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 <lucam1...@gmail.com>
*Date: *Monday, April 30, 2018 at 8:08 AM
*To: *Rui Barradas <ruipbarra...@sapo.pt>
*Cc: *"MacQueen, Don" <macque...@llnl.gov>, array R-help
<r-help@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
<ruipbarra...@sapo.pt<mailto:ruipbarra...@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
<macque...@llnl.gov<mailto:macque...@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$X0) > 0,
1, 0)
}
This may not a complete solution, since I have no idea
what the contents
or structure of d1 are, or what the regexpr() is
expected to return.
And notice the use of double brackets, [[ and ]]. This
is one way to
reference a column of a data frame when you have the
column's name stored
in a variable. Another way is d0[ , nm]
A couple of additional comments:
"t" is a poor choice of object name, because it is
one of R's built-in
functions (immediately after starting a fresh session of
R, with nothing
left over from any previous session, type help("r") and
see what you get).
ifelse() is intended for use on vectors, not scalars,
and it looks like
maybe you're using it on a scalar (can't be sure about
this, though)
For example, ifelse() is designed for this kind of usage:
ifelse( c(TRUE, FALSE, TRUE) , 1:3, 11:13)
[1] 1 12 3
Although it works ok for these
ifelse(TRUE, 3, 4)
[1] 3
ifelse(FALSE, 3, 4)
[1] 4
They are not really what it is intended for.
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
Lab cell 925-724-7509
On 4/24/18, 12:30 AM, "R-help on behalf of Luca Meyer" <
r-help-boun...@r-project.org<mailto:r-help-boun...@r-project.org>on
behalf of
lucam1...@gmail.com<mailto:lucam1...@gmail.com>> wrote:
Hi,
I am trying to debug the following code:
for (i in 1:10){
t <- paste("d0$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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