Bert Gunter
2016-Dec-05 21:46 UTC
[R] Write a function that allows access to columns of a passed dataframe.
Sorry, hit "Send" by mistake. Inline. On Mon, Dec 5, 2016 at 1:34 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote:> Inline. > > -- Bert > > > Bert Gunter > > "The trouble with having an open mind is that people keep coming along > and sticking things into it." > -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) > > > On Mon, Dec 5, 2016 at 9:53 AM, Rui Barradas <ruipbarradas at sapo.pt> wrote: >> Hello, >> >> Inline. >> >> Em 05-12-2016 17:09, David Winsemius escreveu: >>> >>> >>>> On Dec 5, 2016, at 7:29 AM, John Sorkin <jsorkin at grecc.umaryland.edu> >>>> wrote: >>>> >>>> Rui, >>>> I appreciate your suggestion, but eliminating the deparse statement does >>>> not solve my problem. Do you have any other suggestions? See code below. >>>> Thank you, >>>> John >>>> >>>> >>>> mydf <- >>>> data.frame(id=c(1,2,3,4,5),sex=c("M","M","M","F","F"),age=c(20,34,43,32,21)) >>>> mydf >>>> class(mydf) >>>> >>>> >>>> myfun <- function(frame,var){ >>>> call <- match.call() >>>> print(call) >>>> >>>> >>>> indx <- match(c("frame","var"),names(call),nomatch=0) >>>> print(indx) >>>> if(indx[1]==0) stop("Function called without sufficient arguments!") >>>> >>>> >>>> cat("I can get the name of the dataframe as a text string!\n") >>>> #xx <- deparse(substitute(frame)) >>>> print(xx) >>>> >>>> >>>> cat("I can get the name of the column as a text string!\n") >>>> #yy <- deparse(substitute(var)) >>>> print(yy) >>>> >>>> >>>> # This does not work. >>>> print(frame[,var]) >>>> >>>> >>>> # This does not work. >>>> print(frame[,"var"]) >>>> >>>> >>>> >>>> >>>> # This does not work. >>>> col <- xx[,"yy"] >>>> >>>> >>>> # Nor does this work. >>>> col <- xx[,yy] >>>> print(col) >>>> } >>>> >>>> >>>> myfun(mydf,age) >>> >>> >>> >>> When you use that calling syntax, the system will supply the values of >>> whatever the `age` variable contains. (And if there is no `age`-named >>> object, you get an error at the time of the call to `myfun`. >> >> >> Actually, no, which was very surprising to me but John's code worked (not >> the function, the call). And with the change I've proposed, it worked >> flawlessly. No errors. Why I don't know.See ?substitute and in particular the example highlighted there. The technical details are explained in the R Language Definition manual. The key here is the use of promises for lay evaluations. In fact, the expression in the call *is* available within the functions, as is (a pointer to) the environment in which to evaluate the expression. That is how substitute() works. Specifically, quoting from the manual, ***** It is possible to access the actual (not default) expressions used as arguments inside the function. The mechanism is implemented via promises. When a function is being evaluated the actual expression used as an argument is stored in the promise together with a pointer to the environment the function was called from. When (if) the argument is evaluated the stored expression is evaluated in the environment that the function was called from. Since only a pointer to the environment is used any changes made to that environment will be in effect during this evaluation. The resulting value is then also stored in a separate spot in the promise. Subsequent evaluations retrieve this stored value (a second evaluation is not carried out). Access to the unevaluated expression is also available using substitute. ******** -- Bert>> >> Rui Barradas >> >> You need either to call it as: >>> >>> >>> myfun( mydf , "age") >>> >>> >>> # Or: >>> >>> age <- "age" >>> myfun( mydf, age) >>> >>> Unless your value of the `age`-named variable was "age" in the calling >>> environment (and you did not give us that value in either of your postings), >>> you would fail. >>> >> >> ______________________________________________ >> 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 Gunter
2016-Dec-05 21:49 UTC
[R] Write a function that allows access to columns of a passed dataframe.
Typo: "lazy evaluation" not "lay evaluation." -- Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Mon, Dec 5, 2016 at 1:46 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote:> Sorry, hit "Send" by mistake. > > Inline. > > > > On Mon, Dec 5, 2016 at 1:34 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: >> Inline. >> >> -- Bert >> >> >> Bert Gunter >> >> "The trouble with having an open mind is that people keep coming along >> and sticking things into it." >> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) >> >> >> On Mon, Dec 5, 2016 at 9:53 AM, Rui Barradas <ruipbarradas at sapo.pt> wrote: >>> Hello, >>> >>> Inline. >>> >>> Em 05-12-2016 17:09, David Winsemius escreveu: >>>> >>>> >>>>> On Dec 5, 2016, at 7:29 AM, John Sorkin <jsorkin at grecc.umaryland.edu> >>>>> wrote: >>>>> >>>>> Rui, >>>>> I appreciate your suggestion, but eliminating the deparse statement does >>>>> not solve my problem. Do you have any other suggestions? See code below. >>>>> Thank you, >>>>> John >>>>> >>>>> >>>>> mydf <- >>>>> data.frame(id=c(1,2,3,4,5),sex=c("M","M","M","F","F"),age=c(20,34,43,32,21)) >>>>> mydf >>>>> class(mydf) >>>>> >>>>> >>>>> myfun <- function(frame,var){ >>>>> call <- match.call() >>>>> print(call) >>>>> >>>>> >>>>> indx <- match(c("frame","var"),names(call),nomatch=0) >>>>> print(indx) >>>>> if(indx[1]==0) stop("Function called without sufficient arguments!") >>>>> >>>>> >>>>> cat("I can get the name of the dataframe as a text string!\n") >>>>> #xx <- deparse(substitute(frame)) >>>>> print(xx) >>>>> >>>>> >>>>> cat("I can get the name of the column as a text string!\n") >>>>> #yy <- deparse(substitute(var)) >>>>> print(yy) >>>>> >>>>> >>>>> # This does not work. >>>>> print(frame[,var]) >>>>> >>>>> >>>>> # This does not work. >>>>> print(frame[,"var"]) >>>>> >>>>> >>>>> >>>>> >>>>> # This does not work. >>>>> col <- xx[,"yy"] >>>>> >>>>> >>>>> # Nor does this work. >>>>> col <- xx[,yy] >>>>> print(col) >>>>> } >>>>> >>>>> >>>>> myfun(mydf,age) >>>> >>>> >>>> >>>> When you use that calling syntax, the system will supply the values of >>>> whatever the `age` variable contains. (And if there is no `age`-named >>>> object, you get an error at the time of the call to `myfun`. >>> >>> >>> Actually, no, which was very surprising to me but John's code worked (not >>> the function, the call). And with the change I've proposed, it worked >>> flawlessly. No errors. Why I don't know. > > See ?substitute and in particular the example highlighted there. > > The technical details are explained in the R Language Definition > manual. The key here is the use of promises for lay evaluations. In > fact, the expression in the call *is* available within the functions, > as is (a pointer to) the environment in which to evaluate the > expression. That is how substitute() works. Specifically, quoting from > the manual, > > ***** > It is possible to access the actual (not default) expressions used as > arguments inside the function. The mechanism is implemented via > promises. When a function is being evaluated the actual expression > used as an argument is stored in the promise together with a pointer > to the environment the function was called from. When (if) the > argument is evaluated the stored expression is evaluated in the > environment that the function was called from. Since only a pointer to > the environment is used any changes made to that environment will be > in effect during this evaluation. The resulting value is then also > stored in a separate spot in the promise. Subsequent evaluations > retrieve this stored value (a second evaluation is not carried out). > Access to the unevaluated expression is also available using > substitute. > ******** > > -- Bert > > > > >>> >>> Rui Barradas >>> >>> You need either to call it as: >>>> >>>> >>>> myfun( mydf , "age") >>>> >>>> >>>> # Or: >>>> >>>> age <- "age" >>>> myfun( mydf, age) >>>> >>>> Unless your value of the `age`-named variable was "age" in the calling >>>> environment (and you did not give us that value in either of your postings), >>>> you would fail. >>>> >>> >>> ______________________________________________ >>> 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.
Rui Barradas
2016-Dec-06 08:17 UTC
[R] Write a function that allows access to columns of a passed dataframe.
Hello,
Just to say that I wouldn't write the function as John did. I would get
rid of all the deparse/substitute stuff and instinctively use a quoted
argument as a column name. Something like the following.
myfun <- function(frame, var){
[...]
col <- frame[, var] # or frame[[var]]
[...]
}
myfun(mydf, "age") # much better, simpler, no promises.
Rui Barradas
Em 05-12-2016 21:49, Bert Gunter escreveu:> Typo: "lazy evaluation" not "lay evaluation."
>
> -- Bert
>
>
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip
)
>
>
> On Mon, Dec 5, 2016 at 1:46 PM, Bert Gunter <bgunter.4567 at
gmail.com> wrote:
>> Sorry, hit "Send" by mistake.
>>
>> Inline.
>>
>>
>>
>> On Mon, Dec 5, 2016 at 1:34 PM, Bert Gunter <bgunter.4567 at
gmail.com> wrote:
>>> Inline.
>>>
>>> -- Bert
>>>
>>>
>>> Bert Gunter
>>>
>>> "The trouble with having an open mind is that people keep
coming along
>>> and sticking things into it."
>>> -- Opus (aka Berkeley Breathed in his "Bloom County"
comic strip )
>>>
>>>
>>> On Mon, Dec 5, 2016 at 9:53 AM, Rui Barradas <ruipbarradas at
sapo.pt> wrote:
>>>> Hello,
>>>>
>>>> Inline.
>>>>
>>>> Em 05-12-2016 17:09, David Winsemius escreveu:
>>>>>
>>>>>
>>>>>> On Dec 5, 2016, at 7:29 AM, John Sorkin <jsorkin at
grecc.umaryland.edu>
>>>>>> wrote:
>>>>>>
>>>>>> Rui,
>>>>>> I appreciate your suggestion, but eliminating the
deparse statement does
>>>>>> not solve my problem. Do you have any other
suggestions? See code below.
>>>>>> Thank you,
>>>>>> John
>>>>>>
>>>>>>
>>>>>> mydf <-
>>>>>>
data.frame(id=c(1,2,3,4,5),sex=c("M","M","M","F","F"),age=c(20,34,43,32,21))
>>>>>> mydf
>>>>>> class(mydf)
>>>>>>
>>>>>>
>>>>>> myfun <- function(frame,var){
>>>>>> call <- match.call()
>>>>>> print(call)
>>>>>>
>>>>>>
>>>>>> indx <-
match(c("frame","var"),names(call),nomatch=0)
>>>>>> print(indx)
>>>>>> if(indx[1]==0) stop("Function called without
sufficient arguments!")
>>>>>>
>>>>>>
>>>>>> cat("I can get the name of the dataframe as a
text string!\n")
>>>>>> #xx <- deparse(substitute(frame))
>>>>>> print(xx)
>>>>>>
>>>>>>
>>>>>> cat("I can get the name of the column as a text
string!\n")
>>>>>> #yy <- deparse(substitute(var))
>>>>>> print(yy)
>>>>>>
>>>>>>
>>>>>> # This does not work.
>>>>>> print(frame[,var])
>>>>>>
>>>>>>
>>>>>> # This does not work.
>>>>>> print(frame[,"var"])
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>> # This does not work.
>>>>>> col <- xx[,"yy"]
>>>>>>
>>>>>>
>>>>>> # Nor does this work.
>>>>>> col <- xx[,yy]
>>>>>> print(col)
>>>>>> }
>>>>>>
>>>>>>
>>>>>> myfun(mydf,age)
>>>>>
>>>>>
>>>>>
>>>>> When you use that calling syntax, the system will supply
the values of
>>>>> whatever the `age` variable contains. (And if there is no
`age`-named
>>>>> object, you get an error at the time of the call to
`myfun`.
>>>>
>>>>
>>>> Actually, no, which was very surprising to me but John's
code worked (not
>>>> the function, the call). And with the change I've proposed,
it worked
>>>> flawlessly. No errors. Why I don't know.
>>
>> See ?substitute and in particular the example highlighted there.
>>
>> The technical details are explained in the R Language Definition
>> manual. The key here is the use of promises for lay evaluations. In
>> fact, the expression in the call *is* available within the functions,
>> as is (a pointer to) the environment in which to evaluate the
>> expression. That is how substitute() works. Specifically, quoting from
>> the manual,
>>
>> *****
>> It is possible to access the actual (not default) expressions used as
>> arguments inside the function. The mechanism is implemented via
>> promises. When a function is being evaluated the actual expression
>> used as an argument is stored in the promise together with a pointer
>> to the environment the function was called from. When (if) the
>> argument is evaluated the stored expression is evaluated in the
>> environment that the function was called from. Since only a pointer to
>> the environment is used any changes made to that environment will be
>> in effect during this evaluation. The resulting value is then also
>> stored in a separate spot in the promise. Subsequent evaluations
>> retrieve this stored value (a second evaluation is not carried out).
>> Access to the unevaluated expression is also available using
>> substitute.
>> ********
>>
>> -- Bert
>>
>>
>>
>>
>>>>
>>>> Rui Barradas
>>>>
>>>> You need either to call it as:
>>>>>
>>>>>
>>>>> myfun( mydf , "age")
>>>>>
>>>>>
>>>>> # Or:
>>>>>
>>>>> age <- "age"
>>>>> myfun( mydf, age)
>>>>>
>>>>> Unless your value of the `age`-named variable was
"age" in the calling
>>>>> environment (and you did not give us that value in either
of your postings),
>>>>> you would fail.
>>>>>
>>>>
>>>> ______________________________________________
>>>> 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.