Re: [R] date
try to use difftime() instead of as.difftime(). On Thu, Sep 2, 2010 at 10:32 PM, Dunia Scheid dunia.sch...@gmail.com wrote: Hello all, I've 2 strings that representing the start and end values of a date and time. For example, time1 - c(21/04/2005,23/05/2005,11/04/2005) time2 - c(15/07/2009, 03/06/2008, 15/10/2005) as.difftime(time1,time2) Time differences in secs [1] NA NA NA attr(,tzone) [1] How can i calculate the difference between this 2 string? Regards, Dunia [[alternative HTML version deleted]] __ 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. __ 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] , Updating Table
If I am not wrong, it seems that you want to get factor counts in the whole scale of data.raw. Maybe you can do that just like this: table(data.raw) On Sat, Jul 24, 2010 at 1:44 AM, Marcus Liu marcusliu...@yahoo.com wrote: Hi everyone, Is there any command for updating table withing a loop? For instance, at i, I have a table as ZZ = table(data.raw[1:ind[i]]) where ind = c(10, 20, 30, ...). Then , ZZ will be as follow A B C 3 10 2 At (i + 1), ZZ = table(data.raw[(ind[i]+1):ind[i+1]]) A B D 4 7 8 Is there any command that can update the table ZZ for each time so that in the above example, ZZ will be A B C D 7 17 2 8 Thanks. liu [[alternative HTML version deleted]] __ 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. __ 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] getting data from table command
d - 1:4 f - c(2,3,3,1) rep(d,f) [1] 1 1 2 2 2 3 3 3 4 On Sat, Jul 10, 2010 at 10:55 PM, nn roh nn.r...@gmail.com wrote: Hi, I have a question relating to R package If i have the frequencies of data how i can get the data as following : 1 2 3 4 data 2 3 3 1 frequency Which command i can use to get the data set in the above example should be 1 1 2 2 2 3 3 3 4 Thanks Nada [[alternative HTML version deleted]] __ 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. __ 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] Creating named lists
Did you mean this:
n - c('a', 'b')
structure(list(1, 2), names = n)
$a
[1] 1
$b
[1] 2
On Fri, Mar 12, 2010 at 5:28 PM, Rune Schjellerup Philosof
rphilo...@health.sdu.dk wrote:
I often find myself making lists similar to this
list(var1=var1, var2=var2)
It doesn't seem list has an option, to make it use the name of the
variable as name in the list.
Is there another function that does this?
--
Med venlig hilsen
Rune Schjellerup Philosof
Ph.d-stipendiat, Forskningsenheden for Biostatistik
Telefon: 6550 3607
E-mail: rphilo...@health.sdu.dk
Adresse: J.B. Winsløwsvej 9, 5000 Odense C
SYDDANSK UNIVERSITET
___
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Re: [R] Group by
DF
V1 V2 V3
1 10:03:13 3.4 1002
2 10:03:14 5.6 1001
3 10:05:27 7.2 999
4 10:05:33 8.2 998
DF2 - t(sapply(split(DF[,-1], gsub('(.{5}).*', '\\1:00', DF$V1)), colSums))
data.frame(V1 = rownames(DF2), DF2)
V1 V2 V3
10:03:00 10:03:00 9.0 2003
10:05:00 10:05:00 15.4 1997
On Thu, Mar 11, 2010 at 8:16 PM, Henrique Dallazuanna www...@gmail.com wrote:
Try this:
DF - read.csv(textConnection(V1,V2,V3
10:03:13,3.4,1002
10:03:14,5.6,1001
10:05:27,7.2,999
10:05:33,8.2,998), header = TRUE)
tDates - strptime(DF$V1, %H:%M:%S)
tDates[[2]] - floor(strptime(DF$V1, %H:%M:%S)[[2]] / 5) * 5
aggregate(DF[,c('V2', 'V3')],
list(format(tDates, %H:%M:00)),
FUN = sum)
On Thu, Mar 11, 2010 at 6:48 AM, ManInMoon xmoon2...@googlemail.com wrote:
I have a matrix with a POSIXct as a numeric in the first column.
I would like to create a new matrix that is grouped by my chosed time
bars.
i.e. So I would like to group by hour or day or 5 days, and have all my
columns be summed or averaged or counted..
mydata:
V1,V2,V3
10:03:13,3.4,1002
10:03:14,5.6,1001
10:05:27,7.2,999
10:05:33,8.2,998
I want to convert this into say 5 minutes bars and sum columns
mynewdata:
V1,V2,V3
10:00:00,9.0,2003
10:05:00,15.4,1997
How can I do this?
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Re: [R] Define column names to a series of data.frames
It seems that the names of original data frames have not changed in
this way. I guess textConnection() could help, like this:
for (name in objects(pattern = df[0-9]))
eval(parse(textConnection(paste('names(', name, ') -
column_names'
On Thu, Mar 11, 2010 at 9:25 PM, Henrique Dallazuanna www...@gmail.com wrote:
You can try this:
lapply(lapply(ls(pattern = 'DF[0-9]'), get),
'names-', c(SDev,PC1, PC2, PC3, PC4, PC5, PC6))
On Thu, Mar 11, 2010 at 9:33 AM, Nikos Alexandris
nikos.alexand...@felis.uni-freiburg.de wrote:
Greets to the list!
I am aware that this topic has been discussed several times. And I've
read quite some related posts [1]. Yet, can't seem to give a solution to
my problem.
I have 6 data frames consisting of 6 rows x 7 columns put together from
other data.frames.
Something like:
a b c d e f g
v1 # # # # # # #
v2 # # # # # # #
v3 # # # # # # #
v4 # # # # # # #
v5 # # # # # # #
v6 # # # # # # #
I want to give the following column names to each data.frame: (SDev,
PC1, PC2, PC3, PC4, PC5, PC6)
Works fine for one data.frame:
column_names - c(SDev, PC1, PC2, PC3, PC4, PC5, PC6)
names( df1 ) - column_names
How is it to be done at once for all data.frames that the function
objects(pattern = SomePattern) can find?
I've tried several things with assign, get, paste, for but I am not
getting anywhere. I need to integrate this in a function that can handle
lot's of data.frames and not only 6.
Thank you, Nikos
---
[1] I am stuck on this post:
http://www.mail-archive.com/r-h...@stat.math.ethz.ch/msg32063.html.
It reads:
dframes - c(a,b,c)
cols - c(one,two)
df - data.frame(11:20, 21:30)
names(df) - cols
assign(dframes[2], df)
Can't understand the logic behind the [2].
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Re: [R] Define column names to a series of data.frames
Amazing! I haven't seen usage of calling `names-` like this before. Thanks so much! On Thu, Mar 11, 2010 at 9:50 PM, Henrique Dallazuanna www...@gmail.com wrote: Yes, just in the list. If they want change the name in Environment GlobalEnv: for(i in ls(pattern = DF[0-9])) assign(i, `names-`(get(i), c(SDev,PC1, PC2, PC3, PC4, PC5, PC6)), globalenv()) -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ 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] How to match vector with a list ?
Maybe you can create a helper vector first:
helper - structure(names = unlist(j), rep(names(j), sapply(j, length)))
helper
acbd
j1 j1 j2 j2
helper[i]
aabbbccd
j1 j1 j2 j2 j2 j1 j1 j2
On Sat, Mar 6, 2010 at 1:42 AM, Carlos Petti carlos.pe...@gmail.com wrote:
Dear list,
I have a vector of characters and a list of two named elements :
i - c(a,a,b,b,b,c,c,d)
j - list(j1 = c(a,c), j2 = c(b,d))
I'm looking for a fast way to obtain a vector with names, as follows :
[1] j1 j1 j2 j2 j2 j1 j1 j2
I used :
match - lapply(j, function (x) {which(i %in% x)})
k - vector()
for (y in 1:length(match)) {
k[match[[y]]] - names(match[y])}
k
[1] j1 j1 j2 j2 j2 j1 j1 j2
But, I think a better way exists ...
Thanks in advance,
Carlos
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Re: [R] Average regions of non-zeros
Nice shot of cumsum(). Just improve it a little:
x - c(0,0,1,2,3,0,0,4,5,6)
x.groups - split(x, (x != 0) * cumsum(x == 0))[-1]
x.groups
$`2`
[1] 1 2 3
$`4`
[1] 4 5 6
lapply(x.groups, mean)
$`2`
[1] 2
$`4`
[1] 5
On Mon, Mar 8, 2010 at 11:02 AM, jim holtman jholt...@gmail.com wrote:
Try this:
x - c(0,0,1,2,3,0,0,4,5,6)
# partition the data
x.p - split(x, cumsum(x == 0))
# now only process groups 1
x.mean - lapply(x.p, function(a){
+ if (length(a) == 1) return(NULL)
+ return(list(grp=tail(a, -1), mean=mean(tail(a, -1
+ })
# now only return the real values
x.mean[unlist(lapply(x.mean, length) != 0)]
$`2`
$`2`$grp
[1] 1 2 3
$`2`$mean
[1] 2
$`4`
$`4`$grp
[1] 4 5 6
$`4`$mean
[1] 5
On Sun, Mar 7, 2010 at 9:48 PM, Daren Tan dare...@hotmail.com wrote:
x - c(0,0,1,2,3,0,0,4,5,6)
How to identify the regions of non-zeros and average c(1,2,3) and c(4,5,6)
to get 2 and 5.
Thanks
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What is the problem that you are trying to solve?
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Re: [R] Reducing a matrix
Try this: df[!duplicated(df[,'x']),] On Sun, Feb 28, 2010 at 8:56 AM, Juliet Ndukum jpnts...@yahoo.com wrote: I wish to rearrange the matrix, df, such that all there are not repeated x values. Particularly, for each value of x that is reated, the corresponded y value should fall under the appropriate column. For example, the x value 3 appears 4 times under the different columns of y, i.e. y1,y2,y3,y4. The output should be such that for the lone value of 3 selected for x, the corresponding row entries with be 7 under column y1, 16 under column y2, 12 under column y3 and 18 under column y4. This should work for the other rows of x with repeated values. df x y1 y2 y3 y4 1 3 7 NA NA NA 2 3 NA 16 NA NA 3 3 NA NA 12 NA 4 3 NA NA NA 18 5 6 8 NA NA NA 6 10 NA NA 2 NA 7 10 NA 11 NA NA 8 14 NA NA NA 8 9 14 NA 9 NA NA 10 15 NA NA NA 11 11 50 NA NA 13 NA 12 50 20 NA NA NA The output should be: x y1 y2 y3 y4 1 3 7 16 12 18 2 6 8 NA NA NA 3 10 NA 11 2 NA 4 14 NA 9 NA 8 5 15 NA NA NA 11 6 50 20 NA 13 NA Can any write for me a code that would produce these results. Thank you in advance for your help. JN [[alternative HTML version deleted]] __ 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. __ 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] question to make a vector without loop
For general purpose of recursion formula, you could do it like this: make.vector - function(w, n, a, b) c(w, sapply(1:(n-1), function(x) w - w * a / (b + x))) make.vector(w = 1, n = 4, a = 24, b = 1) [1] 1 12 96 576 On Fri, Feb 26, 2010 at 11:23 PM, khaz...@ceremade.dauphine.fr wrote: Hello all, I want to define a vector like w[k+1]=w[k]*a/(b+k) for k=1,...,N-1 without use loop. Is it posible to do in R? Regards khazaei __ 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. __ 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] replicate matrix
You can just rep() it, and c() with extra data, and then matrix() it again: m - matrix(c(1,4,3,6),2) matrix(c(rep(m, 3), c(2, 5)), nrow(m)) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [1,]1313132 [2,]4646465 On Sun, Feb 21, 2010 at 10:58 AM, wendy wendy.q...@utoronto.ca wrote: Hi all, I have a matrix, for example [,1] [,2] [1,] 1 3 [2,] 4 6 I want to replicate the matrix twice and add an extra column at the end, which is [,1] [,2] [,3] [,4] [,5] [,6] [,7] [1,] 1 3 1 3 1 3 2 [2,] 4 6 4 6 4 6 5 I found 'rep' only works for vector. Does anyone know how to replicate a matrix, and append the matrix? Thank you in advance, Wendy -- View this message in context: http://n4.nabble.com/replicate-matrix-tp1563337p1563337.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. __ 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] How to store an array like this?
It's not an array, but a list:
lapply(2:n, function(x) matrix(1:x^2,x))
On Sat, Feb 20, 2010 at 3:17 PM, song song rprojecth...@gmail.com wrote:
maybe its not an array
like
m[1]=matrix(1:4,2,2)
m[2]=matrix(1:9,3,3)
.
.
.
m[k]=matrix(1:k^2,k,k)
likes
for (k in 1:n){
s=matrix(1:k^2,k,k)
}
how to store s like s[1], s[2] ... so that I can use it in other place?
thanks!
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Re: [R] Printing 2 digits after decimal point
format(x, nsmall = 2) On Mon, Feb 15, 2010 at 5:41 PM, t.wun...@stud.uni-heidelberg.de wrote: Hi there, i'm not getting along with the following problem. I'd like to print a real number, e.g. x - 12.3 with exactly two digits after the decimal point, e.g. 12.30 I've tried the whole format(), formatC() and prettyNum() functions but did not have any success with it. This should work with all real numbers, in case even with 0.0 (- 0.00). For cracks this thing is pretty sure obvious, but I've spent the whole morning with it. Please help me! Thanks a lot, Tom __ 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. __ 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] How to: highlight R syntax on webpages ?
To achieve this goal, it seems there are several ways, such as WP-Syntax (http://wordpress.org/extend/plugins/wp-syntax/), SyntaxHighlighter2 (http://mohanjith.com/2009/03/syntaxhighlighter2.html), etc. However, these plugins seem not support R language by default, so you may have to write some code to declare the keywords and functions for R. And also you can write another plugin by yourself if you like, it's not so difficult. On Sat, Nov 21, 2009 at 2:26 AM, Tal Galili tal.gal...@gmail.com wrote: My question if in the Subject, but if to extend: I am specifically curious about WordPress blogs. But any solution will give me a lead. Thanks, Tal -- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com/ (English) [[alternative HTML version deleted]] __ 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. __ 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] How to: highlight R syntax on webpages ?
I have saw it now. Thank you for your excellent works. On Fri, Feb 12, 2010 at 9:15 PM, Tal Galili tal.gal...@gmail.com wrote: Hi Linlin, I am afraid I wasn't clear. In my post I fixed the current WP-Syntax plugin so it WILL support the R syntax :) The link to the article is: http://www.r-statistics.com/2010/02/r-syntax-highlight-on-your-blog-a-wordpress-plugin/ Best, Tal Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) - __ 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] apply command
I guess that the matrix dimension changed because matrix in R are filled by columns. Since you try: apply(b, 1, function(y) sort(y, na.last=F)) The second parameter make it scan matrix b row by row but store result by columns, which make the result be a matrix transposed. If you try: apply(b, 2, function(y) sort(y, na.last=F)) The second parameter means scan column by column, and the result matrix will have the same dimension with origin. On Tue, Jan 19, 2010 at 6:31 PM, Tal Galili tal.gal...@gmail.com wrote: Hello Marco What I would do, is use t to transpose the matrix. Why it is that apply switches the matrix, is beyond my knowledge - and I would love to read more informed replies. Tal Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com/ (English) -- On Tue, Jan 19, 2010 at 12:27 PM, marco salvini marco.salv...@gmail.comwrote: Can you please help on the issue? I using the apply command on a matrix below the example: Create a vector x =c(5, 3, 2:4, NA, 7, 3, 9, 2, 1, 5) create a matrix of 2 rows by 6 columns b=matrix(x, 2,6) print(b) [,1] [,2] [,3] [,4] [,5] [,6] [1,] 5 2 4 7 9 1 [2,] 3 3 NA 3 2 5 using the command apply print(apply(b, 1, function(y) sort(y, na.last=F))) the output is a matrix of 6 rows by 2 columns. [,1] [,2] [1,] 1 NA [2,] 2 2 [3,] 4 3 [4,] 5 3 [5,] 7 3 [6,] 9 5 As you can see in the example I start with a matrix of (2 by 6) and the output of apply is a mtraxi of (6 by 2). This is very strange because I was expecting as output a matrix of the same dim (2 by 6) of the input matrix. I can solve this issues using an if statment on the dim of the matrix but if I am using a square matrix I am not able to control if the result of the apply is correct. Do anyone find a solution to this issue? thanks Marco [[alternative HTML version deleted]] __ 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. [[alternative HTML version deleted]] __ 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. __ 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] as.Date question
I am afraid that although in same literally, they are indeed different functions: as.Date.POSIXct and as.Date.POSIXlt. But I am not sure why they are designed like this, which causes the confusion as you mentioned. On Thu, Dec 24, 2009 at 11:02 PM, MAL diver...@univecom.ch wrote: Mark, not sure that's the answer. Usually one has x=y -- f(x)=f(y) which doesn't seem to hold here (put x=zzz1, y=zzz2, f=as.Date()). Or do I overlook something? - Original Message - From: Marek Janad marek.ja...@gmail.com To: r-help@r-project.org Sent: Thursday, December 24, 2009 00:08 Subject: Re: [R] as.Date question Look at documentation ?as.Date as.Date first represents time in UTC, what gives: as.POSIXlt(zzz1, tz=UTC) HTH 2009/12/20 MAL diver...@univecom.ch: All! This piece of code: zzz1 - as.POSIXct(1999-03-18, tz=CET) zzz2 - as.POSIXlt(1999-03-18, tz=CET) zzz1 == zzz2 as.Date(zzz1) as.Date(zzz2) yields TRUE for zzz1==zzz2, but the two dates returned by as.Date are different: as.Date(zzz1) [1] 1999-03-17 as.Date(zzz2) [1] 1999-03-18 I'm using R 2.10.0. Would be glad for any clarifications. Thanks! __ 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. -- Marek __ 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. __ 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. __ 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] How to find the significant digits of a number?
Try this: f - function(x) length(gregexpr([[:digit:]], as.character(x))[[1]]) f(3.14) [1] 3 f(3.1415) [1] 5 f(3.14159265) [1] 9 On Wed, Dec 16, 2009 at 1:39 PM, Xiang Wu xiang@gmail.com wrote: Is there a function in R that could find the significant digit of a specific number? Such as for 3.1415, return '5'? Thanks in advance. __ 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 there lazy copy in R?
It means that R does have the lazy copy mechanism, which I didn't know, and I think it can be very useful to make R running more quickly. On Tue, Dec 15, 2009 at 12:15 PM, Peng Yu pengyu...@gmail.com wrote: a=1:10 b=a a=1:10 tracemem(a)# I assume the following is address 'a' points to [1] 0x05cf2798 b=a b[1]=1 tracemem[0x05cf2798 - 0x05cf2750]: tracemem[0x05cf2750 - 0x05ed8ba0]: I don't understand what these addresses mean. Would you please help me understand it? __ 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] Removing objects from a list based on nrow
Try these: sapply(lst, nrow) # get row numbers which(sapply(lst, nrow) 3) # get the index of rows which has less than 3 rows lst - lst[-which(sapply(lst, nrow) 3)] # remove the rows from the list On Sun, Nov 29, 2009 at 4:36 PM, Tim Clark mudiver1...@yahoo.com wrote: Dear List, I have a list containing data frames of various numbers of rows. I need to remove any data frame that has less than 3 rows. For example: df1-data.frame(letter=c(A,B,C,D,E),number=c(1,2,3,4,5)) df2-data.frame(letter=c(A,B),number=c(1,2)) df3-data.frame(letter=c(A,B,C,D,E),number=c(1,2,3,4,5)) df4-data.frame(letter=c(A,B,C,D,E),number=c(1,2,3,4,5)) lst-list(df1,df2,df3,df4) How can I determine that the second object (df2) has less than 3 rows and remove it from the list? Thanks! Tim Tim Clark Department of Zoology University of Hawaii __ 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. __ 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] column of dates into time series
There is no year() function. Maybe you can try format() instead. On Sun, Nov 29, 2009 at 8:44 PM, DispersionMap frenc...@btinternet.com wrote: i have a column of dates in this format: data[,Raised.Date] - as.Date(data[,Raised.Date], %d/%m/%Y); data[1:10,Raised.Date] [1] 2006-07-07 2006-07-07 2006-04-03 2006-04-03 2006-04-03 2006-04-03 2006-04-03 2006-04-03 2006-04-03 2006-04-03 I can turn them into months like this... Month-months(data[,Raised.Date]) Month[1:10] [1] July July April April April April April April April April But i also want to turn them into years (and also weeks later on), so tried this... Year-year(data[,Raised.Date]) Error: could not find function year thanks. -- View this message in context: http://n4.nabble.com/column-of-dates-into-time-series-tp930699p930699.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. __ 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] Removing objects from a list based on nrow
Thank Jim! You are right. I didn't notice the case of none of rows match the condition. On Sun, Nov 29, 2009 at 10:10 PM, jim holtman jholt...@gmail.com wrote: One thing to be careful of is if no dataframe have less than 3 rows: df1-data.frame(letter=c(A,B,C,D,E),number=c(1,2,3,4,5)) df2-data.frame(letter=c(A,B),number=c(1,2)) df3-data.frame(letter=c(A,B,C,D,E),number=c(1,2,3,4,5)) df4-data.frame(letter=c(A,B,C,D,E),number=c(1,2,3,4,5)) lst-list(df1,df3,df4) lst [[1]] letter number 1 A 1 2 B 2 3 C 3 4 D 4 5 E 5 [[2]] letter number 1 A 1 2 B 2 3 C 3 4 D 4 5 E 5 [[3]] letter number 1 A 1 2 B 2 3 C 3 4 D 4 5 E 5 lst[-which(sapply(lst, nrow) 3)] list() Notice the list is now empty. Instead use: lst[sapply(lst, nrow) =3] [[1]] letter number 1 A 1 2 B 2 3 C 3 4 D 4 5 E 5 [[2]] letter number 1 A 1 2 B 2 3 C 3 4 D 4 5 E 5 [[3]] letter number 1 A 1 2 B 2 3 C 3 4 D 4 5 E 5 On Sun, Nov 29, 2009 at 3:43 AM, Linlin Yan yanlinli...@gmail.com wrote: Try these: sapply(lst, nrow) # get row numbers which(sapply(lst, nrow) 3) # get the index of rows which has less than 3 rows lst - lst[-which(sapply(lst, nrow) 3)] # remove the rows from the list On Sun, Nov 29, 2009 at 4:36 PM, Tim Clark mudiver1...@yahoo.com wrote: Dear List, I have a list containing data frames of various numbers of rows. I need to remove any data frame that has less than 3 rows. For example: df1-data.frame(letter=c(A,B,C,D,E),number=c(1,2,3,4,5)) df2-data.frame(letter=c(A,B),number=c(1,2)) df3-data.frame(letter=c(A,B,C,D,E),number=c(1,2,3,4,5)) df4-data.frame(letter=c(A,B,C,D,E),number=c(1,2,3,4,5)) lst-list(df1,df2,df3,df4) How can I determine that the second object (df2) has less than 3 rows and remove it from the list? Thanks! Tim Tim Clark Department of Zoology University of Hawaii __ 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. __ 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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ 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] List Name help
On Tue, Nov 24, 2009 at 1:01 AM, Henrique Dallazuanna www...@gmail.com wrote: Try this: test[grep(d2, names(test))] Or it can be simply like this: test[names(test) == d2] On Mon, Nov 23, 2009 at 2:53 PM, Rajasekaramya ramya.vict...@gmail.com wrote: Hi There, I have a named List object.I want to access all the list elements that has the same name for example The List test - list() $d2 v1 v2 v3 v4 1 2 3 4 5 $d2 v1 v2 v3 v4 1 2 3 4 5 $d3 v1 v2 v3 v4 8 9 19 10 $d1 v1 v2 v3 v4 12 14 15 16 so if i say test[[d2]] or test[d2] i should get the first two that matches the name rite but i am not getting them jus getting the first one. Any suggestions on it Ramya -- View this message in context: http://old.nabble.com/List-Name-help-tp26481873p26481873.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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ 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. __ 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] Removing + and ? signs
Try this: gsub([?], , x) On Mon, Nov 23, 2009 at 7:01 AM, Steven Kang stochastick...@gmail.com wrote: Hi all, I get an error message when trying to replace *+* or *?* signs (with empty space) from a string. x - asdf+,jkl? gsub(?, , x) Error message: Error in gsub(?, , x) : invalid regular expression '?' In addition: Warning message: In gsub(?, , x) : regcomp error: 'Invalid preceding regular expression' Your expertise in resolving this issue would be appreciated. Thanks. Steven [[alternative HTML version deleted]] __ 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. __ 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] [Rd] How to generate dependency file that can be used by gnu make?
I don't think this function is same as gcc's option -MM. Because gcc
checks pre-compile command #include, in which the filename can be
fetched definitely. But in your scenario, the filename may be from
some variables, which can not be determined by the R script only.
Maybe you can write a tool by yourself to parse the R syntax to
resolve your problem.
On Tue, Nov 17, 2009 at 11:51 AM, Peng Yu pengyu...@gmail.com wrote:
On Sun, Nov 15, 2009 at 8:45 PM, Peng Yu pengyu...@gmail.com wrote:
gcc has options like -MM, which can generate the dependence files for
a C/C++ file that I can be used by gnu make. I'm wondering if there is
a tool that can generate dependence file for an R script.
For example, I have an R script test.R
#test.R
load('input.RData')
save.image('output.RData')
I want to generate a dependence file like the following. Is there a
tool to do so?
output.RData:test.R input.RData
Is there a way to automatically generate the output files that depends
on an R script and the input files and sourced files that are depended
by an R script? I don't see this option in R. But I wish this can be
implemented in future version of R.
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Re: [R] How to get the string '\'?
Regular expression needs double the '\' again, so try this:
gsub('/','',string)
On Mon, Nov 16, 2009 at 7:35 AM, Peng Yu pengyu...@gmail.com wrote:
My question was from replacing a pattern by '\\'. How to replace '/'
in string by '\'?
string='abc/efg'
gsub('/','\\',string)
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Re: [R] pairs
Hope this help: m - matrix(c(2,1,3,9,5,7,7,8,1,8,6,5,6,2,2,7),4,4) p - c(2, 6) apply(m == p[1], 1, any) apply(m == p[2], 1, any) [1] TRUE FALSE TRUE FALSE If you want the number of rows which contain the pair, sum() could be used: sum(apply(m == p[1], 1, any) apply(m == p[2], 1, any)) [1] 2 On Mon, Nov 16, 2009 at 6:26 AM, cindy Guo cindy.g...@gmail.com wrote: Hi, All, I have an n by m matrix with each entry between 1 and 15000. I want to know the frequency of each pair in 1:15000 that occur together in rows. So for example, if the matrix is 2 5 1 6 1 7 8 2 3 7 6 2 9 8 5 7 Pair (2,6) (un-ordered) occurs together in rows 1 and 3. I want to return the value 2 for this pair as well as that for all pairs. Is there a fast way to do this avoiding loops? Loops take too long. Thank you, Cindy [[alternative HTML version deleted]] __ 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. __ 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] How to strip everything after second whitespace?
Try this: gsub(([a-z]*\\s[a-z]*).*, \\1, nam) [1] Smith John Smith David Smith Ryan On Fri, Nov 6, 2009 at 4:11 PM, johannes rara johannesr...@gmail.com wrote: How to split everything after second whitespace char using regular expression? I want to remove A, B, C and D from these names: nam - c(Smith John A, Smith David B C, Smith Ryan C D) Thanks, Johannes __ 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. __ 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] How to exclude certain columns by column names?
Try this:
x[, colnames(x) != 'a']
[1] 3 4
On Tue, Nov 3, 2009 at 9:31 AM, Peng Yu pengyu...@gmail.com wrote:
I can exclude columns by column number using '-'. But I wondering if
there is an easy way to exclude some columns by column names.
x=cbind(c(1,2),c(3,4))
x
[,1] [,2]
[1,] 1 3
[2,] 2 4
colnames(x)=c('a','b')
x
a b
[1,] 1 3
[2,] 2 4
x[,-'a']
Error in -a : invalid argument to unary operator
x[,-1]
[1] 3 4
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Re: [R] exact string match?
How about using operator == On Sat, Oct 31, 2009 at 5:00 AM, bamsel benam...@gmail.com wrote: Dear R users: I need to compare character strings stored in 2 separate data frames. I need an exact match, so finding a in animal is no good. I've tried regexpr, match, and grepl, but to no avail. Anybody know how to accomplish this? Ben -- View this message in context: http://old.nabble.com/exact-string-match--tp26136677p26136677.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. __ 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] how to do this?
Try this: y[matrix(c(seq_along(x), x), ncol = 2)] [1] 2 16 12 On Fri, Sep 11, 2009 at 4:17 PM, Luca Braglia brag...@poleis.eu wrote: Hello R-users I have a situation like this x=c(1,3,2) y=data.frame(a=1:3, b=4:6, c=7:9)*2 So we have t(t(x)) [,1] [1,] 1 [2,] 3 [3,] 2 And y a b c 1 2 8 14 2 4 10 16 3 6 12 18 I would like to obtain a vector with number taken from the data.frame: x is needed as row index With c(1,3,2), in this case the ouput should be 2 16 12 I've tried a little bit with apply, but unsuccessfully. Thank you __ 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. __ 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] NotePad++ Syntax file
I think NppToR may be a good choice.
http://sourceforge.net/projects/npptor/
On Tue, Aug 11, 2009 at 6:37 AM, Farley, Robertfarl...@metro.net wrote:
Does anyone have an R Syntax Highlighting file {userDefineLang.xml} for
NotePad++?? I've started one, but I'm not so happy with it.
Robert Farley
Metro
1 Gateway Plaza
Mail Stop 99-23-7
Los Angeles, CA 90012-2952
Voice: (213)922-2532
Fax: (213)922-2868
www.Metro.net
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Re: [R] vector
rep(A, each=2) On Thu, Jul 30, 2009 at 12:15 AM, Inchallah Yarabinchallahya...@yahoo.fr wrote: Hi , i have a vector A=(a1,a2,a3,a4) and i want to create another vector B=(a1,a1,a2,a2,a3,a3,a4,a4) !!! i know that it is simple but i begin with R so i nned your help!! thank you for your help !!! [[alternative HTML version deleted]] __ 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. __ 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] Function to standardize matriz?
Did you mean this: m - matrix(1:12, 3, 4) m / max(m) [,1] [,2] [,3] [,4] [1,] 0.0833 0.333 0.583 0.833 [2,] 0.1667 0.417 0.667 0.917 [3,] 0.2500 0.500 0.750 1.000 On Thu, Jul 30, 2009 at 12:52 PM, Samsamanta.fernan...@gmail.com wrote: Hi, this is probably a very basic question but I'm just learning R and i cannot find a function to standardize a data matrix.. I'll be grateful for any tips or help you can provide me. Thank you very much! __ 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. __ 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] Lists
How about like this:
for (i in seq_along(a)) {
result - as.list(a[1:i])
cat(iterator, i, :\n)
print(result)
}
On Sat, Jul 25, 2009 at 6:48 AM, Alberto Lora Malbertolo...@gmail.com wrote:
Hi Everybody
I have the following problem
suppose that we
a-c(uno,dos,tres)
I am working with a while cycle and the idea is in each iteration adding an
item to a list
In the first iteration the resultshould be:
[[1]]
[1] uno
In the second
[[1]]
[1] uno
[[2]]
[1] dos
And the final result
[[1]]
[1] uno
[[2]]
[1] dos
[[3]]
[1] tres
How can I do that
Thx again
--
Alberto
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Re: [R] How to append to a list dynamically?
Function parameters in R are passed by value, not by reference. In
order to resolve it, just remove clusters from the parameter list,
and use clusters[i] - ... to change the value of global variable.
On Wed, Jun 17, 2009 at 7:52 PM, Nick Angelounikola...@yahoo.com wrote:
Hi,
I have a problem with dynamic appending to a list. Here is the list
variable:
clusters - vector(list, 0)
I extended in the function below:
cluster - function (pair, clusters)
{
found - FALSE
for (i in length(clusters))
{
if (length(intersect(pair, clusters[i])) 0)
{
clusters[i] - union(clusters[i], pair)
found - TRUE
}
}
if (!found)
{
clusters - list(clusters, as.vector(pair))
}
}
The function is executed in a loop:
for (i in 1:nrow(adjMatrix))
{
for (j in 1:nrow(adjMatrix))
{
if ((i != j) adjMatrix[i,j] 0) # the matrix element has to be non-zero
in order to be clustered
{
cat(rownames(adjMatrix)[i], colnames(adjMatrix)[j], \n)
cluster(as.vector(c(rownames(adjMatrix)[i], colnames(adjMatrix)[j])),
clusters)
}
}
}
But the list variable remains empty (i.e. length(clusters) = 0) even though
it should not. Somehow the dynamic extension of the list does not work in
this case. Any suggestions?
Best regards,
Nick
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Re: [R] how to get output from a nested loop
How about like this:
t1 - data.frame(row.names=c('c1','c2','c3','c4'), mk1=c(1,1,0,0),
mk2=c(0,0,0,1), mk3=c(1,1,1,1), mk4=c(0,0,0,0), mk5=c(0,0,0,1), S=c(4,5,3,2))
t1
mk1 mk2 mk3 mk4 mk5 S
c1 1 0 1 0 0 4
c2 1 0 1 0 0 5
c3 0 0 1 0 0 3
c4 0 1 1 0 1 2
apply(combn(1:5, 2), 2, function(x) t1[,c(x[1], 6, x[2])])
[[1]]
mk1 S mk2
c1 1 4 0
c2 1 5 0
c3 0 3 0
c4 0 2 1
... ...
On Wed, Jun 10, 2009 at 1:10 PM, Scott Hermannsherm...@bses.org.au wrote:
Dear all,
I imagine that this is a trival question, but it has perplexed for most of
the day. Any help would be greatly appreciated.
Below is an example of what I'm trying to do.
Essentially I want to produce all unique 1 x 1 combinations of certain
columns from a dataframe, and join these to other columns from the same
dataframe. I'm having problems with the nested loop as I can only output
data from the last cycle of the loop. I realise that the problem is with
the st1[[i]] but I'm not sure how to define it differently.
##I want to make a list file of all 1x1 combinations of mk columns, and add
clone and S to these combinations.
clone-c(c1,c2,c3,c4)
mk1-c(1,1,0,0)
mk2-c(0,0,0,1)
mk3-c(1,1,1,1)
mk4-c(0,0,0,0)
mk5-c(0,0,0,1)
S-c(4,5,3,2)
t1-as.data.frame(cbind(clone,mk1,mk2,mk3,mk4, mk5,S))
row.names(t1)-t1$clone
t1-t1[,-1]
t1
###A nested loop. I'm trying to get all combinations of columns 1:5 and join
each of them with column 7.
st1 - list()
for(i in 1:4) {
for(j in (i+1):5){
st1[[i]] - cbind(t1[,c(i,6)],t1[,j])
}
}
st1
Thanks for your help,
Scott
BSES Limited Disclaimer
This email and any files transmitted with it are confide...{{dropped:13}}
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Re: [R] Help with if statements
Try this:
for (i in 1:dim(ALLRESULTS)[1]) {
ALLRESULTS[i,23] - length(ALLRESULTS[i,][ALLRESULTS[i,16:22] = 0.05])
}
On Wed, Jun 10, 2009 at 12:17 AM, Amit Patelamitrh...@yahoo.co.uk wrote:
Hi
I am trying to create a column in a data frame which gives a sigificane score
from 0-7. It should read values from 7 different colums and add 1 to the
counter if the value is =0.05. I get an error message saying
Error in if (ALLRESULTS[i, 16] = 0.05) significance_count =
significance_count + :
missing value where TRUE/FALSE needed
The script is included below
it works if i convert the NA values to zero but this is not appropriate as it
includes the zero as significant.
ANY SUGGESTIONS
#SCRIPT STARTS
for (i in 1:length(ALLRESULTS[,1])) {
significance_count = 0
if (ALLRESULTS[i,16] = 0.05 ) significance_count = significance_count +1
else significance_count = significance_count
if (ALLRESULTS[i,17] = 0.05 ) significance_count = significance_count +1
else significance_count = significance_count
if (ALLRESULTS[i,18] = 0.05 ) significance_count = significance_count +1
else significance_count = significance_count
if (ALLRESULTS[i,19] = 0.05 ) significance_count = significance_count +1
else significance_count = significance_count
if (ALLRESULTS[i,20] = 0.05 ) significance_count = significance_count +1
else significance_count = significance_count
if (ALLRESULTS[i,21] = 0.05 ) significance_count = significance_count +1
else significance_count = significance_count
if (ALLRESULTS[i,22] = 0.05 ) significance_count = significance_count +1
else significance_count = significance_count
ALLRESULTS[i,23] - significance_count}
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Re: [R] seq(...) strange logical value
How about this:
%==% - function(x, y) {
if (length(x) 1) {
sapply(x, function(z) isTRUE(all.equal(z, y)));
} else {
sapply(y, function(z) isTRUE(all.equal(z, x)));
}
}
seq(0, 1, by=0.1) %==% 0.1
[1] FALSE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
seq(0, 1, by=0.1) %==% 0.2
[1] FALSE FALSE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
seq(0, 1, by=0.1) %==% 0.3
[1] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
seq(0, 1, by=0.1) %==% 0.4
[1] FALSE FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE FALSE FALSE
0.3 %==% seq(0, 1, by=0.1)
[1] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
On Mon, Jun 8, 2009 at 4:45 PM, Grześgregori...@gmail.com wrote:
Do you heve any idea why I get after this instruction everywhere false?
seq (0, 1, by=0.1) == 0.3
[1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
But after different step it's ok:
seq(0, 1, by=0.1) == 0.4
[1] FALSE FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE FALSE FALSE
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Re: [R] how to sort data frame order by column?
e.g. dat[ order(dat$a), ] On Sun, May 31, 2009 at 2:34 PM, Угодай n/a ugo...@gmail.com wrote: I have a data frame, for exampe dat - data.frame(a=rnorm(5),b=rnorm(5),c=rnorm(5)) a b c 1 -0.1731141 0.002453991 0.1180976 2 1.2142024 -0.413897606 0.7617472 3 -0.9428484 -0.609312786 0.5132441 4 0.1343336 0.178208961 0.7509650 5 -0.1402286 -0.333476839 -0.4959459 How to make dat2 from dat, where source data frame be ordered by any column? [[alternative HTML version deleted]] __ 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. __ 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] How to set a filter during reading tables
I think you can use readLines(n=1) in loop to skip unwanted rows.
On Mon, Jun 1, 2009 at 12:56 AM, g...@ucalgary.ca wrote:
Thanks, Juliet.
It works for filtering columns.
I am also wondering if there is a way to filter rows.
Thanks again.
-james
One can use colClasses to set which columns get read in. For the
columns you don't
want you can set those to NULL. For example,
cc - c(NULL,rep(numeric,9))
myData -
read.table(myFile.txt,header=TRUE,colClasses=cc,nrow=numRows).
On Wed, May 27, 2009 at 12:27 PM, g...@ucalgary.ca wrote:
We are reading big tables, such as,
Chemicals -
read.table('ftp://ftp.bls.gov/pub/time.series/wp/wp.data.7.Chemicals',header
= TRUE, sep = '\t', as.is =T)
I was wondering if it is possible to set a filter during loading so
that
we just load what we want not the whole table each time. Thanks,
-james
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Re: [R] logical vector as a matrix
On Sat, May 30, 2009 at 2:48 AM, Grześ gregori...@gmail.com wrote: I have a vector like this: h - c(4, 6, NA, 12) and I create the secound logical vector like this: g - c(TRUE, TRUE, FALSE, TRUE) Why don't you create vector g like this: g - ! is.na(h) And my problem is that I would like to get a new m vector as a rasult h and m but with missed NA value, for example: m = (4,6,12) Do you have any idea? As what you tried to do: m - h[g] # which got (4,6,12) you can directly use: m - h[ ! is.na(h) ] __ 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] How to write a loop?
Why did you use different variable names rather than index of list/data.frame? On Wed, May 27, 2009 at 6:34 PM, Maithili Shiva maithili_sh...@yahoo.com wrote: Dear R helpers, Following is a R script I am using to run the Fast Fourier Transform. The csv files has 10 columns with titles m1, m2, m3 .m10. When I use the following commands, I am getting the required results. The probelm is if there are 100 columns, it is not wise to define 100 commands as fk - ONS$mk and so on. Thus, I need some guidance to write the loop for the STEP A and STEP B. Thanking in advance Regards Maithili My R Script --- ONS - read.csv(fast fourier transform.csv, header = TRUE) # STEP A f1 - ONS$m1 f2 - ONS$m2 f3 - ONS$m3 f4 - ONS$m4 f5 - ONS$m5 f6 - ONS$m6 f7 - ONS$m7 f8 - ONS$m8 f9 - ONS$m9 f10 - ONS$m10 # # STEP B g1 - fft(f1) g2 - fft(f2) g3 - fft(f3) g4 - fft(f4) g5 - fft(f5) g6 - fft(f6) g7 - fft(f7) g8 - fft(f8) g9 - fft(f9) g10 - fft(f10) # h - g1*g2*g3*g4*g5*g6*g7*g8*g9*g10 j - fft((h), inverse = TRUE)/length(h) # Cricket on your mind? Visit the ultimate cricket website. Enter http://beta.cricket.yahoo.com [[alternative HTML version deleted]] __ 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. __ 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] How to exclude a column by name?
Hope this helps: df - data.frame(matrix(1:10,2)) df X1 X2 X3 X4 X5 1 1 3 5 7 9 2 2 4 6 8 10 df[,-2] X1 X3 X4 X5 1 1 5 7 9 2 2 6 8 10 df[,-which(names(df)==X2)] X1 X3 X4 X5 1 1 5 7 9 2 2 6 8 10 On Wed, May 27, 2009 at 6:37 PM, Zeljko Vrba zv...@ifi.uio.no wrote: Given an arbitrary data frame, it is easy to exclude a column given its index: df[,-2]. How to do the same thing given the column name? A naive attempt df[,-name] did not work :) __ 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] Sort matrix by column 1 ascending then by column 2 decending
It's a very interesting problem. I just wrote a function for it:
order.matrix - function(m, columnsDecreasing = c('1'=FALSE), rows = 1:nrow(m))
{
if (length(columnsDecreasing) 0)
{
col - as.integer(names(columnsDecreasing[1]));
values - sort(unique(m[rows, col]), decreasing=columnsDecreasing[1]);
unlist(sapply(values, function(x) order.matrix(m,
columnsDecreasing[-1], which((1:nrow(m) %in% rows) (m[,
col]==x);
}
else
{
rows;
}
}
For instance:
m - matrix( c(2, 1, 1, 3, .5, .3, .5, .2), 4)
m
[,1] [,2]
[1,]2 0.5
[2,]1 0.3
[3,]1 0.5
[4,]3 0.2
m[order.matrix(m),]
[,1] [,2]
[1,]1 0.3
[2,]1 0.5
[3,]2 0.5
[4,]3 0.2
m[order.matrix(m, c(1=FALSE, 2=TRUE)),]
[,1] [,2]
[1,]1 0.5
[2,]1 0.3
[3,]2 0.5
[4,]3 0.2
Any comment is welcome! ;)
On Wed, May 27, 2009 at 11:04 PM, Linlin Yan yanlinli...@gmail.com wrote:
m - matrix( c(2, 1, 1, 3, .5, .3, .5, .2), 4)
m
[,1] [,2]
[1,] 2 0.5
[2,] 1 0.3
[3,] 1 0.5
[4,] 3 0.2
m[unlist(sapply(sort(unique(m[,1])), function(x)
which(m[,1]==x)[order(m[(m[,1]==x),2], decreasing=TRUE)])),]
[,1] [,2]
[1,] 1 0.5
[2,] 1 0.3
[3,] 2 0.5
[4,] 3 0.2
On Wed, May 27, 2009 at 8:39 PM, Paul Geeleher paulgeele...@gmail.com wrote:
I've got a matrix with 2 columns and n rows. I need to sort it first
by the values in column 1 ascending. Then for values which are the
same in column 1, sort by column 2 decending. For example:
2 .5
1 .3
1 .5
3 .2
Goes to:
1 .5
1 .3
2 .5
3 .2
This is easy to do in spreadsheet programs but I can't seem to work
out how to do it in R and haven't been able to find a solution
anywhere.
Thanks!
-Paul.
--
Paul Geeleher
School of Mathematics, Statistics and Applied Mathematics
National University of Ireland
Galway
Ireland
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Re: [R] Help needed on R output
Did you mean this: write.table(t, eol=,\n, row.names=FALSE, col.names=FALSE) , 01001001011011101100, 1001001011010101, 1101110100000011, 000100100101001001011001, 000101101101101001101001, Try ?write.table to get the detail of the function please. On Tue, May 26, 2009 at 9:33 PM, peng chen rogerchan2...@gmail.com wrote: Thanks. I am sorry that I did not clearly put my question. I need to output the array like t - c( + , + 01001001011011101100, + 1001001011010101, + 1101110100000011, + 000100100101001001011001, + 000101101101101001101001) to a datafile(e.g., .txt file) where each line is for a binary number in this format: , 01001001011011101100, I was trying to use R-function write.table, however, I wasn't able to get the trailing comma for each line(although I can get the double quotation marks). Thanks. __ 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] Help needed on R output
t - c(
+ ,
+ 01001001011011101100,
+ 1001001011010101,
+ 1101110100000011,
+ 000100100101001001011001,
+ 000101101101101001101001)
{
+ cat ('rom_array := (\n');
+ for (i in 1:length(t)) {
+ cat('', t[i], '',
+ ifelse(i == length(t), '', ',\n'), sep='')
+ };
+ cat(')\n');
+ }
rom_array := (
,
01001001011011101100,
1001001011010101,
1101110100000011,
000100100101001001011001,
000101101101101001101001)
On Tue, May 26, 2009 at 12:30 PM, peng chen rogerchan2...@gmail.com wrote:
Hi, R experts:
I am trying to generate data output in the following format:
rom_array := (
,
01001001011011101100,
1001001011010101,
1101110100000011,
000100100101001001011001,
000101101101101001101001)
I have all the necessary data line, however, I am having trouble generating
the double quotation marks along with the trailing comma for each line.
Anyone can help?
Thanks.
[[alternative HTML version deleted]]
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Re: [R] exists function on list objects gives always a FALSE
SmoothData$span is not an object which can be checked by exists(), but part of an object which can be checked by is.null(). On Wed, May 20, 2009 at 12:07 AM, Žroutík zrou...@gmail.com wrote: Dear R-users, in a minimal example exists() gives FALSE on an object which obviously does exist. How can I check on that list object anyway else, please? SmoothData - list(exists=TRUE, span=0.001) SmoothData $exists [1] TRUE $span [1] 0.001 exists(SmoothData) TRUE exists(SmoothData$span) FALSE exists(SmoothData[[2]]) FALSE Thank you for any opinion regarding this topic. Zroutik [[alternative HTML version deleted]] __ 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. __ 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] Concatenating two vectors into one
z - paste(x, y, sep = '') z [1] A1 B2 C3 D4 E5 F6 On Mon, May 18, 2009 at 7:09 PM, Henning Wildhagen hwildha...@gmx.de wrote: Dear users, a very simple question: Given two vectors x and y x-as.character(c(A,B,C,D,E,F)) y-as.factor(c(1,2,3,4,5,6)) i want to combine them into a single vector z as A1, B2, C3 and so on. z-x*y is not working, i tried several others function, but did not get to the solution. Thanks for your help, Henning -- [[alternative HTML version deleted]] __ 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. __ 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] Concatenating two vectors into one
It seems that c(x,y) is not correct: z-c(x,y) z [1] A B C D E F 1 2 3 4 5 6 On Mon, May 18, 2009 at 7:17 PM, Simon Pickett simon.pick...@bto.org wrote: z-c(x,y) cheers, Simon. - Original Message - From: Henning Wildhagen hwildha...@gmx.de To: r-help@r-project.org Sent: Monday, May 18, 2009 12:09 PM Subject: [R] Concatenating two vectors into one Dear users, a very simple question: Given two vectors x and y x-as.character(c(A,B,C,D,E,F)) y-as.factor(c(1,2,3,4,5,6)) i want to combine them into a single vector z as A1, B2, C3 and so on. z-x*y is not working, i tried several others function, but did not get to the solution. Thanks for your help, Henning -- [[alternative HTML version deleted]] __ 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. __ 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. __ 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] converting numeric to integer
On Sun, May 17, 2009 at 5:00 PM, Thomas Mang thomas.m...@fiwi.at wrote: Hi, Well, also not quite. Suppose x = 5.001 (in finite binary represenation). Then I want x == 5, but that is of course an integer which is now less than the original numeral. Sorry for that I haven't got what your mean exactly about the finite binary representation, because my English is not very good. How much is the difference of x and 5 when result must be 5? Your code would turn it into 6; moreover, your code would still work on numeric-values, so once it gets to the interger-conversion we are back at where the problem started, namely at the issue of finite floating-point representations, which might, or might not, yield the desired integer after the decimal digits have been truncated. BTW, do you know why sometimes replys do not appear in the newsgroups - mine to yours is not there, equally not your last response (althought the later might of course be the resuld because I made a mistak. in the first step) The reason our last replys did not appear in the newsgroup is that you replied to me only rather than to the maillist addres r-h...@stat.math.ethz.ch. cheers and thanks, Thomas Linlin Yan wrote: I see. What you want is the integer with same sign as the original numeral, and whose absolute value is the least integer which is not less than absolute value of the original numeral. Am I right? I am afraid that there may not be any single function could work it out. But I could give the following expression: sign(x) * ceiling(abs(x)), which may be a little clearer. On Sun, May 17, 2009 at 2:02 PM, Thomas Mang thomas.m...@fiwi.at wrote: Hi, ceiling would do the wrong thing for negative values. If x = -4.999, the wanted result would be -5, but ceiling makes a -4 out of it. bye, Thomas Linlin Yan wrote: How about ceiling(x), which return the smallest integer not less than x? On Sun, May 17, 2009 at 2:49 AM, Thomas Mang thomas.m...@fiwi.at wrote: Hello, Suppose I have x, which is a variable of class numeric. The calculations performed to yield x imply that mathematically it should be an integer , but due to round-off errors, it might not be (and so in either direction). The error is however small, so round(x) will yield the appropriate integer value. Moreover, this integer values is guaranteed to be representable by an 'integer' class, that is -2^31 x 2^31, and logically it is an integer anyway. So I want to convert x from class 'numeric' to 'integer'. What is the most elegant, but always correct way, to achieve this conversion ? What comes to mind is of course something along: x = as.integer(round(x)) I am, however, not sure if this always works, because I do not know if the round-function is guaranteed to return a numeric value which, in finite binary representation, is always = the underlying mathematical integer. If that is however guaranteed, that would of course be a simple + elegant one. An alternative I came up with is: x = as.integer(round(x) + ifelse(x = 0, 0.5, -0.5)) Where I explicitly add a bit to ensure the finite binary representation must be = the underlying integer, and then truncate the decimal digits. IMO, this one is always guaranteed to work, at least within the numerical range of what integers are limited to anyway. What's your opinion on the issue ? Any other solution ? Thanks a lot in advance and cheers, Thomas __ 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. __ 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] (no subject)
On Sat, May 16, 2009 at 12:05 PM, Debbie Zhang debbie0...@hotmail.com wrote:
Dear R users,
Does anyone know how to write a function involving derivative?
i.e. I want to implementing Newton's method in R, so my function is something
like
x- x-y/y'
I am not sure how to write y' in my function. Can anyone help?
In addition, if I want to implementing newton's method several times, what
code should I use?
Currently, I use the following code.
-1
Is this line x - 1?
for (i in 1:20){
+ x-x-(x^3-2*x^2+3*x-5)/(3*x^2-4*x+3)}
x
[1] 1.843734
However, when I typed in the following code, I would yield the same answer.
-1
And is this line x - 1, too?
for (i in 1:1){
+ x-x-(x^3-2*x^2+3*x-5)/(3*x^2-4*x+3)}
x
[1] 1.843734
Are you sure about this result?
What I got is like this:
x - 1; for (i in 1:20) { x - x-(x^3-2*x^2+3*x-5)/(3*x^2-4*x+3); }; x
[1] 1.843734
x - 1; for (i in 1:1) { x - x-(x^3-2*x^2+3*x-5)/(3*x^2-4*x+3); }; x
[1] 2.5
There are different!
Can anyone suggest me what the problem is?
Please help.
Thanks so much.
Debbie
_
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Re: [R] converting numeric to integer
How about ceiling(x), which return the smallest integer not less than x? On Sun, May 17, 2009 at 2:49 AM, Thomas Mang thomas.m...@fiwi.at wrote: Hello, Suppose I have x, which is a variable of class numeric. The calculations performed to yield x imply that mathematically it should be an integer , but due to round-off errors, it might not be (and so in either direction). The error is however small, so round(x) will yield the appropriate integer value. Moreover, this integer values is guaranteed to be representable by an 'integer' class, that is -2^31 x 2^31, and logically it is an integer anyway. So I want to convert x from class 'numeric' to 'integer'. What is the most elegant, but always correct way, to achieve this conversion ? What comes to mind is of course something along: x = as.integer(round(x)) I am, however, not sure if this always works, because I do not know if the round-function is guaranteed to return a numeric value which, in finite binary representation, is always = the underlying mathematical integer. If that is however guaranteed, that would of course be a simple + elegant one. An alternative I came up with is: x = as.integer(round(x) + ifelse(x = 0, 0.5, -0.5)) Where I explicitly add a bit to ensure the finite binary representation must be = the underlying integer, and then truncate the decimal digits. IMO, this one is always guaranteed to work, at least within the numerical range of what integers are limited to anyway. What's your opinion on the issue ? Any other solution ? Thanks a lot in advance and cheers, Thomas __ 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. __ 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] Duplicates and duplicated
On Thu, May 14, 2009 at 2:16 PM, christiaan pauw cjp...@gmail.com wrote: Hi everybody. I want to identify not only duplicate number but also the original number that has been duplicated. Example: x=c(1,2,3,4,4,5,6,7,8,9) y=duplicated(x) rbind(x,y) gives: [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] x 1 2 3 4 4 5 6 7 8 9 y 0 0 0 0 1 0 0 0 0 0 i.e. the second 4 [,5] is a duplicate. What I want is the first and second 4. i.e [,4] and [,5] to be TRUE [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] x 1 2 3 4 4 5 6 7 8 9 y 0 0 0 1 1 0 0 0 0 0 How about rbind(x, duplicated(x) | duplicated(x, fromLast=TRUE)) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] x123445678 9 000110000 0 I assume it can be done by sorting the vector and then checking is the next or the previous entry matches using identical() . I am just unsure on how to write such a loop the logic of which (I think) is as follows: sort x for every value of x check if the next value is identical and return TRUE (or 1) if it is and FALSE (or 0) if it is not AND check is the previous value is identical and return TRUE (or 1) if it is and FALSE (or 0) if it is not Im i thinking correct and can some help to write such a function regards Christiaan [[alternative HTML version deleted]] __ 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. __ 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] Duplicates and duplicated
The operator %in% is very good! And that can be simpler like this: x %in% x[duplicated(x)] [1] FALSE FALSE FALSE TRUE TRUE FALSE FALSE FALSE FALSE FALSE On Thu, May 14, 2009 at 4:43 PM, Andrej Blejec andrej.ble...@nib.si wrote: Try this x%in%x[which(y)] From your example x=c(1,2,3,4,4,5,6,7,8,9) y=duplicated(x) rbind(x,y) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] x 1 2 3 4 4 5 6 7 8 9 y 0 0 0 0 1 0 0 0 0 0 which(y) [1] 5 x[which(y)] [1] 4 x%in%x[which(y)] [1] FALSE FALSE FALSE TRUE TRUE FALSE FALSE FALSE FALSE FALSE Andrej -- Andrej Blejec National Institute of Biology Vecna pot 111 POB 141 SI-1000 Ljubljana SLOVENIA e-mail: andrej.ble...@nib.si URL: http://ablejec.nib.si tel: + 386 (0)59 232 789 fax: + 386 1 241 29 80 -- Organizer of Applied Statistics 2009 conference http://conferences.nib.si/AS2009 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of christiaan pauw Sent: Thursday, May 14, 2009 8:17 AM To: r-help@r-project.org Subject: [R] Duplicates and duplicated Hi everybody. I want to identify not only duplicate number but also the original number that has been duplicated. Example: x=c(1,2,3,4,4,5,6,7,8,9) y=duplicated(x) rbind(x,y) gives: [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] x 1 2 3 4 4 5 6 7 8 9 y 0 0 0 0 1 0 0 0 0 0 i.e. the second 4 [,5] is a duplicate. What I want is the first and second 4. i.e [,4] and [,5] to be TRUE [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] x 1 2 3 4 4 5 6 7 8 9 y 0 0 0 1 1 0 0 0 0 0 I assume it can be done by sorting the vector and then checking is the next or the previous entry matches using identical() . I am just unsure on how to write such a loop the logic of which (I think) is as follows: sort x for every value of x check if the next value is identical and return TRUE (or 1) if it is and FALSE (or 0) if it is not AND check is the previous value is identical and return TRUE (or 1) if it is and FALSE (or 0) if it is not Im i thinking correct and can some help to write such a function regards Christiaan [[alternative HTML version deleted]] __ 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. __ 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. __ 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] Simulation
Since you got the most suitable way to get x, why can't you get the
variances in the same way? Just like:
v = vector()
for (i in 1:length(x)) v[i] = var(x[[i]])
BTW, it is much better to use lapply, like this:
lapply(x, var)
On Thu, May 14, 2009 at 8:26 PM, Debbie Zhang debbie0...@hotmail.com wrote:
Thanks for everyone.
I think the approach below is most suitable for me, as a beginner.
x=list()
for(i in 1:n){
x[[i]]=rnorm(i,0,1)
}
Now, I am trying to obtain the sample variance (S^2) of the 1000 samples that
I have generated before.
I am wondering what command I should use in order to get the sample variance
for all the 1000 samples.
What I am capable of doing now is just typing in
var(z[[1]])
var(z[[2]]).
Thanks for help.
Debbie
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Re: [R] Simulation
Does every 100 numbers in rnorm(100 * 1000, 0, 1) have the N(0,1) distribution? On Wed, May 13, 2009 at 11:13 PM, Debbie Zhang debbie0...@hotmail.com wrote: Dear R users, Can anyone please tell me how to generate a large number of samples in R, given certain distribution and size. For example, if I want to generate 1000 samples of size n=100, with a N(0,1) distribution, how should I proceed? (Since I dont want to do rnorm(100,0,1) in R for 1000 times) Thanks for help Debbie _ Looking to change your car this year? Find car news, reviews and more e%2Ecom%2Fcgi%2Dbin%2Fa%2Fci%5F450304%2Fet%5F2%2Fcg%5F801459%2Fpi%5F1004813%2Fai%5F859641_t=762955845_r=tig_OCT07_m=EXT [[alternative HTML version deleted]] __ 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. __ 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.
