Thank everyone for the answers.
I looked into ggmap but could not find how to plot a map that would like
this:
http://www.google.ca/trends/explore#q=flu
I looked into get_map but it can only be of this type:
maptype = c("terrain", "satellite", "roadmap", "hybrid", "toner",
"watercolor")
Any su
Thank you Michael.
I'll try that. Maybe I can find a solution to convert the HTML file into a
let's say SVG file.
Regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/googleVis-plot-and-knitr-sweave-tp4651341p4651463.html
Sent from the R help mailing list archive at Na
Hi and thank you for your answer.
In fact, I don't really need the interactive part of googleVis map. So if
there's a way to plot a static version of the map it would be perfect for
me.
Regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/googleVis-plot-and-knitr-swea
Dear R users.
I'm currently making a report with knitr (RStudio) where I would like to
plot a googleVis map. However, the map generated is an HTML file which I
don't know how to integrate it in my report.
So my question is how to include a map generated with googleVis in a PDF
created with knitr
Thanks for the head-up Don.
Regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Re-projecting-geotiff-tp4646616p4646736.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https
Dear R users,
I'm currently trying to re-project a geotiff in another coordinate system.
For instance, I have a tif image in UTM 19 zone which I would like to
reproject into UTM 18. I was wondering if it was possible in R.
Furthermore, I looked into 'rgdal' package, but I can't really find out if
Thank you Ruben.
You were absolutly right. I'm using trend option now to specify my model.
Thank for the help,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/geoRglm-with-factor-variable-as-covariable-tp4645067p4645310.html
Sent from the R help mailing list archive at Na
Dear R users.
I'm trying to fit a generalised linear spatial mode using the geoRglm
package. To do so, I'm preparing my data (geodata) as follow:
geoData9093 = as.geodata(data9093, coords.col= 17:18, data.col=15,*
covar.col=16*)
where covar.col is a factor variable (years in this case 90-91-92-
Try a look at this:
http://stat.ethz.ch/R-manual/R-patched/library/MASS/html/stepAIC.html
Regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/All-combinations-possible-in-a-mutliple-regression-tp4639762p4639782.html
Sent from the R help mailing list archive at Nabble.
Hi everyone.
I'm trying to customize a biplot (RDA). I would like to remove all axes
ticks. Therefore, I was using *ann=FALSE* (see following example). However,
it only clear the axis 1 and 2 leaving ticks on axis 3 and 4. Any
suggestions to get rdi of the remaining ticks?
data(varespec)
data(va
Hi everyone.
I have these data :
myData = data.frame(Name = c('a', 'a', 'b', 'b'), length = c(1,2,3,4), type
= c('x','x','y','z'))
which gives me:
Name length type
1a 1x
2a 2x
3b 3y
4b 4 z
I would group (mean) this DF using 'Name' as grouping
Thank you, works fine.
Regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Row-count-on-criteria-tp4632100p4632116.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat
Hi.
I would take a look to the /forward.se/l function in the /packfor/ package.
http://r-forge.r-project.org/R/?group_id=195
Good luck,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/regsubsets-Leaps-tp4632083p4632102.html
Sent from the R help mailing list archive at Nabbl
Hi everyone.
I would like to count number of row higher than 0 for each column in a
dataframe.
For instance, I was using something like:
YY = replicate(5, rnorm(10))
apply(YY,2,">",0)
This give me a boolean matrix, but how to have the count of row with numbers
> 0 by column?
Thank in advance
Thank you sire, this is exactly what I was looking for.
Many thanks,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Matrix-transposition-tp4599219p4599310.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-p
Hi everyone.
I want to transpose a data frame. Lets say the following DF:
df = data.frame(matrix(ncol=4, nrow = 10))
df[,1] = c(rep(1,5),rep(2,5))
df[,2] = c(rep('a',4),rep('b',3),rep('c',3))
df[,3] = c(letters[5:14])
df[,4] = runif(10)
I would like to form a data frame with each line correspo
Hi everyone.
I performing a simple PCA using the princomp function. Then, I use the
biplot function to show it. However, the function use line number to
represent samples. I would like to know if there's a way to use a dot
(point) instead of the line number when using the biplot function.
With r
Thank you all.
Problem solved.
Regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Help-with-vectorization-tp4552638p4554565.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
h
Hi and thank you for your time.
I got this error when trying your function.
mouter(wl, k1, k2, k3, FUN = function(w, k1, k2, k3) k1 *exp(k2 / (w + k3)))
"Error in k3/(w + k3) : 'k3' is missing"
Regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Help-with-vectorization
Hi every one. I have a exponential function (3 fitting parameters) that I
would like to use to produce data (6 series) without having to use a loop.
Here
wl = seq(300,500,1)
k1 = c(1.2e-6, 4.9e-6, 9.6e-6, 2.7e-10, 6.7e-8, 7.44e-6)
k2 = c(726, 352, 128, 5232, 1538, 128)
k3 = c(-176, -224, -257, 88
Thank you everyone for your precious advice.
I'll take time to look at it and try to make it work.
Regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Generation-of-correlated-variables-tp4475799p4476346.html
Sent from the R help mailing list archive at Nabble.com.
Hi there.
This is not really working. Here what I have for the moment.
library(ecodist)
x <- 1:100
y1 <- corgen(x=x, r=.85, epsilon=.01)$y
y2 <- corgen(x=x, r=.5, epsilon=.01)$y
y3 <- corgen(x=x, r=.25, epsilon=.01)$y
a = poly(cbind(y1, y2, y3), degree=1)
cor(a[,1], a[,2])
In that case, the c
Hi everyone.
Based on a dependent variable (y), I'm trying to generate some independent
variables with a specified correlation. For this there's no problems.
However, I would like that have all my "regressors" to be orthogonal (i.e.
no correlation among them.
For example,
y = x1 + x2 + x3 where
Thank you Vito for your help.
Works very nice.
Have a nice day,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Help-with-segmented-package-tp4435550p4438589.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@
Hi everyone.
I'm using segmented package to find break point in a bi-linear relationship.
In a particular case, I find 1 pointcut (so 2 slopes).
I would like to know if it is possible to retrieve information in the
segmented object that could let me to plot 1 particular segment with a
different c
Thank everyone for your help.
Problem solved. I'm getting more used with vectorization with your help,
Regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/linear-regression-by-column-tp4432564p4434937.html
Sent from the R help mailing list archive at Nabble.com.
__
Hi everyone.
I have a DF with the first column being my independant variable and all
other columns the dependent variables.
Something like:
x y1 y2 y3
... ... ... ...
... ... ... ...
What I'm trying to do is to perform a linear model for each of my "y". I
Thank you for your help.
I'll simply remove leap days from my serie.
Regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Creating-time-series-ts-object-tp4362762p4364664.html
Sent from the R help mailing list archive at Nabble.com.
_
Hi everyone.
I have have a dataset with daily measurement from January 1st of 1966 up to
December 31th of 2011.
Here's the first part of the data:
DateSLEV
1/1/19661.086
1/2/19661.079
1/3/19661.133
1/4/19661.261
1/5/19661.391
1/6/19661.571
1/7/
Thank for your answer.
Problem solved.
Regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Colors-vector-based-on-group-tp4361425p4362746.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org m
Hi there.
I should have thought about that myself (/shame).
Thank you very much for your time.
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Colors-vector-based-on-group-tp4361425p4361686.html
Sent from the R help mailing list archive at Nabble.com.
_
Hi everyone.
I have a vector like this:
x =
c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,2,2,2,2,2,2,2,2)
I would like to associate a colour to each of these entry. For example, I
could use /topo.colors(3)/ (since I have 3 groups).
I know it is easy to do this with /if/else/
Thank you sire.
You explained it very well. This give ma a good point to start using sapply
more frequently.
Cordially,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/sapply-help-tp4355092p4355376.html
Sent from the R help mailing list archive at Nabble.com.
__
Hi every one.
I'm learning how to use sapply (and other function of this family).
Here's what I'm trying to do.
I have a vector of lets say 5 elements. I also have a matrix of nX5. I would
like to know how many element by column are inferior to each element of my
vector.
On this example:
v = c(
Hi there.
Here's the error message.
Error in seg.lm.fit(y, XREG, Z, PSI, weights, offs, opz) :
(Some) estimated psi out of its range
I have tried many ways to specify the arguments, but apparently the error
message is related to the estimated break point being invalid. However, my
estimation
x = runif(10)
x[4] = NA
sum(x, na.rm = T)
--
View this message in context:
http://r.789695.n4.nabble.com/Sum-of-a-couple-of-variables-of-which-a-few-have-NA-values-tp4282448p4282483.html
Sent from the R help mailing list archive at Nabble.com.
__
R-hel
Hi everyone.
I'm trying to use the segmented function with the following data:
For instance, I use segmented package as follow:
myreg2 = lm(xy$y ~ xy$x)
mysegmented = segmented(myreg2, seg.Z=~x, psi=c(245000), control =
seg.control(display=FALSE))
Which get me to the following error :
As a
Thank you sire. It worked fine.
I was using the parameter in the hist(...) function. I now use it with par
and it works perfectly.
Regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Hist-and-extra-space-tp4169121p4170452.html
Sent from the R help mailing list archive a
Hi everyone.
I have an histogram like this one:
http://imageshack.us/photo/my-images/442/dfsdfsdj.jpg/
I would like to remove the extra space under the histogram so it "touch" the
x axis. Is it possible? Here's my code.
hist(X, freq = F, col = 'gray', axes = FALSE)
abline(v = cyano_euk_min, col
Thank you sire for your help. Thank also for sharing the reference, I'll take
a look at it.
Regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Legend-tp4103799p4108587.html
Sent from the R help mailing list archive at Nabble.com.
___
Thank you for your precious help. It works fine.
However, what about if I have two entries in the legend?
I tryed:
legend('topright', inset = .05, title = 'light ratios', pch = c(21,22),
legend = c(substitute('Green/Red' ~~ R^2 == r2,
list(r2=r2)),substitute('Green/Red' ~~ R^2 == r2, list(
Hi everyone.
I have a linear regression where I retrieve the R2 like this:
r2 = sprintf('%4.2f %s',(summary(reg1)$r.squared))
In my figure I have a legend where I would like to add that R2 value to the
legend text.
Something like: My text R^2 = r2
legend('topright', inset = .05, title='light
Hi everyone.
I'm trying to use small letter phi in a graph produced in R. However, the
small letter phi does not look as it should.
In fact, it looks like this:
http://r.789695.n4.nabble.com/file/n3963311/Untitled.png
instead of what is here http://en.wikipedia.org/wiki/Phi
Here's the code I
Thank you Mark for your help. It worked perfectly with 2 expression and
mtext.
Regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Text-wrap-tp3747574p3747884.html
Sent from the R help mailing list archive at Nabble.com.
__
R-
Hi everyone.
I have a long label that I would like to split. I found that I could use
"strwrap" for simple text. However, this is not working with this label:
str = expression(paste("< 20 µm phytoplankton ","(cells · ",mL^-1,")"))
plot(, ylab = strwrap(str,20),...)
I suspect this is because
Thank you sir for your help.
I'll try to reset cex =1.
Regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Font-size-R-tp3745762p3747271.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org m
Thank you for your answer.
According to the help, they are suggesting to use "layout" which I did.
However, font size is still reduced by the same factor.
Here's my code:
pdf(file="Fig5.pdf", width = 4.86, height = 6.29, family ='Times')
nf = layout(matrix(c(1:4), 2, 2, byrow = TRUE))
par(ps =
Hi everyone.
I'm using this following code:
pdf(file="Fig5.pdf", width = 4.86, height = 6.29, pointsize = 10, family
='Times')
par(mfcol = c(3,2), mai = c(0.4,0.8,0,0), omi = c(0.7, 0, 0.7, 0.1));
hist( rnorm(100) )
dev.off()
When I open this in any vectorial software (like Illustrator), it say
Hi sire.
This is exactly what I was looking for, thank you.
With regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Avoiding-loop-tp3457963p3458152.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-pr
Hi everyone.
I'm using matrix product such as :
#Generate some data
NCols = 5
NRows = 5
A = matrix(runif(NCols*NRows), ncol=NCols)
B = matrix(runif(NCols*NRows), ncol=NCols)
#First calculation
R = A%*%B
for(i in 1:100)
{
R = R%*%B
}
I would like to know if it was possible to avoi
Hi and ty for the answer.
However, it's not working. It will print "expression(d13C Station 1)".
Thank for any help,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Greek-character-and-R-tp304p467.html
Sent from the R help mailing list archive at Nabble.com.
___
Dear R users.
In a loop, I set the title of my graph with :
mytitle = expression(paste(delta^13,'C Station ', i)
title(mytitle)
However, instead of using value of i, it will literally use "i" character.
Any one know the way to concatenate the value of i to the mathematical
expre
Hi everyone.
I would like to modify the axis on my plot.
First, I would like to make a plot without the box. so I use :
plot(x,y, axes = FALSE, type = 'l')
Then, I call :
axis(1, tck = -0.02)
axis(2, tck = -0.02)
to have X and Y axis appear. However, I would like them to join at the
origin
Thank you sir for your answer. I'll take a look at the original legend
function and modify it for my own purpose.
With regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Question-regarding-legend-look-tp3067466p3067894.html
Sent from the R help mailing list archive at
David Winsemius wrote:
>
>
> ...BUT NOT SHOWN US CODE
> OR SAMPLE DATA.
>
Hi and thank you for your help.
For instance, here's my code:
legend("topleft", inset = .05, title="Water masses", pch = c(22,25,21), lty
= c(4,1,2), lwd = 1, c("North","Central","South"), horiz = F, pt.bg
=c("gray
Hi everyone.
I have a quick question regarding the look of my legend in my plot. As you
can see in the next figure, I have 3 series.
http://r.789695.n4.nabble.com/file/n3067466/legend.png
However, I find rather difficult to differentiate the series 1 and 3
according to their line type (lty).
Thank you everyone for your answers.
Regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/List-or-matrix-of-object-tp2992101p2992304.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing
Hi again everyone.
I found I could use a list with
l = list()
l[[1]] = myObj
instead of
l[1] = myObj
Anyone can explain me why the use of double [] is required?
Regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/List-or-matrix-of-object-tp2992101p2992121.html
Sen
Hi everyone.
Is it possible in R to create a matrix or a list (vector) or R object. For
instance, I have
f1 <- function(x) sqrt(x%*%x);
f2 <- function(x) (2x+1);
I would like to do something like
L <- List();
L[1] = f1;
L[2] = f2;
So, is there a way to create matrix or vector that can contai
Thank you for your help!
--
View this message in context:
http://r.789695.n4.nabble.com/loess-and-NA-tp2956235p2956518.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/list
Hi there.
One more thing you could try is to avoid evaluation of length(CRX) every
loop.
Assign the length to a variable before the loop.
Regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Make-a-loop-more-efficient-tp2955912p2956274.html
Sent from the R help mailing
Hi everyone.
I'm trying to do a loess with missing value on independant variable.
doc = c(2.27904, 2.59536, 7.44696, NA, 6.24264, 4.58400, 5.79192, 5.39502,
7.41216, 4.09440, 4.22868, 4.24620, 5.43804, 1.95528);
distance = c(26.5,56.5, 90.3, 123.0, 147.5, 176.0, 215.7, 229.3, 252.0,
325.3, 362
Hi Greg.
Thank you for the time you took to look at my problem.
I'll try your solution until CRAN is updated.
With regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Inserting-a-plot-into-another-tp2720936p2954769.html
Sent from the R help mailing list archive at Na
Hi everyone.
First, thank you for your answers, it helped me alot.
I have 1 more question regarding subplot. I'm using this function in a
graph that contains 6 plots (3 x 2). This is producing strange behavior.
http://img545.imageshack.us/img545/9233/fig6.png See here for the resulting
figure.
Ty sire.
I'll take a look at it.
With regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Inserting-a-plot-into-another-tp2720936p2802966.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org
Hi everyone.
I would like to know if it was possible to insert a plot into another one.
For example I have a plot and I would like to add a smaller plot in the top
right corner.
Best regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Inserting-a-plot-into-another-tp27
I found what was the problem.
Instead of using xpd = NA in par(..) I use it in the plot.
With regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Help-with-graphic-margin-tp2542469p2546395.html
Sent from the R help mailing list archive at Nabble.com.
_
Thank you Greg.
xpd=NA is working as intended. I got 1 more problem with thought. When I'm
using xpd=NA, abline function plots all my regression lines outside the plot
boxes. Any solutions for that strange behavior?
With regards,
--
View this message in context:
http://r.789695.n4.nabble.com/
Hi there and thank you for reply.
I have been trying that, using mar to set margin around each plot and oma to
specify bottom space. However, it wont display the xlab.
Here's my code :
pdf(file="/home/persican/Desktop/test.pdf", width = 3.5, height = 5.6,
pointsize = 8);
par(mfcol = c(3,1), m
Hi everyone.
I would like to make a graphic with 2 subplots (lets say A and B). For
instance, I use :
par(mfcol = c(2,1));
The 2 subplots use the same X axis. So I provide no tics labels and no xlab
for the top plot. I would like the second plot to be right bellow the first
one (small margin).
Thank you sir.
Regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Axis-break-with-gap-plot-tp2533027p2537660.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.eth
Hi again everyone.
Anyone know if there's any limitation with gap.plot concerning the fill
color of plotted markers? I would like to fill the circles with a color :
library(plotrix);
gap.plot(c(1,2,3,4,10), c(1,2,3,4,10), c(5,9), pch = 21, col = "red");
However, it only change the color of the
Thank Jim for your answer.
I actually did my own function to plot with the loess. I just calculated the
intersection between the first and second horizontal gap lines with the line
formed by the 2 points before and after the gap. So I can now plot the loess
from both sides of the gap section.
Th
Hi everyone.
I'm trying to break the y axis on a plot. For instance, I have 2 series
(points and a loess). Since the loess is a "continuous" set of points, it
passes in the break section. However, with gap.plot I cant plot the loess
because of this (I got the message "some values of y will not be
Hhi there.
Here's the example.
pdf(file="C:/Users/Modelisation/Desktop/BubblePlot.pdf", width = 15, height
= 15, family = "Times", pointsize = 25);
par(mfcol = c(3,3), mar = c(3,3,3,3), oma = c(0,0,0,0));
for (i in 5:33)
{
print(bubble(temp,i, col = "cornflowerblue"));
Hi all.
I have been trying to plot 9 bubble plots on each page of a PDF file. For
this, I'm using :
par(mfcol = c(3,3), mar = c(3,3,3,3), oma = c(0,0,0,0));
However, when plotting the bubble plot (gstats package), it plots the graph
on a full page. I tried to plot something else with plot(x,y)
Hi.
This is working fine.
Thank you for your help.
Best regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Mathematical-expression-tp2299253p2299364.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help
Hi everyone.
I'm trying to display the r^2 of a linear regression on a plot using
text(...). I first build the string to display. However, I can't find why
this is not working (it display literally r^2 instead of r superscript 2).
r2string = expression(paste(r^2)," = ", r2);
Any help would be
Ty sir!
--
View this message in context:
http://r.789695.n4.nabble.com/Y-axis-break-tp2295499p2296347.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
PLEAS
Hi.
Thank for your reply. However, the link you provided don't works (for me at
least).
Regards,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Y-axis-break-tp2295499p2296202.html
Sent from the R help mailing list archive at Nabble.com.
___
Hi there.
I bet this question have been answered many times, but I cant find a good
solution for my problem.
I would like to put a axis break (on y axis) on that flowing plot (let's say
between 20 and 80 in y).
y = c(runif(10, 1, 10), 100);
x = y + runif(11, 10, 50);
plot(x,y)
Is there any so
for (i in 1:1)
{
if( (i - 100) %%100 == 0)
print(i);
}
--
View this message in context:
http://r.789695.n4.nabble.com/Displaying-Iteration-Count-tp2264088p2264124.html
Sent from the R help mailing list archive at Nabble.com.
__
Thank you all for your quick answers.
Best,
Phil
--
View this message in context:
http://r.789695.n4.nabble.com/Same-function-name-tp2262633p2262955.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
ht
Hi everyone.
I want to use 2 different functions (in 2 packages) that have same name. for
instance, if I call the function, it will use the one in the last called
package. Is there a way to specify the package to use for a function?
Something like
Package1:myfunction(...)
Package2:myfunction(x
84 matches
Mail list logo