this looks similar:
do.call(expand.grid,split(t(replicate(3,c(0,1,NA))),1:3))
Adrian DUSA a écrit :
Dear R-helpers,
I'm trying to develop a function which specifies all possible expressions that
can be formed using a certain number of variables. For example, with three
variables A, B and C
st - function(data, x, y){
attach(data)
rcc - coef(lm(y~x))
plot(x,y)
abline(rcc)
detach(data)}
st(data=dats, x=visual24, y=visual52)
Pryseley Assam a écrit :
Hello R-users
I am new to R and trying to write some functions. I have problems writing
functions that
sorry if it has already been discussed but i can't understand this:
seq(0.1,1,by=0.1)
[1] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
match(0.1,seq(0.1,1,by=0.1))
[1] 1
match(0.2,seq(0.1,1,by=0.1))
[1] 2
match(0.3,seq(0.1,1,by=0.1))
[1] NA
match(0.4,seq(0.1,1,by=0.1))
[1] 4
R.version
at the article mentionned in FAQ 7.31 since
everything is not clear for me yet .
identical(0.4-0.3,0.1)
[1] FALSE
all.equal(0.4-0.3,0.1)
[1] TRUE
jacques
Roger Bivand a écrit :
On Mon, 30 Jan 2006, Jacques VESLOT wrote:
sorry if it has already been discussed but i can't understand
try:
as.vector(as.dist(data))
Taka Matzmoto a écrit :
Hi R users
I like to extract lower diagonal elements of a matrix in such a way like,
data[1,2], data[1,3],
, data[5,6] are extracted from a matrix called 'data'
This short script below is what I have written so far.
lapply(a,[[,s)
Thomas Steiner a écrit :
I have a list of (same type) lists. I want to retrieve the same
entries of all my objects in the outer list.
eg:
a-list(2006-01-23=list(r=5,s=c(7,12,12,11,4)),
2006-01-24=list(r=6,s=c(3,8,8,9,12)))
a[][[s]]
gives NULL, but I am looking for all the
if you have a list of data frames without the attribute and a
vector/list of the values for this attributes:
mapply(function(x,y) {attr(x,names) - y ; x},
list(data.frame(1:10), data.frame(11:20)), c(something,
anything), SIMPLIFY=F)
Christian Bieli a écrit :
Dear all
I want to generate
try:
DF2 - as.data.frame(matrix(vec, nr=nrow(DF),nc=ncol(DF))==
matrix(1:ncol(DF),nr=nrow(DF),nc=ncol(DF),byrow=T))
DF3 - data.frame(mapply(function(z,x,y) { x[y] - 0 ; x },
names(DF), DF, DF2, SIMPLIFY=F))
but there must be an easier way...
Kenneth Cabrera a écrit :
Hi, R
i am not sure i clearly understood...
do you want to coerce a vector into a list of its elements ?! like this:
split(1:5,1:5)
Norman Goodacre a écrit :
Dear group,
I am nearly beside myself. After an entire night spent on a niggling
little detail, I am no closer to to the truth. I
of course unlist()/as.list() do that !!! (sorry, i am tired)
Norman Goodacre a écrit :
Dear group,
I am nearly beside myself. After an entire night spent on a niggling
little detail, I am no closer to to the truth. I loaded an Excel file in
.csv form into R. It apparentely loads as
use xpd argument in par(), as follows:
?par
par(xpd=T, mar=par()$mar+c(0,0,0,4))
plot(1,1)
legend(1.5,1,point,pch=1)
Abd Rahman Kassim a écrit :
Dear All,
I'm trying to attach a legend outside the plot (Inside plot OK), but failed.
Any help is very much appreciated.
Thanks.
Abd.
with apply loop, x is a vector you are indexing in the first sample code
with its names.
so it failed when you tried indexing it with $ which is dedicated to
data frames.
Fernando Saldanha a écrit :
I have a program where the following code works fine:
df.opt$Delta - apply(df.opt, 1,
(1,1,2,3,6)
Y - c(1,1,9,7,8)
toto - data.frame(Num,Date,Place,X,Y)
Any suggestion is welcome.
Florent Bonneu.
Jacques VESLOT wrote:
OK ! so try this:
merge(toto[1:3], unique(na.omit(toto[3:5])),by=Place,all.x=T)
Florent Bonneu a écrit :
Indeed,
X - c(1,Na,2,3,3,3,6,6)
Y - c(1,Na
something wrong in X and Y definitions... but this could work:
do.call(rbind, lapply(split(toto, toto$Num),
function(x) x[which.min(as.POSIXct(strptime(toto$Date, %d/%m/%y
%H:%M))),]))
i don't understand the second query; do you want to keep the first line
when there are several lines for
in my line Num=2 for my columns X and Y.
Jacques VESLOT wrote:
something wrong in X and Y definitions... but this could work:
do.call(rbind, lapply(split(toto, toto$Num),
function(x) x[which.min(as.POSIXct(strptime(toto$Date, %d/%m/%y
%H:%M))),]))
i don't understand the second query; do
I don't know how to keep factors' levels with :
data.frame(mapply(function(x,y,z) ifelse(is.na(y), z, y),
names(D), D, D2, SIMPLIFY=FALSE))
but in that way it's ok :
data.frame(mapply(function(z,x,y) { y[is.na(y)] - x[is.na(y)] ; y },
names(D), D, D2, SIMPLIFY=F))
(?)
Uwe
?Control
but there are alternative ways :
sapply(), apply(), lapply()...
ifelse()
etc.
Barbora Kocúrová a écrit :
Hallo.
Could you please tell me if it is possible in R use something
like for-cycle or conditions with if and then.
I would need to index z from 0 to m and repeat some operations
image() is fine to plot the cells of a matrix.
i think balloonplot() in gplots and mosaicplot() can do it too.
i sometimes do it with bubble() in gstat.
Jaydutt Bhalshankar a écrit :
Hello Everyone,
I am Naive user of 'R' statistical environment and still i'm in learning
mode.
I would like to
try:
cbind.data.frame(do.call(rbind, strsplit(as.character(yourdf$ID),
-)), yourdf$data)
Henrik Andersson a écrit :
Hello fellow R people,
I can not figure out a pretty way to use strplit with vectors
Imagine that I got the following data from someone with ID's
representing several factors
Please give an example of your data.
set.seed(231)
morp - rnorm(20)
range(morp)
[1] -2.311664 1.650254
You can plot 2 histograms, one of them with the extreme value:
par(mfrow=c(2,1))
hist(morp, breaks=10, freq=F)
lines(density(morp))
par(mfrow=c(1,2))
hist(morp, breaks=10, freq=F)
d - data.frame(x1 = rnorm(100), x2 = rnorm(100), x3 = rnorm(100), cls =
rnorm(100))
dd - subset(d, sel=-cls)
lapply(paste(lm(cls ~, names(dd), ,data=d)), function(x)
eval(parse(text=x)))
Juan Daniel López Serna a écrit :
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Hello:
I'm having
It seems that Xi is a matrix; so it can't be included in a vector.
You need to create a list for the result of your for loop.
res - list()
for ( i in ...) {
res[[i]] - Xi }
Vieilledent Ghislain a écrit :
Dear statisticians,
I would like to save results for a for loop in a vector
you can try :
plot(hclust(dist(ach, manh), ward), hang=-1)
you can choose distance method and clustering method.
you can find a lot of useful functions for multivariate analysis in the
ade4 package ;
but, for clustering, the cluster package is quite nice.
Birgit Kessler a écrit :
I have
try with merge() :
table1 = data.frame(CGID=c(CG_1,CG_3,CG_2, CG_4, CG_5),
diff=c(3,5,6,4,3))
table2 = data.frame(CGID=c(CG_2,CG_3,CG_4, CG_1, CG_5),
diff=c(4,6,3,9,10))
newtable- merge(table1, table2, by=CGID)
matplot(merge(table1,
unique(x[,-2])
Sérgio Nunes a écrit :
Hi again,
the previous answers were great. I was able to do what was planned.
Now, I would like to do the following to a matrix:
|| year | event | team | total ||
where I can have multiple event per team, but each team only has
a year and a total. Thus,
do.call(rbind, strsplit(as.character(date.vector), -))
Richard van Wingerden a écrit :
Hi,
Given a frame with calendar date's:
2005-07-01, 2005-07-02,2005-07-03,2005-07-04,2005-07-05,etc.
I want to extract the following from these dates:
week number
month number
year number
Any ideas how
try:
y - sapply(x, function(x) eval(parse(text=x)))
zhihua li a écrit :
hi netters,
suppose i have a series of objects X1, X2, B1,C1... they all
have the same dimensions. i want to combine into one by using cbind:
y-cbind(X1,X2,B1,C1.)
but i don't want to type the names of
you probably have a dataframe like this :
z - data.frame(y1=c(1,2,3),y2=c(4,5,6),y3=c(4,z,5))
you can do :
z - as.matrix(z)
mode(z) - numeric
zhihua li a écrit :
hi netters,
i have a dataframe TEST like this:
Y1 Y2 Y3
X1 4 7 8
X2 6 2 Z
X3 8 0 1
i would like to change it to a
try :
barplot(do.call(rbind,lapply(list(x,y), function(x) table(cut(x,
breaks =c(0,5,10,20,25,30),beside=T)
Subhabrata a écrit :
Hello R-users,
I am new to R-commands.
I have two sets of data:
x - c(7, 7 , 8, 9, 15, 17, 18)
y - c(7, 8, 9, 15, 17, 19, 20, 20, 25, 23, 22)
I have used
standard deviations of random effects (I don't know how to find
them) but I can't find F
tests for random effects.
I only want to know if there is an easy way (a function ?) to do F tests for
random effects in
random effect models.
Thanks in advance.
Best regards,
Jacques VESLOT
Data
) + (1 | Lignee) + (1 |
Pollinisateur:Lignee), data=mca2))
but I can't get F tests.
Thanks in advance.
Best regards,
Jacques VESLOT
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PLEASE do read the posting
-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Thanks,
Jacques VESLOT
Prof Brian Ripley a écrit :
Nothing was enclosed, nor was the output from summary.aov, so we are
left guessing.
On Thu, 27 Oct 2005, Jacques VESLOT wrote:
Dear R-users,
My question is how
effect
models.
Jacques VESLOT
Doran, Harold a écrit :
I think what you're looking for is in anova()
fm1 - lmer(dv ~ IV ...)
anova(fm1)
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Jacques VESLOT
Sent: Thursday, October 27, 2005 2:22 AM
if it is not necessary.
I hoped there might be an heavier function able to change coordinates in
an let's say intelligent or surgical manner...
jacques
Renaud Lancelot a écrit :
Jacques VESLOT a écrit :
Dear R-users,
Is there an easy way to avoid points one upon another when ploting
rows and columns
of
these points to get a readable plot ?).
Thanks in advance.
Best regards,
Jacques VESLOT
__
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PLEASE do read the posting guide! http://www.R-project.org/posting
:
(...) some devices do not implement line widths less than one.
What I am looking for is how to get readable plots (with not too wide
lines) after size reduction...
Best regards,
Jacques VESLOT
Paul Murrell a écrit :
Hi
Jacques VESLOT wrote:
Dear R-users,
Someone, who uses R under Mac, wants to insert
replaced in Chi2 distribution function to get p-value:
1-pchisq(486.199, df=63)
[1] 0
Is there a better way to perform this or a more appropriate function
dedicated to tests on large-dimensioned contingency tables ?
Thanks in advance,
Jacques VESLOT
+ pl, data = z)
Statistics:
X^2 df P( X^2)
Likelihood Ratio 286.1236 490
Pearson 450.5332 490
Jacques
-Message d'origine-
De : Prof Brian Ripley [mailto:[EMAIL PROTECTED]
Envoye : mardi 15 fevrier 2005 17:41
A : Jacques VESLOT
Cc : R-HELP
class(df)
[1] data.frame
names(df)
[1] x y z
df - df[order(df$x, df$y),]# sort df by x and y with priority to x
-Message d'origine-
De : [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] la part de Luis Ridao Cruz
Envoye : vendredi 11 fevrier 2005 16:03
A :
if * in x1[1:50,]*x2[1:50,] means times, try :
matplot(matrix(x1,ncol=4)*matrix(x2,ncol=4))
if * means vs, try :
matplot(matrix(x1,ncol=4),matrix(x2,ncol=4))
-Message d'origine-
De : [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] la part de NICOLAS DEIG
Envoye : vendredi 28 janvier 2005
;
for that, my first idea was to add this matrix:
X1=X[,rep(1:46,46:1)]
to this one:
res=NULL
for (i in (2:47)) res=c(res,i:47)
X2=X[,res]
(Is it a nice alternative way ?)
Is there a way to create the second matrix X2 without a loop, such as for X1
?
Thanks in advance,
Jacques VESLOT
Y and counting the number of 0, 1 and 2 in it for
inclusion in other operations.
I read your posting...
Thanks for helping,
Jacques VESLOT
-Message d'origine-
De : Adaikalavan Ramasamy [mailto:[EMAIL PROTECTED]
Envoyé : mercredi 26 janvier 2005 15:57
À : [EMAIL PROTECTED]
Cc : R-help
Dear all,
Could someone please make me know if there is a nice script editor available
under Mac, similar to Crimson, that offers R syntax highlighting (and pairs
of parentheses underlining) ?
Thanks in advance,
Jacques VESLOT
Cirad
__
R-help
in if (!any(cond.max.level - cond.current.level 0) (row - 1) * :
missing value where TRUE/FALSE needed
I tried some changes in arguments - notably layout=c(0,1), but anything
works.
Thanks for helping...
Jacques VESLOT
CIRAD Réunion
__
[EMAIL
22 September 2004 09:32, Jacques VESLOT wrote:
Dear all,
I tried to use layout argument in xyplot to get one panel per page.
I have a dataframe named 'data' with the following variables:
x, y = coords,
sub, bloc = 2-level factors,
etat = 5-level factor,
I did :
lset(theme = col.whitebg
if there is another - maybe
better - way to plot spatial marked data.
Jacques VESLOT
CIRAD Réunion
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
- d2004[, -c(concentration, stade)]
Thanks...
Jacques VESLOT / CIRAD
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
a gstat object with two
variables.
Could you please let me know if this is the right way to operate (and what
formula argument means here) ?
Best regards,
Jacques VESLOT
CIRAD
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in order to generate your series, maybe :
format(seq(ISOdate(2003,1,1, 0, 0, 0, tz=), by=min,
length=1440),format=%H:%M)
jacques
-Message d'origine-
De : [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] la part de Bruno Cutayar
Envoyé : jeudi 29 juillet 2004 13:43
À : [EMAIL PROTECTED]
Objet
to have these
two columns sorted in increasing order.
But sort() and order() seem to apply to vectors - or one single column -
only.
Could somedy please let me know how to sort two columns of a dataframe, with
priority to one of them, just like in access ?
Thanks,
Jacques VESLOT
CIRAD
Another related
Sorry sorry,
I didn't pay enough attention to order() help pages indeed...
and it costs me much useless indexing.
As concerns how to put legend in image() plot, is there a piece of
information about somewhere ?
thanks a lot for helping,
Jacques VESLOT
-Message d'origine-
De : Uwe
inside a loop
?
Jacques VESLOT
CIRAD
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It works indeed.
Thanks a lot.
Jacques VESLOT
CIRAD
-Message d'origine-
De : Uwe Ligges [mailto:[EMAIL PROTECTED]
Envoyé : jeudi 15 juillet 2004 11:35
À : [EMAIL PROTECTED]
Cc : [EMAIL PROTECTED]
Objet : Re: [R] About lattice plots and loops
Jacques VESLOT wrote:
I cannot operate
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