Re: [R] avoiding a loop: cumsum-like
Hi
I have not seen any answer yet so I wil try (partly).
I believe that the loop can be vectorised but I am a little bit lost
in your fors and ifs. I found that first part of res is same as
cumsum(tab$x.jour) until about 81st value. However I did not decipher
how to compute the remaining part. I tried to add
cumsum(tab$posit.lat) (after changing NA to 0) what is not correct.
Probably some combination of logical operation and summing can do
what you want. I thought that something like
((cumsum(tab$posit.lat)*0.8)*(cumsum(tab$x.jour)30)+cumsum(tab$x.jour
))
can do it but the result is defferent from your computation.
Not much of help, but maybe you can do better with above suggestion.
Petr
On 2 Nov 2006 at 11:15, GOUACHE David wrote:
Date sent: Thu, 2 Nov 2006 11:15:49 +0100
From: GOUACHE David [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Subject:[R] avoiding a loop: cumsum-like
Hello Rhelpers,
I need to run the following loop over a large number of data-sets, and
was wondering if it could somehow be vectorized. It's more or less a
cumulative sum, but slightly more complex. Here's the code, and an
example dataset (called tab in my code) follows. Thanks in advance for
any suggestions!
res-0
for (i in min(tab$Date):max(tab$Date))
{
if (is.na(tab$posit.lat[tab$Date==i])==T)
{
res-c(res,res[length(res)]+tab$x.jour[tab$Date==i])
}
else
{
if (res[tab$posit.lat[tab$Date==i]+1]30)
{
res-c(res,res[length(res)]+tab$x.jour[tab$Date==i])
}
else
{
res-c(res,res[length(res)]+tab$x.jour[tab$Date==i]+0.8*res[tab$pos
it.lat[tab$Date==i]+1])
}
}
}
res[-1]
Date x.jour posit.lat
35804 0 NA
35805 0 NA
35806 0 NA
35807 0 NA
35808 0 NA
35809 2.97338883 NA
35810 2.796389915 NA
35811 0 NA
35812 0 NA
35813 1.000711886 NA
35814 0.894422571 NA
35815 0 NA
35816 0 NA
35817 0 NA
35818 0 NA
35819 0 NA
35820 0 NA
35821 0 NA
35822 0 NA
35823 0 NA
35824 0 NA
35825 0 NA
35826 0 NA
35827 0 NA
35828 0 NA
35829 0 NA
35830 0 NA
35831 0 NA
35832 0 NA
35833 0 NA
35834 0 NA
35835 0 NA
35836 0 NA
35837 0 NA
35838 0 NA
35839 0 NA
35840 2.47237455 NA
35841 0 2
35842 0 3
35843 0 4
35844 0 5
35845 0 6
35846 0 7
35847 4.842160488 8
35848 2.432125036 9
35849 0 10
35850 0 12
35851 0 14
35852 0 16
35853 3.739683882 18
35854 1.980214421 20
35855 0 22
35856 0 24
35857 5.953444078 27
35858 6.455722475 29
35859 0 31
35860 3.798690334 32
35861 6.222993364 34
35862 3.746243098 35
35863 0 35
35864 0 36
35865 0 37
35866 0 38
35867 0 38
35868 0 39
35869 0 40
35870 0 41
35871 0 42
35872 0 43
35873 0 44
35874 0 45
35875 0 46
35876 0 47
35877 1.951774892 48
35878 0 49
35879 0 50
35880 1.702837643 50
35881 0 52
35882 0 53
35883 0 54
35884 0 55
35885 5.953444078 57
35886 0 58
35887 5.737515358 59
35888 0 61
35889 6.215941227 63
35890 4.731576675 64
35891 0 66
35892 2.255448314 66
35893 3.782283008 67
35894 3.244474546 68
35895 1.808553193 69
35896 2.622680002 70
35897 0 71
35898 0 72
35899 0 72
35900 1.7084177 73
35901 1.28455982 74
35902 2.320013736 76
35903 0 77
35904 0 78
35905 0 79
35906 0 79
35907 0 80
35908 6.716812458 81
35909 0 82
35910 6.796571531 84
35911 5.573668337 85
35912 5.42513958 86
35913 3.774513877 86
35914 0 87
35915 0 89
35916 0 90
35917 4.208252725 91
35918 0 92
35919 0 93
35920 0 95
35921 5.70023661 97
35922 0 98
35923 0 100
35924 0 102
35925 0 103
35926 0 104
David Gouache
Arvalis - Institut du Végétal
Station de La Miničre
78280 Guyancourt
Tel: 01.30.12.96.22 / Port: 06.86.08.94.32
__
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[EMAIL PROTECTED]
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[R] avoiding a loop: cumsum-like
Hello Rhelpers,
I need to run the following loop over a large number of data-sets, and was
wondering if it could somehow be vectorized. It's more or less a cumulative
sum, but slightly more complex.
Here's the code, and an example dataset (called tab in my code) follows.
Thanks in advance for any suggestions!
res-0
for (i in min(tab$Date):max(tab$Date))
{
if (is.na(tab$posit.lat[tab$Date==i])==T)
{
res-c(res,res[length(res)]+tab$x.jour[tab$Date==i])
}
else
{
if (res[tab$posit.lat[tab$Date==i]+1]30)
{
res-c(res,res[length(res)]+tab$x.jour[tab$Date==i])
}
else
{
res-c(res,res[length(res)]+tab$x.jour[tab$Date==i]+0.8*res[tab$posit.lat[tab$Date==i]+1])
}
}
}
res[-1]
Datex.jour posit.lat
35804 0 NA
35805 0 NA
35806 0 NA
35807 0 NA
35808 0 NA
35809 2.97338883 NA
35810 2.796389915 NA
35811 0 NA
35812 0 NA
35813 1.000711886 NA
35814 0.894422571 NA
35815 0 NA
35816 0 NA
35817 0 NA
35818 0 NA
35819 0 NA
35820 0 NA
35821 0 NA
35822 0 NA
35823 0 NA
35824 0 NA
35825 0 NA
35826 0 NA
35827 0 NA
35828 0 NA
35829 0 NA
35830 0 NA
35831 0 NA
35832 0 NA
35833 0 NA
35834 0 NA
35835 0 NA
35836 0 NA
35837 0 NA
35838 0 NA
35839 0 NA
35840 2.47237455 NA
35841 0 2
35842 0 3
35843 0 4
35844 0 5
35845 0 6
35846 0 7
35847 4.842160488 8
35848 2.432125036 9
35849 0 10
35850 0 12
35851 0 14
35852 0 16
35853 3.739683882 18
35854 1.980214421 20
35855 0 22
35856 0 24
35857 5.953444078 27
35858 6.455722475 29
35859 0 31
35860 3.798690334 32
35861 6.222993364 34
35862 3.746243098 35
35863 0 35
35864 0 36
35865 0 37
35866 0 38
35867 0 38
35868 0 39
35869 0 40
35870 0 41
35871 0 42
35872 0 43
35873 0 44
35874 0 45
35875 0 46
35876 0 47
35877 1.951774892 48
35878 0 49
35879 0 50
35880 1.702837643 50
35881 0 52
35882 0 53
35883 0 54
35884 0 55
35885 5.953444078 57
35886 0 58
35887 5.737515358 59
35888 0 61
35889 6.215941227 63
35890 4.731576675 64
35891 0 66
35892 2.255448314 66
35893 3.782283008 67
35894 3.244474546 68
35895 1.808553193 69
35896 2.622680002 70
35897 0 71
35898 0 72
35899 0 72
35900 1.7084177 73
35901 1.28455982 74
35902 2.320013736 76
35903 0 77
35904 0 78
35905 0 79
35906 0 79
35907 0 80
35908 6.716812458 81
35909 0 82
35910 6.796571531 84
35911 5.573668337 85
35912 5.42513958 86
35913 3.774513877 86
35914 0 87
35915 0 89
35916 0 90
35917 4.208252725 91
35918 0 92
35919 0 93
35920 0 95
35921 5.70023661 97
35922 0 98
35923 0 100
35924 0 102
35925 0 103
35926 0 104
David Gouache
Arvalis - Institut du Végétal
Station de La Minière
78280 Guyancourt
Tel: 01.30.12.96.22 / Port: 06.86.08.94.32
__
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[R] avoiding 'for' loop
Hi R users
I have a 1000 * 11 matrix (or data.frame). The first 10 columns consist of 0
or 1 (i.e., binary data) and the last column consists of real values.
I would like to run a logistic regression with the first binary column as a
dependent variable, and the second binary column and the last real value
column as independent variables. After this, I save 'Z' value for the binary
independent variable. Then, with the first binary column as a dependent
variable, and the third binary column and the last real value column as
independent variables. After this, I saved 'Z' value for the binary
independent variable, keeping doing this ...
Here is my simple R codes
logistic.z.value -c()
for (i in 1:9)
{
for (j in (i+1):10)
{
model - glm(X[,i] ~ X[,11] + X[,j], family=binomial(logit))
logistic.z.value -
append(logistic.z.value,coef(summary(model))[3,3])
}
}
I would like to avoid using 'for' loop command. Is there any way for me to
use some fast codes instead of using 'for' loop?
Many thanks
Taka,
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Re: [R] avoiding a loop
Mark,
This is hardly a one-liner. It will only help you if you call it with large
objects. If you have 0 in your data it needs even more extending.
Here it is:
myconstruct - function(aa) {
aa - c(1,1,aa,1)
bb - cumprod(aa)
taa - aa1
difftaa - diff(taa)
starts - which(c(FALSE,difftaa==1))
ends - which(c(FALSE,difftaa==-1))
correction - rep(0,length(aa))
correction[starts] - 1/bb[starts-2]
correction[ends] - -1/bb[starts-2]
correction - cumsum(correction)
correction[correction==0] - 1
correction
bbcor - bb*correction
bbcor[!taa] - aa[!taa]
bbcor[c(-1,-2,-length(bbcor))]
}
aa - runif(1,min=.5,max=1.5)
system.time(target - constructLt(aa))
system.time(found - myconstruct(aa))
identical(target,found)
max(abs(target-found))
system.time(target - constructLt(aa))
[1] 0.059 0.001 0.059 0.000 0.000
system.time(found - myconstruct(aa))
[1] 0.013 0.000 0.014 0.000 0.000
identical(target,found)
[1] FALSE
max(abs(target-found))
[1] 4.440892e-16
Appearently the outcome is different is some low decimal, due to other
calculation method.
Kees
On Tuesday 24 October 2006 20:36, Leeds, Mark (IED) wrote:
I think I asked a similar question 3 years ago to the Splus list and I
think the answer was no or noone answered so noone should spend more
than 5 minutes on this
because it could definitely be a waste of time.
My question is whether the function below can be rewritten without a for
loop. apply is fine if it can be done that way but i doubt it. I call it
a lot and would
prefer to not loop.
#---
--
constructLt-function(invector) {
outvector-invector
for ( i in 2:length(invector) ) {
if ( invector[i] 1 ) {
outvector[i]-invector[i]*outvector[i-1]
}
}
return(outvector)
}
#---
-
This is not an offer (or solicitation of an offer) to buy/sell the
securities/instruments mentioned or an official confirmation. Morgan
Stanley may deal as principal in or own or act as market maker for
securities/instruments mentioned or may advise the issuers. This is not
research and is not from MS Research but it may refer to a research
analyst/research report. Unless indicated, these views are the author's
and may differ from those of Morgan Stanley research or others in the Firm.
We do not represent this is accurate or complete and we may not update
this. Past performance is not indicative of future returns. For
additional information, research reports and important disclosures, contact
me or see https://secure.ms.com/servlet/cls. You should not use e-mail to
request, authorize or effect the purchase or sale of any security or
instrument, to send transfer instructions, or to effect any other
transactions. We cannot guarantee that any such requests received via !
e-mail will be processed in a timely manner. This communication is solely
for the addressee(s) and may contain confidential information. We do not
waive confidentiality by mistransmission. Contact me if you do not wish to
receive these communications. In the UK, this communication is directed in
the UK to those persons who are market counterparties or intermediate
customers (as defined in the UK Financial Services Authority's rules).
[[alternative HTML version deleted]]
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Re: [R] avoiding a loop
On Wednesday 25 October 2006 07:36, Leeds, Mark (IED) wrote:
I think I asked a similar question 3 years ago to the Splus list and I
think the answer was no or noone answered so noone should spend more
than 5 minutes on this
because it could definitely be a waste of time.
My question is whether the function below can be rewritten without a for
loop. apply is fine if it can be done that way but i doubt it. I call it
a lot and would
prefer to not loop.
#---
--
constructLt-function(invector) {
outvector-invector
for ( i in 2:length(invector) ) {
if ( invector[i] 1 ) {
outvector[i]-invector[i]*outvector[i-1]
}
}
return(outvector)
}
Depending on the nature of your data, there is a faster way. It still
involves looping, but not over the entire vector.
Try the following:
constructLt - function(invector) {
outvector - invector
cs - cumsum(rle(invector 1)$lengths)
if (invector[1] 1)
cs - c(1, cs)
for (i in 0:(length(cs)%/%2 - 1)){
starti - cs[2*i + 1]
stopi - cs[starti + 1]
outvector[starti:stopi] - cumprod(invector[starti:stopi])
}
return(outvector)
}
It is in the order of 3 times as fast for random vectors of considerable
length ( 1000). For random vectors of length 50 it is about the same speed
as the full looping algorithm. However if the data is such that there are
longer runs (than N(1, 1)), then you might expect a better speedup.
HTH
Ray Brownrigg
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[R] avoiding a loop
I think I asked a similar question 3 years ago to the Splus list and I
think the answer was no or noone answered so noone should spend more
than 5 minutes on this
because it could definitely be a waste of time.
My question is whether the function below can be rewritten without a for
loop. apply is fine if it can be done that way but i doubt it. I call it
a lot and would
prefer to not loop.
#---
--
constructLt-function(invector) {
outvector-invector
for ( i in 2:length(invector) ) {
if ( invector[i] 1 ) {
outvector[i]-invector[i]*outvector[i-1]
}
}
return(outvector)
}
#---
-
This is not an offer (or solicitation of an offer) to buy/sell the
securities/instruments mentioned or an official confirmation. Morgan Stanley
may deal as principal in or own or act as market maker for
securities/instruments mentioned or may advise the issuers. This is not
research and is not from MS Research but it may refer to a research
analyst/research report. Unless indicated, these views are the author's and
may differ from those of Morgan Stanley research or others in the Firm. We do
not represent this is accurate or complete and we may not update this. Past
performance is not indicative of future returns. For additional information,
research reports and important disclosures, contact me or see
https://secure.ms.com/servlet/cls. You should not use e-mail to request,
authorize or effect the purchase or sale of any security or instrument, to send
transfer instructions, or to effect any other transactions. We cannot
guarantee that any such requests received via !
e-mail will be processed in a timely manner. This communication is solely for
the addressee(s) and may contain confidential information. We do not waive
confidentiality by mistransmission. Contact me if you do not wish to receive
these communications. In the UK, this communication is directed in the UK to
those persons who are market counterparties or intermediate customers (as
defined in the UK Financial Services Authority's rules).
[[alternative HTML version deleted]]
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Re: [R] avoiding a loop
On Tue, 2006-10-24 at 14:36 -0400, Leeds, Mark (IED) wrote:
I think I asked a similar question 3 years ago to the Splus list and I
think the answer was no or noone answered so noone should spend more
than 5 minutes on this
because it could definitely be a waste of time.
My question is whether the function below can be rewritten without a for
loop. apply is fine if it can be done that way but i doubt it. I call it
a lot and would
prefer to not loop.
#---
--
constructLt-function(invector) {
outvector-invector
for ( i in 2:length(invector) ) {
if ( invector[i] 1 ) {
outvector[i]-invector[i]*outvector[i-1]
}
}
return(outvector)
}
You sure can vectorize this. Try this below... I haven't tested, but it
should be close to your solution. There's also
a - invector[-1]
outvector - invector
wh - which(a1)+1
outvector[wh] - a[wh] * invector[-length(invector)][wh-1]
outvector
HTH,
Jerome
--
Jerome Asselin, M.Sc., Agent de recherche, RHCE
CHUM -- Centre de recherche
3875 rue St-Urbain, 3e etage // Montreal QC H2W 1V1
Tel.: 514-890-8000 Poste 15914; Fax: 514-412-7106
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] avoiding a loop
On 10/24/2006 2:53 PM, Jerome Asselin wrote:
On Tue, 2006-10-24 at 14:36 -0400, Leeds, Mark (IED) wrote:
I think I asked a similar question 3 years ago to the Splus list and I
think the answer was no or noone answered so noone should spend more
than 5 minutes on this
because it could definitely be a waste of time.
My question is whether the function below can be rewritten without a for
loop. apply is fine if it can be done that way but i doubt it. I call it
a lot and would
prefer to not loop.
#---
--
constructLt-function(invector) {
outvector-invector
for ( i in 2:length(invector) ) {
if ( invector[i] 1 ) {
outvector[i]-invector[i]*outvector[i-1]
}
}
return(outvector)
}
You sure can vectorize this. Try this below... I haven't tested, but it
should be close to your solution. There's also
a - invector[-1]
outvector - invector
wh - which(a1)+1
outvector[wh] - a[wh] * invector[-length(invector)][wh-1]
outvector
I think this misses the fact that if a sequence of invector values are
less than 1, then the multiplication accumulates: e.g. with invector
equal to rep(0.5, 5) the output should be
constructLt(rep(0.5,5))
[1] 0.5 0.25000 0.12500 0.06250 0.03125
whereas your function gives
asselin(rep(0.5,5))
[1] 0.50 0.25 0.25 0.25 NA
A function like filter() might be able to do this, but in general R
doesn't give a way to write recursive formulae in a simple vector form.
In this particular case, I'd just use the for loop; if it turned out to
be too slow, I'd rewrite it in C, in a nearly literal translation.
Duncan Murdoch
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Re: [R] avoiding a loop
Try this (essentially the trick is to shift the invector to get the y[i-i]
effect):
constructLt-function(invector, a=1) {
invector[invectora] - c(0,invector)[invectora] * invector[invectora]
invector
}
aa - c(1,1,0.5,2,3,0.4,4,5)
aa
[1] 1.0 1.0 0.5 2.0 3.0 0.4 4.0 5.0
constructLt(aa)
[1] 1.0 1.0 0.5 2.0 3.0 1.2 4.0 5.0
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Leeds, Mark (IED)
Sent: Tuesday, October 24, 2006 2:36 PM
To: R-help@stat.math.ethz.ch
Subject: [R] avoiding a loop
I think I asked a similar question 3 years ago to the Splus list and I
think the answer was no or noone answered so noone should spend more than 5
minutes on this because it could definitely be a waste of time.
My question is whether the function below can be rewritten without a for
loop. apply is fine if it can be done that way but i doubt it. I call it a
lot and would prefer to not loop.
#---
--
constructLt-function(invector) {
outvector-invector
for ( i in 2:length(invector) ) {
if ( invector[i] 1 ) {
outvector[i]-invector[i]*outvector[i-1]
}
}
return(outvector)
}
#---
-
This is not an offer (or solicitation of an offer) to buy/sell the
securities/instruments mentioned or an official confirmation. Morgan
Stanley may deal as principal in or own or act as market maker for
securities/instruments mentioned or may advise the issuers. This is not
research and is not from MS Research but it may refer to a research
analyst/research report. Unless indicated, these views are the author's and
may differ from those of Morgan Stanley research or others in the Firm. We
do not represent this is accurate or complete and we may not update this.
Past performance is not indicative of future returns. For additional
information, research reports and important disclosures, contact me or see
https://secure.ms.com/servlet/cls. You should not use e-mail to request,
authorize or effect the purchase or sale of any security or instrument, to
send transfer instructions, or to effect any other transactions. We cannot
guarantee that any such requests received via !
e-mail will be processed in a timely manner. This communication is solely
for the addressee(s) and may contain confidential information. We do not
waive confidentiality by mistransmission. Contact me if you do not wish to
receive these communications. In the UK, this communication is directed in
the UK to those persons who are market counterparties or intermediate
customers (as defined in the UK Financial Services Authority's rules).
[[alternative HTML version deleted]]
__
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] avoiding a loop
I appreciate your effort but I'm fairly certain that won't work. The
loop I have written is pseudo recursive in the sense that
outvector depends on its previous value and this previous value may have
changed because it depnds on the previous value of itself.
The fact that there is no outvector on the right side of your last line
of code tells me that it's probably not what I want. I'll
Try it though. It can't hurt. I do appreciate the help.
-Original Message-
From: Jerome Asselin [mailto:[EMAIL PROTECTED]
Sent: Tuesday, October 24, 2006 2:54 PM
To: Leeds, Mark (IED)
Cc: R-help@stat.math.ethz.ch
Subject: Re: [R] avoiding a loop
On Tue, 2006-10-24 at 14:36 -0400, Leeds, Mark (IED) wrote:
I think I asked a similar question 3 years ago to the Splus list and
I think the answer was no or noone answered so noone should spend more
than 5 minutes on this because it could definitely be a waste of time.
My question is whether the function below can be rewritten without a
for loop. apply is fine if it can be done that way but i doubt it. I
call it a lot and would prefer to not loop.
#-
--
--
constructLt-function(invector) {
outvector-invector
for ( i in 2:length(invector) ) {
if ( invector[i] 1 ) {
outvector[i]-invector[i]*outvector[i-1]
}
}
return(outvector)
}
You sure can vectorize this. Try this below... I haven't tested, but it
should be close to your solution. There's also
a - invector[-1]
outvector - invector
wh - which(a1)+1
outvector[wh] - a[wh] * invector[-length(invector)][wh-1] outvector
HTH,
Jerome
--
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recherche
3875 rue St-Urbain, 3e etage // Montreal QC H2W 1V1
Tel.: 514-890-8000 Poste 15914; Fax: 514-412-7106
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Re: [R] avoiding a loop
thanks duncan. I think you are right. There are very smart people on
this list but there were very smart people on
The splus list 3 years ago so that's why I presented my question with
caution.
I'll just keep it as is for now. Thanks again.
-Original Message-
From: Duncan Murdoch [mailto:[EMAIL PROTECTED]
Sent: Tuesday, October 24, 2006 3:05 PM
To: Jerome Asselin
Cc: Leeds, Mark (IED); R-help@stat.math.ethz.ch
Subject: Re: [R] avoiding a loop
On 10/24/2006 2:53 PM, Jerome Asselin wrote:
On Tue, 2006-10-24 at 14:36 -0400, Leeds, Mark (IED) wrote:
I think I asked a similar question 3 years ago to the Splus list and
I think the answer was no or noone answered so noone should spend
more than 5 minutes on this because it could definitely be a waste of
time.
My question is whether the function below can be rewritten without a
for loop. apply is fine if it can be done that way but i doubt it. I
call it a lot and would prefer to not loop.
#
---
--
constructLt-function(invector) {
outvector-invector
for ( i in 2:length(invector) ) {
if ( invector[i] 1 ) {
outvector[i]-invector[i]*outvector[i-1]
}
}
return(outvector)
}
You sure can vectorize this. Try this below... I haven't tested, but
it should be close to your solution. There's also
a - invector[-1]
outvector - invector
wh - which(a1)+1
outvector[wh] - a[wh] * invector[-length(invector)][wh-1] outvector
I think this misses the fact that if a sequence of invector values are
less than 1, then the multiplication accumulates: e.g. with invector
equal to rep(0.5, 5) the output should be
constructLt(rep(0.5,5))
[1] 0.5 0.25000 0.12500 0.06250 0.03125
whereas your function gives
asselin(rep(0.5,5))
[1] 0.50 0.25 0.25 0.25 NA
A function like filter() might be able to do this, but in general R
doesn't give a way to write recursive formulae in a simple vector form.
In this particular case, I'd just use the for loop; if it turned out to
be too slow, I'd rewrite it in C, in a nearly literal translation.
Duncan Murdoch
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Re: [R] avoiding a loop
christos : change your vector to
1.0,1.0,0.5,2.0,0.2,0.4,4.0,5.0
You should get
1.00 1.00 0.50 2.00 0.40 0.16 4.00 5.00
But your function gives :
1.00 1.00 0.50 2.00 0.40 0.08 4.00 5.00
I think it doesn't work when there are two in a row elements that are
less than 1.0.
Again, I do appreciate your help.
-Original Message-
From: Christos Hatzis [mailto:[EMAIL PROTECTED]
Sent: Tuesday, October 24, 2006 3:01 PM
To: Leeds, Mark (IED); R-help@stat.math.ethz.ch
Subject: RE: [R] avoiding a loop
Try this (essentially the trick is to shift the invector to get the
y[i-i]
effect):
constructLt-function(invector, a=1) {
invector[invectora] - c(0,invector)[invectora] *
invector[invectora]
invector
}
aa - c(1,1,0.5,2,3,0.4,4,5)
aa
[1] 1.0 1.0 0.5 2.0 3.0 0.4 4.0 5.0
constructLt(aa)
[1] 1.0 1.0 0.5 2.0 3.0 1.2 4.0 5.0
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Leeds, Mark (IED)
Sent: Tuesday, October 24, 2006 2:36 PM
To: R-help@stat.math.ethz.ch
Subject: [R] avoiding a loop
I think I asked a similar question 3 years ago to the Splus list and I
think the answer was no or noone answered so noone should spend more
than 5 minutes on this because it could definitely be a waste of time.
My question is whether the function below can be rewritten without a for
loop. apply is fine if it can be done that way but i doubt it. I call it
a lot and would prefer to not loop.
#---
--
constructLt-function(invector) {
outvector-invector
for ( i in 2:length(invector) ) {
if ( invector[i] 1 ) {
outvector[i]-invector[i]*outvector[i-1]
}
}
return(outvector)
}
#---
-
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[R] avoiding a loop?
I have a vector, (not a list)
repeated.measures.FACTOR.names
[1] Insp1 Insp2 Insp3 Insp4 Insp5 Insp6 Insp7 Insp8 Insp9
and would like to convert this into a single string
Insp1,Insp2,Insp3,Insp4,Insp5,Insp6,Insp7,Insp8,Insp9
I can do that with a loop, but isn't there a more elegant way?
result - repeated.measures.FACTOR.names[[1]]
for(i in 2:length(repeated.measures.FACTOR.names)) {
result - paste(result, repeated.measures.FACTOR.names[[i]], sep=,) }
result
[1] Insp1,Insp2,Insp3,Insp4,Insp5,Insp6,Insp7,Insp8,Insp9
Thanks.
Charles Annis, P.E.
[EMAIL PROTECTED]
phone: 561-352-9699
eFax: 614-455-3265
http://www.StatisticalEngineering.com
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Re: [R] avoiding a loop?
On Thu, 2006-10-12 at 12:43 -0400, Charles Annis, P.E. wrote:
I have a vector, (not a list)
repeated.measures.FACTOR.names
[1] Insp1 Insp2 Insp3 Insp4 Insp5 Insp6 Insp7 Insp8 Insp9
and would like to convert this into a single string
Insp1,Insp2,Insp3,Insp4,Insp5,Insp6,Insp7,Insp8,Insp9
I can do that with a loop, but isn't there a more elegant way?
result - repeated.measures.FACTOR.names[[1]]
for(i in 2:length(repeated.measures.FACTOR.names)) {
result - paste(result, repeated.measures.FACTOR.names[[i]], sep=,) }
result
[1] Insp1,Insp2,Insp3,Insp4,Insp5,Insp6,Insp7,Insp8,Insp9
paste() is vectorized and note the use of 'collapse' in lieu of 'sep':
paste(repeated.measures.FACTOR.names, collapse = ,)
[1] Insp1,Insp2,Insp3,Insp4,Insp5,Insp6,Insp7,Insp8,Insp9
HTH,
Marc Schwartz
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Re: [R] avoiding a loop?
Le Jeudi 12 Octobre 2006 12:43, Charles Annis, P.E. a écrit :
I have a vector, (not a list)
repeated.measures.FACTOR.names
[1] Insp1 Insp2 Insp3 Insp4 Insp5 Insp6 Insp7 Insp8 Insp9
and would like to convert this into a single string
Insp1,Insp2,Insp3,Insp4,Insp5,Insp6,Insp7,Insp8,Insp9
I can do that with a loop, but isn't there a more elegant way?
result - repeated.measures.FACTOR.names[[1]]
for(i in 2:length(repeated.measures.FACTOR.names)) {
result - paste(result, repeated.measures.FACTOR.names[[i]], sep=,) }
result
[1] Insp1,Insp2,Insp3,Insp4,Insp5,Insp6,Insp7,Insp8,Insp9
Thanks.
Charles Annis, P.E.
[EMAIL PROTECTED]
phone: 561-352-9699
eFax: 614-455-3265
http://www.StatisticalEngineering.com
paste() will do what you want.
--
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École d'actuariat
Université Laval, Québec
[EMAIL PROTECTED] http://vgoulet.act.ulaval.ca
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Re: [R] avoiding a loop?
On Thu, 2006-10-12 at 12:07 -0500, Marc Schwartz wrote:
On Thu, 2006-10-12 at 12:43 -0400, Charles Annis, P.E. wrote:
I have a vector, (not a list)
repeated.measures.FACTOR.names
[1] Insp1 Insp2 Insp3 Insp4 Insp5 Insp6 Insp7 Insp8 Insp9
and would like to convert this into a single string
Insp1,Insp2,Insp3,Insp4,Insp5,Insp6,Insp7,Insp8,Insp9
I can do that with a loop, but isn't there a more elegant way?
result - repeated.measures.FACTOR.names[[1]]
for(i in 2:length(repeated.measures.FACTOR.names)) {
result - paste(result, repeated.measures.FACTOR.names[[i]], sep=,) }
result
[1] Insp1,Insp2,Insp3,Insp4,Insp5,Insp6,Insp7,Insp8,Insp9
paste() is vectorized and note the use of 'collapse' in lieu of 'sep':
paste(repeated.measures.FACTOR.names, collapse = ,)
[1] Insp1,Insp2,Insp3,Insp4,Insp5,Insp6,Insp7,Insp8,Insp9
Before I forget, you can also do the following to reconstruct the
initial sequence and the final result in a single step:
paste(Insp, 1:9, sep = , collapse = ,)
[1] Insp1,Insp2,Insp3,Insp4,Insp5,Insp6,Insp7,Insp8,Insp9
In this case, we use 'sep' to indicate that there should be no space
between each occurrence of 'Insp' and the integers and then use
'collapse' to indicate (as above) that each alphanum construct is to be
joined by a comma into a single element.
HTH,
Marc
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[R] Avoiding for loop
I understand that in R, for loops are not used as often as other
languages, and am trying to learn how to avoid them. I am wondering
if there is a more efficient way to write a certain piece of code,
which right now I can only envision as a for loop. I have a data file
that basically looks like:
1,55
1,23
2,12
...
that defines a matrix. Each row of the data file corresponds to a row
of the matrix, where each number in the row tells me what column a 1
or -1 should go into. So the first row in the data snippet above
means that the first row of my matrix needs to have a 1 in the 1st
column, and a -1 in the 55nd column. (And 0 elsewhere, which is
already there as I've created the matrix filled with 0s beforehand.)
So my current code looks like:
if(nrow(rawdata) = 1) for(i in 1:nrow(rawdata)) {
X[i, rawdata[i, 1]] - 1
X[i, rawdata[i, 2]] - -1
}
where rawdata is the original data file.
This sort of assignment happens many times in my program so any
improvement would be much appreciated. Thanks.
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Re: [R] Avoiding for loop
On Thu, 4 Aug 2005, Matt Crawford wrote: I understand that in R, for loops are not used as often as other languages, and am trying to learn how to avoid them. I am wondering if there is a more efficient way to write a certain piece of code, which right now I can only envision as a for loop. I have a data file that basically looks like: 1,55 1,23 2,12 ... that defines a matrix. Each row of the data file corresponds to a row of the matrix, where each number in the row tells me what column a 1 or -1 should go into. So the first row in the data snippet above means that the first row of my matrix needs to have a 1 in the 1st column, and a -1 in the 55nd column. (And 0 elsewhere, which is already there as I've created the matrix filled with 0s beforehand.) This may be a job for matrix indexes ii-1:nrow(rawdata) X[cbind(ii,rawdata[,1])] - 1 X[cbind(ii,rawdata[,2])] - -1 -thomas __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Avoiding for loop
On 8/4/05, Matt Crawford [EMAIL PROTECTED] wrote:
I understand that in R, for loops are not used as often as other
languages, and am trying to learn how to avoid them. I am wondering
if there is a more efficient way to write a certain piece of code,
which right now I can only envision as a for loop. I have a data file
that basically looks like:
1,55
1,23
2,12
...
that defines a matrix. Each row of the data file corresponds to a row
of the matrix, where each number in the row tells me what column a 1
or -1 should go into. So the first row in the data snippet above
means that the first row of my matrix needs to have a 1 in the 1st
column, and a -1 in the 55nd column. (And 0 elsewhere, which is
already there as I've created the matrix filled with 0s beforehand.)
So my current code looks like:
if(nrow(rawdata) = 1) for(i in 1:nrow(rawdata)) {
X[i, rawdata[i, 1]] - 1
X[i, rawdata[i, 2]] - -1
}
where rawdata is the original data file.
This sort of assignment happens many times in my program so any
improvement would be much appreciated. Thanks.
idx - seq(length = max(rawdata, 0))
X - outer(rawdata[,1], idx, ==) - outer(rawdata[,2], idx, ==)
Note that we did not have to predefine X and it also works if rawdata
has zero rows:
rawdata - matrix(0, nr = 0, nc = 2)
in which case it gives a 0 by 0 matrix.
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[R] Avoiding a Loop?
Dear R-Helpers,
I have a matrix where the first column is known. The second column is
the result of multiplying this first column with a constant const. The
third column is the result of multiplying the second column with
const.
So far, I did it like this (as a simplified example):
nr.of.columns - 4
myconstant - 27.5
mymatrix - matrix(numeric(0), nrow=5, ncol=nr.of.columns)
mymatrix[,1] - 1:5
for (i in 2:nr.of.columns) {
mymatrix[,i] - myconstant * mymatrix[,i-1]
}
Can anyone give me some advice whether it is possible to avoid this loop
(and if yes: how)?
Any suggestions are welcome!
Thanks,
Roland
+
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RE: [R] Avoiding a Loop?
If `v' the the vector in the first column and `k' is the constant, then the
matrix has k^(j-1) * v in the jth column, right?
k - 27.5^seq(0, length=4)
k
[1] 1.0027.50 756.25 20796.88
outer(1:5, k, *)
[,1] [,2][,3] [,4]
[1,]1 27.5 756.25 20796.88
[2,]2 55.0 1512.50 41593.75
[3,]3 82.5 2268.75 62390.63
[4,]4 110.0 3025.00 83187.50
[5,]5 137.5 3781.25 103984.38
Andy
From: Rau, Roland
Dear R-Helpers,
I have a matrix where the first column is known. The second column is
the result of multiplying this first column with a constant
const. The
third column is the result of multiplying the second column with
const.
So far, I did it like this (as a simplified example):
nr.of.columns - 4
myconstant - 27.5
mymatrix - matrix(numeric(0), nrow=5, ncol=nr.of.columns)
mymatrix[,1] - 1:5
for (i in 2:nr.of.columns) {
mymatrix[,i] - myconstant * mymatrix[,i-1]
}
Can anyone give me some advice whether it is possible to
avoid this loop
(and if yes: how)?
Any suggestions are welcome!
Thanks,
Roland
+
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Rese...{{dropped}}
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Re: [R] Avoiding a Loop?
Does this do what you want?
nr.of.columns - 4
myconstant - 27.5
mymatrix - matrix(myconstant, nrow=5, ncol=nr.of.columns)
mymatrix[,1] - 1:5
t(apply(mymatrix, 1, function(x) cumprod(x)))
__
James HoltmanWhat is the problem you are trying to solve?
Executive Technical Consultant -- Office of Technology, Convergys
[EMAIL PROTECTED]
+1 (513) 723-2929
Rau, Roland
[EMAIL PROTECTED] To:
r-help@stat.math.ethz.ch
Sent by: cc:
[EMAIL PROTECTED]Subject: [R] Avoiding a Loop?
ath.ethz.ch
01/21/2005 07:31
Dear R-Helpers,
I have a matrix where the first column is known. The second column is
the result of multiplying this first column with a constant const. The
third column is the result of multiplying the second column with
const.
So far, I did it like this (as a simplified example):
nr.of.columns - 4
myconstant - 27.5
mymatrix - matrix(numeric(0), nrow=5, ncol=nr.of.columns)
mymatrix[,1] - 1:5
for (i in 2:nr.of.columns) {
mymatrix[,i] - myconstant * mymatrix[,i-1]
}
Can anyone give me some advice whether it is possible to avoid this loop
(and if yes: how)?
Any suggestions are welcome!
Thanks,
Roland
+
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RE: [R] Avoiding a Loop?
Dear R-List,
thank you very much for the fast answers.
Andy Liaw and James Holtman both gave me some example code which does
exactly what I want.
I will now try to check which one is better suited for my actual matrix
(in terms of speed).
Thanks,
Roland
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Sent: Friday, January 21, 2005 3:57 PM
To: Rau, Roland
Cc: r-help@stat.math.ethz.ch; [EMAIL PROTECTED]
Subject: Re: [R] Avoiding a Loop?
Does this do what you want?
nr.of.columns - 4
myconstant - 27.5
mymatrix - matrix(myconstant, nrow=5, ncol=nr.of.columns)
mymatrix[,1] - 1:5
t(apply(mymatrix, 1, function(x) cumprod(x)))
__
James HoltmanWhat is the problem you are trying to solve?
Executive Technical Consultant -- Office of Technology, Convergys
[EMAIL PROTECTED]
+1 (513) 723-2929
Rau, Roland
[EMAIL PROTECTED] To:
r-help@stat.math.ethz.ch
Sent by: cc:
[EMAIL PROTECTED]Subject: [R]
Avoiding a Loop?
ath.ethz.ch
01/21/2005 07:31
Dear R-Helpers,
I have a matrix where the first column is known. The second column is
the result of multiplying this first column with a constant const. The
third column is the result of multiplying the second column with
const.
So far, I did it like this (as a simplified example):
nr.of.columns - 4
myconstant - 27.5
mymatrix - matrix(numeric(0), nrow=5, ncol=nr.of.columns)
mymatrix[,1] - 1:5
for (i in 2:nr.of.columns) {
mymatrix[,i] - myconstant * mymatrix[,i-1]
}
Can anyone give me some advice whether it is possible to avoid this loop
(and if yes: how)?
Any suggestions are welcome!
Thanks,
Roland
+
This mail has been sent through the MPI for Demographic
Rese...{{dropped}}
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+
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Re: [R] Avoiding a Loop?
Rau, Roland [EMAIL PROTECTED] writes:
Dear R-Helpers,
I have a matrix where the first column is known. The second column is
the result of multiplying this first column with a constant const. The
third column is the result of multiplying the second column with
const.
So far, I did it like this (as a simplified example):
nr.of.columns - 4
myconstant - 27.5
mymatrix - matrix(numeric(0), nrow=5, ncol=nr.of.columns)
mymatrix[,1] - 1:5
for (i in 2:nr.of.columns) {
mymatrix[,i] - myconstant * mymatrix[,i-1]
}
Can anyone give me some advice whether it is possible to avoid this loop
(and if yes: how)?
How about:
myconstant - 27.5
a - 1:5
myconstant - myconstant^((1:nr.of.columns)-1)
myconstant
[1] 1.0027.50 756.25 20796.88
mymatrix - outer(a,myconstant)
mymatrix
[,1] [,2][,3] [,4]
[1,]1 27.5 756.25 20796.88
[2,]2 55.0 1512.50 41593.75
[3,]3 82.5 2268.75 62390.62
[4,]4 110.0 3025.00 83187.50
[5,]5 137.5 3781.25 103984.38
HTH
Zoltan
Any suggestions are welcome!
Thanks,
Roland
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--
Z. Barta
Dept. of Evol. Zool., Univ. of Debrecen, Debrecen, H-4010, Hungary
Phone: 36 52 316 666 ext. 2334, Fax: 36 52 533 677
E-mail: [EMAIL PROTECTED], http://puma.unideb.hu/~zbarta/
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Re: [R] Avoiding a Loop?
Rau, Roland wrote:
Dear R-Helpers,
I have a matrix where the first column is known. The second column is
the result of multiplying this first column with a constant const. The
third column is the result of multiplying the second column with
const.
So far, I did it like this (as a simplified example):
nr.of.columns - 4
myconstant - 27.5
mymatrix - matrix(numeric(0), nrow=5, ncol=nr.of.columns)
mymatrix[,1] - 1:5
for (i in 2:nr.of.columns) {
mymatrix[,i] - myconstant * mymatrix[,i-1]
}
nr.of.columns - 4
myconstant - 27.5
mycolumn - 1:5
outer(mycolumn, myconstant^(0:(nr.of.columns-1)))
Uwe Ligges
Can anyone give me some advice whether it is possible to avoid this loop
(and if yes: how)?
Any suggestions are welcome!
Thanks,
Roland
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RE: [R] Avoiding a Loop?
Roland:
Andy Liaw and others have already given perfectly good answers to this (but
note: Using apply() type functions does **not** avoid loops; apply's **are**
loops). However, mostly as an illustration to reinforce Uwe Ligges's
comments (in the dim vs length thread) about the usefulness of sometimes
treating arrays as vectors, I offer the following:
nc-ncol(yourmatrix)
matrix(yourmatrix[,1]*rep(k^seq(0,length=nc),e=nrow(yourmatrix)),ncol=nc)
Alteratively, one could use matrix multiplication:
yourmatrix[,1]%*%matrix(k^seq(0,length=nc),nrow=1)
Both of these will be very fast (although unlikely to make a noticeable
difference without a lot of rows or columns) and I think use less memory
than outer() (again, unlikely to make a noticeable difference).
-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Rau, Roland
Sent: Friday, January 21, 2005 4:32 AM
To: r-help@stat.math.ethz.ch
Subject: [R] Avoiding a Loop?
Dear R-Helpers,
I have a matrix where the first column is known. The second column is
the result of multiplying this first column with a constant
const. The
third column is the result of multiplying the second column with
const.
So far, I did it like this (as a simplified example):
nr.of.columns - 4
myconstant - 27.5
mymatrix - matrix(numeric(0), nrow=5, ncol=nr.of.columns)
mymatrix[,1] - 1:5
for (i in 2:nr.of.columns) {
mymatrix[,i] - myconstant * mymatrix[,i-1]
}
Can anyone give me some advice whether it is possible to
avoid this loop
(and if yes: how)?
Any suggestions are welcome!
Thanks,
Roland
+
This mail has been sent through the MPI for Demographic
Rese...{{dropped}}
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
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http://www.R-project.org/posting-guide.html
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