[Rd] eval(parse()) within mutate() returning same value for all rows

[Mateo Obregón]

Hi all-

Looking through stackoverflow for R string combining examples, I found the 
following from 3 years ago:



The top answer suggests to use eval(parse(sprintf())). I tried the suggestion 
and it did not return the expected combines strings. I thought that this might 
be an issue with some leftover values being reused, so I explicitly eval() 
with a new.env():

> library(dplyr)
> df <- tibble(words=c("%s plus %s equals %s"), 
args=c("1,1,2","2,2,4","3,3,6"))
> df |> mutate(combined = eval(parse(text=sprintf("sprintf('%s', %s)", words, 
args)), envir=new.env()))

# A tibble: 3 × 3
  wordsargs  combined 
   
1 %s plus %s equals %s 1,1,2 3 plus 3 equals 6
2 %s plus %s equals %s 2,2,4 3 plus 3 equals 6
3 %s plus %s equals %s 3,3,6 3 plus 3 equals 6

The `combined`  is not what I was expecting, as the same last eval() is 
returned for all three rows.
 
Am I missing something? What has changed in the past three years?

Mateo.
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[Rd] The case for a pipe assignment operator + database of R code

[David Hugh-Jones]

Hi all,

I wrote up the case for having a pipe assignment operator in R here:
https://hughjonesd.github.io/case-for-pipe-assignment.html

A pipe assignment operator would expand e.g.

obj <|> do_something()

to

obj <- obj |> do_something()

and therefore to

obj <- do_something(obj)

Just for fun, I made a patch to the R source at
https://hughjonesd.github.io/pipe-assignment.patch. It is highly imperfect,
and made against github rather than svn, so it is just a proof of concept.

Maybe more interesting to list readers is the dataset of R code I created
to learn how people use assignments in real world code. It has about 26000
code snippets, from github, Stackoverflow questions, and R package
examples. It might be useful if you are researching real world usage
patterns in R. I didn't know of any existing resources:

https://github.com/hughjonesd/codesamples

Happy new year!

David

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Re: [Rd] eval(parse()) within mutate() returning same value for all rows

[Dirk Eddelbuettel]

On 29 December 2023 at 14:13, Mateo Obregón wrote:
| Hi all-
| 
| Looking through stackoverflow for R string combining examples, I found the 
| following from 3 years ago:
| 
| 

| 
| The top answer suggests to use eval(parse(sprintf())). I tried the suggestion 

Well:

   > fortunes::fortune(106)

   If the answer is parse() you should usually rethink the question.
  -- Thomas Lumley
 R-help (February 2005)
   
   > 

| and it did not return the expected combines strings. I thought that this 
might 
| be an issue with some leftover values being reused, so I explicitly eval() 
| with a new.env():
| 
| > library(dplyr)
| > df <- tibble(words=c("%s plus %s equals %s"), 
| args=c("1,1,2","2,2,4","3,3,6"))
| > df |> mutate(combined = eval(parse(text=sprintf("sprintf('%s', %s)", words, 
| args)), envir=new.env()))
| 
| # A tibble: 3 × 3
|   wordsargs  combined 
|
| 1 %s plus %s equals %s 1,1,2 3 plus 3 equals 6
| 2 %s plus %s equals %s 2,2,4 3 plus 3 equals 6
| 3 %s plus %s equals %s 3,3,6 3 plus 3 equals 6
| 
| The `combined`  is not what I was expecting, as the same last eval() is 
| returned for all three rows.
|  
| Am I missing something? What has changed in the past three years?

Nothing if you use the first answer which relies only on base R and still
works: 

   > words <- c('%s + %s equal %s', '%s + %s equal %s')
   > arguments <- c('1,1,2', '2,2,4')
   > df <- data.frame(words, arguments)
   > df
words arguments
   1 %s + %s equal %s 1,1,2
   2 %s + %s equal %s 2,2,4
   > df$combined <- apply(df, 1, function(x) do.call(sprintf, 
c(as.list(strsplit(x[2], ',')[[1]]), fmt = x[[1]])))
   > df
words arguments  combined
   1 %s + %s equal %s 1,1,2 1 + 1 equal 2
   2 %s + %s equal %s 2,2,4 2 + 2 equal 4
   > 

I am not the best person to answer what may have changed in `dplyr` in those
three years -- and neither is this list which is primarily concerned with
developing R itself.  

Dirk

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Re: [Rd] eval(parse()) within mutate() returning same value for all rows

[Jan Gorecki]

Unless you are able to reproduce the problem without dplyr then you should
submit your question to dplyr issues tracker. R-devel is for issues in R,
not in a third party packages.

On Fri, Dec 29, 2023, 15:13 Mateo Obregón  wrote:

> Hi all-
>
> Looking through stackoverflow for R string combining examples, I found the
> following from 3 years ago:
>
> <
> https://stackoverflow.com/questions/63881854/how-to-format-strings-using-values-from-other-column-in-r
> >
>
> The top answer suggests to use eval(parse(sprintf())). I tried the
> suggestion
> and it did not return the expected combines strings. I thought that this
> might
> be an issue with some leftover values being reused, so I explicitly eval()
> with a new.env():
>
> > library(dplyr)
> > df <- tibble(words=c("%s plus %s equals %s"),
> args=c("1,1,2","2,2,4","3,3,6"))
> > df |> mutate(combined = eval(parse(text=sprintf("sprintf('%s', %s)",
> words,
> args)), envir=new.env()))
>
> # A tibble: 3 × 3
>   wordsargs  combined
>
> 1 %s plus %s equals %s 1,1,2 3 plus 3 equals 6
> 2 %s plus %s equals %s 2,2,4 3 plus 3 equals 6
> 3 %s plus %s equals %s 3,3,6 3 plus 3 equals 6
>
> The `combined`  is not what I was expecting, as the same last eval() is
> returned for all three rows.
>
> Am I missing something? What has changed in the past three years?
>
> Mateo.
> --
> Mateo Obregón
>
> __
> R-devel@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-devel
>

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Re: [Rd] eval(parse()) within mutate() returning same value for all rows

[Mateo Obregón]

Thanks Dirk for your response.

The "base" R code is just too unwieldy to be easily understandable. The 
eval(parse(sprintf())) solution provided is ungainly too, but at least its 
more understandable from a casual reading of it.

I'll punt this case over to the dplyr people then.

Cheers.

Mateo.
--  
Mateo Obregón
On Friday, 29 December 2023 15:35:20 GMT Dirk Eddelbuettel wrote:
> On 29 December 2023 at 14:13, Mateo Obregón wrote:
> | Hi all-
> | 
> | Looking through stackoverflow for R string combining examples, I found the
> | following from 3 years ago:
> | 
> |  
> | values-from-other-column-in-r>
> | 
> | The top answer suggests to use eval(parse(sprintf())). I tried the
> | suggestion
> Well:
>> fortunes::fortune(106)
> 
>If the answer is parse() you should usually rethink the question.
>   -- Thomas Lumley
>  R-help (February 2005)
> 
> | and it did not return the expected combines strings. I thought that this
> | might be an issue with some leftover values being reused, so I explicitly
> | eval()| 
> | with a new.env():
> | > library(dplyr)
> | > df <- tibble(words=c("%s plus %s equals %s"),
> | 
> | args=c("1,1,2","2,2,4","3,3,6"))
> | 
> | > df |> mutate(combined = eval(parse(text=sprintf("sprintf('%s', %s)",
> | > words,
> | 
> | args)), envir=new.env()))
> | 
> | # A tibble: 3 × 3
> | 
> |   wordsargs  combined
> |
> | 
> | 1 %s plus %s equals %s 1,1,2 3 plus 3 equals 6
> | 2 %s plus %s equals %s 2,2,4 3 plus 3 equals 6
> | 3 %s plus %s equals %s 3,3,6 3 plus 3 equals 6
> | 
> | The `combined`  is not what I was expecting, as the same last eval() is
> | returned for all three rows.
> | 
> | Am I missing something? What has changed in the past three years?
> 
> Nothing if you use the first answer which relies only on base R and still
> 
> works:
>> words <- c('%s + %s equal %s', '%s + %s equal %s')
>> arguments <- c('1,1,2', '2,2,4')
>> df <- data.frame(words, arguments)
>> df
> 
> words arguments
>1 %s + %s equal %s 1,1,2
>2 %s + %s equal %s 2,2,4
> 
>> df$combined <- apply(df, 1, function(x) do.call(sprintf,
>> c(as.list(strsplit(x[2], ',')[[1]]), fmt = x[[1]]))) df
> 
> words arguments  combined
>1 %s + %s equal %s 1,1,2 1 + 1 equal 2
>2 %s + %s equal %s 2,2,4 2 + 2 equal 4
> 
> 
> I am not the best person to answer what may have changed in `dplyr` in those
> three years -- and neither is this list which is primarily concerned with
> developing R itself.
> 
> Dirk

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Re: [Rd] eval(parse()) within mutate() returning same value for all rows

[Duncan Murdoch]

On 29/12/2023 9:13 a.m., Mateo Obregón wrote:

Hi all-

Looking through stackoverflow for R string combining examples, I found the
following from 3 years ago:



The top answer suggests to use eval(parse(sprintf())). I tried the suggestion
and it did not return the expected combines strings. I thought that this might
be an issue with some leftover values being reused, so I explicitly eval()
with a new.env():


library(dplyr)
df <- tibble(words=c("%s plus %s equals %s"),

args=c("1,1,2","2,2,4","3,3,6"))

df |> mutate(combined = eval(parse(text=sprintf("sprintf('%s', %s)", words,

args)), envir=new.env()))

# A tibble: 3 × 3
   wordsargs  combined

1 %s plus %s equals %s 1,1,2 3 plus 3 equals 6
2 %s plus %s equals %s 2,2,4 3 plus 3 equals 6
3 %s plus %s equals %s 3,3,6 3 plus 3 equals 6

The `combined`  is not what I was expecting, as the same last eval() is
returned for all three rows.
  
Am I missing something? What has changed in the past three years?




I don't know if this is a change, but when `eval()` is passed an 
expression vector, it evaluates the elements in order and returns the 
value of the last one.  This is only partially documented:


"Value:  The result of evaluating the object: for an expression vector 
this is the result of evaluating the last element."



That text has been unchanged in the help page for 13 years.

Duncan Murdoch

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Re: [Rd] eval(parse()) within mutate() returning same value for all rows

[Gabor Grothendieck]

If the question is how to accomplish this as opposed to how to use eval
then we can do it without eval like this provided we can assume that words
contains three %s .

  library(dplyr)
  library(tidyr)
  df <- tibble(words=c("%s plus %s equals %s"),args=c("1,1,2","2,2,4","3,3,6"))

  df |>
separate_wider_delim(args, ",", names = c("a", "b", "c")) |>
mutate(combined = sprintf(words, a, b, c))
  ## # A tibble: 3 × 5
  ## wordsa b c combined
  ##
  ## 1 %s plus %s equals %s 1 1 2 1 plus 1 equals 2
  ## 2 %s plus %s equals %s 2 2 4 2 plus 2 equals 4
  ## 3 %s plus %s equals %s 3 3 6 3 plus 3 equals 6

On Fri, Dec 29, 2023 at 9:14 AM Mateo Obregón  wrote:
>
> Hi all-
>
> Looking through stackoverflow for R string combining examples, I found the
> following from 3 years ago:
>
> 
>
> The top answer suggests to use eval(parse(sprintf())). I tried the suggestion
> and it did not return the expected combines strings. I thought that this might
> be an issue with some leftover values being reused, so I explicitly eval()
> with a new.env():
>
> > library(dplyr)
> > df <- tibble(words=c("%s plus %s equals %s"),
> args=c("1,1,2","2,2,4","3,3,6"))
> > df |> mutate(combined = eval(parse(text=sprintf("sprintf('%s', %s)", words,
> args)), envir=new.env()))
>
> # A tibble: 3 × 3
>   wordsargs  combined
>
> 1 %s plus %s equals %s 1,1,2 3 plus 3 equals 6
> 2 %s plus %s equals %s 2,2,4 3 plus 3 equals 6
> 3 %s plus %s equals %s 3,3,6 3 plus 3 equals 6
>
> The `combined`  is not what I was expecting, as the same last eval() is
> returned for all three rows.
>
> Am I missing something? What has changed in the past three years?
>
> Mateo.
> --
> Mateo Obregón
>
> __
> R-devel@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-devel



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email: ggrothendieck at gmail.com

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Re: [Rd] eval(parse()) within mutate() returning same value for all rows

[Mateo Obregón]

Thanks Gabor, I like your solution that splits the args into separate columns, 
in turn making the sprintf() call more interpretable .

Cheers.

Mateo.
--  
Mateo Obregón
On Friday, 29 December 2023 18:45:06 GMT Gabor Grothendieck wrote:
> If the question is how to accomplish this as opposed to how to use eval
> then we can do it without eval like this provided we can assume that words
> contains three %s .
> 
>   library(dplyr)
>   library(tidyr)
>   df <- tibble(words=c("%s plus %s equals
> %s"),args=c("1,1,2","2,2,4","3,3,6"))
> 
>   df |>
> separate_wider_delim(args, ",", names = c("a", "b", "c")) |>
> mutate(combined = sprintf(words, a, b, c))
>   ## # A tibble: 3 × 5
>   ## wordsa b c combined
>   ##
>   ## 1 %s plus %s equals %s 1 1 2 1 plus 1 equals 2
>   ## 2 %s plus %s equals %s 2 2 4 2 plus 2 equals 4
>   ## 3 %s plus %s equals %s 3 3 6 3 plus 3 equals 6
> 
> On Fri, Dec 29, 2023 at 9:14 AM Mateo Obregón  
wrote:
> > Hi all-
> > 
> > Looking through stackoverflow for R string combining examples, I found the
> > following from 3 years ago:
> > 
> >  
> > > values-from-other-column-in-r>
> > 
> > The top answer suggests to use eval(parse(sprintf())). I tried the
> > suggestion and it did not return the expected combines strings. I thought
> > that this might be an issue with some leftover values being reused, so I
> > explicitly eval()> 
> > with a new.env():
> > > library(dplyr)
> > > df <- tibble(words=c("%s plus %s equals %s"),
> > 
> > args=c("1,1,2","2,2,4","3,3,6"))
> > 
> > > df |> mutate(combined = eval(parse(text=sprintf("sprintf('%s', %s)",
> > > words,
> > 
> > args)), envir=new.env()))
> > 
> > # A tibble: 3 × 3
> > 
> >   wordsargs  combined
> >
> > 
> > 1 %s plus %s equals %s 1,1,2 3 plus 3 equals 6
> > 2 %s plus %s equals %s 2,2,4 3 plus 3 equals 6
> > 3 %s plus %s equals %s 3,3,6 3 plus 3 equals 6
> > 
> > The `combined`  is not what I was expecting, as the same last eval() is
> > returned for all three rows.
> > 
> > Am I missing something? What has changed in the past three years?
> > 
> > Mateo.
> > --
> > Mateo Obregón
> > 
> > __
> > R-devel@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-devel

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Re: [Rd] eval(parse()) within mutate() returning same value for all rows

[Dirk Eddelbuettel]

On 29 December 2023 at 22:31, Mateo Obregón wrote:
| Thanks Gabor, I like your solution that splits the args into separate 
columns, 
| in turn making the sprintf() call more interpretable .

Well you may also like `tstrsplit()`, a gem inside data.table:

> suppressMessages(library(data.table))
>
> D <- data.table(words="%s plus %s equals %s", args=c("1,1,2", "2,2,4", 
> "3,3,6"))
> D
  words   args
  
1: %s plus %s equals %s  1,1,2
2: %s plus %s equals %s  2,2,4
3: %s plus %s equals %s  3,3,6
>
> D[, c('a','b','c') := tstrsplit(args, ",")]
> D
  words   args  a  b  c
 
1: %s plus %s equals %s  1,1,2  1  1  2
2: %s plus %s equals %s  2,2,4  2  2  4
3: %s plus %s equals %s  3,3,6  3  3  6
>
> D[, res := sprintf(words, a, b, c)]
> D
  words   args  a  b  c   res
 
1: %s plus %s equals %s  1,1,2  1  1  2 1 plus 1 equals 2
2: %s plus %s equals %s  2,2,4  2  2  4 2 plus 2 equals 4
3: %s plus %s equals %s  3,3,6  3  3  6 3 plus 3 equals 6
> 

so all we do here is a one-liner in data.table if you're so inclined:


> D <- data.table(words="%s plus %s equals %s", args=c("1,1,2", "2,2,4", 
> "3,3,6"))
> D[, c('a','b','c') := tstrsplit(args, ",")][, res := sprintf(words, a, b, 
> c)][, .(res)]
 res
  
1: 1 plus 1 equals 2
2: 2 plus 2 equals 4
3: 3 plus 3 equals 6
> 

data.table is very powerful and expressive. It is much worth getting into
which I really only did ten or so years into using R.

Dirk

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