Hello,
Sorry, I forgot to Cc the list.
Rui Barradas
Em 18-06-2013 16:29, Rui Barradas escreveu:
Hello,
Inline.
Em 18-06-2013 15:54, Dzu escreveu:
Dear all,
I need to create a for-loop in which I can compute multiple histograms
My code is the following :
#singlefile includes huge csv file
clear.
Rui Barradas
Em 18-06-2013 21:04, Dizem Uerek escreveu:
Hello
Thanks for reply
I want to compute several histograms in a for loop.I am trying to set the
binsize constant in the beginning.
#compute the histograms
for (i in 1:12)
{
binsize <- -20 :20/2
hist(single
ll to hist() that does work, without a for loop.
Rui Barradas
I want to do the following :
#I have created a huge csv.files with 44 colums
#I want to select the specific colums from these files
#CL1 consist data from which I want to compute the histogramms, CL2 is the
cloumn which has numbers
Hello,
Try the following.
aggregate(Project ~ Country + Iso, data = df, FUN = length)
Hope this helps,
Rui Barradas
Em 19-06-2013 15:23, Arnaud Michel escreveu:
Hello
I have the following dataframe
df <- data.frame(
Project=c("Abaco","Abaco","Abac"
Hello,
The Pr(>F) is a regular p-value, with the usual meaning.
Rui Barradas
Em 21-06-2013 11:05, pieter escreveu:
k.
Thanks.
Sorry i'm not the biggest statician, but I still find the output write
out not the clearest of all. Anyway, that is probably a personal
problem. :-)
Just one
Hello,
Just to add to the last post, you can try to get that p-value using what
that output gives you:
df1 <- 1
df2 <- 356
F_stat <- 10.948
pf(F_stat, df1, df2, lower.tail = FALSE) # same value
Hope this helps,
Rui Barradas
Em 21-06-2013 12:11, Rui Barradas escreveu:
Hello,
engths)[order(order(x))]
plot(x, col = cols)
Hope this helps,
Rui Barradas
Em 21-06-2013 16:13, Suparna Mitra escreveu:
Hello R experts,
I want to define desired numbers to a vector based on the present
numbers. Can anybody please help me?
Obviously I found worst ways to do it, but I believe the
Hello,
It's not your first post, you should know to say from which package is
the function you're using.
As for the question the answer is no, those values are hard coded in the
source. What you can do is to download the source and change file hdr.den.R
Hope this helps,
Rui Ba
Hello,
At an R prompt type
?lgamma
?nlminb
and read the respective help pages.
Hope this helps,
Rui Barradas
Em 22-06-2013 10:54, 심정인 escreveu:
Could you tell me what mean is that lgamma, digamma, and nlminb ?
[[alternative HTML version deleted
Hello,
There's a no homework policy. Please ask your question here:
http://stackoverflow.com/questions/tagged/r
Hope this helps,
Rui Barradas
Em 22-06-2013 10:06, 심정인 escreveu:
Hi!
my name is jeong in Sim.
Now I'm studying MLE(Maximum Likelyhood Estimation) and MOM(Method of m
Hello,
Sorry I forgot to Cc the list.
And I had forgotten the case where x == 0.
extract <- function(x){
e <- ifelse(x == 0, 0, floor(log10(x)))
m <- x/10^e
list(mantissa = m, exponent = e)
}
extract(c(0, 1.234e12, 12345678901234, 123e123))
Hope this he
Hello,
Sorry, I forgot to Cc the list.
Rui Barradas
Em 23-06-2013 21:44, Rui Barradas escreveu:
Hello,
See if the following does what you want.
lapply(seq_len(obsv), function(i) adf.test(df[df$ID == i, 3]))
Hope this helps,
Rui Barradas
Em 23-06-2013 19:12, Olga Musayev escreveu:
Short
Hello,
To work with xls files, check out package XLConnect. The vignette
explains how to use it rather well.
Also, don't use attach() It can be confusing and a source for errors.
Hope this helps,
Rui Barradas
Em 23-06-2013 21:54, Sérgio Henrique almeida da silva ju escreveu:
I crea
he suggested.
Hope this helps,
Rui Barradas
Em 24-06-2013 10:34, Stefano Sofia escreveu:
Thank you for your tips.
I tried the code that you suggested, but there is a problem with the length of
fixed.
In the help page it is specified that the length of fixed must be total number
of parameters
Hello,
Try the following.
result <- lapply(unique(dataset$ID), function(uid) dataset[dataset$ID ==
uid, ])
names(result) <- unique(dataset$ID)
Hope this helps,
Rui Barradas
Em 24-06-2013 15:36, matteo escreveu:
Hi guys,
I'm a newby, so sorry for the easy question.
I ha
Hello,
You don't have a sub-data.frame, what you have is a list, with each
element of that list a df. Try to see, for instance, result[[1]]. This
should be a data.frame corresponding to the first ID.
Rui Barradas
Em 24-06-2013 18:03, matteo escreveu:
Hi,
result <- lapply(unique
Hello,
I had forgotten the much simpler solutions. The following should do it.
split(dataset, dataset$ID)
Rui Barradas
Em 24-06-2013 18:13, Bert Gunter escreveu:
First of all, is your data structure a matrix or a data frame? They
are different!
Assuming the latter, a shorter version of
3 residuals, all zero, like seen above.)
Rui Barradas
Em 24-06-2013 20:12, Olga Musayev escreveu:
Rui-- thanks so much for the help!
I'm getting this error though, which is leaving me stumped:
test<-lapply(ids, function(i) {
if(!any(is.na <http://is.na>(df[df$ID==i,3])))
{adf.t
a <- 5
g(x)
}
f(2) # 7
Now R is finding 'a'. Because g() exists in the environment of f() (like
'a' does.)
Hope this helps,
Rui Barradas
Em 24-06-2013 21:27, David Kulp escreveu:
According to
http://cran.r-project.org/doc/contrib/Fox-Companion/appen
Hello,
R is open source. You can download the source code for package sp.
http://cran.r-project.org/web/packages/sp/index.html
Hope this helps,
Rui Barradas
Em 25-06-2013 08:06, Dr. Alireza Zolfaghari escreveu:
Hi list,
I would like to write the function [R_point_in_polygon_sp] in c# as I
Hello,
Maybe ?cat
cat(X, "\n")
(But it doesn't put the vector elements between quotes.)
Hope this helps,
Rui Barradas
Em 25-06-2013 08:14, nevil amos escreveu:
I want to print a vector of strings to the console formatted as if it were
input
X<-c("a","b&
Hello,
Use graphic parameter las = 2.
See the help for ?par.
Hope this helps,
Rui Barradas
Em 25-06-2013 07:03, alfonso.carf...@uniparthenope.it escreveu:
Hi,
I need to generate a barplot in in which the x-labels must to be
perpendiculars to the x-axis
in my data have a list of probability
Hello,
I'm not sure I understand, but it seems as simple as
merge(data1, data)
Hope this helps,
Rui Barradas
Em 25-06-2013 10:34, Nico Met escreveu:
Dear all,
I would like to fetch a list (data1) of entities from a big data file
(data) and merged together. for example: data is the
Hello,
You can write a function gmean and tapply it to your data.
gmean <- function(x, na.rm = FALSE){
if(na.rm) x <- x[!is.na(x)]
n <- length(x)
prod(x)^(1/n)
}
tapply(data$value, data$group, gmean)
Hope this helps,
Rui Barradas
Em 25-06-2013 11:58, Sh
ata[] <- sapply(data, function(x) as.numeric(sub(",", "\\.", x)))
data1 <- as.matrix(data)
heatmap.2(...) # No errors
Hope this helps,
Rui Barradas
Em 25-06-2013 19:49, Gitte Brinch Andersen escreveu:
Hi
I am trying to do a heatmap, but I can't see what I am d
Hello,
Try using ?cut
y2 <- cut(x, breaks)
by(x, y2, sum)
Hope this helps,
Rui Barradas
Em 27-06-2013 09:24, Witold E Wolski escreveu:
Is there a build in function to create an index for tapply or by given a
a numeric vector x an a vector of breaks?
What I want to do is:
x <-
Hello,
After installing the package (just once) you have to load it in the R
session (every time you start a session that will use it.) You do this
with the following command.
library(agricolae) # load the package
Hope this helps,
Rui Barradas
Em 27-06-2013 11:17, bawonpon chonipat
Hello,
Or use ?sapply.
sapply(split(dat[1:2], dat[3]), function(x) cor(x[1], x[2]))
Hope this helps,
Rui Barradas
Em 27-06-2013 20:22, David Carlson escreveu:
You can pass a matrix to by()
set.seed(42)
dat <- data.frame(x=runif(50)*20, y=runif(50)*20,
g=rep(LETTERS[1:2], each
ew = newdat)
There are many other examples like this one. If you are doing data
modeling, use data frames.
Hope this helps,
Rui Barradas
Em 27-06-2013 19:26, Anika Masters escreveu:
When "should" I use a dataframe vs. a matrix? What are the pros and cons?
If I have data of all the
])))
Hope this helps,
Rui Barradas
Em 28-06-2013 09:31, Jannis escreveu:
Yes, I had a look at that function. From the documentation, however, it
did not get clear to me how to split the dataframe into subsets of rows
based on an index argument. Like:
testframe <- data.frame(a=rnorm(100), b = rnorm(
Hello,
The following does what you want but the order of columns is different.
reshape(df1, v.names = c("Cat", "Début", "Fin"),
idvar = "Mat", timevar = "Cat", direction = "wide")
Hope this helps,
Rui Barradas
Em 29-06
Hello,
Try instead
list.files(pattern = "*")
Hope this helps,
Rui Barradas
Em 01-07-2013 19:23, Thomas Grzybowski escreveu:
I am using the "write" function like so (R 3.0.1 on linux):
"wrt" <-
function()
{
write(system("ls *"),file=&qu
o[, -1], fun)
Kgeno
Also, the best way to post data is by using ?dput.
dput(head(Kgeno[, 1:5], 30)) # post the output of this
Hope this helps,
Rui Barradas
Em 02-07-2013 21:46, kathleen askland escreveu:
I'm new to R (previously used SAS primarily) and I have a genetics data
frame consisti
Mm <- paste0(as.character(x[1]), as.character(x[2])) # Major minor
mm <- paste0(as.character(x[2]), as.character(x[2])) # minor minor
x[x == MM] <- 0
x[x == Mm] <- 1
x[x == mm] <- 2
x
}
Rui Barradas
Em 02-07-2013 22:15, Rui Barradas
el = "VA", class =
"factor"),
Value = c(73L, 73L, 76L, 76L, 74L, 75L)), .Names = c("Site",
"State", "Value"), class = "data.frame", row.names = c(NA, -6L))
# Now use the argument 'fromLast'
dfA$dups <- duplicated(dfA) | dupl
bands = FALSE,
centers = ratioBiomass,
[...etc...]
Rui Barradas
Em 10-09-2012 23:07, barbara costa escreveu:
Hi Rui,
I'd really need the plot only in black so I ran the following code using
your suggestion (with col="black)", but I'm getting the color scale at
the right sid
Hello,
See also
http://cran.r-project.org/doc/contrib/Leisch-CreatingPackages.pdf
Hope this helps,
Rui Barradas
Em 11-09-2012 09:53, R. Michael Weylandt escreveu:
Start with ?package.skeleton()
Cheers,
Michael
On Tue, Sep 11, 2012 at 9:47 AM, purushothaman wrote:
Hi
how to make a
Hello,
Try
which("B" == Mylist)
Hope this helps,
Rui Barradas
Em 11-09-2012 13:31, Rantony escreveu:
How can I get index from a list if I know the listitem ?
For eg:- Mylist<-c("A","B","C")
Acually,Here I need to get the index of "B" a
became reproducible.
set.seed(1)
df1 <- data.frame(var1=sample(c('a','b','c',NA), replace=TRUE, size=100),
var2=sample(c('a','b','c',NA), replace=TRUE, size=100))
sum(is.na(df1$var1) & is.na(df1$var2)) # 8
So I suppose this i
Hello,
Try, using data and matrix dims of your choice,
replicate(100, matrix(rnorm(12), ncol = 3))
Hope this helps,
Rui Barradas
Em 11-09-2012 15:17, cesare orsini escreveu:
Dear people,
I need to generate 100 different matrices, without particular
characteristics but using only one command
, 'fmax', 'below')
dat <- data.frame(Depth = sample(depth, n, TRUE), Value = runif(n))
# This is sorted by increasing Depth
agg <- aggregate(Value ~ Depth, data = dat, FUN = mean)
ix <- order(depth) # First step
agg[order(ix), ] # inverse permutation
Hope this helps,
Ru
if (lIf you want 4 column data.frames, uncomment the instruction above (and
comment out the previuos one).
Hope this helps,
Rui Barradas
Em 11-09-2012 19:44, Rui Esteves escreveu:
Hello,
I have 2 functions (a and b)
a = function(n) { matrix (runif(n*2,0.0,1), n) }
b = function (m, matri
;degF",3,line=2)
To the op:
according to Hadley Wickham, "It's not possible in ggplot2 because plots
with multiple y scales are fundamentally flawed." See
http://stackoverflow.com/questions/3099219/how-to-use-ggplot2-make-plot-with-2-y-axes-one-y-axis-on-the-left-and-another
H
NA
ct$p.value
[1] 0.1101842
To the op: you should say which library you are using. Even if Hmisc is
a very popular one.
Hope this helps,
Rui Barradas
Em 12-09-2012 16:10, R. Michael Weylandt escreveu:
On Wed, Sep 12, 2012 at 4:04 PM, Jason Love wrote:
Hello,
I'd like to t
cs = c(4.9268, 2.892, 2.6428, 1.1124, 2.7237
)), .Names = c("D.Prime", "T.statistics"), class = "data.frame",
row.names = c("1", "2", "3", "4", "5"))
library(Hmisc)
rc <- rcorr(as.matrix(Var), type="pearson")
p
value = d2[lower.tri(d2, diag = TRUE)])
res
Hope this helps,
Rui Barradas
Em 12-09-2012 21:45, csmeredith escreveu:
Hello
I have data like this
x1 x2 x3 x4 x5
I want to create a matrix similar to a correlation matrix, but with the
difference between the two values, like this
x1
character(unique(unlist(T80[ycols])))
result <- lapply( 4:2, function(.n)
do.call(rbind, lapply(covers, makeCovers, .n)) )
names(result) <- paste("nYears", 4:2, sep = ".")
str(result)
# See the results for 4 years
result[[ "nYears.4" ]] # or any o
Hello,
Try the following.
Q_use_2 <- split(Q_use, as.Date(Q_use$date))
SC_use_2 <- split(SC_use, as.Date(SC_use$date))
lapply(seq_along(Q_use_2), function(i) cor(Q_use_2[[i]]$Q,
SC_use_2[[i]]$SC))
Hope this helps,
Rui Barradas
Em 14-09-2012 01:35, emorway escreveu:
useRs,
Here is
1:2]
Hope this helps,
Rui Barradas
Em 13-09-2012 23:04, Curtis Burkhalter escreveu:
Hello,
I am having trouble creating a (1000,3,201) array in R from a data set
created within the same script. My problem is that I want to take a matrix
that is 1000x603
(called "landmat") and separ
t the format returned by as.Date uses '-' as separator:
as.Date(x, format = "%d.%b.%Y-%H:%M")
[1] "2012-08-14" "2012-09-03"
Hope this helps,
Rui Barradas
Em 14-09-2012 10:44, Martin Batholdy escreveu:
Hi,
I have two data-frames which I want to match by a date
Hello,
Use ?colorRampPalette
nr <- nrow(yes.matrix_11)
clrs <- colorRampPalette(c("white", "black"))(nr)
barplot(yes.matrix_11, beside =TRUE, ylim=c(0,250), col = clrs)
[...etc...]
Hope this helps,
Rui Barradas
Em 14-09-2012 12:33, Robert Pazur escreveu:
Hi,
i c
-6
[5] mvtnorm_0.9-9992
loaded via a namespace (and not attached):
[1] fortunes_1.5-0 grid_2.15.1tools_2.15.1
Rui Barradas
Em 14-09-2012 19:02, Duncan Murdoch escreveu:
On 14/09/2012 10:44 AM, Jie wrote:
Dear All,
This might be a tiny question but I do not know the reason. On my
desktop
Hello,
Check the packages UScensus2000blkgrp and UScensus2000tract
http://stat.ethz.ch/CRAN/web/packages/UScensus2000blkgrp/index.html
http://stat.ethz.ch/CRAN/web/packages/UScensus2000tract/index.html
Hope this helps,
Rui Barradas
Em 14-09-2012 15:22, Michael Leitson escreveu:
Does anyone
Homework. There's a no homework policy, but see ?aggregate.
Rui Barradas
Em 15-09-2012 10:22, kangam3 escreveu:
Hi Friends
I am new here and have a problem
Year Market Winner BID
1 1990ABC Apple 0.1260
2 1990
Hello,
Since logical values F/T are coded as integers 0/1, you can use this:
set.seed(5712) # make it reproducible
n <- 1e3
x <- data.frame(A = sample(0:1, n, TRUE), B = sample(0:10, n, TRUE))
count <- sum(x$A == 1 & x$B > 5) # 207
Hope this helps,
Rui Barradas
Em
(start), y = peak))
p2 + geom_rect(aes(xmin = as.factor(start), xmax = as.factor(end), ymin
= 0, ymax = peak))
The level 5291926 is place last. Shouldn't it be expected to plot as first?
Rui Barradas
Em 16-09-2012 00:20, John Kane escreveu:
Thanks for the data. It makes things much easi
Hello,
Here's another one.
logic.result <- with(rep_data, know %in% c("very well", "fairly well") &
getalong %in% c(4,5))
rep_data$clo <- 1*logic.result # coerce to numeric
Rui Barradas
Em 16-09-2012 13:29, Stephen Politzer-Ahles escreveu:
Hi Niklas,
I
Hello,
Try the following.
1)
pattern <- "response."
m <- regexpr(pattern, ex) #gregexpr to get all "response"
regmatches(ex, m)
2)
gsub("\\$", "\\.", ex)
Hope this helps,
Rui Barradas
Em 16-09-2012 15:35, Özgür Asar escreveu:
Dear all,
I wan
Maybe a bug in ggplot2::geom_rect?
I'm Cceing this to Hadley Wickham, maybe he has an answer.
Rui Barradas
Em 16-09-2012 17:04, John Kane escreveu:
-Original Message-
From: ruipbarra...@sapo.pt
Sent: Sun, 16 Sep 2012 13:13:47 +0100
To: jrkrid...@inbox.com
Subject: Re: [R]
)
t3 <- system.time(z3 <- f3(z))
identical(z1, z2) #[1] TRUE
identical(z1, z3) #[1] TRUE
rbind(t1, t2, t3)
user.self sys.self elapsed user.child sys.child
t1 2.55 0.473.01 NANA
t2 1.57 0.291.87 NANA
t3 1.51 0.261.78
Hello,
This should do it. You can collapse the first two instructions, but I've
left it like this for clarity.
s <- unlist(strsplit(ex, "[,)[:blank:]]"))
s <- gsub("^.*\\$", "", s)
s[nchar(s) > 0]
Rui Barradas
Em 16-09-2012 17:26, Özgür Asar escre
NULL
res
}
f5 <- function(w){
sp <- split(w, w$V8)
res <- do.call( rbind, lapply(sp, f4) )
res <- data.frame(res)
res
}
T80 <- read.table("sample.txt", header = TRUE, sep = ";")
ycols <- grep("y", names(T80))
ws <
if(10, 0, 10)
Y <- 2 + 3*X
a <- data.frame(X = X, Y = Y)
fun <- function(a){
fit <- lm(Y ~ X, data=a)
return(coef(fit))
}
result <- boot::tsboot(a, statistic = fun, R = 10, sim = "geom", l = 10,
orig.t = TRUE)
Hope this helps,
Rui Barradas
Em 17-09-2012 14:42, Hock Ann L
ev.new()
bwplot(conc~site, data=test,
scales = list(y=list(log=10)),
panel = function(...){
panel.bwplot(..., stats = mystats)
}
)
With a median _line_ it would be perfect.
(Not a follow-up, it was already answered some time ago, use pch = "|"
in panel.bwplot.
s(dat3) <- seq_len(nrow(dat3))
all.equal(dat2, dat3) # only names are different
Hope this helps,
Rui Barradas
Em 17-09-2012 19:32, john james escreveu:
> Dear R users,
>
> I have the following problems. My dataset (dat) is as follows:
>
> a <- c(1,2,3)
> id <- rep(
Google est ton ami:
http://lmgtfy.com/?q=R+statistiques+fran%C3%A7ais
Le deuxième parait prometteur.
Bonne chance,
Rui Barradas
Em 17-09-2012 20:06, Seydou Badiane escreveu:
HELLO, I SHALL NEED A HELP. It is my official language(tongue) is French,
even if I manage little in English, I shall
Hello,
Try the following.
sp <- split(mydata, mydata$Sample)
do.call(rbind, lapply(sp, function(x){x$Gain <- x$Mass[3] - x$Mass[2]; x}))
Hope this helps,
Rui Barradas
Em 18-09-2012 00:15, Julie Lee-Yaw escreveu:
> Hi
>
> I have a dataframe similar to:
>
>> Sample<-c
Or diff(x[2:3])
Rui Barradas
Em 18-09-2012 01:05, David Winsemius escreveu:
On Sep 17, 2012, at 5:00 PM, David Winsemius wrote:
On Sep 17, 2012, at 4:15 PM, Julie Lee-Yaw wrote:
Hi
I have a dataframe similar to:
Sample<-c(1,1,1,2,2,2,3,3,3)
Time<-c(1,2,3,1,2,3,1,2,3)
M
Volume.y)
Importance <- Fi*Vi/sum(Fi*Vi)
3) Maybe you can combine both ways and find a use for the data.frame
'm1'. And have
m1$Importance <- ...etc...
Hope this helps,
Rui Barradas
Em 18-09-2012 05:48, Raoni Rodrigues escreveu:
Hello all,
I'm new in R, and I
you do in your post.
With the side effect that any other software package would be able to
access, and get rid of Excel dependency.
But if for some reason you need Excell, then the error message is clear,
only file extensions xls and xlsx are allowed.
Rui Barradas
Em 18-09-2012 13:54, Amir Kasa
st
elements precisely to have simplified access to each combination of Crop
and Period.
(Condition 19, the last one, comes first because split sorts its output
by list element name, and 'crop' is the first.)
Rui Barradas
John Kane
Kingston ON Canada
-Original Message-
F
(dat, as.numeric(sub("%", "", pct_total))/100)
str(dat) # see its STRucture
top <- which(dat$pct_total >= median(dat$pct_total)) # make index vector
sum(dat$pct_total[top])
Hope this helps,
Rui Barradas
Em 18-09-2012 15:41, ramoss escreveu:
Hello,
I am a newbie to R
Hello,
Try the following.
agg <- aggregate(buddleiat ~ samplet + datet, data = traffic, FUN = mean)
mrg <- merge(encounters, agg,
by.x = c("samplec", "datec"),
by.y = c("samplet", "datet"))
mrg$Div <- with(mrg, Bladen/buddleiat)
Hello,
I'm not sure I understand the question, but is it something like this?
set.seed(3648)
x <- rnorm(100)
plot(ecdf(x), cex = 0.5)
# Now create a function; don't use
# uppercase F, it's a symbol for FALSE.
f <- ecdf(x)
f(0) # Should be near 0.5
Hope this helps,
Rui B
re
unlikely to occur. That's what P. Burns is talking about when he says
that "Surprise in movies and novels is good. Surprise in computer code
is bad." To understand a function look at the way it is called, and to
what happens in it. Forget almost all of the rest, the exception b
Hello,
Try ?read.csv.
dat <- read.csv("C:\\Users\\Anthi\\Desktop\\R\\A.csv")
Note that unlike what happens with load() you'll need an explicit
assignment.
Hope this helps,
Rui Barradas
Em 19-09-2012 09:54, Anthi Oikonomou escreveu:
> Hi,
> I am trying to import c
such thing as c:\\\\ ---> c:\\).
Rui Barradas
Sarah
-Oorspronkelijk bericht-
Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
Namens Sarah Goslee
Verzonden: woensdag 19 september 2012 15:44
Aan: Anthi Oikonomou
CC: r-h...@stat.math.ethz.ch
Onderwerp: Re: [R]
%3Cbr%3EMaunga_Whau_Volcano,21
Hope this helps,
Rui Barradas
Em 19-09-2012 16:55, Tinus Sonnekus escreveu:
Hi All,
I am trying to reproduce this using R instead.
[image: Full-size image (38 K)]
I tried using the following code
*SChla <- read.csv("SM_Chla_data.csv")*
*Atlantis <- SChla[16
Hello,
Try the following.
x <- rnorm(100)
hist(x, xaxt = "n")
axis(1, at = pretty(x), pos = 0)
Hope this helps,
Rui Barradas
Em 19-09-2012 18:51, firespot escreveu:
Hi,
So I plot a histogram using the built-in hist function:
hist(rnorm(100), ...).
Now the y-axis starts at
manually create it whenever
this happened. And to issue a print statement after a missed ?function
one. Apparently the error message was waiting to be displayed somewhere.
The problem was corrected with R 2.15.0 so I recommened you update your
installation of R.
Hope this helps,
Rui Barrada
'b'. This time it's a vector, not a
data.frame.
The ifelse corrected is
# 'b' is now 1-dim
ifelse(b[1] < Sys.time()+5*60, rnorm(1,0,5), Sys.sleep)
(Why the '*' and the double quotes?)
Then I've realized that you had restarted the counting, but startin
e epoch of incredulity it was the season of Light it was
the season of Darkness"
fun(x, y)
z <- "It was the era of R"
fun(x, y, z)
Hope this helps,
Rui Barradas
Em 19-09-2012 20:08, mcelis escreveu:
I am new to R and am looking to merge two or more frequency tables into one.
I
s!
?predict.lm
Without 'newdata' it gives you the fitted values. With new data it gives
you predictions.
Beware, 'newdata' must be a data.frame.
Hope this helps,
Rui Barradas
Because I dont have two but six
matrices as independent variables, so it becomes pretty lengthy. I
Maybe it is longer, but it's also more general, it issues an error if
the tables are not 1-dim. That's where most of the function's extra
lines are. Otherwise it's the same as your first solution. The second
one has the problem you've mentioned.
Rui Barradas
Em 20
by=5), lty=2)
abline(v=seq(1, 30, by=1), lty=2)
boxplot (TMAX ~ HOUR, data=soton.df,
xlab="Forecast Hour", ylab="MAX TEMP",
main="GEFS $$MM$DD ${HH}Z FORECAST MAX TEMPS",
whiskcol="red", col="red", outline=TRUE,
ylim=c(0, 1
t(datas, level)
sm <- sapply(sp, function(x) cumsum( x[[column]] ))
as.vector(sm)
}
cut.paste.2(dummy, level3)
cut_paste(dummy, "datas2", level3)
cut.paste.1(dummy, level3)
cut_paste(dummy, "datas1", level3)
Hope this helps,
Rui Barradas
Thanks very much!
Ben
else
*which = NA_INTEGER;
UNPROTECT(1);
return(Which);
}
SEXP firstGreaterDbl(SEXP X, SEXP A){
R_len_t *which, i = 0;
double *x = REAL(X), *a = REAL(A);
SEXP Which;
PROTECT(Which = allocVector(INTSXP, 1));
which = INTEGER(Which);
while(x[i] <= a[0]
. (And why cbind, by the way?)
2. Use something folloing these lines. (Untested, obviously, without a
data example.)
pred <- predict(...etc...)
no_na <- complete.cases( cbind(matrix1, matrix2) )
matrix1[ no_na, ] <- pred[1, ]
matrix2[ no_na, ] <- pred[2, ]
Hope this helps,
Rui Barrad
Hello,
For questions like this,
install.packages('sos') # if not yet installed
library(sos) # load into R session
findFn('mzxml ')
The frist two look promising.
Hope this helps,
Rui Barradas
Em 21-09-2012 20:02, Greg Barding escreveu:
Hi Everyone,
So I've b
lines in ggplot2
using transparency in order to give a visual picture of where there are
more lines.
Hope this helps,
Rui Barradas
Em 22-09-2012 00:11, eliza botto escreveu:
Dear useRs,
my question could be very basic for which i apologize in advance.
Each column of a matrix with dimensions 365
om_line(data = dm, colour = alpha("blue", 1/5)) +
geom_line(data = dm, aes(y = avg), colour = "darkblue")
Rui Barradas
Em 22-09-2012 02:02, arun escreveu:
HI,
Similar graph in xyplot:
set.seed(1)
mat <- matrix(rnorm(100*37), ncol = 37)
mat <- apply(mat, 2, cumsum)
mat1&
here are
three 1s in 'z'). The same goes for 'y'.
Correct:
# Create an index matrix
z.inx <- which(z == 1, arr.ind = TRUE)
z.inx
# Test
x1 <- x2 <- x3 <- x # Use copies to test
x1[z == 1] <- y[z == 1]
x2[z.inx] <- y[z.inx]
# 1 and 0 to T/F
x3[as.logical(z)] <
Hello,
Try diff(v)
Hope this helps,
Rui Barradas
Em 24-09-2012 12:00, Hermann Norpois escreveu:
Hello,
I have a vector (numeric) v-> c(a,b,c,d,e) and I want to create the vector
n->c(b-a,c-b,d-c,e-d). How can I do that?
Thank you
Hermann
[[alternative HTML version d
Hello,
See also ?setdiff
Hope this helps,
Rui Barradas
Em 24-09-2012 08:30, Milan Bouchet-Valat escreveu:
Le lundi 24 septembre 2012 à 13:22 +1000, Chintanu a écrit :
Hi,
I have two dataframes (Dataframe_A, Dataframe_B) with the same no. of
columns. The first column of both the dataframes
t;b", "c", "d",
"e"), class = "factor"), X = 1:5), .Names = c("A", "X"), row.names = c(NA,
-5L), class = "data.frame")
Rui Barradas
Em 24-09-2012 12:20, benrgillespie escreveu:
Hi guys,
It would be great if you could help me
used as to why the logical condition is returning
numbers, are you sure of that?)
Anyway, the right way would be to index 'mPV' using a logical or an
index matrix.
Hope this helps,
Rui Barradas
If so I am still confused as this is not what I thought was supposed to by
happen
I've just reread my answer and it's not very clear. Not at all. Inline.
Em 24-09-2012 18:34, Rui Barradas escreveu:
Hello,
Inline.
Em 24-09-2012 15:31, Bazman76 escreveu:
Thanks Rui Barrudas and Peter Alspach,
I understand better now:
x<-matrix(c(1,0,0,0,2,0,0,0,2),nrow=3)
Hello,
Also, R uses doubles, not floats.
Hope this helps,
Rui Barradas
Em 24-09-2012 23:27, Peter Langfelder escreveu:
Erin,
you seem to confuse R and C syntax a bit, among other things. See below.
On Mon, Sep 24, 2012 at 3:03 PM, Erin Hodgess wrote:
Dear R People:
I'm working
Hello,
Try the following.
countTrait <- function(x) length( unique(x[x != 0]) )
presence <- 1*(!is.na(dat1[-1]))
result <- apply(presence, 2, function(x) apply(dat2[, -1]*x, 2, countTrait))
result <- t(result)
rowSums(result)
#Site1 Site2 Site3
# 5 7 3
Hope this
same mistake, you simply return the function
without ever calling it.
Hope this helps,
Rui Barradas
Em 25-09-2012 13:28, Al Ehan escreveu:
Hi,
I'm trying to compile two functions into one function. the first funtion is
called 'fs' which is self-made function, another f
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