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

Mon Apr 30 17:25:26 CEST 2018

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 using gmail.com>
Date: Monday, April 30, 2018 at 8:08 AM
To: Rui Barradas <ruipbarradas using sapo.pt>
Cc: "MacQueen, Don" <macqueen1 using llnl.gov>, array R-help <r-help using 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 using sapo.pt<mailto:ruipbarradas using 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 using llnl.gov<mailto:macqueen1 using 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 using r-project.org<mailto:r-help-bounces using r-project.org> on behalf of lucam1968 using gmail.com<mailto:lucam1968 using 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

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