[R] Trailing path separators and dirname(), file.exists(), file.path()
I've found something irritating in the behaviour of dirname() and
subsequent interactions with file.path() and file.exists(). The problem is
how it and they deal with trailing path separators. It seems the code is
correct to specification but I think it could be better. Part of the
problem is restricted to Windows and this is alluded to in the docs but the
implications are not fully explicit. Providing a path with a trailing path
separator to dirname() causes the final directory to be removed.
In Linux (R 3.3.1)
> file.path("","p1","p2","p3")
[1] "/p1/p2/p3"
> dirname(file.path("","p1","p2","p3")) # p3 is treated as a filename,
not a directory in the path - fine.
[1] "/p1/p2"
> file.path("","p1","p2","p3","/") # not the "//" by file.path"
[1] "/p1/p2/p3//"
> dirname(file.path("","p1","p2","p3","/"))# p3 is now treated as a
directory - fine.
[1] "/p1/p2/p3"
> dirname("/p1/p2/p3/") # but when I give a path with no file at the end.
What I consider a directory, is chopped off as a final filename.
[1] "/p1/p2"
In Windows (R 3.3.1)
> file.path("","p1","p2","p3") # as per linux
[1] "/p1/p2/p3"
> dirname(file.path("","p1","p2","p3")) # as per linux
[1] "/p1/p2"
> file.path("","p1","p2","p3","/") # N.B. no second "/"
[1] "/p1/p2/p3/"
> dirname(file.path("","p1","p2","p3","/")) # last 'directory' lost
[1] "/p1/p2"
> dirname("/p1/p2/p3/") # last 'directory' lost
[1] "/p1/p2"
This is a problem for me when testing for the existence of directories
(e.g. have I already made an output directory for this analysis)
In Windows (R 3.3.1)
> dir.create("temp")
> file.exists("temp")
[1] TRUE
> file.exists("temp/") # R on windows not recognising and removing the
trailing path separator.
[1] FALSE
> file.exists(dirname("temp/"))
[1] TRUE
> dirname("temp/")
[1] "."
On Linux, I can use a trailing separator, but I still shouldn't use
dirname() :-
> dir.create("temp")
> file.exists("temp")
[1] TRUE
> file.exists("temp/")
[1] TRUE
> file.exists(dirname("temp/")) # false positive
[1] TRUE
> dirname("temp/")# what it is really testing for.
[1] "."
Does anyone else feel this is not as good as it should be?
I don't know how old this behavious is. I've only tested R 3.3.1 but the
docs for file.path() suggest there may have been some recent tinkering:-
"Trailing path separators are invalid for Windows file paths apart from ‘/’
and ‘d:/’ (although some functions/utilities do accept them), so as from R
3.1.0 a trailing / or \ is removed."
Dave Gerrard
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[R] Variable 'A' is not a factor Error message
I am running a DOE with the following code library(Rcmdr)
library(RcmdrMisc)
library(RcmdrPlugin.DoE)
# Define plackett burman experiment
PB.DOE <- pb(nruns= 12 ,n12.taguchi= FALSE ,nfactors= 12 -1, ncenter= 0 ,
replications= 1 ,repeat.only= FALSE ,randomize= TRUE ,seed=
27241 ,
factor.names=list(
A=c(100,1000),B=c(100,200),C=c(1,3),D=c(1,1.7),
E=c(1000,1500),G=c(-2,2) ) )
as.numeric2 <- function(x) as.numeric(as.character(x))
# Calculate response column
IP <-
with(PB.DOE,(as.numeric2(A)*as.numeric2(B)*(5000-3000))/(141.2*as.numeric2(C)*as.numeric2(D)*(log(as.numeric2(E)/0.25)-(1/2)+as.numeric2(G
# Combine response column with exp design table
final_set <- within(PB.DOE, {
IP<-
((as.numeric2(A)*as.numeric2(B)*(5000-3000))/(141.2*as.numeric2(C)*as.numeric2(D)*(log(as.numeric2(E)/0.25)-(1/2)+as.numeric2(G
})I then ran a regression as follows:LinearModel.1 <- lm(IP ~ A + B + C + D
+ E + G,
data=final_set)
summary(LinearModel.1)Following this i wanted to run a predict using specified
values as predictors in a Monte Carlo:n = 1
# Define probability distributions of predictors
A = rnorm(n,450,100)
hist(A,col = "blue",breaks = 50)
B = rnorm(n, 150,10)
hist(B,col = "blue",breaks = 50)
C = rnorm(n, 1.5, 0.5)
hist(C,col = "blue",breaks = 50)
D = runif(n,1.2,1.7)
hist(D,col = "blue",breaks = 50)
E = rnorm(n,1250,50)
hist(E,col = "blue",breaks = 50)
G = rnorm(n,0,0.5)
hist(G,col = "blue",breaks = 50)
MCtable <- data.frame(A=A,B=B,C=C,D=D,E=E,G=G)
for (n in 1:n) {
N=predict(LinearModel.1,MCtable)
}
hist(N,col = "yellow",breaks = 10)I end up getting this error:"Warning in
model.frame.default(Terms, newdata, na.action = na.action, xlev =
object$xlevels) :
variable 'A' is not a factor"Using str() to get some info on the
LinearModel.1 and from what I understand seems to indicates that since the
predictors A,B,C etc are factors with 2 levels I have to convert my data.frame
table to factors aswell. Is that correct?Doing this would mean I would also
need to specify the number of levels which would mean that since I have set my
n to 1 would mean 1 levels for each factor. How would I go about doing
this? Is there a better solution? Any help would be appreciated.
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Re: [R] Average every 4 columns
Hi Miluj,
Perhaps you didn't get my previous email. Let your data frame be named "msdf":
block_col_summ<-function(x,step,block_size,FUN="mean") {
dimx<-dim(x)
return_value<-NA
start<-1
end<-start+block_size-1
block_count<-1
while(end <= dimx[2]) {
return_value[block_count]<-
do.call(FUN,list(as.matrix(x[,start:end]),na.rm=TRUE))
block_count<-block_count+1
start<-start+step
end<-start+block_size-1
}
return(return_value)
}
# use the step and block size that you want
block_col_summ(msdf,4,4)
Jim
On Thu, Nov 10, 2016 at 3:29 AM, Miluji Sb wrote:
> Dear all,
>
> I have a dataset with hundreds of columns, I am only providing only 12
> columns. Is it possible to take the mean of every four (or 12) columns
> (value601, value602, value603, value604 etc.in this case) and repeat for
> the hundreds of columns? Thank you.
>
> Sincerely,
>
> Milu
>
> structure(list(value601 = c(10.1738710403442, 3.54112911224365,
> 12.9192342758179, 3.17447590827942, 11.7332258224487, 7.68282270431519,
> -7.11564493179321, 0.987620949745178, 13.0476207733154, 6.36939525604248
> ), value602 = c(13.0642414093018, 5.53129482269287, 16.0519638061523,
> 2.88946437835693, 14.9204912185669, 9.42428588867188, -6.80674123764038,
> -0.614241063594818, 16.7947769165039, 7.9541072845459), value603 =
> c(22.0399188995361,
> 14.398024559021, 24.9523792266846, 12.0878629684448, 23.6459674835205,
> 18.3277816772461, -2.54092741012573, 10.5550804138184, 25.1016540527344,
> 16.2166938781738), value604 = c(27.7165412902832, 20.3255825042725,
> 30.8430004119873, 16.6856250762939, 29.2485408782959, 24.3775005340576,
> 6.47758340835571, 15.5897912979126, 30.7387924194336, 22.3637084960938
> ), value605 = c(31.6644763946533, 23.4093952178955, 35.1488723754883,
> 19.7132263183594, 33.3924179077148, 29.5846366882324, 10.2083873748779,
> 19.3551616668701, 35.3076629638672, 27.4299201965332), value606 =
> c(33.9698333740234,
> 26.8574161529541, 36.8900833129883, 22.8604583740234, 34.8642921447754,
> 33.8158760070801, 14.7055835723877, 22.1144580841064, 37.0545425415039,
> 32.1087913513184), value607 = c(36.0279846191406, 26.9297180175781,
> 38.2701225280762, 23.2643146514893, 36.7398796081543, 34.1216125488281,
> 17.1387901306152, 24.0419750213623, 37.8542327880859, 32.7677421569824
> ), value608 = c(34.0242347717285, 25.7720966339111, 36.4897193908691,
> 22.0332260131836, 34.8011703491211, 32.6856842041016, 16.6232261657715,
> 21.5571365356445, 36.1491546630859, 31.1716938018799), value609 =
> c(27.5402088165283,
> 21.7590408325195, 30.5214176177979, 18.4252090454102, 29.1156253814697,
> 26.9878330230713, 12.4962501525879, 17.7259578704834, 30.9099159240723,
> 25.4832077026367), value610 = c(23.4706859588623, 17.0126209259033,
> 26.8166942596436, 15.297459602356, 25.1733055114746, 23.5616931915283,
> 8.86995983123779, 13.5793552398682, 27.5732250213623, 22.1691932678223
> ), value611 = c(14.5820417404175, 9.08279132843018, 17.8419170379639,
> 8.36016654968262, 16.5633754730225, 14.8123331069946, 1.32095837593079,
> 5.73408317565918, 18.9752082824707, 13.572542236), value612 =
> c(9.12979793548584,
> 2.79943537712097, 11.6504030227661, 2.21584677696228, 10.5404834747314,
> 7.55471754074097, -5.58141136169434, -0.566209673881531, 12.3264112472534,
> 6.65576601028442)), .Names = c("value601", "value602", "value603",
> "value604", "value605", "value606", "value607", "value608", "value609",
> "value610", "value611", "value612"), row.names = c("1", "2",
> "3", "4", "5", "6", "7", "8", "9", "10"), class = "data.frame")
>
> [[alternative HTML version deleted]]
>
> __
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> and provide commented, minimal, self-contained, reproducible code.
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[R] replace items in vector with character based on logical operator
Hi group,
I have a vector of numeric items and I want to replace based on
logical operator (> or <) with 'up', 'down' or 'nochange'
The way I am doing works sometimes and does not work sometime. I dont
understand the problem. In this example, that works for condition DN
first, gets over-written by UP and then by NC.
I appreciate your help.
Thanks
Adrian
In the example below (dput code given below).
qt = quantile(kx)
kx[kx <= qt[[2]]] <- 'DN'
kx[kx >=qt [[4]]] <- 'UP'
kx[kx < qt[[4]] | kx > qt[[2]]] <- 'NC'
> qt
0% 25% 50% 75% 100%
9.341531 9.995026 10.186305 10.444237 11.013304
> kx[kx <= qt[[2]]] <- 'DN'
> kx
TCGA-3H-AB3K TCGA-3H-AB3L TCGA-3H-AB3M
TCGA-3H-AB3O TCGA-3H-AB3S TCGA-3H-AB3T TCGA-3H-AB3U
TCGA-3H-AB3X
"11.0133035117116" "DN" "10.2976687932597"
"10.080008468093" "10.3661039885332" "DN"
"10.8971723299346" "10.0327245097586"
TCGA-3U-A98D TCGA-3U-A98E TCGA-3U-A98F
TCGA-3U-A98G TCGA-3U-A98H TCGA-3U-A98I TCGA-3U-A98J
TCGA-LK-A4NW
"10.0094206497379" "10.4816534666287" "10.4317651665439"
"DN" "10.2313364037012" "DN" "10.1412738417844"
"DN"
TCGA-LK-A4NY TCGA-LK-A4NZ TCGA-LK-A4O0 TCGA-LK-A4O2
"10.603941260297" "10.7959493941044" "10.0501324788234" "10.3877479259697"
kx <- structure(c(11.0133035117116, 9.95184228858716, 10.2976687932597,
10.080008468093, 10.3661039885332, 9.34153148077152, 10.8971723299346,
10.0327245097586, 10.0094206497379, 10.4816534666287, 10.4317651665439,
9.57711797674944, 10.2313364037012, 9.39562974544228, 10.1412738417844,
9.80706011783492, 10.603941260297, 10.7959493941044, 10.0501324788234,
10.3877479259697), .Names = c("TCGA-3H-AB3K", "TCGA-3H-AB3L",
"TCGA-3H-AB3M", "TCGA-3H-AB3O", "TCGA-3H-AB3S", "TCGA-3H-AB3T",
"TCGA-3H-AB3U", "TCGA-3H-AB3X", "TCGA-3U-A98D", "TCGA-3U-A98E",
"TCGA-3U-A98F", "TCGA-3U-A98G", "TCGA-3U-A98H", "TCGA-3U-A98I",
"TCGA-3U-A98J", "TCGA-LK-A4NW", "TCGA-LK-A4NY", "TCGA-LK-A4NZ",
"TCGA-LK-A4O0", "TCGA-LK-A4O2"))
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Re: [R] replace items in vector with character based on logical operator
Your goal is fundamentally flawed, since all elements of a vector must be if
the same type.
Create another vector to hold your character information.
kc <- rep( 'NC', length( kx ) )
kc[ kx <= qt[[2]] ] <- 'DN'
kc[ kx >=qt [[4]] ] <- 'UP'
--
Sent from my phone. Please excuse my brevity.
On November 12, 2016 2:25:51 PM PST, Adrian Johnson
wrote:
>Hi group,
>
>I have a vector of numeric items and I want to replace based on
>logical operator (> or <) with 'up', 'down' or 'nochange'
>
>The way I am doing works sometimes and does not work sometime. I dont
>understand the problem. In this example, that works for condition DN
>first, gets over-written by UP and then by NC.
>
>
> I appreciate your help.
>Thanks
>Adrian
>
>In the example below (dput code given below).
>
>qt = quantile(kx)
>kx[kx <= qt[[2]]] <- 'DN'
>kx[kx >=qt [[4]]] <- 'UP'
>kx[kx < qt[[4]] | kx > qt[[2]]] <- 'NC'
>
>
>> qt
> 0% 25% 50% 75% 100%
> 9.341531 9.995026 10.186305 10.444237 11.013304
>
>> kx[kx <= qt[[2]]] <- 'DN'
>> kx
> TCGA-3H-AB3K TCGA-3H-AB3L TCGA-3H-AB3M
>TCGA-3H-AB3O TCGA-3H-AB3S TCGA-3H-AB3T TCGA-3H-AB3U
> TCGA-3H-AB3X
>"11.0133035117116" "DN" "10.2976687932597"
>"10.080008468093" "10.3661039885332" "DN"
>"10.8971723299346" "10.0327245097586"
> TCGA-3U-A98D TCGA-3U-A98E TCGA-3U-A98F
>TCGA-3U-A98G TCGA-3U-A98H TCGA-3U-A98I TCGA-3U-A98J
> TCGA-LK-A4NW
>"10.0094206497379" "10.4816534666287" "10.4317651665439"
>"DN" "10.2313364037012" "DN" "10.1412738417844"
> "DN"
> TCGA-LK-A4NY TCGA-LK-A4NZ TCGA-LK-A4O0 TCGA-LK-A4O2
>"10.603941260297" "10.7959493941044" "10.0501324788234"
>"10.3877479259697"
>
>
>
>
>
>
>
>kx <- structure(c(11.0133035117116, 9.95184228858716, 10.2976687932597,
>10.080008468093, 10.3661039885332, 9.34153148077152, 10.8971723299346,
>10.0327245097586, 10.0094206497379, 10.4816534666287, 10.4317651665439,
>9.57711797674944, 10.2313364037012, 9.39562974544228, 10.1412738417844,
>9.80706011783492, 10.603941260297, 10.7959493941044, 10.0501324788234,
>10.3877479259697), .Names = c("TCGA-3H-AB3K", "TCGA-3H-AB3L",
>"TCGA-3H-AB3M", "TCGA-3H-AB3O", "TCGA-3H-AB3S", "TCGA-3H-AB3T",
>"TCGA-3H-AB3U", "TCGA-3H-AB3X", "TCGA-3U-A98D", "TCGA-3U-A98E",
>"TCGA-3U-A98F", "TCGA-3U-A98G", "TCGA-3U-A98H", "TCGA-3U-A98I",
>"TCGA-3U-A98J", "TCGA-LK-A4NW", "TCGA-LK-A4NY", "TCGA-LK-A4NZ",
>"TCGA-LK-A4O0", "TCGA-LK-A4O2"))
>
>__
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>https://stat.ethz.ch/mailman/listinfo/r-help
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>http://www.R-project.org/posting-guide.html
>and provide commented, minimal, self-contained, reproducible code.
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Re: [R] replace items in vector with character based on logical operator
Your first replacement in kx changed it to a character vector and
comparisons
of character strings given different results than comparisons of numbers
> kx <- 1:10
> qt <- quantile(kx)
> kx[kx <= qt[2]] <- "DN"
> kx
[1] "DN" "DN" "DN" "4" "5" "6" "7" "8" "9" "10"
> "10" < "6"
[1] TRUE
>
> # compare to
> kx <- 1:10
> code <- character(length(kx))
> qt <- quantile(kx)
> code[kx <= qt[2]] <- "DN"
> code
[1] "DN" "DN" "DN" "" "" "" "" "" "" ""
> code[kx >= qt[4]] <- "DN"
> code
[1] "DN" "DN" "DN" "" "" "" "" "DN" "DN" "DN"
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Sat, Nov 12, 2016 at 2:25 PM, Adrian Johnson
wrote:
> Hi group,
>
> I have a vector of numeric items and I want to replace based on
> logical operator (> or <) with 'up', 'down' or 'nochange'
>
> The way I am doing works sometimes and does not work sometime. I dont
> understand the problem. In this example, that works for condition DN
> first, gets over-written by UP and then by NC.
>
>
> I appreciate your help.
> Thanks
> Adrian
>
> In the example below (dput code given below).
>
> qt = quantile(kx)
> kx[kx <= qt[[2]]] <- 'DN'
> kx[kx >=qt [[4]]] <- 'UP'
> kx[kx < qt[[4]] | kx > qt[[2]]] <- 'NC'
>
>
> > qt
>0% 25% 50% 75% 100%
> 9.341531 9.995026 10.186305 10.444237 11.013304
>
> > kx[kx <= qt[[2]]] <- 'DN'
> > kx
> TCGA-3H-AB3K TCGA-3H-AB3L TCGA-3H-AB3M
> TCGA-3H-AB3O TCGA-3H-AB3S TCGA-3H-AB3T TCGA-3H-AB3U
> TCGA-3H-AB3X
> "11.0133035117116" "DN" "10.2976687932597"
> "10.080008468093" "10.3661039885332" "DN"
> "10.8971723299346" "10.0327245097586"
> TCGA-3U-A98D TCGA-3U-A98E TCGA-3U-A98F
> TCGA-3U-A98G TCGA-3U-A98H TCGA-3U-A98I TCGA-3U-A98J
> TCGA-LK-A4NW
> "10.0094206497379" "10.4816534666287" "10.4317651665439"
> "DN" "10.2313364037012" "DN" "10.1412738417844"
> "DN"
> TCGA-LK-A4NY TCGA-LK-A4NZ TCGA-LK-A4O0 TCGA-LK-A4O2
> "10.603941260297" "10.7959493941044" "10.0501324788234" "10.3877479259697"
>
>
>
>
>
>
>
> kx <- structure(c(11.0133035117116, 9.95184228858716, 10.2976687932597,
> 10.080008468093, 10.3661039885332, 9.34153148077152, 10.8971723299346,
> 10.0327245097586, 10.0094206497379, 10.4816534666287, 10.4317651665439,
> 9.57711797674944, 10.2313364037012, 9.39562974544228, 10.1412738417844,
> 9.80706011783492, 10.603941260297, 10.7959493941044, 10.0501324788234,
> 10.3877479259697), .Names = c("TCGA-3H-AB3K", "TCGA-3H-AB3L",
> "TCGA-3H-AB3M", "TCGA-3H-AB3O", "TCGA-3H-AB3S", "TCGA-3H-AB3T",
> "TCGA-3H-AB3U", "TCGA-3H-AB3X", "TCGA-3U-A98D", "TCGA-3U-A98E",
> "TCGA-3U-A98F", "TCGA-3U-A98G", "TCGA-3U-A98H", "TCGA-3U-A98I",
> "TCGA-3U-A98J", "TCGA-LK-A4NW", "TCGA-LK-A4NY", "TCGA-LK-A4NZ",
> "TCGA-LK-A4O0", "TCGA-LK-A4O2"))
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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.
>
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[R] Error in Summary.factor(c(24L, 28L, 29L, 30L, 31L, 32L, 33L, 34L, 35L, :
Dear all; I am getting the following error message when I run maSigPro package by using make.design.matrix commend. I could not figure out why this error happens. Thanks for your help Greg Error in Summary.factor(c(24L, 28L, 29L, 30L, 31L, 32L, 33L, 34L, 35L, : ‘min’ not meaningful for factors [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Error in Summary.factor(c(24L, 28L, 29L, 30L, 31L, 32L, 33L, 34L, 35L, :
On 12/11/2016 7:33 PM, greg holly wrote: Dear all; I am getting the following error message when I run maSigPro package by using make.design.matrix commend. I could not figure out why this error happens. Thanks for your help Greg Error in Summary.factor(c(24L, 28L, 29L, 30L, 31L, 32L, 33L, 34L, 35L, : ‘min’ not meaningful for factors You'll need to give a minimal reproducible example (i.e. something others can run, but not containing a lot of unnecessary stuff) if you want help with this. Duncan Murdoch __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Error in Summary.factor(c(24L, 28L, 29L, 30L, 31L, 32L, 33L, 34L, 35L, :
Hi Duncan; Thanks for this. I used followed command. In command line: blood is the name of the data file, degree=11 because I have 12-time point. I need to create the design matrix of dummies (named des_blood) for fitting time series micorarray gene expression experiments using maSigPro program. des_blood=make.design.matrix(blood, degree=11) On Sat, Nov 12, 2016 at 7:54 PM, Duncan Murdoch wrote: > On 12/11/2016 7:33 PM, greg holly wrote: > >> Dear all; >> >> I am getting the following error message when I run maSigPro package by >> using >> >> make.design.matrix commend. I could not figure out why this error happens. >> Thanks for your help >> >> Greg >> >> Error in Summary.factor(c(24L, 28L, 29L, 30L, 31L, 32L, 33L, 34L, 35L, : >> ‘min’ not meaningful for factors >> >> > You'll need to give a minimal reproducible example (i.e. something others > can run, but not containing a lot of unnecessary stuff) if you want help > with this. > > Duncan Murdoch > > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] value matching %in% for a number pair
Hi,
We can match one numerical value as follows
> 3 %in% c(4,5)
[1] FALSE
> 3 %in% c(4,5,3)
[1] TRUE
To see whether value pairs are identical,
> identical(c(3,4), c(3,5))
[1] FALSE
> identical(c(3,4), c(3,4))
[1] TRUE
Is there any way to test whether “A value pair is in a set of value
pairs”? For example, can we test whether the pair (2,3) is identical to one
of the pairs in the set S={(1,2), (4,3), (3,3), (2,3), (4,5)}?
In this case, the answer is yes because the 4th element of S is (2,3).
Is there any simple way to code it? Thanks!
John
[[alternative HTML version deleted]]
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Re: [R] value matching %in% for a number pair
On 12/11/2016 8:36 PM, John wrote:
Hi,
We can match one numerical value as follows
3 %in% c(4,5)
[1] FALSE
3 %in% c(4,5,3)
[1] TRUE
To see whether value pairs are identical,
identical(c(3,4), c(3,5))
[1] FALSE
identical(c(3,4), c(3,4))
[1] TRUE
Is there any way to test whether “A value pair is in a set of value
pairs”? For example, can we test whether the pair (2,3) is identical to one
of the pairs in the set S={(1,2), (4,3), (3,3), (2,3), (4,5)}?
In this case, the answer is yes because the 4th element of S is (2,3).
Is there any simple way to code it? Thanks!
You'll have to type a long expression or write your own function for it.
Here's one way, if you store your set as a list:
inlist <- function(x, thelist)
any(sapply(thelist, identical, x))
For example:
> S <- list(c(1, 2), c(4, 3))
> inlist(c(4, 3), S)
[1] TRUE
> inlist(c(3, 4), S)
[1] FALSE
Duncan Murdoch
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Re: [R] Error in Summary.factor(c(24L, 28L, 29L, 30L, 31L, 32L, 33L, 34L, 35L, :
On 12/11/2016 8:14 PM, greg holly wrote: Hi Duncan; Thanks for this. I used followed command. In command line: blood is the name of the data file, degree=11 because I have 12-time point. I need to create the design matrix of dummies (named des_blood) for fitting time series micorarray gene expression experiments using maSigPro program. Who but you can run this? We don't have "blood". Duncan Murdoch des_blood=make.design.matrix(blood, degree=11) On Sat, Nov 12, 2016 at 7:54 PM, Duncan Murdoch mailto:murdoch.dun...@gmail.com>> wrote: On 12/11/2016 7:33 PM, greg holly wrote: Dear all; I am getting the following error message when I run maSigPro package by using make.design.matrix commend. I could not figure out why this error happens. Thanks for your help Greg Error in Summary.factor(c(24L, 28L, 29L, 30L, 31L, 32L, 33L, 34L, 35L, : ‘min’ not meaningful for factors You'll need to give a minimal reproducible example (i.e. something others can run, but not containing a lot of unnecessary stuff) if you want help with this. Duncan Murdoch __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] value matching %in% for a number pair
This really depends on the way you represent these pairs. Here is some food for
thought:
p <- matrix( c( 1,2, 4,3, 3,3, 2,3, 4,5 ), byrow=TRUE, ncol=2 )
( 2 %in% p[,1] ) & ( 3 %in% p[,2] ) # (2,3)
( c( 2, 1, 5 ) %in% p[,1] ) & ( c( 3, 2, 5 ) %in% p[,2] ) # (2,3) (1,2) (5,5)
--
Sent from my phone. Please excuse my brevity.
On November 12, 2016 5:36:37 PM PST, John wrote:
>Hi,
>
> We can match one numerical value as follows
>> 3 %in% c(4,5)
>[1] FALSE
>> 3 %in% c(4,5,3)
>[1] TRUE
>
> To see whether value pairs are identical,
>> identical(c(3,4), c(3,5))
>[1] FALSE
>> identical(c(3,4), c(3,4))
>[1] TRUE
>
> Is there any way to test whether “A value pair is in a set of value
>pairs”? For example, can we test whether the pair (2,3) is identical to
>one
>of the pairs in the set S={(1,2), (4,3), (3,3), (2,3), (4,5)}?
> In this case, the answer is yes because the 4th element of S is (2,3).
>Is there any simple way to code it? Thanks!
>
>John
>
> [[alternative HTML version deleted]]
>
>__
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>http://www.R-project.org/posting-guide.html
>and provide commented, minimal, self-contained, reproducible code.
__
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Re: [R] value matching %in% for a number pair
> S <- list(c(1,2), c(4,3), c(3,3), c(2,3), c(4,5))
> list(c(1,2), c(3,4), c(2,3)) %in% S # is in S?
[1] TRUE FALSE TRUE
> match(list(c(1,2), c(3,4), c(2,3)), S) # which element of S?
[1] 1 NA 4
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Sat, Nov 12, 2016 at 5:36 PM, John wrote:
> Hi,
>
>We can match one numerical value as follows
> > 3 %in% c(4,5)
> [1] FALSE
> > 3 %in% c(4,5,3)
> [1] TRUE
>
>To see whether value pairs are identical,
> > identical(c(3,4), c(3,5))
> [1] FALSE
> > identical(c(3,4), c(3,4))
> [1] TRUE
>
>Is there any way to test whether “A value pair is in a set of value
> pairs”? For example, can we test whether the pair (2,3) is identical to one
> of the pairs in the set S={(1,2), (4,3), (3,3), (2,3), (4,5)}?
>In this case, the answer is yes because the 4th element of S is (2,3).
> Is there any simple way to code it? Thanks!
>
> John
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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]]
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Re: [R] value matching %in% for a number pair
Sorry, that was a fail. Better to think about:
any( 2 == p[,1] & 3 == p[,2] )
v <- matrix( c( 2,3, 1,2, 5,5 ), byrow=TRUE, ncol=2 )
apply( v, 1, function(x) { any( x[1]==p[,1] & x[2]==p[,2] ) } )
--
Sent from my phone. Please excuse my brevity.
On November 12, 2016 6:11:13 PM PST, Jeff Newmiller
wrote:
>This really depends on the way you represent these pairs. Here is some
>food for thought:
>
>p <- matrix( c( 1,2, 4,3, 3,3, 2,3, 4,5 ), byrow=TRUE, ncol=2 )
>( 2 %in% p[,1] ) & ( 3 %in% p[,2] ) # (2,3)
>( c( 2, 1, 5 ) %in% p[,1] ) & ( c( 3, 2, 5 ) %in% p[,2] ) # (2,3) (1,2)
>(5,5)
>
>--
>Sent from my phone. Please excuse my brevity.
>
>On November 12, 2016 5:36:37 PM PST, John wrote:
>>Hi,
>>
>> We can match one numerical value as follows
>>> 3 %in% c(4,5)
>>[1] FALSE
>>> 3 %in% c(4,5,3)
>>[1] TRUE
>>
>> To see whether value pairs are identical,
>>> identical(c(3,4), c(3,5))
>>[1] FALSE
>>> identical(c(3,4), c(3,4))
>>[1] TRUE
>>
>> Is there any way to test whether “A value pair is in a set of value
>>pairs”? For example, can we test whether the pair (2,3) is identical
>to
>>one
>>of the pairs in the set S={(1,2), (4,3), (3,3), (2,3), (4,5)}?
>> In this case, the answer is yes because the 4th element of S is
>(2,3).
>>Is there any simple way to code it? Thanks!
>>
>>John
>>
>> [[alternative HTML version deleted]]
>>
>>__
>>R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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>>http://www.R-project.org/posting-guide.html
>>and provide commented, minimal, self-contained, reproducible code.
>
>__
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Re: [R] value matching %in% for a number pair
Hi john,
I don't know whether this breaks any rules, but:
target_pair<-c(3,4)
pair_list<-list(c(1,2),c(3,4),c(5,6))
sapply(pair_list,identical,target_pair)
[1] FALSE TRUE FALSE
Jim
On Sun, Nov 13, 2016 at 1:32 PM, Jeff Newmiller
wrote:
> Sorry, that was a fail. Better to think about:
>
> any( 2 == p[,1] & 3 == p[,2] )
> v <- matrix( c( 2,3, 1,2, 5,5 ), byrow=TRUE, ncol=2 )
> apply( v, 1, function(x) { any( x[1]==p[,1] & x[2]==p[,2] ) } )
>
> --
> Sent from my phone. Please excuse my brevity.
>
> On November 12, 2016 6:11:13 PM PST, Jeff Newmiller
> wrote:
>>This really depends on the way you represent these pairs. Here is some
>>food for thought:
>>
>>p <- matrix( c( 1,2, 4,3, 3,3, 2,3, 4,5 ), byrow=TRUE, ncol=2 )
>>( 2 %in% p[,1] ) & ( 3 %in% p[,2] ) # (2,3)
>>( c( 2, 1, 5 ) %in% p[,1] ) & ( c( 3, 2, 5 ) %in% p[,2] ) # (2,3) (1,2)
>>(5,5)
>>
>>--
>>Sent from my phone. Please excuse my brevity.
>>
>>On November 12, 2016 5:36:37 PM PST, John wrote:
>>>Hi,
>>>
>>> We can match one numerical value as follows
3 %in% c(4,5)
>>>[1] FALSE
3 %in% c(4,5,3)
>>>[1] TRUE
>>>
>>> To see whether value pairs are identical,
identical(c(3,4), c(3,5))
>>>[1] FALSE
identical(c(3,4), c(3,4))
>>>[1] TRUE
>>>
>>> Is there any way to test whether “A value pair is in a set of value
>>>pairs”? For example, can we test whether the pair (2,3) is identical
>>to
>>>one
>>>of the pairs in the set S={(1,2), (4,3), (3,3), (2,3), (4,5)}?
>>> In this case, the answer is yes because the 4th element of S is
>>(2,3).
>>>Is there any simple way to code it? Thanks!
>>>
>>>John
>>>
>>> [[alternative HTML version deleted]]
>>>
>>>__
>>>R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>>https://stat.ethz.ch/mailman/listinfo/r-help
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>>>http://www.R-project.org/posting-guide.html
>>>and provide commented, minimal, self-contained, reproducible code.
>>
>>__
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>>and provide commented, minimal, self-contained, reproducible code.
>
> __
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Re: [R] Is this foreach behaviour correct?
I'm still not clear about whether this is a bug in foreach. Should c.Date be
invoked by foreach with .combine='c'?
On 11/06/2016 07:02 PM, William Dunlap wrote:
Note that in the OP's example c.Date is never invoked. c.Date is called if
.combine
calls c rather than if .combine is c:
> library(zoo)
> trace(c.Date, quote(print(sys.call(
Tracing function "c.Date" in package "base"
[1] "c.Date"
> foreach(i=1:10003, .combine=c) %do% { as.Date(i) }
[1] 1 10001 10002 10003
> foreach(i=1:10003, .combine=function(...)c(...)) %do% { as.Date(i) }
Tracing c.Date(...) on entry
eval(expr, envir, enclos)
Tracing c.Date(...) on entry
eval(expr, envir, enclos)
Tracing c.Date(...) on entry
eval(expr, envir, enclos)
[1] "1997-05-19" "1997-05-20" "1997-05-21" "1997-05-22"
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Sun, Nov 6, 2016 at 2:20 PM, Duncan Murdoch
mailto:murdoch.dun...@gmail.com>> wrote:
On 06/11/2016 5:02 PM, Jim Lemon wrote:
hi James,
I think you have to have a starting date ("origin") for as.Date to
convert numbers to dates.
That's true with the function in the base package, but the zoo package also has
an as.Date() function, which defaults the origin to "1970-01-01". If James is
using zoo his code would be okay. If he's not, he would have got an error, so
I think he must have been.
Duncan Murdoch
Jim
On Sun, Nov 6, 2016 at 12:10 PM, James Hirschorn
mailto:james.hirsch...@hotmail.com>> wrote:
This seemed odd so I wanted to check:
> x <- foreach(i=1:10100, .combine='c') %do% { as.Date(i) }
yields a numeric vector for x:
> class(x)
[1] "numeric"
Should it not be a vector of Date?
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