Re: [R] Trouble retrieving the second largest value from each row of a data.frame
Hi,
Here is a little function that will do what you want and return a nice output:
#Function To calculate top two values and return
my.finder - function(mydata) {
my.fun - function(data) {
strongest - which.max(data)
secondstrongest - which.max(data[-strongest])
strongestantenna - names(data)[strongest]
secondstrongantenna - names(data[-strongest])[secondstrongest]
value - matrix(c(data[strongest], data[secondstrongest],
strongestantenna, secondstrongantenna), ncol =4)
return(value)
}
dat - apply(mydata, 1, my.fun)
dat - t(dat)
dat - as.data.frame(dat, stringsAsFactors = FALSE)
colnames(dat) - c(strongest, secondstrongest,
strongestantenna, secondstrongantenna)
dat[ , strongest] - as.numeric(dat[ , strongest])
dat[ , secondstrongest] - as.numeric(dat[ , secondstrongest])
return(dat)
}
#Using your example data:
yourdata - structure(list(value0 = c(-13007L, -12838L, -12880L, -12805L,
-12834L, -11068L, -12807L, -12770L, -12988L, -11779L), value60 = c(-11707L,
-13210L, -11778L, -11653L, -13527L, -11698L, -14068L, -11665L,
-11736L, -12873L), value120 = c(-11072L, -11176L, -3L, -11071L,
-11067L, -12430L, -11092L, -11061L, -11137L, -12973L), value180 = c(-12471L,
-11799L, -12439L, -12385L, -11638L, -12430L, -11709L, -12373L,
-12570L, -12537L), value240 = c(-12838L, -13210L, -13089L, -11561L,
-13527L, -12430L, -11607L, -11426L, -13467L, -12973L), value300 = c(-13357L,
-13845L, -13880L, -13317L, -13873L, -12814L, -13025L, -12805L,
-13739L, -11146L)), .Names = c(value0, value60, value120,
value180, value240, value300), class = data.frame, row.names = c(1,
2, 3, 4, 5, 6, 7, 8, 9, 10))
my.finder(yourdata) #and what you want is in a nicely labeled data frame
#A potential problem is that it is not very efficient
#Here is a test using a matrix of 100,000 rows
#sampled from the same range as your data
#with the same number of columns
data.test - matrix(
sample(seq(min(yourdata),max(yourdata)), size = 50, replace = TRUE),
ncol = 5)
system.time(my.finder(data.test))
#On my system I get
system.time(my.finder(data.test))
user system elapsed
2.890.002.89
Hope that helps,
Josh
On Fri, Jul 23, 2010 at 6:20 PM, mpw...@illinois.edu wrote:
I have a data frame with a couple million lines and want to retrieve the
largest and second largest values in each row, along with the label of the
column these values are in. For example
row 1
strongest=-11072
secondstrongest=-11707
strongestantenna=value120
secondstrongantenna=value60
Below is the code I am using and a truncated data.frame. Retrieving the
largest value was easy, but I have been getting errors every way I have tried
to retrieve the second largest value. I have not even tried to retrieve the
labels for the value yet.
Any help would be appreciated
Mike
data-data.frame(value0,value60,value120,value180,value240,value300)
data
value0 value60 value120 value180 value240 value300
1 -13007 -11707 -11072 -12471 -12838 -13357
2 -12838 -13210 -11176 -11799 -13210 -13845
3 -12880 -11778 -3 -12439 -13089 -13880
4 -12805 -11653 -11071 -12385 -11561 -13317
5 -12834 -13527 -11067 -11638 -13527 -13873
6 -11068 -11698 -12430 -12430 -12430 -12814
7 -12807 -14068 -11092 -11709 -11607 -13025
8 -12770 -11665 -11061 -12373 -11426 -12805
9 -12988 -11736 -11137 -12570 -13467 -13739
10 -11779 -12873 -12973 -12537 -12973 -11146
#largest value in the row
strongest-apply(data,1,max)
#second largest value in the row
n-function(data)(1/(min(1/(data[1,]-max(data[1,]+ (max(data[1,])))
secondstrongest-apply(data,1,n)
Error in data[1, ] : incorrect number of dimensions
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/
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Re: [R] Trouble retrieving the second largest value from each row of a data.frame
Maybe efficiency is less of an issue than I thought, on 2 million
rows, it only took a bit over a minute, and my system only jumped up ~
600 MB of memory so it wasn't much of a strain there either.
data.test - matrix(
+ sample(seq(min(yourdata),max(yourdata)), size = 1000, replace = TRUE),
+ ncol = 5)
nrow(data.test)
[1] 200
system.time(my.finder(data.test))
user system elapsed
74.540.20 75.20
On Fri, Jul 23, 2010 at 11:01 PM, Joshua Wiley jwiley.ps...@gmail.com wrote:
Hi,
Here is a little function that will do what you want and return a nice output:
#Function To calculate top two values and return
my.finder - function(mydata) {
my.fun - function(data) {
strongest - which.max(data)
secondstrongest - which.max(data[-strongest])
strongestantenna - names(data)[strongest]
secondstrongantenna - names(data[-strongest])[secondstrongest]
value - matrix(c(data[strongest], data[secondstrongest],
strongestantenna, secondstrongantenna), ncol =4)
return(value)
}
dat - apply(mydata, 1, my.fun)
dat - t(dat)
dat - as.data.frame(dat, stringsAsFactors = FALSE)
colnames(dat) - c(strongest, secondstrongest,
strongestantenna, secondstrongantenna)
dat[ , strongest] - as.numeric(dat[ , strongest])
dat[ , secondstrongest] - as.numeric(dat[ , secondstrongest])
return(dat)
}
#Using your example data:
yourdata - structure(list(value0 = c(-13007L, -12838L, -12880L, -12805L,
-12834L, -11068L, -12807L, -12770L, -12988L, -11779L), value60 = c(-11707L,
-13210L, -11778L, -11653L, -13527L, -11698L, -14068L, -11665L,
-11736L, -12873L), value120 = c(-11072L, -11176L, -3L, -11071L,
-11067L, -12430L, -11092L, -11061L, -11137L, -12973L), value180 = c(-12471L,
-11799L, -12439L, -12385L, -11638L, -12430L, -11709L, -12373L,
-12570L, -12537L), value240 = c(-12838L, -13210L, -13089L, -11561L,
-13527L, -12430L, -11607L, -11426L, -13467L, -12973L), value300 = c(-13357L,
-13845L, -13880L, -13317L, -13873L, -12814L, -13025L, -12805L,
-13739L, -11146L)), .Names = c(value0, value60, value120,
value180, value240, value300), class = data.frame, row.names = c(1,
2, 3, 4, 5, 6, 7, 8, 9, 10))
my.finder(yourdata) #and what you want is in a nicely labeled data frame
#A potential problem is that it is not very efficient
#Here is a test using a matrix of 100,000 rows
#sampled from the same range as your data
#with the same number of columns
data.test - matrix(
sample(seq(min(yourdata),max(yourdata)), size = 50, replace = TRUE),
ncol = 5)
system.time(my.finder(data.test))
#On my system I get
system.time(my.finder(data.test))
user system elapsed
2.89 0.00 2.89
Hope that helps,
Josh
On Fri, Jul 23, 2010 at 6:20 PM, mpw...@illinois.edu wrote:
I have a data frame with a couple million lines and want to retrieve the
largest and second largest values in each row, along with the label of the
column these values are in. For example
row 1
strongest=-11072
secondstrongest=-11707
strongestantenna=value120
secondstrongantenna=value60
Below is the code I am using and a truncated data.frame. Retrieving the
largest value was easy, but I have been getting errors every way I have
tried to retrieve the second largest value. I have not even tried to
retrieve the labels for the value yet.
Any help would be appreciated
Mike
data-data.frame(value0,value60,value120,value180,value240,value300)
data
value0 value60 value120 value180 value240 value300
1 -13007 -11707 -11072 -12471 -12838 -13357
2 -12838 -13210 -11176 -11799 -13210 -13845
3 -12880 -11778 -3 -12439 -13089 -13880
4 -12805 -11653 -11071 -12385 -11561 -13317
5 -12834 -13527 -11067 -11638 -13527 -13873
6 -11068 -11698 -12430 -12430 -12430 -12814
7 -12807 -14068 -11092 -11709 -11607 -13025
8 -12770 -11665 -11061 -12373 -11426 -12805
9 -12988 -11736 -11137 -12570 -13467 -13739
10 -11779 -12873 -12973 -12537 -12973 -11146
#largest value in the row
strongest-apply(data,1,max)
#second largest value in the row
n-function(data)(1/(min(1/(data[1,]-max(data[1,]+ (max(data[1,])))
secondstrongest-apply(data,1,n)
Error in data[1, ] : incorrect number of dimensions
__
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.
--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/
--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/
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PLEASE
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] union data in column
Fahim Md wrote:
Is there any function/way to merge/unite the following data
GENEID col1 col2 col3col4
G234064 1 0 0 0
G234064 1 0 0 0
G234064 1 0 0 0
G234064 0 1 0 0
G234065 0 1 0 0
G234065 0 1 0 0
G234065 0 1 0 0
G234065 0 0 1 0
G234065 0 0 1 0
G234065 0 0 0 1
into
GENEID col1 col2 col3col4
G234064 1 1 0 0
// 1 appears in col1 and col2 above, rest are zero
G234065 0 1 1 1
// 1 appears in col2 , 3 and 4 above.
Thank
Warning on terminology: there is a merge function in R that lines up
rows from different tables to make a new set of longer rows (more
columns). The usual term for combining column values from multiple rows
is aggregation.
In addition to the example offered by Jim Holtzman, here are some other
options in no particular order:
x - read.table(textConnection( GENEID col1 col2 col3 col4
G234064 1 0 0 0
G234064 1 0 0 0
G234064 1 0 0 0
G234064 0 1 0 0
G234065 0 1 0 0
G234065 0 1 0 0
G234065 0 1 0 0
G234065 0 0 1 0
G234065 0 0 1 0
G234065 0 0 0 1
), header=TRUE, as.is=TRUE, row.names=NULL)
closeAllConnections()
# syntactic repackaging of Jim's basic approach
library(plyr)
ddply( x, .(GENEID), function(df)
{with(as.integer(c(col1=any(col1),col2=any(col2),col3=any(col3),col4=any(col4}
)
# if you are familiar with SQL, this approach allows you merge data
frames and aggregate rows in the same step, but has a somewhat limited
range of aggregating functions available
library(sqldf)
sqldf(SELECT GENEID, max(col1) AS col1, max(col2) AS col2, max(col3) AS
col3, max(col4) AS col4 FROM x GROUP BY GENEID)
# when the data you want to crunch is separate from the key(s) or you
can separate them yourself
crunch - function(tmp){as.integer(any(tmp))}
aggregate( x[,c(col1,col2,col3,col4)], by=list(x$GENEID),
FUN=crunch )
# this is typically used if you want to aggregate many columns with the
same set of operations
library(doBy)
summaryBy( .~GENEID, data=x, FUN=c(sum,crunch) )
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to import simple java/mathematica expression to R
Hello, 2010/7/24 jim holtman jholt...@gmail.com: Well, I took you equation and put the following at the start: Power - function(x,y) x^y EE - function(x) x alp - 2 x - 900*Power(-0.2030178326474623 + 0.23024073983368956*(1 - alp) + ... Actually, there are many functions like this (about 18*2), for each situation with input data. So there is an idea to put them in different files (via Export from Mathematica) and then use them somehow (without modifications) in R program. Thank You for the answer. -- Regards, -Andrey __ 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 import simple java/mathematica expression to R
2010/7/24 David Winsemius dwinsem...@comcast.net:
On Jul 23, 2010, at 10:29 PM, jim holtman wrote:
It has some interesting properties with that identity definition of EE(x).
Local maxima at 1 and 0, blows up beyond -1 and 2, and several local
minima nearby:
Math.Fn - function(alp) { big-long-expression }
plot( seq(-.3,1.5,by=0.01), Math.Fn(seq(-.3,1.5,by=0.01) ), cex=0.2)
Indeed, EE(x) is not a function with real x, but a symbol with
subscript - i.e. EE(1) and EE(2) are two variables e1 and e2.
Thanks for Your interest.
--
Regards,
-Andrey
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[R] Understanding Prototype of class
Hi R gurus, here I am trying to understand how R define and work with the prototype property in class definition. Here I have created a class: setClass(a, representation=list(x=numeric, y=character)) [1] a As I did not explicitly define the prototype in above code, R is supposed to generate default. I wanted to see what R generated: prototype(a) An object of class classPrototypeDef Slot object: [1] a Slot slots: character(0) Slot dataPart: [1] TRUE I really could not understand the meaning which R returned. I understood part of above (please correct me if I am wrong): An object of class classPrototypeDef : I understood R treats the prototype of any class, as an object of class classPrototypeDef Slot object: : I believe under this heading, R shows the name of the underlying class, whose prototype I am looking for Slot slots: : I really could not understand what this meant for Slot dataPart: : again I really could not understand what this meant for Can anyone please help me to understand above clearly? Thanks and regards, [[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.
Re: [R] interpretation of stress in NMDS
On 7/22/2010 2:21 PM, Graves, Gregory wrote: Among those users of Primer, stress values greater than 0.3 are interpreted as questionable. Using both isoMDS and metaMDS (vegan package), the stress values returned are much higher using my own data and using examples provided in R Help. For example Rstress = 8.3, and the stressplot r2 = 0.99 indicating (to me) that the ordination is OK. I am guessing that the stress values across packages are not the same, and googling about has not returned a satisfactory answer ... thus this posting. My concern being that reporting a stress value of 8 for what I consider a satisfactory result may raise a few Primer-user's eyebrows. Dear Gregory, as the help file of ?isoMDS tells us: stress The final stress achieved (in percent) So 8.3 is in reality 0.083. Thomas Petzoldt __ 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 calculate the product of every two elements in two vectors
as.vector(t(outer(A, B))) [1] 9 10 11 12 18 20 22 24 27 30 33 36 HTH, Dennis On Fri, Jul 23, 2010 at 8:11 AM, aegea gche...@gmail.com wrote: Thanks in advance! A=c(1, 2,3) B=c (9, 10, 11, 12) I want to get C=c(1*9, 1*10, 1*11, 1*12, ., 3*9, 3*10, 3*11, 3*12)? C is still a vector with 12 elements Is there a way to do that? -- View this message in context: http://r.789695.n4.nabble.com/how-to-calculate-the-product-of-every-two-elements-in-two-vectors-tp2300299p2300299.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. [[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.
Re: [R] Question regarding panel data diagnostic
My thought is this:
It depends on what you have in the panel. Are your data cross-section data
observed over ten years for, say, 3 countries (or regions within the same
country)? If so, yes you can perform integration properties (what people
usually call unit root test) and then test for cointegration. But if the data
are quarterly or monthly, these techniques are not relevant.
Hope this helps.
Lexi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of amatoallah ouchen
Sent: Friday, July 23, 2010 12:18 AM
To: r-help@r-project.org
Subject: [R] Question regarding panel data diagnostic
Good day R-listers,
I'm currently working on a panel data analysis (N=17, T=5), in order
to check for the spurious regression problem, i have to test for
stationarity but i've read somewhere that i needn't to test for it as
my T10 , what do you think? if yes is there any other test i have
to perform in such case (a kind of cointegration test for small T?)
Any hint would be highly appreciated.
Ama.
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
__
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and provide commented, minimal, self-contained, reproducible code.
DISCLAIMER:\ Sample Disclaimer added in a VBScript.\ ...{{dropped:3}}
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Question regarding panel data diagnostic
Let me correct an omission in my response below. The last sentence
should read But if the data are 10 quarterly or monthly values, these
techniques are not relevant.
Cheers
Lexi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Setlhare Lekgatlhamang
Sent: Saturday, July 24, 2010 12:54 PM
To: amatoallah ouchen; r-help@r-project.org
Subject: Re: [R] Question regarding panel data diagnostic
My thought is this:
It depends on what you have in the panel. Are your data cross-section
data observed over ten years for, say, 3 countries (or regions within
the same country)? If so, yes you can perform integration properties
(what people usually call unit root test) and then test for
cointegration. But if the data are quarterly or monthly, these
techniques are not relevant.
Hope this helps.
Lexi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of amatoallah ouchen
Sent: Friday, July 23, 2010 12:18 AM
To: r-help@r-project.org
Subject: [R] Question regarding panel data diagnostic
Good day R-listers,
I'm currently working on a panel data analysis (N=17, T=5), in order
to check for the spurious regression problem, i have to test for
stationarity but i've read somewhere that i needn't to test for it as
my T10 , what do you think? if yes is there any other test i have
to perform in such case (a kind of cointegration test for small T?)
Any hint would be highly appreciated.
Ama.
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
__
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.
DISCLAIMER:\ Sample Disclaimer added in a VBScript.\ ..{{dropped:14}}
__
R-help@r-project.org mailing list
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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 regarding panel data diagnostic
Let me correct an omission in my response below. The last sentence
should read But if the data are 10 quarterly or monthly values, these
techniques are not relevant.
Cheers
Lexi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Setlhare Lekgatlhamang
Sent: Saturday, July 24, 2010 12:54 PM
To: amatoallah ouchen; r-help@r-project.org
Subject: Re: [R] Question regarding panel data diagnostic
My thought is this:
It depends on what you have in the panel. Are your data cross-section
data observed over ten years for, say, 3 countries (or regions within
the same country)? If so, yes you can perform integration properties
(what people usually call unit root test) and then test for
cointegration. But if the data are quarterly or monthly, these
techniques are not relevant.
Hope this helps.
Lexi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of amatoallah ouchen
Sent: Friday, July 23, 2010 12:18 AM
To: r-help@r-project.org
Subject: [R] Question regarding panel data diagnostic
Good day R-listers,
I'm currently working on a panel data analysis (N=17, T=5), in order
to check for the spurious regression problem, i have to test for
stationarity but i've read somewhere that i needn't to test for it as
my T10 , what do you think? if yes is there any other test i have
to perform in such case (a kind of cointegration test for small T?)
Any hint would be highly appreciated.
Ama.
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
DISCLAIMER:\ Sample Disclaimer added in a VBScript.\ ..{{dropped:14}}
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Trouble retrieving the second largest value from each row of a data.frame
On Jul 23, 2010, at 9:20 PM, mpw...@illinois.edu wrote: I have a data frame with a couple million lines and want to retrieve the largest and second largest values in each row, along with the label of the column these values are in. For example row 1 strongest=-11072 secondstrongest=-11707 strongestantenna=value120 secondstrongantenna=value60 Below is the code I am using and a truncated data.frame. Retrieving the largest value was easy, but I have been getting errors every way I have tried to retrieve the second largest value. I have not even tried to retrieve the labels for the value yet. Any help would be appreciated Mike Using Holtman's extract of your data with x as the name and the order function to generate an index to names of the dataframe: t(apply(x, 1, sort, decreasing=TRUE)[1:3, ]) [,1] [,2] [,3] 1 -11072 -11707 -12471 2 -11176 -11799 -12838 3 -3 -11778 -12439 4 -11071 -11561 -11653 5 -11067 -11638 -12834 6 -11068 -11698 -12430 7 -11092 -11607 -11709 8 -11061 -11426 -11665 9 -11137 -11736 -12570 10 -11146 -11779 -12537 Putting it all together: matrix( paste( t(apply(x, 1, sort, decreasing=TRUE)[1:3, ]), names(x)[ t(apply(x, 1, order, decreasing=TRUE) [1:3, ]) ]), ncol=3) [,1] [,2] [,3] [1,] -11072 value120 -11707 value60 -12471 value180 [2,] -11176 value120 -11799 value180 -12838 value0 [3,] -3 value120 -11778 value60 -12439 value180 [4,] -11071 value120 -11561 value240 -11653 value60 [5,] -11067 value120 -11638 value180 -12834 value0 [6,] -11068 value0 -11698 value60 -12430 value120 [7,] -11092 value120 -11607 value240 -11709 value180 [8,] -11061 value120 -11426 value240 -11665 value60 [9,] -11137 value120 -11736 value60 -12570 value180 [10,] -11146 value300 -11779 value0 -12537 value180 -- David. data-data.frame(value0,value60,value120,value180,value240,value300) data value0 value60 value120 value180 value240 value300 1 -13007 -11707 -11072 -12471 -12838 -13357 2 -12838 -13210 -11176 -11799 -13210 -13845 3 -12880 -11778 -3 -12439 -13089 -13880 4 -12805 -11653 -11071 -12385 -11561 -13317 5 -12834 -13527 -11067 -11638 -13527 -13873 6 -11068 -11698 -12430 -12430 -12430 -12814 7 -12807 -14068 -11092 -11709 -11607 -13025 8 -12770 -11665 -11061 -12373 -11426 -12805 9 -12988 -11736 -11137 -12570 -13467 -13739 10 -11779 -12873 -12973 -12537 -12973 -11146 #largest value in the row strongest-apply(data,1,max) #second largest value in the row n-function(data)(1/(min(1/(data[1,]-max(data[1,]+ (max(data[1,]))) secondstrongest-apply(data,1,n) Error in data[1, ] : incorrect number of dimensions __ 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] local polynomial with differnt kernal functions
Hi, R users I need to use the function (locpoly) to fit a local poynomial regression model, The defult for kernal function is normal , but I need to use different kernal functions such as :Uniform,Triangular,Epanechnikov,.. Could someone help me define these functions to fit local polynomial regression model?. Email:assaed...@yahoo.com Thanks alot [[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] Weights in mixed models
Hello everyone, I wonder if sample size can be used as weight in a weighted mixed model. Or should I use just the inverse of the variance? For example, the class'lmer' in the 'lme4' package have an option 'weights' (as in the class 'lm' of 'stats'). In the help of lme4 there is an example using 'herd size' as weight in a mixed model. So, if I have a measurement data (eg height) of 10 groups (sample size ranging from 30 to 3000 individuals for each group) can I use this number (N, sample size) in the 'weights' option? Is this wrong? Finally, what to do if the results (coefficients) of weighing by 'inverse of the variance' or by 'sample size' are very different, even opposite? Thank you very much in advance David __ 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] Odp: Help me with prediction in linear model
Thanks Murphy and pikal, I need another help,for fitting first fourier transformation ,i used following thing .Please advise on this beer_monthl has 400+ records EXample: head(beer_monthly) beer 1 93.2 2 96.0 3 95.2 4 77.1 5 70.9 6 64.8 time-seq(1956,1995.2,length=length(beer_monthly)) sin.t-sin(2*pi*time) cos.t-cos(2*pi*time) beer_fit_fourier=lm(beer_monthly[,1]~poly(time,2)+sin.t+cos.t) #this is not working beer_fit_fourier=lm(beer_monthly[,1]~time+time2+sin.t+cos.t) #it is working #prediction is not workinng tpred_four - data.frame(time = seq(1995, 1998, length = 20)) predict(beer_fit_fourier, newdata = tpred_four) Is there any way to fit first fourier frequency , Please assist. Thanks in advance -- View this message in context: http://r.789695.n4.nabble.com/Help-me-with-prediction-in-linear-model-tp2297313p2300991.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] union data in column
On Sat, Jul 24, 2010 at 2:23 AM, Jeff Newmiller
jdnew...@dcn.davis.ca.us wrote:
Fahim Md wrote:
Is there any function/way to merge/unite the following data
GENEID col1 col2 col3 col4
G234064 1 0 0 0
G234064 1 0 0 0
G234064 1 0 0 0
G234064 0 1 0 0
G234065 0 1 0 0
G234065 0 1 0 0
G234065 0 1 0 0
G234065 0 0 1 0
G234065 0 0 1 0
G234065 0 0 0 1
into
GENEID col1 col2 col3 col4
G234064 1 1 0 0
// 1 appears in col1 and col2 above, rest are zero
G234065 0 1 1 1
// 1 appears in col2 , 3 and 4 above.
Thank
Warning on terminology: there is a merge function in R that lines up rows
from different tables to make a new set of longer rows (more columns). The
usual term for combining column values from multiple rows is aggregation.
In addition to the example offered by Jim Holtzman, here are some other
options in no particular order:
x - read.table(textConnection( GENEID col1 col2 col3 col4
G234064 1 0 0 0
G234064 1 0 0 0
G234064 1 0 0 0
G234064 0 1 0 0
G234065 0 1 0 0
G234065 0 1 0 0
G234065 0 1 0 0
G234065 0 0 1 0
G234065 0 0 1 0
G234065 0 0 0 1
), header=TRUE, as.is=TRUE, row.names=NULL)
closeAllConnections()
# syntactic repackaging of Jim's basic approach
library(plyr)
ddply( x, .(GENEID), function(df)
{with(as.integer(c(col1=any(col1),col2=any(col2),col3=any(col3),col4=any(col4}
)
You can do this a little more succinctly with colwise:
any_1 - function(x) as.integer(any(x))
ddply(x, GENEID, numcolwise(any_1))
Hadley
--
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
http://had.co.nz/
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Re: [R] UseR! 2010 - my impressions
On 07/23/2010 06:50 PM, Ravi Varadhan wrote: Dear UseRs!, Everything about UseR! 2010 was terrific! I really mean everything - the tutorials, invited talks, kaleidoscope sessions, focus sessions, breakfast, snacks, lunch, conference dinner, shuttle services, and the participants. The organization was fabulous. NIST were gracious hosts, and provided top notch facilities. The rousing speech by Antonio Possolo, who is the chief of Statistical Engineering Division at NIST, set the tempo for the entire conference. Excellent invited lectures by Luke Tierney, Frank Harrell, Mark Handcock, Diethelm Wurtz, Uwe Ligges, and Fritz Leisch. All the sessions that I attended had many interesting ideas and useful contributions. During the whole time that I was there, I could not help but get the feeling that I am a part of something great. Before I end, let me add a few words about a special person. This conference would not have been as great as it was without the tireless efforts of Kate Mullen. The great thing about Kate is that she did so much without ever hogging the limelight. Thank you, Kate and thank you NIST! I cannot wait for UseR!2011! Best, Ravi. Ravi Varadhan, Ph.D. Assistant Professor, Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University Ph. (410) 502-2619 email: rvarad...@jhmi.edu I want to echo what Ravi said. The talks were terrific (thanks to the program committee and the speakers) and Kate Mullen and her team did an extraordinary job in putting the conference together and running it. I am proud to have been a part of it. Thank you all! Frank -- Frank E Harrell Jr Professor and ChairmanSchool of Medicine Department of Biostatistics Vanderbilt University __ 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
On Fri, 23 Jul 2010, Marcus Liu wrote: Hi everyone, Is there any command for updating table withing a loop?? Loops? We don't need no stinking loops! (From 'The Good, the Bad, and the Rgly') tab - table(data.raw, findInterval(seq(along=data.raw), ind+1 ) ) tab %*% upper.tri(tab,diag=T) or tab2 - tapply( factor(data.raw), findInterval(seq(along=data.raw), ind+1 ), table) Reduce( +, tab2, accum=TRUE ) HTH, Chuck p.s. See the posting guide re including a reproducible example with requests like yours. 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]] Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:cbe...@tajo.ucsd.edu UC San Diego http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901 __ 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
On 24/07/2010 11:25 AM, Charles C. Berry wrote: On Fri, 23 Jul 2010, Marcus Liu wrote: Hi everyone, Is there any command for updating table withing a loop?� Loops? We don't need no stinking loops! (From 'The Good, the Bad, and the Rgly') Actually, that quote comes from the TreasR of the SieRa MadRe. Duncan Murdoch tab - table(data.raw, findInterval(seq(along=data.raw), ind+1 ) ) tab %*% upper.tri(tab,diag=T) or tab2 - tapply( factor(data.raw), findInterval(seq(along=data.raw), ind+1 ), table) Reduce( +, tab2, accum=TRUE ) HTH, Chuck p.s. See the posting guide re including a reproducible example with requests like yours. 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]] Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:cbe...@tajo.ucsd.edu UC San Diego http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901 __ 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] glm - prediction of a factor with several levels
blackscorpio olivier.collignon at live.fr writes: I'm currently attempting to predict the occurence of an event (factor) having more than 2 levels with several continuous predictors. The model being ordinal, I was waiting the glm function to return several intercepts, which is not the case when looking to my results (I only have one intercept). I finally managed to perform an ordinal polytomous logisitc regression with the polr function, which gives several intercepts. But does anyone know what was the model performed with glm and why only one intercept was given ? It's not sufficiently clear (to me at least) what you're trying to do. Please provide a minimal reproducible example ... As far as I know, polr is the right way to do ordinal regression; it's not clear how you were trying to use glm to do it. Ben Bolker __ 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] Doubt about a population competition function
Hi,
I'm doing a function that describe two populations in competition.
that's the function that i wrote:
exclusao-function(n10, n20, k1, k2, alfa, beta, t){
n1-k1-(alfa*n20)
n2-k2-(beta*n10)
if(t==0){plot(t, n10, type='b', xlim=range(c(1:t),c
(1:t)), ylim=range(n10, n20), xlab='tempo',
ylab='tamanho populacional')
points(t, n20, type='b', col=red)
points(t,n10,type=b, col=black)
legend(topleft, c(Pop1,Pop2), cex=0.8, col=c
(black,red), pch=21:21, lty=1:1);
}
if(t0){
for (i in 1:t){
n1[i==1]-n1
n2[i==1]-n2
n1[i+1]-k1-alfa*n2[i]
n2[i+1]-k2-beta*n1[i]
if(n1[i]==0){n1[i:t]==0}
if(n2[i]==0){n2[i:t]==0}
}
plot(c(1:t), n1[1:t], type='b', xlim=range(c(1:t),c
(1:t)), ylim=range(n1[1:t], n2[1:t]), xlab='tempo',
ylab='tamanho populacional')
points(c(1:t), n2[1:t], type='b', col=red)
legend(topleft, c(Pop1,Pop2), cex=0.8, col=c
(black,red), pch=21:21, lty=1:1);
}}
Where n10: size population in time 0, n20: size population in time 0, k1:
carrying capacity of the population 1, k2: carrying capacity of the population
2, alfa: competition coefficient of population 2 in population 1, beta:
competition coefficient of population 1 in population 2, t: time.
and when some population becomes 0 (ZERO), i want that population still 0
(ZERO) until the end of t. i have tried to put if(n1[i]==0){n1[i:t]==0}
if(n2[i]==0){n2[i:t]==0} after n2[i+1]-k2-beta*n1[i] in the for function,
but nothing happens. What may i do ?
Thanks
Bruno
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Doubt about a population competition function
Hi,
I'm doing a function that describe two populations in competition.
that's the function that i wrote:
exclusao-function(n10, n20, k1, k2, alfa, beta, t){
n1-k1-(alfa*n20)
n2-k2-(beta*n10)
if(t==0){plot(t, n10, type='b', xlim=range(c(1:t),c
(1:t)), ylim=range(n10, n20), xlab='tempo',
ylab='tamanho populacional')
points(t, n20, type='b', col=red)
points(t,n10,type=b, col=black)
legend(topleft, c(Pop1,Pop2), cex=0.8, col=c
(black,red), pch=21:21, lty=1:1);
}
if(t0){
for (i in 1:t){
n1[i==1]-n1
n2[i==1]-n2
n1[i+1]-k1-alfa*n2[i]
n2[i+1]-k2-beta*n1[i]
if(n1[i]==0){n1[i:t]==0}
if(n2[i]==0){n2[i:t]==0}
}
plot(c(1:t), n1[1:t], type='b', xlim=range(c(1:t),c
(1:t)), ylim=range(n1[1:t], n2[1:t]), xlab='tempo',
ylab='tamanho populacional')
points(c(1:t), n2[1:t], type='b', col=red)
legend(topleft, c(Pop1,Pop2), cex=0.8, col=c
(black,red), pch=21:21, lty=1:1);
}}
Where n10: size population in time 0, n20: size population in time 0, k1:
carrying capacity of the population 1, k2: carrying capacity of the population
2, alfa: competition coefficient of population 2 in population 1, beta:
competition coefficient of population 1 in population 2, t: time.
and when some population becomes 0 (ZERO), i want that population still 0
(ZERO) until the end of t. i have tried to put if(n1[i]==0){n1[i:t]==0}
if(n2[i]==0){n2[i:t]==0} after n2[i+1]-k2-beta*n1[i] in the for function,
but nothing happens. What may i do ?
Thanks
Bruno
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Book on R's Programming Language
Can someone please recommend to me a book on the programming language that R is based on? I'm looking for a foundational book that teaches the logic of the S language. It seems that knowing the underpinnings of the language can only make using R a bit easier. Any leads are greatly appreciated . . . Matt. [[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.
Re: [R] how to calculate the product of every two elements in two vectors
Try this: c(as.matrix(B) %*% A) On Fri, Jul 23, 2010 at 12:11 PM, aegea gche...@gmail.com wrote: Thanks in advance! A=c(1, 2,3) B=c (9, 10, 11, 12) I want to get C=c(1*9, 1*10, 1*11, 1*12, ., 3*9, 3*10, 3*11, 3*12)? C is still a vector with 12 elements Is there a way to do that? -- View this message in context: http://r.789695.n4.nabble.com/how-to-calculate-the-product-of-every-two-elements-in-two-vectors-tp2300299p2300299.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 [[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.
Re: [R] Book on R's Programming Language
Hi Matt, http://www.r-project.org/doc/bib/R-books.html Lists a variety of books, and seems to include most (i.e., my searches through google, amazon, and barnes and noble, didn't really turn up others) books dedicated to R. I have always been under the impression that Programming with Data (the Green Book) is a classic. http://cran.r-project.org/manuals.html has the official manuals Similar questions have been asked several times on this list so you can also search for previous threads (e.g., http://tolstoy.newcastle.edu.au/R/help/04/06/0063.html ) Best regards, Josh On Sat, Jul 24, 2010 at 9:39 AM, Matt Stati mattst...@yahoo.com wrote: Can someone please recommend to me a book on the programming language that R is based on? I'm looking for a foundational book that teaches the logic of the S language. It seems that knowing the underpinnings of the language can only make using R a bit easier. Any leads are greatly appreciated . . . Matt. [[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. -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to calculate the product of every two elements in two vectors
On Fri, Jul 23, 2010 at 11:11 AM, aegea gche...@gmail.com wrote: Thanks in advance! A=c(1, 2,3) B=c (9, 10, 11, 12) I want to get C=c(1*9, 1*10, 1*11, 1*12, ., 3*9, 3*10, 3*11, 3*12)? C is still a vector with 12 elements Is there a way to do that? Here are yet a few more. The first one is the only one so far that uses a single function and the last two are slight variations of ones already posted. kronecker(A, B) c(tcrossprod(B, A)) c(outer(B, A)) c(B %o% A) Here is a speed comparison. The fastest are as.matrix, %outer% and %o% . They are so close that random fluctuations might easily change their order and since %o% involves the least keystrokes that one might be a good overall choice. Although not among the fastest the kronecker solution is the simplest since it only involves a single function call so it might be preferred on that count. A - B - 1:400 out - benchmark( + as.matrix = c(as.matrix(B) %*% A), + crossprod = c(tcrossprod(B, A)), + outer = c(outer(B, A)), + o = c(B %o% A), + kronecker = kronecker(A, B), + touter = as.vector(t(outer(A, B out[order(out$relative), ] test replications elapsed relative user.self sys.self user.child sys.child 1 as.matrix 1000.92 1.00 0.62 0.28 NANA 3 outer 1000.93 1.010870 0.59 0.35 NANA 4 o 1000.94 1.021739 0.66 0.28 NANA 2 crossprod 1001.11 1.206522 0.67 0.43 NANA 5 kronecker 1001.45 1.576087 1.25 0.21 NANA 6touter 1001.84 2.00 1.40 0.43 NANA __ 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] Book on R's Programming Language
Matt, you might want to check out programming with data by John Chambers. http://www.amazon.com/Programming-Data-Guide-S-Language/dp/0387985034/ref=sr_1_1?ie=UTF8s=booksqid=1279990404sr=8-1 Best Joe On Sat, Jul 24, 2010 at 11:39 AM, Matt Stati mattst...@yahoo.com wrote: Can someone please recommend to me a book on the programming language that R is based on? I'm looking for a foundational book that teaches the logic of the S language. It seems that knowing the underpinnings of the language can only make using R a bit easier. Any leads are greatly appreciated . . . Matt. [[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.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joseph C. Magagnoli Doctoral Student Department of Political Science University of North Texas 1155 Union Circle #305340 Denton, Texas 76203-5017 Email: jcm0...@unt.edu [[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.
Re: [R] Constrain density to 0 at 0?
Look at the logspline package. This is a different approach to density estimation from the kernel densities used by 'density', but does allow you to set fixed boundaries. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Farley, Robert Sent: Monday, July 19, 2010 7:57 PM To: r-help@r-project.org Subject: [R] Constrain density to 0 at 0? I'm plotting some trip length frequencies using the following code: plot( density(zTestData$Distance, weights=zTestData$Actual), xlim=c(0,10), main=Test TLFD, xlab=Distance, col=6 ) lines(density(zTestData$Distance, weights=zTestData$FlatWeight), col=2) lines(density(zTestData$Distance, weights=zTestData$BrdWeight ), col=3) which works fine except the distances are all positive, but the densities don't drop to 0 until around -2 or -3. Is there a way for me to force the density plot to 0 at 0? Thanks 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 [[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] matest function for multiple factors
Hi, I am using maanova package for analysis. In my dataset, two fixed factors time and treatment and sample as random factor is there. Am able to get madata object and fitmaanova object. But, am unable to do f-test with two factors, but i have done f-test seperately for two factors. fit.full.mix - fitmaanova(madata, formula = ~Sample+Time+Treatment, random = ~Sample) ftest.all = *matest*(madata, fit.full.mix, test.method=c(1,1), shuffle.method=sample, *term=Time+Treatment*, n.perm= 100) Can u please suggest me, how to represent multiple factors in the above function simultaneously in term. [[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.
Re: [R] How to deal with more than 6GB dataset using R?
You may want to look at the biglm package as another way to regression models on very large data sets. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of babyfoxlo...@sina.com Sent: Friday, July 23, 2010 10:10 AM To: r-help@r-project.org Subject: [R] How to deal with more than 6GB dataset using R? nbsp;Hi there, Sorry to bother those who are not interested in this problem. I'm dealing with a large data set, more than 6 GB file, and doing regression test with those data. I was wondering are there any efficient ways to read those data? Instead of just using read.table()? BTW, I'm using a 64bit version desktop and a 64bit version R, and the memory for the desktop is enough for me to use. Thanks. --Gin [[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] Using R to fill ETM+ data gaps?
Hi, I was wondering if anyone knows of a method (or if it's possible) to use R in interpolating Landsat ETM+ data gaps? Regards, Hakim Abdi _ Abdulhakim Abdi, M.Sc. Conservation GIS/Remote Sensing Lab Smithsonian Conservation Biology Institute 1500 Remount Road Front Royal, VA 22630 phone: +1 540 635 6578 mobile: +1 747 224 7006 fax: +1 540 635 6506 (Attn: ABDI/GIS Lab) email: ab...@si.edu http://nationalzoo.si.edu/SCBI/ConservationGIS/ __ 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] How to generate a sequence of dates without hardcoding the year
Hi: I have a dataframe named 'spring' and I am trying to add a new variable named 'IdDate' This line of code works fine: spring$idDate - seq(as.Date(2008-07-01),as.Date(2009-06-30),by=week) But I don't want to hardcode the year because it will be used again the following year Is it possible to just generate dates with the month and day? I tried the code below: seq(as.Date(7-1,%B%d),as.Date(6-30,%B%d),by=week) and got this error message: Error in seq.int(0, to - from, by) : 'to' must be finite Thanks for any pointers Felipe D. Carrillo Supervisory Fishery Biologist Department of the Interior US Fish Wildlife Service California, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to generate a sequence of dates without hardcoding the year
Try this: format(seq(as.Date(2008-07-01),as.Date(2009-06-30),by=week), %d/%m) On Sat, Jul 24, 2010 at 6:07 PM, Felipe Carrillo mazatlanmex...@yahoo.comwrote: Hi: I have a dataframe named 'spring' and I am trying to add a new variable named 'IdDate' This line of code works fine: spring$idDate - seq(as.Date(2008-07-01),as.Date(2009-06-30),by=week) But I don't want to hardcode the year because it will be used again the following year Is it possible to just generate dates with the month and day? I tried the code below: seq(as.Date(7-1,%B%d),as.Date(6-30,%B%d),by=week) and got this error message: Error in seq.int(0, to - from, by) : 'to' must be finite Thanks for any pointers Felipe D. Carrillo Supervisory Fishery Biologist Department of the Interior US Fish Wildlife Service California, USA __ 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 [[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.
Re: [R] (no subject)
2010/7/23 w 霍 hw_joyce...@hotmail.com:
I use the constrOptim to maximize a function with four constriants but the
answer does not leave from the starting value and there is only one outer
iteration. The function is defined as follows:
tm-function(p){
p1-p[1]; p2-p[2]; p3-p[3];
p4-1-p1-p2-p3;
p1*p2*p3*p4}
##the constraints are p1=0; p2=0; p3=0 and p4=0 i.e. p1+p2+p3=1
start-c(0.991,0.001,0.001)
dev-rbind(diag(3),-diag(3),rep(-1,3))
bvec-c(rep(0,3),rep(-1,4))
constrOptim(start,tm,NULL,ui=dev,ci=bvec,control=list(maxit=1))
Am i missing something obviously that cause the problem or there is some bugs
in constrOptim. Could you please help me out
Wenwen,
I believe that the reason why constrOptim behaves as described is
related to the fact that
(p1, p2, p3) = (1,0,0)
is a stationary point and you use it as a starting point. Try a
different starting point.
If the objective is to maximize, then you should use the following command:
constrOptim(start,tm,NULL,ui=dev,ci=bvec,control=list(maxit=1,fnscale=-1))
(Notice fnscale=-1.)
Finally, whenever you ask something on this list, please use a
meaningful title for your message, as it will dramatically increase
the chances of you getting an answer.
Good luck,
Paul
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Re: [R] How to generate a sequence of dates without hardcoding the year
Is this what you want if you want to assume that the date without a year is this year: seq(as.Date(7-1,%m-%d),by=week, length=52) [1] 2010-07-01 2010-07-08 2010-07-15 2010-07-22 2010-07-29 2010-08-05 2010-08-12 2010-08-19 [9] 2010-08-26 2010-09-02 2010-09-09 2010-09-16 2010-09-23 2010-09-30 2010-10-07 2010-10-14 [17] 2010-10-21 2010-10-28 2010-11-04 2010-11-11 2010-11-18 2010-11-25 2010-12-02 2010-12-09 [25] 2010-12-16 2010-12-23 2010-12-30 2011-01-06 2011-01-13 2011-01-20 2011-01-27 2011-02-03 [33] 2011-02-10 2011-02-17 2011-02-24 2011-03-03 2011-03-10 2011-03-17 2011-03-24 2011-03-31 [41] 2011-04-07 2011-04-14 2011-04-21 2011-04-28 2011-05-05 2011-05-12 2011-05-19 2011-05-26 [49] 2011-06-02 2011-06-09 2011-06-16 2011-06-23 On Sat, Jul 24, 2010 at 5:07 PM, Felipe Carrillo mazatlanmex...@yahoo.com wrote: Hi: I have a dataframe named 'spring' and I am trying to add a new variable named 'IdDate' This line of code works fine: spring$idDate - seq(as.Date(2008-07-01),as.Date(2009-06-30),by=week) But I don't want to hardcode the year because it will be used again the following year Is it possible to just generate dates with the month and day? I tried the code below: seq(as.Date(7-1,%B%d),as.Date(6-30,%B%d),by=week) and got this error message: Error in seq.int(0, to - from, by) : 'to' must be finite Thanks for any pointers Felipe D. Carrillo Supervisory Fishery Biologist Department of the Interior US Fish Wildlife Service California, USA __ 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] glm - prediction of a factor with several levels
As far as I know, glm only works with dichotomous or count data. polr in
the MASS package works and so does lrm {Design} for ordinal dependent
variables. I would assume that the model produced by glm is a dichotomous
version of your model but not sure. Only one intercept would be given
because if you used the log link then it would have produced a dichotomous
model instead of an ordered logistic regression. My suggestion is to use
polr or lrm.
On Fri, Jul 23, 2010 at 6:15 PM, blackscorpio [via R]
ml-node+2300793-1751019155-246...@n4.nabble.comml-node%2b2300793-1751019155-246...@n4.nabble.com
wrote:
Dear community,
I'm currently attempting to predict the occurence of an event (factor)
having more than 2 levels with several continuous predictors. The model
being ordinal, I was waiting the glm function to return several intercepts,
which is not the case when looking to my results (I only have one
intercept). I finally managed to perform an ordinal polytomous logisitc
regression with the polr function, which gives several intercepts.
But does anyone know what was the model performed with glm and why only one
intercept was given ?
Thanks a lot for your help !
--
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Re: [R] Trouble retrieving the second largest value from each row of a data.frame
On Jul 24, 2010, at 4:54 PM, mpw...@illinois.edu wrote:
THANKS, but I have one issue and one question.
For some reason the secondstrongest value for row 3 and 6 are
incorrect (they are the strongest) the remaining 10 are correct??
In my run of Wiley's code I instead get identical values for rows
2,5,6. Holtman's and my solutions did not suffer from that defect,
although mine suffered from my misreading of your request, thinking
that you wanted the top 3. The fix is trivial
These data are being used to track radio-tagged birds, they are from
automated radio telemetry receivers. I will applying the following
formula
diff - ((strongest- secondstrongest)/100)
bearingdiff -30-(-0.0624*(diff**2))-(2.8346*diff)
vals - c(value0, value60, value120, value180, value240,
value300)
value.str2 - (match(yourdata$secondstrongestantenna, vals)-1)*60
value.str1 - (match(yourdata$strongestantenna, vals)-1)*60
change.ind - abs(match(yourdata, vals) - which(match(yourdata, vals) )
A) Then the bearing diff is added to strongestantenna (value0 =
0degrees) if the secondstrongestatenna is greater (eg value0 and
value60),
B) or if the secondstrongestantenna is smaller than the
strongestantenna,
then the bearingdiff is substracted from the strongestantenna.
C) The only exception is that if value0 (0degrees) is strongest and
value300(360degrees) is the secondstrongestantenna then the bearing
is 360-bearingdiff.
D) Also the strongestantenna and secondstrongestantenna have to be
next to each other (e.g. value0 with value60, value240 with
value300, value0 with value300) or the results should be NA.
After setting finalbearing with A, B, and C then:
yourdata$finalbearing - with(yourdata, ifelse (
change.ind 5 change.ind 1 ,
NA, finalbearing) )
I have been trying to use a series of if,else statements to produce
these bearing, but all I am producing is errors. Any suggestion
would be appreciated.
Again THANKS for you efforts.
Mike
Original message
Date: Fri, 23 Jul 2010 23:01:56 -0700
From: Joshua Wiley jwiley.ps...@gmail.com
Subject: Re: [R] Trouble retrieving the second largest value from
each row of a data.frame
To: mpw...@illinois.edu
Cc: r-help@r-project.org
Hi,
Here is a little function that will do what you want and return a
nice output:
#Function To calculate top two values and return
my.finder - function(mydata) {
my.fun - function(data) {
strongest - which.max(data)
secondstrongest - which.max(data[-strongest])
strongestantenna - names(data)[strongest]
secondstrongantenna - names(data[-strongest])[secondstrongest]
value - matrix(c(data[strongest], data[secondstrongest],
strongestantenna, secondstrongantenna), ncol =4)
return(value)
}
dat - apply(mydata, 1, my.fun)
dat - t(dat)
dat - as.data.frame(dat, stringsAsFactors = FALSE)
colnames(dat) - c(strongest, secondstrongest,
strongestantenna, secondstrongantenna)
dat[ , strongest] - as.numeric(dat[ , strongest])
dat[ , secondstrongest] - as.numeric(dat[ , secondstrongest])
return(dat)
}
#Using your example data:
yourdata - structure(list(value0 = c(-13007L, -12838L, -12880L,
-12805L,
-12834L, -11068L, -12807L, -12770L, -12988L, -11779L), value60 =
c(-11707L,
-13210L, -11778L, -11653L, -13527L, -11698L, -14068L, -11665L,
-11736L, -12873L), value120 = c(-11072L, -11176L, -3L, -11071L,
-11067L, -12430L, -11092L, -11061L, -11137L, -12973L), value180 =
c(-12471L,
-11799L, -12439L, -12385L, -11638L, -12430L, -11709L, -12373L,
-12570L, -12537L), value240 = c(-12838L, -13210L, -13089L, -11561L,
-13527L, -12430L, -11607L, -11426L, -13467L, -12973L), value300 =
c(-13357L,
-13845L, -13880L, -13317L, -13873L, -12814L, -13025L, -12805L,
-13739L, -11146L)), .Names = c(value0, value60, value120,
value180, value240, value300), class = data.frame,
row.names = c(1,
2, 3, 4, 5, 6, 7, 8, 9, 10))
my.finder(yourdata) #and what you want is in a nicely labeled data
frame
#A potential problem is that it is not very efficient
#Here is a test using a matrix of 100,000 rows
#sampled from the same range as your data
#with the same number of columns
data.test - matrix(
sample(seq(min(yourdata),max(yourdata)), size = 50, replace =
TRUE),
ncol = 5)
system.time(my.finder(data.test))
#On my system I get
system.time(my.finder(data.test))
user system elapsed
2.890.002.89
Hope that helps,
Josh
On Fri, Jul 23, 2010 at 6:20 PM, mpw...@illinois.edu wrote:
I have a data frame with a couple million lines and want to
retrieve the largest and second largest values in each row, along
with the label of the column these values are in. For example
row 1
strongest=-11072
secondstrongest=-11707
strongestantenna=value120
secondstrongantenna=value60
Below is the code I am using and a truncated data.frame.
Retrieving the largest value was easy, but
Re: [R] Trouble retrieving the second largest value from each row of a data.frame
On Sat, Jul 24, 2010 at 5:09 PM, David Winsemius dwinsem...@comcast.net wrote:
On Jul 24, 2010, at 4:54 PM, mpw...@illinois.edu wrote:
THANKS, but I have one issue and one question.
For some reason the secondstrongest value for row 3 and 6 are incorrect
(they are the strongest) the remaining 10 are correct??
In my run of Wiley's code I instead get identical values for rows 2,5,6.
Yes, my apologies; I neglected a [-strongest] when extracting the
second highest value. I included a corrected form below; however,
Winsemius' code is cleaner, not to mention easier to generalize, so I
see no reason not to use that option. You might consider using a
different object name than 'diff' since it is also the name of a
function.
Josh
###
my.finder - function(mydata) {
my.fun - function(data) {
strongest - which.max(data)
secondstrongest - which.max(data[-strongest])
strongestantenna - names(data)[strongest]
secondstrongantenna - names(data[-strongest])[secondstrongest]
value - matrix(c(data[strongest], data[-strongest][secondstrongest],
strongestantenna, secondstrongantenna), ncol =4)
return(value)
}
dat - apply(mydata, 1, my.fun)
dat - t(dat)
dat - as.data.frame(dat, stringsAsFactors = FALSE)
colnames(dat) - c(strongest, secondstrongest,
strongestantenna, secondstrongantenna)
dat[ , strongest] - as.numeric(dat[ , strongest])
dat[ , secondstrongest] - as.numeric(dat[ , secondstrongest])
return(dat)
}
Holtman's and my solutions did not suffer from that defect, although mine
suffered from my misreading of your request, thinking that you wanted the
top 3. The fix is trivial
These data are being used to track radio-tagged birds, they are from
automated radio telemetry receivers. I will applying the following formula
diff - ((strongest- secondstrongest)/100)
bearingdiff -30-(-0.0624*(diff**2))-(2.8346*diff)
vals - c(value0, value60, value120, value180, value240,
value300)
value.str2 - (match(yourdata$secondstrongestantenna, vals)-1)*60
value.str1 - (match(yourdata$strongestantenna, vals)-1)*60
change.ind - abs(match(yourdata, vals) - which(match(yourdata, vals) )
A) Then the bearing diff is added to strongestantenna (value0 = 0degrees)
if the secondstrongestatenna is greater (eg value0 and value60),
B) or if the secondstrongestantenna is smaller than the strongestantenna,
then the bearingdiff is substracted from the strongestantenna.
C) The only exception is that if value0 (0degrees) is strongest and
value300(360degrees) is the secondstrongestantenna then the bearing is
360-bearingdiff.
D) Also the strongestantenna and secondstrongestantenna have to be next to
each other (e.g. value0 with value60, value240 with value300, value0 with
value300) or the results should be NA.
After setting finalbearing with A, B, and C then:
yourdata$finalbearing - with(yourdata, ifelse (
change.ind 5 change.ind 1 ,
NA, finalbearing) )
I have been trying to use a series of if,else statements to produce these
bearing, but all I am producing is errors. Any suggestion would be
appreciated.
Again THANKS for you efforts.
Mike
Original message
Date: Fri, 23 Jul 2010 23:01:56 -0700
From: Joshua Wiley jwiley.ps...@gmail.com
Subject: Re: [R] Trouble retrieving the second largest value from each
row of a data.frame
To: mpw...@illinois.edu
Cc: r-help@r-project.org
Hi,
Here is a little function that will do what you want and return a nice
output:
#Function To calculate top two values and return
my.finder - function(mydata) {
my.fun - function(data) {
strongest - which.max(data)
secondstrongest - which.max(data[-strongest])
strongestantenna - names(data)[strongest]
secondstrongantenna - names(data[-strongest])[secondstrongest]
value - matrix(c(data[strongest], data[secondstrongest],
strongestantenna, secondstrongantenna), ncol =4)
return(value)
}
dat - apply(mydata, 1, my.fun)
dat - t(dat)
dat - as.data.frame(dat, stringsAsFactors = FALSE)
colnames(dat) - c(strongest, secondstrongest,
strongestantenna, secondstrongantenna)
dat[ , strongest] - as.numeric(dat[ , strongest])
dat[ , secondstrongest] - as.numeric(dat[ , secondstrongest])
return(dat)
}
#Using your example data:
yourdata - structure(list(value0 = c(-13007L, -12838L, -12880L, -12805L,
-12834L, -11068L, -12807L, -12770L, -12988L, -11779L), value60 =
c(-11707L,
-13210L, -11778L, -11653L, -13527L, -11698L, -14068L, -11665L,
-11736L, -12873L), value120 = c(-11072L, -11176L, -3L, -11071L,
-11067L, -12430L, -11092L, -11061L, -11137L, -12973L), value180 =
c(-12471L,
-11799L, -12439L, -12385L, -11638L, -12430L, -11709L, -12373L,
-12570L, -12537L), value240 = c(-12838L, -13210L, -13089L, -11561L,
-13527L, -12430L, -11607L, -11426L, -13467L, -12973L), value300 =
c(-13357L,
Re: [R] Trouble retrieving the second largest value from each row of a data.frame
On Jul 24, 2010, at 8:09 PM, David Winsemius wrote:
On Jul 24, 2010, at 4:54 PM, mpw...@illinois.edu wrote:
THANKS, but I have one issue and one question.
For some reason the secondstrongest value for row 3 and 6 are
incorrect (they are the strongest) the remaining 10 are correct??
In my run of Wiley's code I instead get identical values for rows
2,5,6. Holtman's and my solutions did not suffer from that defect,
although mine suffered from my misreading of your request, thinking
that you wanted the top 3. The fix is trivial
These data are being used to track radio-tagged birds, they are
from automated radio telemetry receivers. I will applying the
following formula
diff - ((strongest- secondstrongest)/100)
bearingdiff -30-(-0.0624*(diff**2))-(2.8346*diff)
vals - c(value0, value60, value120, value180, value240,
value300)
value.str2 - (match(yourdata$secondstrongestantenna, vals)-1)*60
value.str1 - (match(yourdata$strongestantenna, vals)-1)*60
change.ind - abs(match(yourdata, vals) - which(match(yourdata,
vals) )
OOOPs should have been
change.ind - abs(match(yourdata, vals) - match(yourdata, vals) )
A) Then the bearing diff is added to strongestantenna (value0 =
0degrees) if the secondstrongestatenna is greater (eg value0 and
value60),
B) or if the secondstrongestantenna is smaller than the
strongestantenna,
then the bearingdiff is substracted from the strongestantenna.
yourdata$finalbearing - with(yourdata, ifelse (value.str2value.str1,
bearingdiff+value.str1, value.str1-bearingdiff) )
C) The only exception is that if value0 (0degrees) is strongest and
value300(360degrees) is the secondstrongestantenna then the bearing
is 360-bearingdiff.
yourdata$finalbearing - with(yourdata, ifelse (strongestantenna ==
value0 secondstrongantenna == value300, 360- bearingdiff,
finalbearing) );
D) Also the strongestantenna and secondstrongestantenna have to be
next to each other (e.g. value0 with value60, value240 with
value300, value0 with value300) or the results should be NA.
After setting finalbearing with A, B, and C then:
yourdata$finalbearing - with(yourdata, ifelse (
change.ind 5 change.ind 1 ,
NA, finalbearing) )
yourdata
strongest secondstrongest strongestantenna secondstrongantenna
finalbearing
1 -11072 -11707 value120 value60
-11086.52
2 -11176 -11799 value120value180
-11190.76
3 -3 -11778 value120 value60
-11126.91
4 -11071 -11561 value120
value240 NA
5 -11067 -11638 value120value180
-11082.85
6 -11068 -11698 value0 value60
-11082.62
7 -11092 -11607 value120
value240 NA
8 -11061 -11426 value120
value240 NA
9 -11137 -11736 value120 value60
-11152.26
10-11146 -11779 value300 value0
-11160.56
I have been trying to use a series of if,else statements to produce
these bearing,
ifelse is the correct construct for processing vectors
--
David.
but all I am producing is errors. Any suggestion would be
appreciated.
Again THANKS for you efforts.
Mike
Original message
Date: Fri, 23 Jul 2010 23:01:56 -0700
From: Joshua Wiley jwiley.ps...@gmail.com
Subject: Re: [R] Trouble retrieving the second largest value from
each row of a data.frame
To: mpw...@illinois.edu
Cc: r-help@r-project.org
Hi,
Here is a little function that will do what you want and return a
nice output:
#Function To calculate top two values and return
my.finder - function(mydata) {
my.fun - function(data) {
strongest - which.max(data)
secondstrongest - which.max(data[-strongest])
strongestantenna - names(data)[strongest]
secondstrongantenna - names(data[-strongest])[secondstrongest]
value - matrix(c(data[strongest], data[secondstrongest],
strongestantenna, secondstrongantenna), ncol =4)
return(value)
}
dat - apply(mydata, 1, my.fun)
dat - t(dat)
dat - as.data.frame(dat, stringsAsFactors = FALSE)
colnames(dat) - c(strongest, secondstrongest,
strongestantenna, secondstrongantenna)
dat[ , strongest] - as.numeric(dat[ , strongest])
dat[ , secondstrongest] - as.numeric(dat[ , secondstrongest])
return(dat)
}
#Using your example data:
yourdata - structure(list(value0 = c(-13007L, -12838L, -12880L,
-12805L,
-12834L, -11068L, -12807L, -12770L, -12988L, -11779L), value60 =
c(-11707L,
-13210L, -11778L, -11653L, -13527L, -11698L, -14068L, -11665L,
-11736L, -12873L), value120 = c(-11072L, -11176L, -3L, -11071L,
-11067L, -12430L, -11092L, -11061L, -11137L, -12973L), value180 =
c(-12471L,
-11799L, -12439L,
Re: [R] Trouble retrieving the second largest value from each row of a data.frame
On Jul 24, 2010, at 9:27 PM, David Winsemius wrote:
On Jul 24, 2010, at 8:09 PM, David Winsemius wrote:
On Jul 24, 2010, at 4:54 PM, mpw...@illinois.edu wrote:
THANKS, but I have one issue and one question.
For some reason the secondstrongest value for row 3 and 6 are
incorrect (they are the strongest) the remaining 10 are correct??
In my run of Wiley's code I instead get identical values for rows
2,5,6. Holtman's and my solutions did not suffer from that defect,
although mine suffered from my misreading of your request, thinking
that you wanted the top 3. The fix is trivial
These data are being used to track radio-tagged birds, they are
from automated radio telemetry receivers. I will applying the
following formula
diff - ((strongest- secondstrongest)/100)
bearingdiff -30-(-0.0624*(diff**2))-(2.8346*diff)
vals - c(value0, value60, value120, value180, value240,
value300)
value.str2 - (match(yourdata$secondstrongestantenna, vals)-1)*60
Had a misspelling ... rather:
match(yourdata$secondstrongantenna, vals)
value.str1 - (match(yourdata$strongestantenna, vals)-1)*60
change.ind - abs(match(yourdata, vals) - which(match(yourdata,
vals) )
OOOPs should have been
change.ind - abs(match(yourdata, vals) - match(yourdata, vals) )
A) Then the bearing diff is added to strongestantenna (value0 =
0degrees) if the secondstrongestatenna is greater (eg value0 and
value60),
B) or if the secondstrongestantenna is smaller than the
strongestantenna,
then the bearingdiff is substracted from the strongestantenna.
yourdata$finalbearing - with(yourdata, ifelse
(value.str2value.str1, bearingdiff+value.str1, value.str1-
bearingdiff) )
C) The only exception is that if value0 (0degrees) is strongest
and value300(360degrees) is the secondstrongestantenna then the
bearing is 360-bearingdiff.
yourdata$finalbearing - with(yourdata, ifelse (strongestantenna ==
value0 secondstrongantenna == value300, 360- bearingdiff,
finalbearing) );
D) Also the strongestantenna and secondstrongestantenna have to be
next to each other (e.g. value0 with value60, value240 with
value300, value0 with value300) or the results should be NA.
After setting finalbearing with A, B, and C then:
yourdata$finalbearing - with(yourdata, ifelse (
change.ind 5 change.ind 1 ,
NA, finalbearing) )
Better result with proper creation of value.str2:
yourdata
strongest secondstrongest strongestantenna secondstrongantenna
finalbearing
1 -11072 -11707 value120 value60
105.48359
2 -11176 -11799 value120value180
134.76237
3 -3 -11778 value120 value60
106.09061
4 -11071 -11561 value120
value240 NA
5 -11067 -11638 value120value180
135.84893
6 -11068 -11698 value0 value60
14.61868
7 -11092 -11607 value120
value240 NA
8 -11061 -11426 value120
value240 NA
9 -11137 -11736 value120 value60
104.74034
10-11146 -11779 value300 value0
285.44272
I have been trying to use a series of if,else statements to
produce these bearing,
ifelse is the correct construct for processing vectors
--
David.
but all I am producing is errors. Any suggestion would be
appreciated.
Again THANKS for you efforts.
Mike
Original message
Date: Fri, 23 Jul 2010 23:01:56 -0700
From: Joshua Wiley jwiley.ps...@gmail.com
Subject: Re: [R] Trouble retrieving the second largest value from
each row of a data.frame
To: mpw...@illinois.edu
Cc: r-help@r-project.org
Hi,
Here is a little function that will do what you want and return a
nice output:
#Function To calculate top two values and return
my.finder - function(mydata) {
my.fun - function(data) {
strongest - which.max(data)
secondstrongest - which.max(data[-strongest])
strongestantenna - names(data)[strongest]
secondstrongantenna - names(data[-strongest])[secondstrongest]
value - matrix(c(data[strongest], data[secondstrongest],
strongestantenna, secondstrongantenna), ncol =4)
return(value)
}
dat - apply(mydata, 1, my.fun)
dat - t(dat)
dat - as.data.frame(dat, stringsAsFactors = FALSE)
colnames(dat) - c(strongest, secondstrongest,
strongestantenna, secondstrongantenna)
dat[ , strongest] - as.numeric(dat[ , strongest])
dat[ , secondstrongest] - as.numeric(dat[ , secondstrongest])
return(dat)
}
#Using your example data:
yourdata - structure(list(value0 = c(-13007L, -12838L, -12880L,
-12805L,
-12834L, -11068L, -12807L, -12770L, -12988L, -11779L), value60 =
c(-11707L,
-13210L, -11778L, -11653L, -13527L, -11698L, -14068L, -11665L,
[R] c-statiscs 95% CI for cox regression model
Dear all, how can i do the calculate the C-statistics 95% confidences interval for the cox regression model. Thanks very much for your any help. Paaveenthan _ [[elided Hotmail spam]] [[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.
Re: [R] UseR! 2010 - my impressions
On Sat, Jul 24, 2010 at 08:55:01AM -0500, Frank E Harrell Jr wrote: On 07/23/2010 06:50 PM, Ravi Varadhan wrote: I want to echo what Ravi said. The talks were terrific (thanks to the program committee and the speakers) and Kate Mullen and her team did an extraordinary job in putting the conference together and running it. I am proud to have been a part of it. Thank you all! Not much to add to this, so I just leave it at Yup!. Thanks for useR! 2010. A job well done, and then some. -- Dirk Eddelbuettel | e...@debian.org | http://dirk.eddelbuettel.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] c-statiscs 95% CI for cox regression model
On 07/24/2010 09:51 PM, paaventhan jeyaganth wrote:
Dear all,
how can i do the calculate the C-statistics
95% confidences interval for the cox regression model.
Thanks very much for your any help.
Paaveenthan
install.packages('Hmisc')
require(Hmisc
?rcorr.cens (there is an example at the bottom)
Frank
--
Frank E Harrell Jr Professor and ChairmanSchool of Medicine
Department of Biostatistics Vanderbilt University
__
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] Equivalent to go-to statement
Dear all,
I'm working with a code that consists of two parts: In Part 1 I'm generating
a random graph using the igraph library (which represents the relationships
between different nodes) and a vector (which represents a certain
characteristic for each node):
library(igraph)
g - watts.strogatz.game(1,100,5,0.05)
z - rlnorm(100,0,1)
In Part 2 I'm iteratively changing the elements of z in order to reach a
certain value of a certain target variable. I'm doing this using a while
statement:
while (target_variable threshold) {## adapt z}
The problem is that in some rare cases this iterative procedure can take
very long (a couple of million of iterations), depending on the specific
structure of the graph generated in Part 1. I therefore would like to change
Part 2 of my code in the sense that once a certain threshold number of
iterations has been achieved, the iterative process in Part 2 stops and goes
back to Part 1 to generate a new graph structure. So my idea is as follows:
- Run Part 1 and generate g and z
- Run Part 2 and iteratively modify z to maximize the target variable
- If Part 2 can be obtained in less than X steps, then go to Part 3
- If Part 2 takes more than X steps then go back to Part 1 and start again
I think that R does not have a function like go-to or go-back.
Does anybody know of a convenient way of doing this?
Thanks very much for your help,
Michael
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and provide commented, minimal, self-contained, reproducible code.
