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MacQueen, Don

2018-Apr-30 15:25 UTC

[R] How to visualise what code is processed within a for loop

Luca,

If speed is important, you might improve performance by making d0 into a true
matrix, rather than a data frame (assuming d0 is indeed a data frame at this
point). Although data frames may look like matrices, they aren?t, and they have
some overhead that matrices don?t.  I don?t think you would be able to use the
[[nm]] syntax with a matrix, but [ , nm] should work, provided the matrix has
column names. Or you could perhaps index by column number.

I had a project some years ago in which I reduced calculation time a lot by
extracting the numeric columns of a data frame and working with them, then
recombining them with the character columns. R?s performance working with data
frames has improved since then, so I really don?t know if it would make a
difference for your task.

-Don

--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
Lab cell 925-724-7509


From: Luca Meyer <lucam1968 at gmail.com>
Date: Monday, April 30, 2018 at 8:08 AM
To: Rui Barradas <ruipbarradas at sapo.pt>
Cc: "MacQueen, Don" <macqueen1 at llnl.gov>, array R-help
<r-help at r-project.org>
Subject: Re: [R] How to visualise what code is processed within a for loop

Hi Rui
Thank you for your suggestion,

I have tested the code suggested by you against that supplied by Don in terms of
timing and results are very much aligned: to populate a 5954x899 0/1 matrix on
my machine your procedure took 79 secs, while the one with ifelse employed 80
secs, hence unfortunately not really any significant time saved there.
Nevertheless thank you for your contribution.
Kind regards,

Luca

2018-04-28 23:18 GMT+02:00 Rui Barradas <ruipbarradas at
sapo.pt<mailto:ruipbarradas at sapo.pt>>:
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<mailto: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<mailto:r-help-bounces at r-project.org> on
behalf of lucam1968 at gmail.com<mailto: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<mailto: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<mailto: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<mailto: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]]

Rui Barradas

2018-Apr-30 15:56 UTC

head link

[R] How to visualise what code is processed within a for loop

Hello,

Another thing to note is that regexpr is likely to take (much) more time 
than ifelse or as.integer.
And the code will therefore not be very optimizable.

Rui Barradas

On 4/30/2018 4:25 PM, MacQueen, Don wrote:> Luca,
> 
> If speed is important, you might improve performance by making d0 into a 
> true matrix, rather than a data frame (assuming d0 is indeed a data 
> frame at this point). Although data frames may look like matrices, they 
> aren?t, and they have some overhead that matrices don?t.? I don?t think 
> you would be able to use the [[nm]] syntax with a matrix, but [ , nm] 
> should work, provided the matrix has column names. Or you could perhaps 
> index by column number.
> 
> I had a project some years ago in which I reduced calculation time a lot 
> by extracting the numeric columns of a data frame and working with them, 
> then recombining them with the character columns. R?s performance 
> working with data frames has improved since then, so I really don?t know 
> if it would make a difference for your task.
> 
> -Don
> 
> --
> 
> Don MacQueen
> 
> Lawrence Livermore National Laboratory
> 
> 7000 East Ave., L-627
> 
> Livermore, CA 94550
> 
> 925-423-1062
> 
> Lab cell 925-724-7509
> 
> *From: *Luca Meyer <lucam1968 at gmail.com>
> *Date: *Monday, April 30, 2018 at 8:08 AM
> *To: *Rui Barradas <ruipbarradas at sapo.pt>
> *Cc: *"MacQueen, Don" <macqueen1 at llnl.gov>, array R-help
> <r-help at r-project.org>
> *Subject: *Re: [R] How to visualise what code is processed within a for
loop
> 
> Hi Rui
> 
> Thank you for your suggestion,
> 
> I have tested the code suggested by you against that supplied by Don in 
> terms of timing and results are very much aligned: to populate a 
> 5954x899 0/1 matrix on my machine your procedure took 79 secs, while the 
> one with ifelse employed 80 secs, hence unfortunately not really any 
> significant time saved there.
> 
> Nevertheless thank you for your contribution.
> 
> Kind regards,
> 
> Luca
> 
> 2018-04-28 23:18 GMT+02:00 Rui Barradas 
> <ruipbarradas at sapo.pt<mailto:ruipbarradas at sapo.pt>>:
> 
>     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<mailto: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<mailto:r-help-bounces at
r-project.org>on
>                 behalf of
>                 lucam1968 at gmail.com<mailto: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<mailto: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<mailto: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<mailto: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.
>

Luca Meyer

2018-Apr-30 16:25 UTC

head link

[R] How to visualise what code is processed within a for loop

Thank you for both replies Don & Rui,

The very issue here is that there is a search that needs to be done within
a text field and I agree with Rui later comment that regexpr might indeed
be the time consuming piece of code.

I might try to optimise this piece of code later on, but for the time being
I am working on the following part of building a neural network to try
indeed classifying some text.

Again, thanks,

Luca

2018-04-30 17:25 GMT+02:00 MacQueen, Don <macqueen1 at llnl.gov>:
> Luca,
>
>
>
> If speed is important, you might improve performance by making d0 into a
> true matrix, rather than a data frame (assuming d0 is indeed a data frame
> at this point). Although data frames may look like matrices, they aren?t,
> and they have some overhead that matrices don?t.  I don?t think you would
> be able to use the [[nm]] syntax with a matrix, but [ , nm] should work,
> provided the matrix has column names. Or you could perhaps index by column
> number.
>
>
>
> I had a project some years ago in which I reduced calculation time a lot
> by extracting the numeric columns of a data frame and working with them,
> then recombining them with the character columns. R?s performance working
> with data frames has improved since then, so I really don?t know if it
> would make a difference for your task.
>
>
>
> -Don
>
>
>
> --
>
> Don MacQueen
>
> Lawrence Livermore National Laboratory
>
> 7000 East Ave., L-627
>
> Livermore, CA 94550
>
> 925-423-1062
>
> Lab cell 925-724-7509
>
>
>
>
>
> *From: *Luca Meyer <lucam1968 at gmail.com>
> *Date: *Monday, April 30, 2018 at 8:08 AM
> *To: *Rui Barradas <ruipbarradas at sapo.pt>
> *Cc: *"MacQueen, Don" <macqueen1 at llnl.gov>, array R-help
<
> r-help at r-project.org>
> *Subject: *Re: [R] How to visualise what code is processed within a for
> loop
>
>
>
> Hi Rui
>
> Thank you for your suggestion,
>
>
>
> I have tested the code suggested by you against that supplied by Don in
> terms of timing and results are very much aligned: to populate a 5954x899
> 0/1 matrix on my machine your procedure took 79 secs, while the one with
> ifelse employed 80 secs, hence unfortunately not really any significant
> time saved there.
>
> Nevertheless thank you for your contribution.
>
> Kind regards,
>
>
>
> Luca
>
>
>
> 2018-04-28 23:18 GMT+02:00 Rui Barradas <ruipbarradas at sapo.pt>:
>
> 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.
>
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Bert Gunter

2018-Apr-30 17:28 UTC

head link

[R] How to visualise what code is processed within a for loop

"I am working on the following part of building a neural network to try
indeed classifying some text."

... and so you are most likely trying to reinvent wheels. There are
already many such tools available here:
https://cran.r-project.org/web/views/NaturalLanguageProcessing.html

Some of these are interfaces to C language code, and so are probably
much more efficient than anything you (or I) can do at the R level.

Of course, if this is mainly a programming exercise for you, than this
is largely irrelevant.

Cheers,
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, Apr 30, 2018 at 9:25 AM, Luca Meyer <lucam1968 at gmail.com>
wrote:> Thank you for both replies Don & Rui,
>
> The very issue here is that there is a search that needs to be done within
> a text field and I agree with Rui later comment that regexpr might indeed
> be the time consuming piece of code.
>
> I might try to optimise this piece of code later on, but for the time being
> I am working on the following part of building a neural network to try
> indeed classifying some text.
>
> Again, thanks,
>
> Luca
>
> 2018-04-30 17:25 GMT+02:00 MacQueen, Don <macqueen1 at llnl.gov>:
>
>> Luca,
>>
>>
>>
>> If speed is important, you might improve performance by making d0 into
a
>> true matrix, rather than a data frame (assuming d0 is indeed a data
frame
>> at this point). Although data frames may look like matrices, they
aren?t,
>> and they have some overhead that matrices don?t.  I don?t think you
would
>> be able to use the [[nm]] syntax with a matrix, but [ , nm] should
work,
>> provided the matrix has column names. Or you could perhaps index by
column
>> number.
>>
>>
>>
>> I had a project some years ago in which I reduced calculation time a
lot
>> by extracting the numeric columns of a data frame and working with
them,
>> then recombining them with the character columns. R?s performance
working
>> with data frames has improved since then, so I really don?t know if it
>> would make a difference for your task.
>>
>>
>>
>> -Don
>>
>>
>>
>> --
>>
>> Don MacQueen
>>
>> Lawrence Livermore National Laboratory
>>
>> 7000 East Ave., L-627
>>
>> Livermore, CA 94550
>>
>> 925-423-1062
>>
>> Lab cell 925-724-7509
>>
>>
>>
>>
>>
>> *From: *Luca Meyer <lucam1968 at gmail.com>
>> *Date: *Monday, April 30, 2018 at 8:08 AM
>> *To: *Rui Barradas <ruipbarradas at sapo.pt>
>> *Cc: *"MacQueen, Don" <macqueen1 at llnl.gov>, array
R-help <
>> r-help at r-project.org>
>> *Subject: *Re: [R] How to visualise what code is processed within a for
>> loop
>>
>>
>>
>> Hi Rui
>>
>> Thank you for your suggestion,
>>
>>
>>
>> I have tested the code suggested by you against that supplied by Don in
>> terms of timing and results are very much aligned: to populate a
5954x899
>> 0/1 matrix on my machine your procedure took 79 secs, while the one
with
>> ifelse employed 80 secs, hence unfortunately not really any significant
>> time saved there.
>>
>> Nevertheless thank you for your contribution.
>>
>> Kind regards,
>>
>>
>>
>> Luca
>>
>>
>>
>> 2018-04-28 23:18 GMT+02:00 Rui Barradas <ruipbarradas at
sapo.pt>:
>>
>> 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.
>>
>>
>>
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>>
>> ______________________________________________
>> 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.
>>
>>
>>
>
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>
> ______________________________________________
> 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
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> and provide commented, minimal, self-contained, reproducible code.

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