 & C == 21"           "B == 30 & C == 21"          
#> [13] "A == 11 & B == 22 & C == 88" "A == 11 & B == 22 & C == 47"
#> [15] "A == 2 & B == 30 & C == 21"  "A == 3 & B == 30 & C == 21"

edited Nov 17, 2017 at 11:38

answered Nov 16, 2017 at 13:21

Romain Francois

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3 Comments

Citizen Over a year ago

Thank you very much, Romain. I have run the code but I have got "Error in as_function(.f, ...) : object 'f_name' not found". I have done something wrong, I guess. Could you show me the way to solve this?

Romain Francois Over a year ago

I probably should have loaded rlang too. I have added library(rlang) just now.

Citizen Over a year ago

Thank you ever so much, Romain. It works very well. Now, I have to think and work about doing this with any number of columns, i.e. for any big dataframe input, but this would be another potencial question. I have selected your answer as the best right answer. Please, let me know if I have to do something else.

Milan Valášek · Accepted Answer · 2017-11-15 11:33:25Z

0

Is this output what you are after?

DF <- data.frame(A = c(11, 11, 2, 3),
                 B = c(22, 22, 30, 30),
                 C = c(88, 47, 21, 21))

r <- c("A==A[i] & B==B[i]", "A==A[i] & C==C[i]",
       "B==B[i] & C==C[i]", "A==A[i] & B==B[i] & C==C[i]")

output=list()
for (j in r){
  for (i in 1:nrow(DF))
    output[[j]][[i]] <- DF[with(DF, eval(parse(text = j))), ]
}

output

$`A==A[i] & B==B[i]`
$`A==A[i] & B==B[i]`[[1]]
   A  B  C
1 11 22 88
2 11 22 47

$`A==A[i] & B==B[i]`[[2]]
   A  B  C
1 11 22 88
2 11 22 47

$`A==A[i] & B==B[i]`[[3]]
  A  B  C
3 2 30 21

$`A==A[i] & B==B[i]`[[4]]
  A  B  C
4 3 30 21


$`A==A[i] & C==C[i]`
$`A==A[i] & C==C[i]`[[1]]
   A  B  C
1 11 22 88
...

edited Nov 15, 2017 at 11:33

answered Nov 15, 2017 at 11:14

Milan Valášek

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2 Comments

Romain Francois Over a year ago

Or perhaps surround the expression with a with( DF ) to avoid the risky gsub, e.g. output[[j]][i] <- DF[ with(DF, eval(parse(text=j))), ]

Citizen Over a year ago

Thank you very much. I am afraid my question was not asked in the right way, so that I have explained it better.

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