R help - Apr 2018 - How to visualise what code is processed within a for loop

Thanks Don,

    for (i in 1:10){
      nm <- paste0("V", i)
      d0[[nm]] <- ifelse( regexpr(d1[i,1], d0$X0) > 0, 1, 0)
    }

is exaclty what I needed.

Best regards,

Luca


2018-04-25 23:03 GMT+02:00 MacQueen, Don <macqueen1 at llnl.gov>:
> Your code doesn't make sense to me in a couple of ways.
>
> Inside the loop, the first line assigns a value to an object named
"t".
> Then, the second line does the same thing, assigns a value to an object
> named "t".
>
> The value of the object named "t" after the second line will be
the output
> of the ifelse() expression, whatever that is. This has the effect of making
> the first line irrelevant. Whatever value t has after the first line is
> replaced by whatever it gets from the second line.
>
> It looks like the first line inside the loop is constructing the name of a
> data frame column, and storing that name as a character string. However,
> the second line doesn't use that name at all. If your goal is to update
the
> contents of a column, you need to assign something to that column in the
> next line. Instead you assign it to the object named "t".
>
> What you're looking for will be more along the lines of this:
>
>     for (i in 1:10){
>       nm <- paste0("V", i)
>       d0[[nm]] <- ifelse( regexpr(d1[i,1], d0$X0) > 0, 1, 0)
>     }
>
> This may not a complete solution, since I have no idea what the contents
> or structure of d1 are, or what the regexpr() is expected to return.
>
> And notice the use of double brackets, [[ and ]]. This is one way to
> reference a column of a  data frame when you have the column's name
stored
> in a variable. Another way is d0[ , nm]
>
>
> A couple of additional comments:
>
>  "t" is a poor choice of object name, because it is one of
R's built-in
> functions (immediately after starting a fresh session of R, with nothing
> left over from any previous session, type help("r") and see what
you get).
>
>  ifelse() is intended for use on vectors, not scalars, and it looks like
> maybe you're using it on a scalar (can't be sure about this,
though)
>
> For example, ifelse() is designed for this kind of usage:
> > ifelse( c(TRUE, FALSE, TRUE) , 1:3, 11:13)
> [1]  1 12  3
>
> Although it works ok for these
> > ifelse(TRUE, 3, 4)
> [1] 3
> > ifelse(FALSE, 3, 4)
> [1] 4
> They are not really what it is intended for.
>
> --
> Don MacQueen
> Lawrence Livermore National Laboratory
> 7000 East Ave., L-627
> Livermore, CA 94550
> 925-423-1062
> Lab cell 925-724-7509
>
>
> ?On 4/24/18, 12:30 AM, "R-help on behalf of Luca Meyer" <
> r-help-bounces at r-project.org on behalf of lucam1968 at gmail.com>
wrote:
>
>     Hi,
>
>     I am trying to debug the following code:
>
>     for (i in 1:10){
>       t <- paste("d0$V",i,sep="")
>       t <- ifelse(regexpr(d1[i,1],d0$X0)>0,1,0)
>     }
>
>     and I would like to see what code is actually processing R, how can I
> do
>     that?
>
>     More to the point, I am trying to update my variables d0$V1 to d0$V10
>     according to the presence or absence of some text (contained in the
> file
>     d1) within the d0$X0 variable.
>
>     The code seem to run ok, if I add print(table(t)) within the loop I
> can see
>     that the ifelse procedure is working and to some cases within the
> d0$V1 to
>     d0$V10 variable range a 1 is assigned. But when checking my d0$V1 to
> d0$V10
>     after the for loop they are all still equal to zero...
>
>     Thanks,
>
>     Luca
>
>         [[alternative HTML version deleted]]
>
>     ______________________________________________
>     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.
>
>
>
	[[alternative HTML version deleted]]

Hello,

instead of ifelse, the following is exactly the same and much more 
efficient.

d0[[nm]] <- as.integer(regexpr(d1[i,1], d0$X0) > 0)


Hope this helps,

Rui Barradas

On 4/28/2018 8:45 PM, Luca Meyer wrote:
> Thanks Don,
> 
>      for (i in 1:10){
>        nm <- paste0("V", i)
>        d0[[nm]] <- ifelse( regexpr(d1[i,1], d0$X0) > 0, 1, 0)
>      }
> 
> is exaclty what I needed.
> 
> Best regards,
> 
> Luca
> 
> 
> 2018-04-25 23:03 GMT+02:00 MacQueen, Don <macqueen1 at llnl.gov>:
> 
>> Your code doesn't make sense to me in a couple of ways.
>>
>> Inside the loop, the first line assigns a value to an object named
"t".
>> Then, the second line does the same thing, assigns a value to an object
>> named "t".
>>
>> The value of the object named "t" after the second line will
be the output
>> of the ifelse() expression, whatever that is. This has the effect of
making
>> the first line irrelevant. Whatever value t has after the first line is
>> replaced by whatever it gets from the second line.
>>
>> It looks like the first line inside the loop is constructing the name
of a
>> data frame column, and storing that name as a character string.
However,
>> the second line doesn't use that name at all. If your goal is to
update the
>> contents of a column, you need to assign something to that column in
the
>> next line. Instead you assign it to the object named "t".
>>
>> What you're looking for will be more along the lines of this:
>>
>>      for (i in 1:10){
>>        nm <- paste0("V", i)
>>        d0[[nm]] <- ifelse( regexpr(d1[i,1], d0$X0) > 0, 1, 0)
>>      }
>>
>> This may not a complete solution, since I have no idea what the
contents
>> or structure of d1 are, or what the regexpr() is expected to return.
>>
>> And notice the use of double brackets, [[ and ]]. This is one way to
>> reference a column of a  data frame when you have the column's name
stored
>> in a variable. Another way is d0[ , nm]
>>
>>
>> A couple of additional comments:
>>
>>   "t" is a poor choice of object name, because it is one of
R's built-in
>> functions (immediately after starting a fresh session of R, with
nothing
>> left over from any previous session, type help("r") and see
what you get).
>>
>>   ifelse() is intended for use on vectors, not scalars, and it looks
like
>> maybe you're using it on a scalar (can't be sure about this,
though)
>>
>> For example, ifelse() is designed for this kind of usage:
>>> ifelse( c(TRUE, FALSE, TRUE) , 1:3, 11:13)
>> [1]  1 12  3
>>
>> Although it works ok for these
>>> ifelse(TRUE, 3, 4)
>> [1] 3
>>> ifelse(FALSE, 3, 4)
>> [1] 4
>> They are not really what it is intended for.
>>
>> --
>> Don MacQueen
>> Lawrence Livermore National Laboratory
>> 7000 East Ave., L-627
>> Livermore, CA 94550
>> 925-423-1062
>> Lab cell 925-724-7509
>>
>>
>> ?On 4/24/18, 12:30 AM, "R-help on behalf of Luca Meyer" <
>> r-help-bounces at r-project.org on behalf of lucam1968 at gmail.com>
wrote:
>>
>>      Hi,
>>
>>      I am trying to debug the following code:
>>
>>      for (i in 1:10){
>>        t <- paste("d0$V",i,sep="")
>>        t <- ifelse(regexpr(d1[i,1],d0$X0)>0,1,0)
>>      }
>>
>>      and I would like to see what code is actually processing R, how
can I
>> do
>>      that?
>>
>>      More to the point, I am trying to update my variables d0$V1 to
d0$V10
>>      according to the presence or absence of some text (contained in
the
>> file
>>      d1) within the d0$X0 variable.
>>
>>      The code seem to run ok, if I add print(table(t)) within the loop
I
>> can see
>>      that the ifelse procedure is working and to some cases within the
>> d0$V1 to
>>      d0$V10 variable range a 1 is assigned. But when checking my d0$V1
to
>> d0$V10
>>      after the for loop they are all still equal to zero...
>>
>>      Thanks,
>>
>>      Luca
>>
>>          [[alternative HTML version deleted]]
>>
>>      ______________________________________________
>>      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.
>>
>>
>>
> 
> 	[[alternative HTML version deleted]]
> 
> ______________________________________________
> 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.
>

I forgot to explain why my suggestion.

The logical condition returns FALSE/TRUE that in R are coded as 0/1.
So all you have to do is coerce to integer.

This works because the ifelse will return a 1 or a 0 depending on the 
condition. Meaning exactly the same values. And is more efficient since 
ifelse creates both vectors, the true part and the false part, and then 
indexes those vectors in order to return the appropriate values. This is 
the double of the trouble and a great deal of memory used.

Rui Barradas

On 4/28/2018 10:12 PM, Rui Barradas wrote:
> Hello,
> 
> instead of ifelse, the following is exactly the same and much more 
> efficient.
> 
> d0[[nm]] <- as.integer(regexpr(d1[i,1], d0$X0) > 0)
> 
> 
> Hope this helps,
> 
> Rui Barradas
> 
> On 4/28/2018 8:45 PM, Luca Meyer wrote:
>> Thanks Don,
>>
>> ???? for (i in 1:10){
>> ?????? nm <- paste0("V", i)
>> ?????? d0[[nm]] <- ifelse( regexpr(d1[i,1], d0$X0) > 0, 1, 0)
>> ???? }
>>
>> is exaclty what I needed.
>>
>> Best regards,
>>
>> Luca
>>
>>
>> 2018-04-25 23:03 GMT+02:00 MacQueen, Don <macqueen1 at llnl.gov>:
>>
>>> Your code doesn't make sense to me in a couple of ways.
>>>
>>> Inside the loop, the first line assigns a value to an object named
"t".
>>> Then, the second line does the same thing, assigns a value to an
object
>>> named "t".
>>>
>>> The value of the object named "t" after the second line
will be the
>>> output
>>> of the ifelse() expression, whatever that is. This has the effect
of
>>> making
>>> the first line irrelevant. Whatever value t has after the first
line is
>>> replaced by whatever it gets from the second line.
>>>
>>> It looks like the first line inside the loop is constructing the
name
>>> of a
>>> data frame column, and storing that name as a character string.
However,
>>> the second line doesn't use that name at all. If your goal is
to
>>> update the
>>> contents of a column, you need to assign something to that column
in the
>>> next line. Instead you assign it to the object named "t".
>>>
>>> What you're looking for will be more along the lines of this:
>>>
>>> ???? for (i in 1:10){
>>> ?????? nm <- paste0("V", i)
>>> ?????? d0[[nm]] <- ifelse( regexpr(d1[i,1], d0$X0) > 0, 1, 0)
>>> ???? }
>>>
>>> This may not a complete solution, since I have no idea what the
contents
>>> or structure of d1 are, or what the regexpr() is expected to
return.
>>>
>>> And notice the use of double brackets, [[ and ]]. This is one way
to
>>> reference a column of a? data frame when you have the column's
name
>>> stored
>>> in a variable. Another way is d0[ , nm]
>>>
>>>
>>> A couple of additional comments:
>>>
>>> ? "t" is a poor choice of object name, because it is one
of R's built-in
>>> functions (immediately after starting a fresh session of R, with
nothing
>>> left over from any previous session, type help("r") and
see what you
>>> get).
>>>
>>> ? ifelse() is intended for use on vectors, not scalars, and it
looks
>>> like
>>> maybe you're using it on a scalar (can't be sure about
this, though)
>>>
>>> For example, ifelse() is designed for this kind of usage:
>>>> ifelse( c(TRUE, FALSE, TRUE) , 1:3, 11:13)
>>> [1]? 1 12? 3
>>>
>>> Although it works ok for these
>>>> ifelse(TRUE, 3, 4)
>>> [1] 3
>>>> ifelse(FALSE, 3, 4)
>>> [1] 4
>>> They are not really what it is intended for.
>>>
>>> -- 
>>> Don MacQueen
>>> Lawrence Livermore National Laboratory
>>> 7000 East Ave., L-627
>>> Livermore, CA 94550
>>> 925-423-1062
>>> Lab cell 925-724-7509
>>>
>>>
>>> ?On 4/24/18, 12:30 AM, "R-help on behalf of Luca Meyer"
<
>>> r-help-bounces at r-project.org on behalf of lucam1968 at
gmail.com> wrote:
>>>
>>> ???? Hi,
>>>
>>> ???? I am trying to debug the following code:
>>>
>>> ???? for (i in 1:10){
>>> ?????? t <- paste("d0$V",i,sep="")
>>> ?????? t <- ifelse(regexpr(d1[i,1],d0$X0)>0,1,0)
>>> ???? }
>>>
>>> ???? and I would like to see what code is actually processing R,
how
>>> can I
>>> do
>>> ???? that?
>>>
>>> ???? More to the point, I am trying to update my variables d0$V1 to
>>> d0$V10
>>> ???? according to the presence or absence of some text (contained
in the
>>> file
>>> ???? d1) within the d0$X0 variable.
>>>
>>> ???? The code seem to run ok, if I add print(table(t)) within the
loop I
>>> can see
>>> ???? that the ifelse procedure is working and to some cases within
the
>>> d0$V1 to
>>> ???? d0$V10 variable range a 1 is assigned. But when checking my 
>>> d0$V1 to
>>> d0$V10
>>> ???? after the for loop they are all still equal to zero...
>>>
>>> ???? Thanks,
>>>
>>> ???? Luca
>>>
>>> ???????? [[alternative HTML version deleted]]
>>>
>>> ???? ______________________________________________
>>> ???? 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.
>>>
>>>
>>>
>>
>> ????[[alternative HTML version deleted]]
>>
>> ______________________________________________
>> 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.
>>
> 
> ______________________________________________
> 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.