Re: [R] Is it possible to remove this loop? [SEC=UNCLASSIFIED]
Hello,
Try:
a1$h2 - 0
a1$h2[a1$h1==H] - 1
Regards
Le 12/07/03 16:18, jin...@ga.gov.au a écrit :
Hi all,
I would like create a new column in a data.frame (a1) to store 0, 1 data
converted from a factor as below.
a1$h2-NULL
for (i in 1:dim(a1)[1]) {
if (a1$h1[i]==H) a1$h2[i]-1 else a1$h2[i]-0
}
My question: is it possible to remove the loop from above code to achieve the
desired result?
Thanks in advance,
Jin
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Re: [R] Is it possible to remove this loop? [SEC=UNCLASSIFIED]
Hi
Hi all,
I would like create a new column in a data.frame (a1) to store 0, 1 data
converted from a factor as below.
a1$h2-NULL
for (i in 1:dim(a1)[1]) {
if (a1$h1[i]==H) a1$h2[i]-1 else a1$h2[i]-0
}
My question: is it possible to remove the loop from above code to
achieve
the desired result?
Untested
a1$h2 - (a1$h1==H)*1
Regards
Petr
Thanks in advance,
Jin
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Re: [R] Is it possible to remove this loop? [SEC=UNCLASSIFIED]
On 07/03/2012 05:18 PM, jin...@ga.gov.au wrote:
Hi all,
I would like create a new column in a data.frame (a1) to store 0, 1 data
converted from a factor as below.
a1$h2-NULL
for (i in 1:dim(a1)[1]) {
if (a1$h1[i]==H) a1$h2[i]-1 else a1$h2[i]-0
}
My question: is it possible to remove the loop from above code to achieve the
desired result?
Hi Jin,
Just to provide you with an embarrassment of alternatives:
a1$h2-ifelse(a1$h1==H,1,0)
Jim
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Re: [R] Is it possible to remove this loop? [SEC=UNCLASSIFIED]
On Jul 3, 2012, at 5:08 AM, Jim Lemon wrote:
On 07/03/2012 05:18 PM, jin...@ga.gov.au wrote:
Hi all,
I would like create a new column in a data.frame (a1) to store 0, 1
data converted from a factor as below.
a1$h2-NULL
for (i in 1:dim(a1)[1]) {
if (a1$h1[i]==H) a1$h2[i]-1 else a1$h2[i]-0
}
My question: is it possible to remove the loop from above code to
achieve the desired result?
Hi Jin,
Just to provide you with an embarrassment of alternatives:
a1$h2-ifelse(a1$h1==H,1,0)
One more. Similar to Petr's, but perhaps a bit more accessible to a
new R user:
a1$h2 - as.numeric(a1$h1==H)
I wasn't sure whether NA's would be handled in the same manner by
these two methods so I tested:
ifelse( factor(c(H, h, NA))==H, 1, 0)
[1] 1 0 NA
as.numeric( factor(c(H, h, NA))==H)
[1] 1 0 NA
--
David Winsemius, MD
West Hartford, CT
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Re: [R] Is it possible to remove this loop? [SEC=UNCLASSIFIED]
And one more alternative: a1$h2 - apply(a1,1, function(x) if (x[h1]==H) 1 else 0 ) -- View this message in context: http://r.789695.n4.nabble.com/Is-it-possible-to-remove-this-loop-SEC-UNCLASSIFIED-tp4635250p4635271.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list 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.
Re: [R] Is it possible to remove this loop? [SEC=UNCLASSIFIED]
Thank you all for providing various alternatives. They are all pretty fast. Great help! Based on a test of a dataset with 800,000 rows, the time used varies from 0.04 to 11.56 s. The champion is: a1$h2 - 0 a1$h2[a1$h1==H] - 1 Regards, Jin Geoscience Australia Disclaimer: This e-mail (and files transmitted with it) is intended only for the person or entity to which it is addressed. If you are not the intended recipient, then you have received this e-mail by mistake and any use, dissemination, forwarding, printing or copying of this e-mail and its file attachments is prohibited. The security of emails transmitted cannot be guaranteed; by forwarding or replying to this email, you acknowledge and accept these risks. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is it possible to remove this loop? [SEC=UNCLASSIFIED]
On 2012-07-03 17:23, jin...@ga.gov.au wrote:
Thank you all for providing various alternatives. They are all pretty fast.
Great help! Based on a test of a dataset with 800,000 rows, the time used
varies from 0.04 to 11.56 s. The champion is:
a1$h2 - 0
a1$h2[a1$h1==H] - 1
Interesting. My testing shows that Petr's solution is about
twice as fast. Not that it matters much - the time is pretty
small in any case.
a0 - data.frame(h1 = sample(c(H,J,K), 1e7, replace = TRUE),
stringsAsFactors = FALSE)
a1 - a0
system.time({a1$h2 - 0; a1$h2[a1$h1 == H] - 1})
# user system elapsed
# 1.470.481.96
a11 - a1
a1 - a0
system.time(a1$h2 - (a1$h1 == H) * 1)
# user system elapsed
# 0.370.170.56
a12 - a1
all.equal(a11,a12)
#[1] TRUE
Peter Ehlers
Regards,
Jin
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Re: [R] Is it possible to remove this loop? [SEC=UNCLASSIFIED]
Le 04/07/2012 12:43, Peter Ehlers a écrit :
On 2012-07-03 17:23, jin...@ga.gov.au wrote:
Thank you all for providing various alternatives. They are all pretty
fast. Great help! Based on a test of a dataset with 800,000 rows, the
time used varies from 0.04 to 11.56 s. The champion is:
a1$h2 - 0
a1$h2[a1$h1==H] - 1
Interesting. My testing shows that Petr's solution is about
twice as fast. Not that it matters much - the time is pretty
small in any case.
a0 - data.frame(h1 = sample(c(H,J,K), 1e7, replace = TRUE),
stringsAsFactors = FALSE)
a1 - a0
system.time({a1$h2 - 0; a1$h2[a1$h1 == H] - 1})
# user system elapsed
# 1.470.481.96
a11 - a1
a1 - a0
system.time(a1$h2 - (a1$h1 == H) * 1)
# user system elapsed
# 0.370.170.56
a12 - a1
all.equal(a11,a12)
#[1] TRUE
Peter Ehlers
I got the same result. Petr's solution is the fastest. Good to know it.
Pascal Oettli
Regards,
Jin
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Re: [R] Is it possible to remove this loop? [SEC=UNCLASSIFIED]
Thanks for your validation. Yes Peter's solution is the fastest, faster than
the previous one by saving 25% time. It was missed out in my previous testing.
Jin
-Original Message-
From: Pascal Oettli [mailto:kri...@ymail.com]
Sent: Wednesday, 4 July 2012 2:07 PM
To: Li Jin
Cc: r-help@r-project.org
Subject: Re: [R] Is it possible to remove this loop? [SEC=UNCLASSIFIED]
Le 04/07/2012 12:43, Peter Ehlers a écrit :
On 2012-07-03 17:23, jin...@ga.gov.au wrote:
Thank you all for providing various alternatives. They are all pretty
fast. Great help! Based on a test of a dataset with 800,000 rows, the
time used varies from 0.04 to 11.56 s. The champion is:
a1$h2 - 0
a1$h2[a1$h1==H] - 1
Interesting. My testing shows that Petr's solution is about
twice as fast. Not that it matters much - the time is pretty
small in any case.
a0 - data.frame(h1 = sample(c(H,J,K), 1e7, replace = TRUE),
stringsAsFactors = FALSE)
a1 - a0
system.time({a1$h2 - 0; a1$h2[a1$h1 == H] - 1})
# user system elapsed
# 1.470.481.96
a11 - a1
a1 - a0
system.time(a1$h2 - (a1$h1 == H) * 1)
# user system elapsed
# 0.370.170.56
a12 - a1
all.equal(a11,a12)
#[1] TRUE
Peter Ehlers
I got the same result. Petr's solution is the fastest. Good to know it.
Pascal Oettli
Regards,
Jin
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