This is precisely what I needed; I can't believe how simple it is. Thanks!
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nal Message-
> From: vashchyshy...@gmail.com
> Sent: Tue, 5 Jun 2012 07:48:51 -0700 (PDT)
> To: r-help@r-project.org
> Subject: [R] rowSums problem
>
> I'm having a very frustrating problem, trying to find the inverse
> distance
> squared weighted interpolants of some
Hello,
The files you've uploaded are the weights file and the results file, not
the original temp.csv.
So this is untested but it seems you have a standard matrix multiply
problem.
temp3880W <- temp[, 3:50] %*% weight3880
Hope this helps,
Rui Barradas
Em 05-06-2012 15:48, alonis10 escreveu
http://r.789695.n4.nabble.com/file/n4632406/temp3880.csv temp3880.csv
http://r.789695.n4.nabble.com/file/n4632406/weight3880.csv weight3880.csv
Here are the files I promised to upload.
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Sent fr
I'm having a very frustrating problem, trying to find the inverse distance
squared weighted interpolants of some weather data.
I have a data frame of weights, which sum to 1. I have attached the weights
data. I also have a data frame of temperatures at 48 grid points, which I
have also attached.
he first commandment of floating point programming is
THOU SHALT NOT TEST WHETHER TWO FP NUMBERS ARE EQUAL
HTH
Rex
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of thomas.salve...@syngenta.com
Sent: Monday, March 07, 2011 2:09 AM
To: r-h
Hi Tom,
That's once again the floating point number issue: see FAQ 7.31.
Look at this:
sum(m[161,])
[1] 1
sum(m[161,])==1
[1] FALSE
sum(m[161,])-1
[1] 2.220446e-16
So 0.6+0.3+0.1 is indeed greater than 1
Try this instead:
round(sum(m[161,]))==1
[1] TRUE
HTH,
Ivan
Le 3/7/2011 08:08, thomas.sa
I am trying to construct a data set with some sequences for example:
a = seq(0,1,0.1)
m = matrix(nrow = 1331, ncol = 3)
m[,1] = rep(a,121)
m[,2] = rep(a,11,each = 11)
m[,3] = rep(a,1,each = 121)
I realize that there may be better ways of doing this, but this approach
demonstrates the problem I'
Thanks Jorge
It works great. I solved it by using loop, but i like your way better.
Thanks again
Cameron
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Hi Cameron,
May be this (untested)?
rowSums(is.na(tsObj), na.rm = TRUE)
HTH,
Jorge
On Thu, Nov 18, 2010 at 2:38 PM, cameron <> wrote:
>
> thanks Henrique
>
> I have another question.
>
> Lets say i have a timeSeries table
>AB C
> 1/1/90 NA 1 2
> 1/2/90
thanks Henrique
I have another question.
Lets say i have a timeSeries table
AB C
1/1/90 NA 1 2
1/2/90 NA 1 1
1/3/90 NA 1 -1
1/4/90 NA -1 1
1/5/901 1 1
1/6/901 51 1
I want to
Try this:
rowSums(tsObj, na.rm = TRUE)
On Thu, Nov 18, 2010 at 3:58 PM, cameron wrote:
>
>
> I have a question on RowSums.
>
> Lets say i have a timeSeries table
>A B C
> 1/1/90 NA 1 1
> 1/2/90 NA 1 1
> 1/3/90 NA 1 1
> 1/4/90 NA 1
I have a question on RowSums.
Lets say i have a timeSeries table
A B C
1/1/90 NA 1 1
1/2/90 NA 1 1
1/3/90 NA 1 1
1/4/90 NA 1 1
1/5/901 1 1
1/6/901 1 1
if i use RowSums, i will get
1/5/903
1/
On 9/24/2008 10:38 AM, Marc Schwartz wrote:
> on 09/24/2008 09:06 AM Doran, Harold wrote:
>> Say I have the following data:
>>
>> testDat <- data.frame(A = c(1,NA,3), B = c(NA, NA, 3))
>>
>>> testDat
>>A B
>> 1 1 NA
>> 2 NA NA
>> 3 3 3
>>
>> rowsums() with na.rm=TRUE generates the following
on 09/24/2008 09:06 AM Doran, Harold wrote:
> Say I have the following data:
>
> testDat <- data.frame(A = c(1,NA,3), B = c(NA, NA, 3))
>
>> testDat
>A B
> 1 1 NA
> 2 NA NA
> 3 3 3
>
> rowsums() with na.rm=TRUE generates the following, which is not desired:
>
>> rowSums(testDat[, c('A',
try the following:
testDat <- data.frame(A = c(1,NA,3), B = c(NA, NA, 3))
ind <- rowSums(is.na(testDat)) == length(testDat)
out <- rowSums(testDat, na.rm = TRUE)
out[ind] <- NA
out
I hope it helps.
Best,
Dimitris
Doran, Harold wrote:
Say I have the following data:
testDat <- data.frame(A =
On 9/24/2008 10:06 AM, Doran, Harold wrote:
> Say I have the following data:
>
> testDat <- data.frame(A = c(1,NA,3), B = c(NA, NA, 3))
>
>> testDat
>A B
> 1 1 NA
> 2 NA NA
> 3 3 3
>
> rowsums() with na.rm=TRUE generates the following, which is not desired:
>
>> rowSums(testDat[, c('A',
I guess this would be the fastest way would be:
rs <- rowSums( testDat, na.rm=T)
rs[ which( rowMeans(is.na(testDat)) == 1 ) ] <- NA
since both rowSums and rowMeans are internally coded in C.
Regards, Adai
Doran, Harold wrote:
Say I have the following data:
testDat <- data.frame(A = c(1,N
Say I have the following data:
testDat <- data.frame(A = c(1,NA,3), B = c(NA, NA, 3))
> testDat
A B
1 1 NA
2 NA NA
3 3 3
rowsums() with na.rm=TRUE generates the following, which is not desired:
> rowSums(testDat[, c('A', 'B')], na.rm=T)
[1] 1 0 6
rowsums() with na.rm=F generates the fol
On Wed, 21 Nov 2007, Robin Hankin wrote:
>
> On 21 Nov 2007, at 08:30, Prof Brian Ripley wrote:
>
>> On Tue, 20 Nov 2007, Tim Hesterberg wrote:
>>
>>> I wrote the original rowSums (in S-PLUS).
>>> There, rowSums() does not coerce integer to double.
>>
>> Actaully, neither does R. It computes a
On 21 Nov 2007, at 08:30, Prof Brian Ripley wrote:
> On Tue, 20 Nov 2007, Tim Hesterberg wrote:
>
>> I wrote the original rowSums (in S-PLUS).
>> There, rowSums() does not coerce integer to double.
>
> Actaully, neither does R. It computes a double answer but does no
> coercion per se.
>
>> Ho
On Tue, 20 Nov 2007, Tim Hesterberg wrote:
> I wrote the original rowSums (in S-PLUS).
> There, rowSums() does not coerce integer to double.
Actaully, neither does R. It computes a double answer but does no
coercion per se.
> However, one advantage of coercion is to avoid integer overflow.
In
I wrote the original rowSums (in S-PLUS).
There, rowSums() does not coerce integer to double.
However, one advantage of coercion is to avoid integer overflow.
Tim Hesterberg
>... So, why does rowSums() coerce to double (behaviour
>that is undesirable for me)?
__
On 10 Nov 2007, at 07:32, Prof Brian Ripley wrote:
> On Fri, 9 Nov 2007, Robin Hankin wrote:
>
>> Hi
>>
>> [R-2.6.0, macOSX 10.4.10].
>>
>> The helppage says that rowSums() and colSums()
>> are equivalent to 'apply' with 'FUN = sum'.
>>
>> But I came across this:
>>
>> > a <- matrix(1:30,5,6)
>>
On Fri, 9 Nov 2007, Robin Hankin wrote:
> Hi
>
> [R-2.6.0, macOSX 10.4.10].
>
> The helppage says that rowSums() and colSums()
> are equivalent to 'apply' with 'FUN = sum'.
>
> But I came across this:
>
> > a <- matrix(1:30,5,6)
> > is.integer(apply(a,1,sum))
> [1] TRUE
> > is.integer(rowSums(a))
Hi
[R-2.6.0, macOSX 10.4.10].
The helppage says that rowSums() and colSums()
are equivalent to 'apply' with 'FUN = sum'.
But I came across this:
> a <- matrix(1:30,5,6)
> is.integer(apply(a,1,sum))
[1] TRUE
> is.integer(rowSums(a))
[1] FALSE
>
so rowSums() returns a float.
Why is this?
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