R help - Jul 2024 - Using the pipe, |>, syntax with "names<-"

With further fooling around, I realized that explicitly assigning my
last "solution" 'works'; i.e.

names(z)[2] <- "foo"

can be piped as:

 z <- z |>(\(x) "names<-"(x,value =
"[<-"(names(x),2,'foo')))()
> z
  a foo
1 1   a
2 2   b
3 3   c

This is even awfuller than before. So my query still stands.

-- Bert

On Sat, Jul 20, 2024 at 1:14?PM Bert Gunter <bgunter.4567 at gmail.com>
wrote:
>
> Nope, I still got it wrong: None of my approaches work.  :(
>
> So my query remains: how to do it via piping with |> ?
>
> Bert
>
>
> On Sat, Jul 20, 2024 at 1:06?PM Bert Gunter <bgunter.4567 at
gmail.com> wrote:
> >
> > This post is likely pretty useless;  it is motivated by a recent post
> > from "Val" that was elegantly answered using Tidyverse
constructs, but
> > I wondered how to do it using base R only. Along the way, I ran into
> > the following question to which I think my answer (below) is pretty
> > awful. I would be interested in more elegant base R approaches. So...
> >
> > z <- data.frame(a = 1:3, b = letters[1:3])
> > > z
> >   a h
> > 1 1 a
> > 2 2 b
> > 3 3 c
> >
> > Suppose I want to change the name of the second column of z from
'b'
> > to 'foo' . This is very easy using nested function syntax by:
> >
> > names(z)[2] <- "foo"
> > > z
> >   a foo
> > 1 1   a
> > 2 2   b
> > 3 3   c
> >
> > Now suppose I wanted to do this using |> syntax, along the lines
of:
> >
> > z |> names()[2] <- "foo"  ## throws an error
> >
> > Slightly fancier is:
> >
> > z |> (\(x)names(x)[2] <- "b")()
> > ## does nothing, but does not throw an error.
> >
> > However, the following, which resulted from a more careful read of
> > ?names works (after changing the name of the second column back to
"b"
> > of course):
> >
> > z |>(\(x) "names<-"(x,value =
"[<-"(names(x),2,'foo')))()
> > >z
> >   a foo
> > 1 1   a
> > 2 2   b
> > 3 3   c
> >
> > This qualifies to me as "pretty awful." I'm sure there
are better ways
> > to do this using pipe syntax, so I would appreciate any better
> > approaches.
> >
> > Best,
> > Bert

?s 21:46 de 20/07/2024, Bert Gunter escreveu:
> With further fooling around, I realized that explicitly assigning my
> last "solution" 'works'; i.e.
> 
> names(z)[2] <- "foo"
> 
> can be piped as:
> 
>   z <- z |>(\(x) "names<-"(x,value =
"[<-"(names(x),2,'foo')))()
>> z
>    a foo
> 1 1   a
> 2 2   b
> 3 3   c
> 
> This is even awfuller than before. So my query still stands.
> 
> -- Bert
> 
> On Sat, Jul 20, 2024 at 1:14?PM Bert Gunter <bgunter.4567 at
gmail.com> wrote:
>>
>> Nope, I still got it wrong: None of my approaches work.  :(
>>
>> So my query remains: how to do it via piping with |> ?
>>
>> Bert
>>
>>
>> On Sat, Jul 20, 2024 at 1:06?PM Bert Gunter <bgunter.4567 at
gmail.com> wrote:
>>>
>>> This post is likely pretty useless;  it is motivated by a recent
post
>>> from "Val" that was elegantly answered using Tidyverse
constructs, but
>>> I wondered how to do it using base R only. Along the way, I ran
into
>>> the following question to which I think my answer (below) is pretty
>>> awful. I would be interested in more elegant base R approaches.
So...
>>>
>>> z <- data.frame(a = 1:3, b = letters[1:3])
>>>> z
>>>    a h
>>> 1 1 a
>>> 2 2 b
>>> 3 3 c
>>>
>>> Suppose I want to change the name of the second column of z from
'b'
>>> to 'foo' . This is very easy using nested function syntax
by:
>>>
>>> names(z)[2] <- "foo"
>>>> z
>>>    a foo
>>> 1 1   a
>>> 2 2   b
>>> 3 3   c
>>>
>>> Now suppose I wanted to do this using |> syntax, along the lines
of:
>>>
>>> z |> names()[2] <- "foo"  ## throws an error
>>>
>>> Slightly fancier is:
>>>
>>> z |> (\(x)names(x)[2] <- "b")()
>>> ## does nothing, but does not throw an error.
>>>
>>> However, the following, which resulted from a more careful read of
>>> ?names works (after changing the name of the second column back to
"b"
>>> of course):
>>>
>>> z |>(\(x) "names<-"(x,value =
"[<-"(names(x),2,'foo')))()
>>>> z
>>>    a foo
>>> 1 1   a
>>> 2 2   b
>>> 3 3   c
>>>
>>> This qualifies to me as "pretty awful." I'm sure
there are better ways
>>> to do this using pipe syntax, so I would appreciate any better
>>> approaches.
>>>
>>> Best,
>>> Bert
> 
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
Hello,

This is not exactly the same but in one of your attempts all you have to 
do is to return x.
The following works and does something.


z |> (\(x){names(x)[2] <- "foo";x})()
#   a foo
# 1 1   a
# 2 2   b
# 3 3   c


Hope this helps,

Rui Barradas



-- 
Este e-mail foi analisado pelo software antiv?rus AVG para verificar a presen?a
de v?rus.
www.avg.com

I suspect that you would want to define a function which was aware of 
the limitations of piping to handle this.  For example:

rename <- function(x, col, newname) {
   names(x)[col] <- newname
   x
}

Then

z |> rename(2, "foo")

would be fine.

Duncan Murdoch

On 2024-07-20 4:46 p.m., Bert Gunter wrote:
> With further fooling around, I realized that explicitly assigning my
> last "solution" 'works'; i.e.
> 
> names(z)[2] <- "foo"
> 
> can be piped as:
> 
>   z <- z |>(\(x) "names<-"(x,value =
"[<-"(names(x),2,'foo')))()
>> z
>    a foo
> 1 1   a
> 2 2   b
> 3 3   c
> 
> This is even awfuller than before. So my query still stands.
> 
> -- Bert
> 
> On Sat, Jul 20, 2024 at 1:14?PM Bert Gunter <bgunter.4567 at
gmail.com> wrote:
>>
>> Nope, I still got it wrong: None of my approaches work.  :(
>>
>> So my query remains: how to do it via piping with |> ?
>>
>> Bert
>>
>>
>> On Sat, Jul 20, 2024 at 1:06?PM Bert Gunter <bgunter.4567 at
gmail.com> wrote:
>>>
>>> This post is likely pretty useless;  it is motivated by a recent
post
>>> from "Val" that was elegantly answered using Tidyverse
constructs, but
>>> I wondered how to do it using base R only. Along the way, I ran
into
>>> the following question to which I think my answer (below) is pretty
>>> awful. I would be interested in more elegant base R approaches.
So...
>>>
>>> z <- data.frame(a = 1:3, b = letters[1:3])
>>>> z
>>>    a h
>>> 1 1 a
>>> 2 2 b
>>> 3 3 c
>>>
>>> Suppose I want to change the name of the second column of z from
'b'
>>> to 'foo' . This is very easy using nested function syntax
by:
>>>
>>> names(z)[2] <- "foo"
>>>> z
>>>    a foo
>>> 1 1   a
>>> 2 2   b
>>> 3 3   c
>>>
>>> Now suppose I wanted to do this using |> syntax, along the lines
of:
>>>
>>> z |> names()[2] <- "foo"  ## throws an error
>>>
>>> Slightly fancier is:
>>>
>>> z |> (\(x)names(x)[2] <- "b")()
>>> ## does nothing, but does not throw an error.
>>>
>>> However, the following, which resulted from a more careful read of
>>> ?names works (after changing the name of the second column back to
"b"
>>> of course):
>>>
>>> z |>(\(x) "names<-"(x,value =
"[<-"(names(x),2,'foo')))()
>>>> z
>>>    a foo
>>> 1 1   a
>>> 2 2   b
>>> 3 3   c
>>>
>>> This qualifies to me as "pretty awful." I'm sure
there are better ways
>>> to do this using pipe syntax, so I would appreciate any better
>>> approaches.
>>>
>>> Best,
>>> Bert
> 
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

It should be written more like this:

```R
z <- data.frame(a = 1:3, b = letters[1:3])
z |> names() |> _[2] <- "foo"
z
```

Regards,
    Iris

On Sat, Jul 20, 2024 at 4:47?PM Bert Gunter <bgunter.4567 at gmail.com>
wrote:
>
> With further fooling around, I realized that explicitly assigning my
> last "solution" 'works'; i.e.
>
> names(z)[2] <- "foo"
>
> can be piped as:
>
>  z <- z |>(\(x) "names<-"(x,value =
"[<-"(names(x),2,'foo')))()
> > z
>   a foo
> 1 1   a
> 2 2   b
> 3 3   c
>
> This is even awfuller than before. So my query still stands.
>
> -- Bert
>
> On Sat, Jul 20, 2024 at 1:14?PM Bert Gunter <bgunter.4567 at
gmail.com> wrote:
> >
> > Nope, I still got it wrong: None of my approaches work.  :(
> >
> > So my query remains: how to do it via piping with |> ?
> >
> > Bert
> >
> >
> > On Sat, Jul 20, 2024 at 1:06?PM Bert Gunter <bgunter.4567 at
gmail.com> wrote:
> > >
> > > This post is likely pretty useless;  it is motivated by a recent
post
> > > from "Val" that was elegantly answered using Tidyverse
constructs, but
> > > I wondered how to do it using base R only. Along the way, I ran
into
> > > the following question to which I think my answer (below) is
pretty
> > > awful. I would be interested in more elegant base R approaches.
So...
> > >
> > > z <- data.frame(a = 1:3, b = letters[1:3])
> > > > z
> > >   a h
> > > 1 1 a
> > > 2 2 b
> > > 3 3 c
> > >
> > > Suppose I want to change the name of the second column of z from
'b'
> > > to 'foo' . This is very easy using nested function syntax
by:
> > >
> > > names(z)[2] <- "foo"
> > > > z
> > >   a foo
> > > 1 1   a
> > > 2 2   b
> > > 3 3   c
> > >
> > > Now suppose I wanted to do this using |> syntax, along the
lines of:
> > >
> > > z |> names()[2] <- "foo"  ## throws an error
> > >
> > > Slightly fancier is:
> > >
> > > z |> (\(x)names(x)[2] <- "b")()
> > > ## does nothing, but does not throw an error.
> > >
> > > However, the following, which resulted from a more careful read
of
> > > ?names works (after changing the name of the second column back
to "b"
> > > of course):
> > >
> > > z |>(\(x) "names<-"(x,value =
"[<-"(names(x),2,'foo')))()
> > > >z
> > >   a foo
> > > 1 1   a
> > > 2 2   b
> > > 3 3   c
> > >
> > > This qualifies to me as "pretty awful." I'm sure
there are better ways
> > > to do this using pipe syntax, so I would appreciate any better
> > > approaches.
> > >
> > > Best,
> > > Bert
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

Bert,

You need to consider LHS vs RHS functionality.

Before I start, I would have done your example setup lie this:

trio <- 1:3
z <- data.frame(a = trio, b = letters[trio])

Just kidding!

Syntactic sugar means you are calling this function:
> `names<-`
function (x, value)  .Primitive("names<-")

Not particularly documented is a gimmick I just tried of supplying a second
argument to the more routine function version:
> `names<-`(z, c("one","two"))
  one two
1   1   a
2   2   b
3   3   c

The above does not change z, but returns a new DF with new names.

In a pipeline, try this:

z <-
  z |>
  `names<-`( c("one","two"))
> z
  a b
1 1 a
2 2 b
3 3 c
> z <-
+   z |>
+   `names<-`( c("one","two"))
> z
  one two
1   1   a
2   2   b
3   3   c





-----Original Message-----
From: R-help <r-help-bounces at r-project.org> On Behalf Of Bert Gunter
Sent: Saturday, July 20, 2024 4:47 PM
To: R-help <R-help at r-project.org>
Subject: Re: [R] Using the pipe, |>, syntax with "names<-"

With further fooling around, I realized that explicitly assigning my
last "solution" 'works'; i.e.

names(z)[2] <- "foo"

can be piped as:

 z <- z |>(\(x) "names<-"(x,value =
"[<-"(names(x),2,'foo')))()
> z
  a foo
1 1   a
2 2   b
3 3   c

This is even awfuller than before. So my query still stands.

-- Bert

On Sat, Jul 20, 2024 at 1:14?PM Bert Gunter <bgunter.4567 at gmail.com>
wrote:
>
> Nope, I still got it wrong: None of my approaches work.  :(
>
> So my query remains: how to do it via piping with |> ?
>
> Bert
>
>
> On Sat, Jul 20, 2024 at 1:06?PM Bert Gunter <bgunter.4567 at
gmail.com> wrote:
> >
> > This post is likely pretty useless;  it is motivated by a recent post
> > from "Val" that was elegantly answered using Tidyverse
constructs, but
> > I wondered how to do it using base R only. Along the way, I ran into
> > the following question to which I think my answer (below) is pretty
> > awful. I would be interested in more elegant base R approaches. So...
> >
> > z <- data.frame(a = 1:3, b = letters[1:3])
> > > z
> >   a h
> > 1 1 a
> > 2 2 b
> > 3 3 c
> >
> > Suppose I want to change the name of the second column of z from
'b'
> > to 'foo' . This is very easy using nested function syntax by:
> >
> > names(z)[2] <- "foo"
> > > z
> >   a foo
> > 1 1   a
> > 2 2   b
> > 3 3   c
> >
> > Now suppose I wanted to do this using |> syntax, along the lines
of:
> >
> > z |> names()[2] <- "foo"  ## throws an error
> >
> > Slightly fancier is:
> >
> > z |> (\(x)names(x)[2] <- "b")()
> > ## does nothing, but does not throw an error.
> >
> > However, the following, which resulted from a more careful read of
> > ?names works (after changing the name of the second column back to
"b"
> > of course):
> >
> > z |>(\(x) "names<-"(x,value =
"[<-"(names(x),2,'foo')))()
> > >z
> >   a foo
> > 1 1   a
> > 2 2   b
> > 3 3   c
> >
> > This qualifies to me as "pretty awful." I'm sure there
are better ways
> > to do this using pipe syntax, so I would appreciate any better
> > approaches.
> >
> > Best,
> > Bert
______________________________________________
R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.