[R] reorder() in the latticeExtra library
Hello all, thanks for your time and help. Below are my commands, and it
generates a really nice plot, however I am not happy with the reorder()
function. I would like the order to be the same as they appear in the
genotype variable genotype - c(CJ1450 NW 4/25/12,CJ1450 BAL
4/25/12,CJ1450 NW 4/27/12,CJ1450 BAL 4/27/12,CJ1721 NW
4/27/12,CJ1721 BAL 4/27/12,CJ1721 NW 4/29/12,CJ1721 BAL 4/29/12 )
and not as it is currently coded.
Is there any way to turn off the reorder, or set it up so the values appear
in the order above, thank you again!
(I am open to all suggestions)
JD
genotype - c(CJ1450 NW 4/25/12,CJ1450 BAL 4/25/12,CJ1450 NW
4/27/12,CJ1450 BAL 4/27/12,CJ1721 NW 4/27/12,CJ1721 BAL
4/27/12,CJ1721 NW 4/29/12,CJ1721 BAL 4/29/12 )
#paste(Animal, as.roman(1:8), sep = -)
plant.height - c(0.001173003, 0.001506127, 0.001361596, 0.001922572,
0.034272147, 0.030466017, 0.001654299, 0.001071724)
SE - c(0.000444123, 0.000290096, 0.000372844, 0.00197687, 0.033945128,
0.035231568, 0.001094518, 0.000423545)
lower - plant.height - SE; upper - plant.height + SE
x - data.frame(group = genotype, lower = lower, est = plant.height, upper =
upper)
library(latticeExtra)
segplot(reorder(genotype, est) ~ lower + upper, data = x, draw.bands =
FALSE, centers = est, segments.fun = panel.arrows, ends = both, angle =
90, length = 0, par.settings = simpleTheme(pch = 19, col = 1), xlab =
expression(nucleotide diversity %+-% sd), panel = function(x, y, z,
...) {
panel.abline(h = z, col = grey, lty = dashed)
panel.abline(v = 14.20, col = grey)
panel.segplot(x, y, z, ...)})
--
View this message in context:
http://r.789695.n4.nabble.com/reorder-in-the-latticeExtra-library-tp4648299.html
Sent from the R help mailing list archive at Nabble.com.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] reorder() in the latticeExtra library
On Nov 2, 2012, at 8:04 PM, JDINIS wrote:
Hello all, thanks for your time and help. Below are my commands, and it
generates a really nice plot, however I am not happy with the reorder()
function. I would like the order to be the same as they appear in the
genotype variable genotype - c(CJ1450 NW 4/25/12,CJ1450 BAL
4/25/12,CJ1450 NW 4/27/12,CJ1450 BAL 4/27/12,CJ1721 NW
4/27/12,CJ1721 BAL 4/27/12,CJ1721 NW 4/29/12,CJ1721 BAL 4/29/12 )
and not as it is currently coded.
Is there any way to turn off the reorder, or set it up so the values appear
in the order above, thank you again!
(I am open to all suggestions)
JD
genotype - c(CJ1450 NW 4/25/12,CJ1450 BAL 4/25/12,CJ1450 NW
4/27/12,CJ1450 BAL 4/27/12,CJ1721 NW 4/27/12,CJ1721 BAL
4/27/12,CJ1721 NW 4/29/12,CJ1721 BAL 4/29/12 )
#paste(Animal, as.roman(1:8), sep = -)
plant.height - c(0.001173003, 0.001506127, 0.001361596, 0.001922572,
0.034272147, 0.030466017, 0.001654299, 0.001071724)
SE - c(0.000444123, 0.000290096, 0.000372844, 0.00197687, 0.033945128,
0.035231568, 0.001094518, 0.000423545)
lower - plant.height - SE; upper - plant.height + SE
x - data.frame(group = genotype, lower = lower, est = plant.height, upper =
upper)
library(latticeExtra)
segplot(reorder(genotype, est) ~ lower + upper, data = x, draw.bands =
FALSE, centers = est, segments.fun = panel.arrows, ends = both, angle =
90, length = 0, par.settings = simpleTheme(pch = 19, col = 1), xlab =
expression(nucleotide diversity %+-% sd), panel = function(x, y, z,
...) {
panel.abline(h = z, col = grey, lty = dashed)
panel.abline(v = 14.20, col = grey)
panel.segplot(x, y, z, ...)})
Wouldn't you just define genotype as a factor with the desired sequence of
levels and remove the call to reorder?
--
David Winsemius, MD
Alameda, CA, USA
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Re: [R] mergeing a large number of large .csvs
Jeff,
If you're willing to educate, I'd be happy to learn what wide vs long
format means. I'll give rbind a shot in the meantime.
Ben
On Nov 2, 2012 4:31 PM, Jeff Newmiller jdnew...@dcn.davis.ca.us wrote:
I would first confirm that you need the data in wide format... many
algorithms are more efficient in long format anyway, and rbind is way more
efficient than merge.
If you feel this is not negotiable, you may want to consider sqldf. Yes,
you need to learn a bit of SQL, but it is very well integrated into R.
---
Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live
Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
Benjamin Caldwell btcaldw...@berkeley.edu wrote:
Dear R help;
I'm currently trying to combine a large number (about 30 x 30) of large
.csvs together (each at least 1 records). They are organized by
plots,
hence 30 X 30, with each group of csvs in a folder which corresponds to
the
plot. The unmerged csvs all have the same number of columns (5). The
fifth
column has a different name for each csv. The number of rows is
different.
The combined csvs are of course quite large, and the code I'm running
is
quite slow - I'm currently running it on a computer with 10 GB ram,
ssd,
and quad core 2.3 ghz processor; it's taken 8 hours and it's only 75%
of
the way through (it's hung up on one of the largest data groupings now
for
an hour, and using 3.5 gigs of RAM.
I know that R isn't the most efficient way of doing this, but I'm not
familiar with sql or C. I wonder if anyone has suggestions for a
different
way to do this in the R environment. For instance, the key function now
is
merge, but I haven't tried join from the plyr package or rbind from
base.
I'm willing to provide a dropbox link to a couple of these files if
you'd
like to see the data. My code is as follows:
#multmerge is based on code by Tony cookson,
http://www.r-bloggers.com/merging-multiple-data-files-into-one-data-frame/
;
The function takes a path. This path should be the name of a folder
that
contains all of the files you would like to read and merge together and
only those files you would like to merge.
multmerge = function(mypath){
filenames=list.files(path=mypath, full.names=TRUE)
datalist = try(lapply(filenames,
function(x){read.csv(file=x,header=T)}))
try(Reduce(function(x,y) {merge(x, y, all=TRUE)}, datalist))
}
#this function renames files using a fixed list and outputs a .csv
merepk - function (path, nf.name) {
output-multmerge(mypath=path)
name - list(x, y, z, depth, amplitude)
try(names(output) - name)
write.csv(output, nf.name)
}
#assumes all folders are in the same directory, with nothing else there
merge.by.folder - function (folderpath){
foldernames-list.files(path=folderpath)
n- length(foldernames)
setwd(folderpath)
for (i in 1:n){
path-paste(folderpath,foldernames[i], sep=\\)
nf.name - as.character(paste(foldernames[i],.csv, sep=))
merepk (path,nf.name)
}
}
folderpath - yourpath
merge.by.folder(folderpath)
Thanks for looking, and happy friday!
*Ben Caldwell*
PhD Candidate
University of California, Berkeley
[[alternative HTML version deleted]]
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[[alternative HTML version deleted]]
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Re: [R] finding global variables in a function containing formulae
On Thu, Nov 1, 2012 at 2:04 PM, Hafen, Ryan P ryan.ha...@pnnl.gov wrote:
I need to find all global variables being used in a function and
findGlobals() in the codetools package works quite nicely. However, I am not
able to find variables that are used in formulae. Simply avoiding formulae
in functions is not an option because I do not have control over what
functions this will be applied to.
Here is an example to illustrate:
library(codetools)
xGlobal - rnorm(10)
yGlobal - rnorm(10)
plotFn1 - function() {
plot(yGlobal ~ xGlobal)
}
plotFn2 - function() {
y - yGlobal
x - xGlobal
plot(y ~ x)
}
plotFn3 - function() {
plot(xGlobal, yGlobal)
}
findGlobals(plotFn1, merge=FALSE)$variables
# character(0)
findGlobals(plotFn2, merge=FALSE)$variables
# [1] xGlobal yGlobal
findGlobals(plotFn3, merge=FALSE)$variables
# [1] xGlobal yGlobal
I would like to find that plotFn1 also uses globals xGlobal and yGlobal. Any
suggestions on how I might do this?
If this is only being applied to your own functions then we can have a
convention when writing them to help it in which we declare such
variables so that findGlobals can locate them:
plotFn1 - function() {
xGlobal; yGlobal
plot(yGlobal ~ xGlobal)
}
findGlobals(plotFn1)
--
Statistics Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
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Re: [R] rgl package and animation
On 12-11-02 7:47 PM, Robert Baer wrote:
I am trying to figure out how to use rgl package for animation. It
appears that this is done using the play3d() function. Below I have
some sample code that plots a 3D path and puts a sphere at the point
farthest from the origin (which in this case also appears to be at the
end of the path). What I would like to do is animate the movement of
another sphere along the length of the path while simultaneously
rotating the viewport.
Duncan Murdock's (wonderful) Braided Knot YouTube video:
(http://www.youtube.com/watch?v=prdZWQD7L5c)
makes it clear that such things can be done, but I am having trouble
understanding how to construct the f(time) function that gets passed to
play3d(). The demo(flag) example is a little helpful, but I still can't
quite translate it to my problem.
Can anyone point to some some simple f(time) function examples that I
could use for reference or give me a little hint as to how to construct
f(time) for movement along the path while simultaneously rotating the
viewport?
Thanks,
Rob
library(rgl)
# Generate a 3D path
dat -
structure(list(X = c(0, 0.06181308, 0.002235635,
-0.03080658, -0.1728054, -0.372467, -0.5877065,
-0.8814848, -1.103668, -1.366157, -1.625862, -1.948066,
-2.265388, -2.689826, -3.095001, -3.49749, -3.946068,
-4.395653, -4.772034, -5.111259, -5.410515, -5.649475, -5.73439,
-5.662201, -5.567145, -5.390334, -5.081581, -4.796631,
-4.496559, -4.457024, -4.459564, -4.641746, -4.849105,
-5.0899430001, -5.43129, -5.763724, -6.199448, -6.517578,
-6.864234, -6.907439), Y = c(0, -0.100724,
-0.1694719,
0.036505999886, -0.09299519, -0.222977, -0.3557596,
-0.3658229, -0.3299489, -0.2095574,
-0.08041446,
0.02013388, 0.295372, 0.1388314, 0.2811047,
0.2237614, 0.1419052, 0.06029464,
-0.09330875,
-0.2075969, -0.3286296, -0.4385684,
-0.4691093,
-0.6235059, -0.5254676, -0.568444, -0.6388859,
-0.727356, -1.073769, -1.0321350001, -1.203461, -1.438637,
-1.6502310001, -1.861351, -2.169083, -2.4314730001,
-2.6991430001,
-2.961258, -3.239381, -3.466103), Z = c(0, 0.1355290002,
0.40106200024, 1.216374, 1.5539550003, 1.7308050003,
1.8116760003, 2.185124, 2.5260320004, 3.034794,
3.4265440004, 3.822512, 4.7449040002, 4.644837,
5.4184880002, 5.8586730001, 6.378356, 6.8339540001,
7.216339, 7.5941160004, 7.9559020004, 8.352936, 8.709319,
9.0166930003, 9.4855350003, 9.9000550001, 10.397003,
10.932068, 11.025726, 12.334595, 13.177887, 13.741852, 14.61142,
15.351013, 16.161255, 16.932831, 17.897186, 18.826691, 19.776001,
20.735596), time = c(0, 0.0116, 0.0196,
0.0311, 0.0391, 0.0507,
0.0623,
0.0703, 0.0818, 0.0899,
0.101,
0.109, 0.121, 0.129, 0.141,
0.152, 0.16, 0.172, 0.18, 0.191,
0.199, 0.211, 0.222, 0.23,
0.242, 0.25, 0.262, 0.27, 0.281,
0.289, 0.301, 0.312, 0.32,
0.332, 0.34, 0.351, 0.359,
0.371, 0.379, 0.391)), .Names = c(X,
Y, Z, time), row.names = c(1844, 1845, 1846, 1847,
1848, 1849, 1850, 1851, 1852, 1853, 1854, 1855,
1856, 1857, 1858, 1859, 1860, 1861, 1862, 1863,
1864, 1865, 1866, 1867, 1868, 1869, 1870, 1871,
1872, 1873, 1874, 1875, 1876, 1877, 1878, 1879,
1880, 1881, 1882, 1883), class = data.frame)
# Plot 3d path
with(dat, plot3d(X,Y,Z, type = 'l', col = 'blue', lty = 1))
# get absolute distance from origin
dat$r = sqrt(dat$X ^ 2 + dat$Y ^ 2 + dat$Z ^ 2)
mxpnt = dat[dat$r == mr,] # Coordinates of furthest point
# Plot a blue sphere at max distance
plot3d(mxpnt$X, mxpnt$Y, mxpnt$Z, type = 's', radius = 1, col = 'blue',
add = TRUE)
Your code didn't include the mr variable, but I assume it's just
max(dat$r). With that assumption, I'd do the animation function as follows:
First, draw the new sphere at the first point and save the object id:
sphereid - sphere3d(dat[1,c(X, Y, Z)], col=red, radius=1)
# Also save the spinner that you like:
spin - spin3d( ) #maybe with different parms
# Now, the animation function:
f - function(time) {
par3d(skipRedraw = TRUE) # stops intermediate redraws
on.exit(par3d(skipRedraw=FALSE)) # redraw at the end
rgl.pop(id=sphereid) # delete the old sphere
pt - time %% 40 + 1 # compute which one to draw
pnt - dat[pt, c(X, Y, Z)] # maybe interpolate instead?
sphereid - spheres3d(pnt, radius=1,
Re: [R] How to make pch symbols thicker?
On Nov 2, 2012, at 10:06 PM, 21rosit wrote: Hi I need to know how to make pch symbols like pch=3 (+) or pch=4(x) or even the border of squares or triangles thicker without changing the size. I have a lot of symbols of different colors but you can't see the colors clearly and I don't want to change the symbol. Thanks! Hi, Have you tried the lwd= argument? Like this... plot(1:10, pch = 3, lwd = 4) Cheers, Ben Ben Tupper Bigelow Laboratory for Ocean Sciences 180 McKown Point Rd. P.O. Box 475 West Boothbay Harbor, Maine 04575-0475 http://www.bigelow.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Bioconductor, merging annotation with list of probeids
i will sorry! anyway it's a data.frame object. isn't that good? -- View this message in context: http://r.789695.n4.nabble.com/Bioconductor-merging-annotation-with-list-of-probeids-tp4648251p4648305.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] reorder() in the latticeExtra library
Thanks David, I used you suggestion and it worked fine, please see below for what I did. segplot(reorder(factor(genotype), genotype) ~ lower + upper On Nov 3, 2012, at 2:47 AM, David Winsemius wrote: define genotype as a factor [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Contrasts in manova
Hi everybody I am trying to find contrast in MANOVA. I used next code contrasts(ffage)-ctr contrasts(ffage) MANOVA.agec-manova(Y1~ffage,data=vol18.df) summary(MANOVA.agec, split =list (ffage=list(0-17 v over 18=0, 18-25 v over 26=1, 26-31 v over 32=2, 32-42 v over 43=3, 43-65 v 66+=4))) But the output was only the overall fit contrasts(ffage)-ctr contrasts(ffage) [,1] [,2] [,3] [,4] [,5] 050000 1 -14000 2 -1 -1300 3 -1 -1 -120 4 -1 -1 -1 -11 5 -1 -1 -1 -1 -1 MANOVA.agec-manova(Y1~ffage,data=vol18.df) summary(MANOVA.agec, split =list (ffage=list(0-17 v over 18=0, 18-25 v over 26=1, 26-31 v over 32=2, 32-42 v over 43=3, 43-65 v 66+=4))) Df Pillai approx F num Df den Df Pr(F) ffage 5 0.30607 2.1543 20520 0.002681 ** Residuals 130 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 I try to use linearHypothesis, but i couldnt find anything, maybe I didnt use in a corect way, *any advice is welcome* I always get the error Error in solve.default(wcrossprod(model.matrix(model), w = wts)) : Lapack routine dgesv: system is exactly singular Thanks in advance Paola -- View this message in context: http://r.789695.n4.nabble.com/Contrasts-in-manova-tp4648306.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Bioconductor, merging annotation with list of probeids
On 03.11.2012 14:56, Brawni wrote: i will sorry! anyway it's a data.frame object. isn't that good? And what are you referring to? I do not see any citation in this message? A, some Nabble generated mail... Please do read the posting guide to this mailing list. Uwe Ligges -- View this message in context: http://r.789695.n4.nabble.com/Bioconductor-merging-annotation-with-list-of-probeids-tp4648251p4648305.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] mergeing a large number of large .csvs
On the absence of any data examples from you per the posting guidelines, I will
refer you to the help files for the melt function in the reshape2 package.
Note that there can be various mixtures of wide versus long... such as a wide
file with one date column and columns representing all stock prices and all
trade volumes. The longest format would be what melt gives (date, column name,
and value) but an in-between format would have one distinct column each for
dollar values and volume values with a column indicating ticker label and of
course another for date.
If your csv files can be grouped according to those with similar column
types, then as you read them in you can use cbind( csvlabel=somelabel,
csvdf) to distinguish it and then rbind those data frames together to create an
intermediate-width data frame. When dealing with large amounts of data you will
want to minimize the amount of reshaping you do, but it would require knowledge
of your data and algorithms to say any more.
---
Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
Benjamin Caldwell btcaldw...@berkeley.edu wrote:
Jeff,
If you're willing to educate, I'd be happy to learn what wide vs long
format means. I'll give rbind a shot in the meantime.
Ben
On Nov 2, 2012 4:31 PM, Jeff Newmiller jdnew...@dcn.davis.ca.us
wrote:
I would first confirm that you need the data in wide format... many
algorithms are more efficient in long format anyway, and rbind is way
more
efficient than merge.
If you feel this is not negotiable, you may want to consider sqldf.
Yes,
you need to learn a bit of SQL, but it is very well integrated into
R.
---
Jeff NewmillerThe . . Go
Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live
Go...
Live: OO#.. Dead: OO#..
Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#.
rocks...1k
---
Sent from my phone. Please excuse my brevity.
Benjamin Caldwell btcaldw...@berkeley.edu wrote:
Dear R help;
I'm currently trying to combine a large number (about 30 x 30) of
large
.csvs together (each at least 1 records). They are organized by
plots,
hence 30 X 30, with each group of csvs in a folder which corresponds
to
the
plot. The unmerged csvs all have the same number of columns (5). The
fifth
column has a different name for each csv. The number of rows is
different.
The combined csvs are of course quite large, and the code I'm
running
is
quite slow - I'm currently running it on a computer with 10 GB ram,
ssd,
and quad core 2.3 ghz processor; it's taken 8 hours and it's only
75%
of
the way through (it's hung up on one of the largest data groupings
now
for
an hour, and using 3.5 gigs of RAM.
I know that R isn't the most efficient way of doing this, but I'm
not
familiar with sql or C. I wonder if anyone has suggestions for a
different
way to do this in the R environment. For instance, the key function
now
is
merge, but I haven't tried join from the plyr package or rbind from
base.
I'm willing to provide a dropbox link to a couple of these files if
you'd
like to see the data. My code is as follows:
#multmerge is based on code by Tony cookson,
http://www.r-bloggers.com/merging-multiple-data-files-into-one-data-frame/
;
The function takes a path. This path should be the name of a folder
that
contains all of the files you would like to read and merge together
and
only those files you would like to merge.
multmerge = function(mypath){
filenames=list.files(path=mypath, full.names=TRUE)
datalist = try(lapply(filenames,
function(x){read.csv(file=x,header=T)}))
try(Reduce(function(x,y) {merge(x, y, all=TRUE)}, datalist))
}
#this function renames files using a fixed list and outputs a .csv
merepk - function (path, nf.name) {
output-multmerge(mypath=path)
name - list(x, y, z, depth, amplitude)
try(names(output) - name)
write.csv(output, nf.name)
}
#assumes all folders are in the same directory, with nothing else
there
merge.by.folder - function (folderpath){
foldernames-list.files(path=folderpath)
n- length(foldernames)
setwd(folderpath)
for (i in 1:n){
path-paste(folderpath,foldernames[i],
Re: [R] rgl package and animation
On 11/3/2012 6:47 AM, Duncan Murdoch wrote:
On 12-11-02 7:47 PM, Robert Baer wrote:
I am trying to figure out how to use rgl package for animation. It
appears that this is done using the play3d() function. Below I have
some sample code that plots a 3D path and puts a sphere at the point
farthest from the origin (which in this case also appears to be at the
end of the path). What I would like to do is animate the movement of
another sphere along the length of the path while simultaneously
rotating the viewport.
Duncan Murdock's (wonderful) Braided Knot YouTube video:
(http://www.youtube.com/watch?v=prdZWQD7L5c)
makes it clear that such things can be done, but I am having trouble
understanding how to construct the f(time) function that gets passed to
play3d(). The demo(flag) example is a little helpful, but I still can't
quite translate it to my problem.
Can anyone point to some some simple f(time) function examples that I
could use for reference or give me a little hint as to how to construct
f(time) for movement along the path while simultaneously rotating the
viewport?
Thanks,
Rob
library(rgl)
# Generate a 3D path
dat -
structure(list(X = c(0, 0.06181308, 0.002235635,
-0.03080658, -0.1728054, -0.372467, -0.5877065,
-0.8814848, -1.103668, -1.366157, -1.625862, -1.948066,
-2.265388, -2.689826, -3.095001, -3.49749, -3.946068,
-4.395653, -4.772034, -5.111259, -5.410515, -5.649475, -5.73439,
-5.662201, -5.567145, -5.390334, -5.081581, -4.796631,
-4.496559, -4.457024, -4.459564, -4.641746, -4.849105,
-5.0899430001, -5.43129, -5.763724, -6.199448, -6.517578,
-6.864234, -6.907439), Y = c(0, -0.100724,
-0.1694719,
0.036505999886, -0.09299519, -0.222977, -0.3557596,
-0.3658229, -0.3299489, -0.2095574,
-0.08041446,
0.02013388, 0.295372, 0.1388314, 0.2811047,
0.2237614, 0.1419052, 0.06029464,
-0.09330875,
-0.2075969, -0.3286296, -0.4385684,
-0.4691093,
-0.6235059, -0.5254676, -0.568444, -0.6388859,
-0.727356, -1.073769, -1.0321350001, -1.203461, -1.438637,
-1.6502310001, -1.861351, -2.169083, -2.4314730001,
-2.6991430001,
-2.961258, -3.239381, -3.466103), Z = c(0, 0.1355290002,
0.40106200024, 1.216374, 1.5539550003, 1.7308050003,
1.8116760003, 2.185124, 2.5260320004, 3.034794,
3.4265440004, 3.822512, 4.7449040002, 4.644837,
5.4184880002, 5.8586730001, 6.378356, 6.8339540001,
7.216339, 7.5941160004, 7.9559020004, 8.352936,
8.709319,
9.0166930003, 9.4855350003, 9.9000550001, 10.397003,
10.932068, 11.025726, 12.334595, 13.177887, 13.741852, 14.61142,
15.351013, 16.161255, 16.932831, 17.897186, 18.826691, 19.776001,
20.735596), time = c(0, 0.0116, 0.0196,
0.0311, 0.0391, 0.0507,
0.0623,
0.0703, 0.0818, 0.0899,
0.101,
0.109, 0.121, 0.129,
0.141,
0.152, 0.16, 0.172, 0.18, 0.191,
0.199, 0.211, 0.222, 0.23,
0.242, 0.25, 0.262, 0.27, 0.281,
0.289, 0.301, 0.312, 0.32,
0.332, 0.34, 0.351, 0.359,
0.371, 0.379, 0.391)), .Names =
c(X,
Y, Z, time), row.names = c(1844, 1845, 1846, 1847,
1848, 1849, 1850, 1851, 1852, 1853, 1854, 1855,
1856, 1857, 1858, 1859, 1860, 1861, 1862, 1863,
1864, 1865, 1866, 1867, 1868, 1869, 1870, 1871,
1872, 1873, 1874, 1875, 1876, 1877, 1878, 1879,
1880, 1881, 1882, 1883), class = data.frame)
# Plot 3d path
with(dat, plot3d(X,Y,Z, type = 'l', col = 'blue', lty = 1))
# get absolute distance from origin
dat$r = sqrt(dat$X ^ 2 + dat$Y ^ 2 + dat$Z ^ 2)
mr = max(dat$r) # yes sorry, didn't get copied to original
email code
mxpnt = dat[dat$r == mr,] # Coordinates of furthest point
# Plot a blue sphere at max distance
plot3d(mxpnt$X, mxpnt$Y, mxpnt$Z, type = 's', radius = 1, col = 'blue',
add = TRUE)
Your code didn't include the mr variable, but I assume it's just
max(dat$r). With that assumption, I'd do the animation function as
follows:
First, draw the new sphere at the first point and save the object id:
sphereid - sphere3d(dat[1,c(X, Y, Z)], col=red, radius=1)
# Also save the spinner that you like:
spin - spin3d( ) #maybe with different parms
# Now, the animation function:
f - function(time) {
par3d(skipRedraw = TRUE) # stops intermediate redraws
on.exit(par3d(skipRedraw=FALSE)) # redraw at the end
rgl.pop(id=sphereid) # delete the old sphere
pt - time %% 40
Re: [R] override date in xts time series
Hello Arun, I too am using R 2.15 and am unable to get the same result as you. You will notice in the R code that follows that when I use 'update' the time in the xts object goes haywire. For example, 2004-04-04 01:15:00 EST gets converted to 2004-01-03 22:15:00 PST (see below). Because I can't get the result you showed me in your previous response, and maybe you won't be able to get this result, I've resorted back to your other suggestion using gsub. I don't have a good handle on regular expressions and was wondering if in the last line of code below, the replace month is 'hardwired'? In other words, could \\101\\2 somehow be replaced with unique(month(index(x.1)))in the last line of code below so that x.1 is providing the replacement month, rather than have it fixed? Or perhaps I've misunderstood the regular expression, which is entirely possible. sessionInfo() #R version 2.15.2 (2012-10-26) #Platform: x86_64-w64-mingw32/x64 (64-bit) library(xts) library(lubridate) x.Date - rep(1/1/2004,times=5) x.Times- c(01:15:00, 01:30:00, 01:45:00, 02:00:00, 02:30:00, 03:00:00, 03:15:00) x-paste(x.Date,x.Times) y.Date - rep(4/4/2004,times=4) y.Times- c(01:15:00, 01:30:00, 01:45:00, 02:00:00, 02:30:00, 03:30:00) y-paste(y.Date,y.Times) fmt - %m/%d/%Y %H:%M:%S x.1-xts(1:7, as.POSIXct(x, format=fmt, tz = EST)) y.1-xts(1:6, as.POSIXct(y, format=fmt, tz = EST)) y.1 #[,1] #2004-04-04 01:15:001 #2004-04-04 01:30:002 #2004-04-04 01:45:003 #2004-04-04 02:00:004 #2004-04-04 02:30:005 #2004-04-04 03:30:006 #Warning message: #timezone of object (EST) is different than current timezone (). index(y.1) # 2004-04-04 01:15:00 EST 2004-04-04 01:30:00 EST # 2004-04-04 01:45:00 EST 2004-04-04 02:00:00 EST # 2004-04-04 02:30:00 EST 2004-04-04 03:30:00 EST index(y.1)-update(index(y.1),month=unique(month(index(x.1 index(y.1) # 2004-01-03 22:15:00 PST 2004-01-03 22:30:00 PST # 2004-01-03 22:45:00 PST 2004-01-03 23:00:00 PST # 2004-01-03 23:30:00 PST 2004-01-04 00:30:00 PST index(y.1)-as.POSIXct(gsub((.*\\-).*(\\-.*),\\101\\2,index(y.1))) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] AUTO: Alan Chalk has left RSA (returning 30/11/2012)
I am out of the office until 30/11/2012. Please send work related emails to laura.jor...@uk.rsagroup.com or personal emails to alanch...@gmail.com. Note: This is an automated response to your message R-help Digest, Vol 117, Issue 3 sent on 03/11/2012 11:00:07. This is the only notification you will receive while this person is away. Please consider the environment - Think before you print RSA -The UK's first carbon neutral insurer *** Royal Sun Alliance Insurance plc (No. 93792). Registered in England Wales at St. Mark's Court, Chart Way, Horsham, West Sussex, RH12 1XL. Authorised and Regulated by the Financial Services Authority. For your protection, telephone calls may be recorded and monitored. The information in this e-mail is confidential and may be read, copied or used only by the intended recipients. If you have received it in error please contact the sender immediately by returning the e-mail. Please delete the e-mail and do not disclose any of its contents to anyone. No responsibility is accepted for loss or damage arising from viruses or changes made to this message after it was sent. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] override date in xts time series
Sys.setenv(TZ=GMT) did the trick! Thank you very much. I'll continue to work the larger problem with this option. Out of curiosity, however, can the following code be modified so that the replacement argument is informed by the month of x.1?: index(y.1)-as.POSIXct(gsub((.*\\-).*(\\-.*),\\101\\2,index(y.1))) Something to the tune of the following seems to work, but is it robust?: txt-paste(\\10,as.character(unique(month(index(x.1,\\2,sep=) index(y.1)-as.POSIXct(gsub((.*\\-).*(\\-.*),txt,index(y.1))) index(y.1) # 2004-01-04 01:15:00 PST 2004-01-04 01:30:00 PST # 2004-01-04 01:45:00 PST 2004-01-04 02:00:00 PST # 2004-01-04 02:30:00 PST 2004-01-04 03:30:00 PST What would the gsub 'pattern' string be to replace the day, if I may ask? I'm not trying to push my luck, but the gsub approach is new to me and don't quite follow everything that is going on. -Eric [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] override date in xts time series
Hi, Sorry, I forgot to answer the second question. txt-paste(\\10,unique(month(index(x.1))),\\2,sep=) #without the as.character() also should work #because str(paste(\\10,unique(month(index(x.1))),\\2,sep=)) # it returns a character # chr \\101\\2 #Here too: str(paste(10,unique(month(index(x.1))),2,sep=)) # chr 1012 #According to the description in paste() Concatenate vectors after converting to character. as.POSIXct(gsub((.*\\-).*(\\-.*),txt,index(y.1))) #[1] 2004-01-04 01:15:00 EST 2004-01-04 01:30:00 EST #[3] 2004-01-04 01:45:00 EST 2004-01-04 02:00:00 EST #[5] 2004-01-04 02:30:00 EST 2004-01-04 03:30:00 EST #Now, suppose if I want to change both the month and day from the original y.1 index(y.1)-as.POSIXct(gsub((.*\\-).*(\\-).*(\\s.*),\\101\\207\\3,index(y.1))) #Here, the month will be 01 and day 07 y.1 # [,1] #2004-01-07 01:15:00 1 #2004-01-07 01:30:00 2 #2004-01-07 01:45:00 3 #2004-01-07 02:00:00 4 #2004-01-07 02:30:00 5 #2004-01-07 03:30:00 6 Hope it helps. A.K. From: Eric Morway emor...@usgs.gov To: arun smartpink...@yahoo.com Cc: R help r-help@r-project.org Sent: Saturday, November 3, 2012 12:32 PM Subject: Re: [R] override date in xts time series Sys.setenv(TZ=GMT) did the trick! Thank you very much. I'll continue to work the larger problem with this option. Out of curiosity, however, can the following code be modified so that the replacementargument is informed by the month of x.1?: index(y.1)-as.POSIXct(gsub((.*\\-).*(\\-.*),\\101\\2,index(y.1))) Something to the tune of the following seems to work, but is it robust?: txt-paste(\\10,as.character(unique(month(index(x.1,\\2,sep=) index(y.1)-as.POSIXct(gsub((.*\\-).*(\\-.*),txt,index(y.1))) index(y.1) # 2004-01-04 01:15:00 PST 2004-01-04 01:30:00 PST # 2004-01-04 01:45:00 PST 2004-01-04 02:00:00 PST # 2004-01-04 02:30:00 PST 2004-01-04 03:30:00 PST What would the gsub 'pattern' string be to replace the day, if I may ask? I'm not trying to push my luck, but the gsub approach is new to me and don't quite follow everything that is going on. -Eric __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Replacing NAs in long format
Hi, I have the following data: data[1:20,c(1,2,20)] idr schyear year 1 80 1 91 1 10 NA 2 4 NA 2 5 -1 2 60 2 71 2 82 2 93 2 104 2 11 NA 2 126 3 4 NA 3 5 -2 3 6 -1 3 70 3 81 3 92 3 103 3 11 NA What I want to do is replace the NAs in the year variable with the following: idr schyear year 1 80 1 91 1 10 2 2 4 -2 2 5 -1 2 60 2 71 2 82 2 93 2 104 2 11 5 2 126 3 4 -3 3 5 -2 3 6 -1 3 70 3 81 3 92 3 103 3 11 4 I have no idea how to do this. What it needs to do is make sure that for each subject (idr) that it either adds a 1 if it is preceded by a value in year or subtracts a 1 if it comes before a year value. Does that make sense? I could do this in Excel but I am at a loss for how to do this in R. Please reply to me as well as the list if you respond. Thanks! Chris [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] reorder() in the latticeExtra library
On Nov 3, 2012, at 6:36 AM, Jorge Dinis wrote: Thanks David, I used you suggestion and it worked fine, please see below for what I did. segplot(reorder(factor(genotype), genotype) ~ lower + upper Perhaps a missing close-paren . ^ Although reading this as a formatted posting such as you sent might cause a registration error. On Nov 3, 2012, at 2:47 AM, David Winsemius wrote: define genotype as a factor -- David Winsemius, MD Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Replacing NAs in long format
Hello,
Try the following. I've called your data.frames 'dat' and 'dat2'
# First your datasets, see ?dput
dput(dat)
structure(list(idr = c(1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), schyear = c(8L, 9L,
10L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 4L, 5L, 6L, 7L, 8L,
9L, 10L, 11L), year = c(0L, 1L, NA, NA, -1L, 0L, 1L, 2L, 3L,
4L, NA, 6L, NA, -2L, -1L, 0L, 1L, 2L, 3L, NA)), .Names = c(idr,
schyear, year), class = data.frame, row.names = c(NA, -20L
))
dput(dat2)
structure(list(idr = c(1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), schyear = c(8L, 9L,
10L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 4L, 5L, 6L, 7L, 8L,
9L, 10L, 11L), year = c(0L, 1L, 2L, -2L, -1L, 0L, 1L, 2L, 3L,
4L, 5L, 6L, -3L, -2L, -1L, 0L, 1L, 2L, 3L, 4L)), .Names = c(idr,
schyear, year), class = data.frame, row.names = c(NA, -20L
))
# Now the code
fun - function(x){
for(i in which(is.na(x$year))){
if(i == 1)
x$year[i] - x$year[i + 1] - 1L
else
x$year[i] - x$year[i - 1] + 1L
}
x
}
result - do.call(rbind, lapply(split(dat, dat$idr), fun))
all.equal(result, dat2)
Hope this helps,
Rui Barradas
Em 03-11-2012 17:14, Christopher Desjardins escreveu:
Hi,
I have the following data:
data[1:20,c(1,2,20)]
idr schyear year
1 80
1 91
1 10 NA
2 4 NA
2 5 -1
2 60
2 71
2 82
2 93
2 104
2 11 NA
2 126
3 4 NA
3 5 -2
3 6 -1
3 70
3 81
3 92
3 103
3 11 NA
What I want to do is replace the NAs in the year variable with the
following:
idr schyear year
1 80
1 91
1 10 2
2 4 -2
2 5 -1
2 60
2 71
2 82
2 93
2 104
2 11 5
2 126
3 4 -3
3 5 -2
3 6 -1
3 70
3 81
3 92
3 103
3 11 4
I have no idea how to do this. What it needs to do is make sure that for
each subject (idr) that it either adds a 1 if it is preceded by a value in
year or subtracts a 1 if it comes before a year value.
Does that make sense? I could do this in Excel but I am at a loss for how
to do this in R. Please reply to me as well as the list if you respond.
Thanks!
Chris
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] finding global variables in a function containing formulae
findGlobals must be explicitly ignoring calls to the ~ function.
You could poke through the source code of codetools and find
where this is happening.
Or, if you have the source code for the package you are investigating,
use sed to change all ~ to %TILDE% and then use findGlobals on
the resulting source code. The messages will be a bit garbled but
should give you a start. E.g., compare the following two, in which y
is defined in the function but x is not:
findGlobals(function(y)lm(y~x))
[1] ~ lm
findGlobals(function(y)lm(y %TILDE% x))
[1] lm %TILDE% x
You will get false alarms, since in a call like lm(y~x+z, data=dat) findGlobals
cannot know if dat includes columns called 'x', 'y', and 'z' and the above
approach errs on the side of reporting the potential problem.
You could use code in codetools to analyze S code instead of source code
to globally replace all calls to ~ with calls to %TILDE% but that is more
work than using sed on the source code.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of Hafen, Ryan P
Sent: Friday, November 02, 2012 4:28 PM
To: Bert Gunter
Cc: r-help@r-project.org
Subject: Re: [R] finding global variables in a function containing formulae
Thanks. That works if I a have the formula expression handy. But suppose
I want a function, findGlobalVars() that takes a function as an argument
and finds globals in it, where I have absolutely no idea what is in the
supplied function:
findGlobalVars - function(f) {
require(codetools)
findGlobals(f, merge=FALSE)$variables
}
findGlobalVars(plotFn1)
I would like findGlobalVars() to be able to find variables in formulae
that might be present in f.
On 11/1/12 1:19 PM, Bert Gunter gunter.ber...@gene.com wrote:
Does
?all.vars
##as in
all.vars(y~x)
[1] y x
help?
-- Bert
On Thu, Nov 1, 2012 at 11:04 AM, Hafen, Ryan P ryan.ha...@pnnl.gov
wrote:
I need to find all global variables being used in a function and
findGlobals() in the codetools package works quite nicely. However, I
am not able to find variables that are used in formulae. Simply
avoiding formulae in functions is not an option because I do not have
control over what functions this will be applied to.
Here is an example to illustrate:
library(codetools)
xGlobal - rnorm(10)
yGlobal - rnorm(10)
plotFn1 - function() {
plot(yGlobal ~ xGlobal)
}
plotFn2 - function() {
y - yGlobal
x - xGlobal
plot(y ~ x)
}
plotFn3 - function() {
plot(xGlobal, yGlobal)
}
findGlobals(plotFn1, merge=FALSE)$variables
# character(0)
findGlobals(plotFn2, merge=FALSE)$variables
# [1] xGlobal yGlobal
findGlobals(plotFn3, merge=FALSE)$variables
# [1] xGlobal yGlobal
I would like to find that plotFn1 also uses globals xGlobal and
yGlobal. Any suggestions on how I might do this?
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Bert Gunter
Genentech Nonclinical Biostatistics
Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-bio
statistics/pdb-ncb-home.htm
__
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__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] some help
Hi People! I have following concern consisting of some steps to do in R: I have an ascii file (table) consisting of many columns and rows. 1. I would like to order all values of the columns one under each other. It will begin with column 1, then column 2 under column 1, column 3 under column 2 etc. until at the end there is only 1 column. How do I do it? 2. Second problem is to make a scatterplot of two variables (I think after further explanation scatter plot itself will not be needed). I have two columns of two different variables (that I produces before), column 1 with variable 1 and column 2 with variable 2. I would like to order them by one variable and 0,01 interval (the varibale values will range between 0 and 1). From each 0,01 interval (100 intervals) i want to pick the maximum and minimum value of variable 2. 3. From the obtained max and min of values of each interval i would like to make a linear least square regression. I hope someone can help me out! Thanks Stefan -- View this message in context: http://r.789695.n4.nabble.com/some-help-tp4648316.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] reorder() in the latticeExtra library
Hi,
Try this:
genotype1-factor(genotype,levels=c(CJ1450 NW 4/25/12,CJ1450 BAL 4/25/12,
CJ1450 NW\n4/27/12,
CJ1450 BAL 4/27/12, CJ1721 NW 4/27/12, CJ1721 BAL\n4/27/12,
CJ1721 NW 4/29/12, CJ1721 BAL 4/29/12) )
segplot(genotype1 ~ lower + upper, data = x, draw.bands =
FALSE, centers = est, segments.fun = panel.arrows, ends = both, angle =
90, length = 0, par.settings = simpleTheme(pch = 19, col = 1), xlab =
expression(nucleotide diversity %+-% sd), panel = function(x, y, z,
...) {
panel.abline(h = z, col = grey, lty = dashed)
panel.abline(v = 14.20, col = grey)
panel.segplot(x, y, z, ...)})
A.K.
- Original Message -
From: JDINIS jorgemdi...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Friday, November 2, 2012 11:04 PM
Subject: [R] reorder() in the latticeExtra library
Hello all, thanks for your time and help. Below are my commands, and it
generates a really nice plot, however I am not happy with the reorder()
function. I would like the order to be the same as they appear in the
genotype variable genotype - c(CJ1450 NW 4/25/12,CJ1450 BAL
4/25/12,CJ1450 NW 4/27/12,CJ1450 BAL 4/27/12,CJ1721 NW
4/27/12,CJ1721 BAL 4/27/12,CJ1721 NW 4/29/12,CJ1721 BAL 4/29/12 )
and not as it is currently coded.
Is there any way to turn off the reorder, or set it up so the values appear
in the order above, thank you again!
(I am open to all suggestions)
JD
genotype - c(CJ1450 NW 4/25/12,CJ1450 BAL 4/25/12,CJ1450 NW
4/27/12,CJ1450 BAL 4/27/12,CJ1721 NW 4/27/12,CJ1721 BAL
4/27/12,CJ1721 NW 4/29/12,CJ1721 BAL 4/29/12 )
#paste(Animal, as.roman(1:8), sep = -)
plant.height - c(0.001173003, 0.001506127, 0.001361596, 0.001922572,
0.034272147, 0.030466017, 0.001654299, 0.001071724)
SE - c(0.000444123, 0.000290096, 0.000372844, 0.00197687, 0.033945128,
0.035231568, 0.001094518, 0.000423545)
lower - plant.height - SE; upper - plant.height + SE
x - data.frame(group = genotype, lower = lower, est = plant.height, upper =
upper)
library(latticeExtra)
segplot(reorder(genotype, est) ~ lower + upper, data = x, draw.bands =
FALSE, centers = est, segments.fun = panel.arrows, ends = both, angle =
90, length = 0, par.settings = simpleTheme(pch = 19, col = 1), xlab =
expression(nucleotide diversity %+-% sd), panel = function(x, y, z,
...) {
panel.abline(h = z, col = grey, lty = dashed)
panel.abline(v = 14.20, col = grey)
panel.segplot(x, y, z, ...)})
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[R] optim .C / Crashing on run
Hello, I am attempting to use optim under the default Nelder-Mead algorithm for model fitting, minimizing a Chi^2 statistic whose value is determined by a .C call to an external shared library compiled from C C++ code. My problem has been that the R session will immediately crash upon starting the simplex run, without it taking a single step. This is strange, as the .C call itself works, is error-free (as far as I can tell!) does not return NAN or Inf under any initial starting parameters that I have tested it with in R. It only ever crashes the R session when the Chi^2 function to be minimized is called from optim, not under any other circumstances. In the interests of reproducibility, I attach R code that reads attached data files attempts a N-M optim run. The required shared library containing the external code (compiled in Ubuntu 12.04 x64 with g++ 4.6.3) is also attached. Calculating an initial Chi^2 value for a starting set of model parameters works, then the R session crashes when the optim call is made. Is there something I'm perhaps doing wrong in the specification of the optim run? Is it inadvisable to use external code with optim? There doesn't seem to be a problem with the external code itself, so I'm very stumped as to the source of the crashes. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] to print system.time always
Hi all; I want to print system.time whenever I execute any command. It takes too much time to type system.time() function to all command. is there any solution on it? And, apply(matrix,1,cumsum) command is too slow to some large matrix. is there any function like rowCumSums ? thank u! -- View this message in context: http://r.789695.n4.nabble.com/to-print-system-time-always-tp4648314.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] override date in xts time series
HI, Could you check whether you are getting the same result with tz=GMT? as.POSIXct(x,format=fmt,tz=GMT) #[1] 2004-01-01 01:15:00 GMT 2004-01-01 01:30:00 GMT #[3] 2004-01-01 01:45:00 GMT 2004-01-01 02:00:00 GMT #[5] 2004-01-01 02:30:00 GMT 2004-01-01 03:00:00 GMT #[7] 2004-01-01 03:15:00 GMT as.POSIXct(y,format=fmt,tz=GMT) #[1] 2004-04-04 01:15:00 GMT 2004-04-04 01:30:00 GMT #[3] 2004-04-04 01:45:00 GMT 2004-04-04 02:00:00 GMT #[5] 2004-04-04 02:30:00 GMT 2004-04-04 03:30:00 GMT x.1-xts(1:7,as.POSIXct(x,format=fmt,tz=GMT)) x.1 # [,1] #2004-01-01 01:15:00 1 #2004-01-01 01:30:00 2 #2004-01-01 01:45:00 3 #2004-01-01 02:00:00 4 #2004-01-01 02:30:00 5 #2004-01-01 03:00:00 6 #2004-01-01 03:15:00 7 #Warning message: #timezone of object (GMT) is different than current timezone (). y.1-xts(1:6,as.POSIXct(y,format=fmt,tz=GMT)) y.1 # [,1] #2004-04-04 01:15:00 1 #2004-04-04 01:30:00 2 #2004-04-04 01:45:00 3 #2004-04-04 02:00:00 4 #2004-04-04 02:30:00 5 #2004-04-04 03:30:00 6 #Warning message: #timezone of object (GMT) is different than current timezone (). update(index(y.1),month=unique(month(index(x.1 #[1] 2004-01-04 01:15:00 GMT 2004-01-04 01:30:00 GMT #[3] 2004-01-04 01:45:00 GMT 2004-01-04 02:00:00 GMT #[5] 2004-01-04 02:30:00 GMT 2004-01-04 03:30:00 GMT #Here is where the problem occurs index(y.1)-update(index(y.1),month=unique(month(index(x.1 y.1 # [,1] #2004-01-03 20:15:00 1 #2004-01-03 20:30:00 2 #2004-01-03 20:45:00 3 #2004-01-03 21:00:00 4 #2004-01-03 21:30:00 5 #2004-01-03 22:30:00 6 #Warning message: #timezone of object (GMT) is different than current timezone (). #So I am going to change the timezone in the system and see what happens Sys.setenv(TZ=GMT) y.1-xts(1:6,as.POSIXct(y,format=fmt,tz=GMT)) update(index(y.1),month=unique(month(index(x.1 #[1] 2004-01-04 01:15:00 GMT 2004-01-04 01:30:00 GMT #[3] 2004-01-04 01:45:00 GMT 2004-01-04 02:00:00 GMT #[5] 2004-01-04 02:30:00 GMT 2004-01-04 03:30:00 GMT index(y.1)-update(index(y.1),month=unique(month(index(x.1 y.1 # [,1] #2004-01-04 01:15:00 1 #2004-01-04 01:30:00 2 #2004-01-04 01:45:00 3 #2004-01-04 02:00:00 4 #2004-01-04 02:30:00 5 #2004-01-04 03:30:00 6 A.K. From: Eric Morway emor...@usgs.gov To: arun smartpink...@yahoo.com Cc: R help r-help@r-project.org Sent: Saturday, November 3, 2012 11:44 AM Subject: Re: [R] override date in xts time series Hello Arun, I too am using R 2.15 and am unable to get the same result as you. You will notice in the R code that follows that when I use 'update' the time in the xts object goes haywire. For example, 2004-04-04 01:15:00 ESTgets converted to 2004-01-03 22:15:00 PST(see below). Because I can't get the result you showed me in your previous response, and maybe you won't be able to get this result, I've resorted back to your other suggestion using gsub. I don't have a good handle on regular expressions and was wondering if in the last line of code below, the replace month is 'hardwired'? In other words, could \\101\\2somehow be replaced with unique(month(index(x.1)))in the last line of code below so that x.1 is providing the replacement month, rather than have it fixed? Or perhaps I've misunderstood the regular expression, which is entirely possible. sessionInfo() #R version 2.15.2 (2012-10-26) #Platform: x86_64-w64-mingw32/x64 (64-bit) library(xts) library(lubridate) x.Date - rep(1/1/2004,times=5) x.Times- c(01:15:00, 01:30:00, 01:45:00, 02:00:00, 02:30:00, 03:00:00, 03:15:00) x-paste(x.Date,x.Times) y.Date - rep(4/4/2004,times=4) y.Times- c(01:15:00, 01:30:00, 01:45:00, 02:00:00, 02:30:00, 03:30:00) y-paste(y.Date,y.Times) fmt - %m/%d/%Y %H:%M:%S x.1-xts(1:7, as.POSIXct(x, format=fmt, tz = EST)) y.1-xts(1:6, as.POSIXct(y, format=fmt, tz = EST)) y.1 # [,1] #2004-04-04 01:15:00 1 #2004-04-04 01:30:00 2 #2004-04-04 01:45:00 3 #2004-04-04 02:00:00 4 #2004-04-04 02:30:00 5 #2004-04-04 03:30:00 6 #Warning message: #timezone of object (EST) is different than current timezone (). index(y.1) # 2004-04-04 01:15:00 EST 2004-04-04 01:30:00 EST # 2004-04-04 01:45:00 EST 2004-04-04 02:00:00 EST # 2004-04-04 02:30:00 EST 2004-04-04 03:30:00 EST index(y.1)-update(index(y.1),month=unique(month(index(x.1 index(y.1) # 2004-01-03 22:15:00 PST 2004-01-03 22:30:00 PST # 2004-01-03 22:45:00 PST 2004-01-03 23:00:00 PST # 2004-01-03 23:30:00 PST 2004-01-04 00:30:00 PST index(y.1)-as.POSIXct(gsub((.*\\-).*(\\-.*),\\101\\2,index(y.1))) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal,
Re: [R] Replacing NAs in long format
x - read.table(text = idr schyear year
+ 1 80
+ 1 91
+ 1 10 NA
+ 2 4 NA
+ 2 5 -1
+ 2 60
+ 2 71
+ 2 82
+ 2 93
+ 2 104
+ 2 11 NA
+ 2 126
+ 3 4 NA
+ 3 5 -2
+ 3 6 -1
+ 3 70
+ 3 81
+ 3 92
+ 3 103
+ 3 11 NA, header = TRUE)
# you did not specify if there might be multiple contiguous NAs,
# so there are a lot of checks to be made
x.l - lapply(split(x, x$idr), function(.idr){
+ # check for all NAs -- just return indeterminate state
+ if (sum(is.na(.idr$year)) == nrow(.idr)) return(.idr)
+ # repeat until all NAs have been fixed; takes care of contiguous ones
+ while (any(is.na(.idr$year))){
+ # find all the NAs
+ for (i in which(is.na(.idr$year))){
+ if ((i == 1L) (!is.na(.idr$year[i + 1L]))){
+ .idr$year[i] - .idr$year[i + 1L] - 1
+ } else if ((i 1L) (!is.na(.idr$year[i - 1L]))){
+ .idr$year[i] - .idr$year[i - 1L] + 1
+ } else if ((i nrow(.idr)) (!is.na(.idr$year[i + 1L]))){
+ .idr$year[i] - .idr$year[i + 1L] -1
+ }
+ }
+ }
+ return(.idr)
+ })
do.call(rbind, x.l)
idr schyear year
1.11 80
1.21 91
1.31 102
2.42 4 -2
2.52 5 -1
2.62 60
2.72 71
2.82 82
2.92 93
2.10 2 104
2.11 2 115
2.12 2 126
3.13 3 4 -3
3.14 3 5 -2
3.15 3 6 -1
3.16 3 70
3.17 3 81
3.18 3 92
3.19 3 103
3.20 3 114
On Sat, Nov 3, 2012 at 1:14 PM, Christopher Desjardins
cddesjard...@gmail.com wrote:
Hi,
I have the following data:
data[1:20,c(1,2,20)]
idr schyear year
1 80
1 91
1 10 NA
2 4 NA
2 5 -1
2 60
2 71
2 82
2 93
2 104
2 11 NA
2 126
3 4 NA
3 5 -2
3 6 -1
3 70
3 81
3 92
3 103
3 11 NA
What I want to do is replace the NAs in the year variable with the
following:
idr schyear year
1 80
1 91
1 10 2
2 4 -2
2 5 -1
2 60
2 71
2 82
2 93
2 104
2 11 5
2 126
3 4 -3
3 5 -2
3 6 -1
3 70
3 81
3 92
3 103
3 11 4
I have no idea how to do this. What it needs to do is make sure that for
each subject (idr) that it either adds a 1 if it is preceded by a value in
year or subtracts a 1 if it comes before a year value.
Does that make sense? I could do this in Excel but I am at a loss for how
to do this in R. Please reply to me as well as the list if you respond.
Thanks!
Chris
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Re: [R] to print system.time always
Here is a faster solution to your 'apply'; use 'sapply' instead: str(x) num [1:100, 1:30] 0.0346 0.4551 0.66 0.8528 0.5494 ... system.time(y - apply(x, 1, cumsum)) user system elapsed 13.240.61 14.02 system.time(ys - sapply(1:col, function(a) cumsum(x[,a]))) user system elapsed 1.400.141.59 On Sat, Nov 3, 2012 at 11:52 AM, mrzung mrzun...@gmail.com wrote: Hi all; I want to print system.time whenever I execute any command. It takes too much time to type system.time() function to all command. is there any solution on it? And, apply(matrix,1,cumsum) command is too slow to some large matrix. is there any function like rowCumSums ? thank u! -- View this message in context: http://r.789695.n4.nabble.com/to-print-system-time-always-tp4648314.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] to print system.time always
On 03.11.2012 16:52, mrzung wrote: Hi all; I want to print system.time whenever I execute any command. It takes too much time to type system.time() function to all command. is there any solution on it? See ?Rprof on how to profile your code. And, apply(matrix,1,cumsum) command is too slow to some large matrix. is there any function like rowCumSums ? You had: result1 - apply(matrix,1,cumsum) This is only slow, if you have lots of rows. Now think in matrices how to to that: b - sapply(1:ncol(matrix), function(i) c(rep(1, i), rep(0, ncol(matrix)-i) result2 - t(x %*% b) This is roughly 10 times faster on a 100 x 10 matrix. Check the results: all.equal(result1, result2) Uwe Ligges thank u! -- View this message in context: http://r.789695.n4.nabble.com/to-print-system-time-always-tp4648314.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] to print system.time always
On 03.11.2012 19:42, jim holtman wrote: Here is a faster solution to your 'apply'; use 'sapply' instead: str(x) num [1:100, 1:30] 0.0346 0.4551 0.66 0.8528 0.5494 ... system.time(y - apply(x, 1, cumsum)) user system elapsed 13.240.61 14.02 system.time(ys - sapply(1:col, function(a) cumsum(x[,a]))) user system elapsed 1.400.141.59 Which solves another problem (cumsum of cols rather than rows). Applying it on rows won't be much faster. Uwe Ligges On Sat, Nov 3, 2012 at 11:52 AM, mrzung mrzun...@gmail.com wrote: Hi all; I want to print system.time whenever I execute any command. It takes too much time to type system.time() function to all command. is there any solution on it? And, apply(matrix,1,cumsum) command is too slow to some large matrix. is there any function like rowCumSums ? thank u! -- View this message in context: http://r.789695.n4.nabble.com/to-print-system-time-always-tp4648314.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] to print system.time always
I use notepad++ on Windows, so it is easy to add a hotkey that will
surround a block of code that you want to execute with:
system.time({..code to run..})
Usually you don't want it around each statement. I use the following
function to have it print out CPU and memory usage at various
locations in my script since it allows me to see the progress and
where time might be going:
my.stats -
local({
# local variables to hold the last times
# first two variables are the elasped and CPU times from the last report
Level- 1
MaxLevel - 30
Stack- matrix(0, ncol=2, nrow=MaxLevel)
function(text = stats, reset=FALSE, oper=)
{
procTime - proc.time()[1:3]# get current metrics
if (reset){ # setup to mark timing from this point
Level - 1 # reset the Level
Stack[Level, ] - c(procTime[3], procTime[1] + procTime[2])
}
if (oper == push){
if (Level MaxLevel) Level - Level + 1
Stack[Level, ] - c(procTime[3], procTime[1] + procTime[2])
}
.caller - sys.calls()
if (length(.caller) == 1) .caller - Rgui
else .caller - as.character(.caller[[length(.caller) - 1]])[1]
cat(sprintf(%s (%d) - %s : %s %.1f %.1f %.1f : %.1fMB\n,
text,
Level,
.caller,
format(Sys.time(), format=%H:%M:%S),
procTime[1] + procTime[2] - Stack[Level, 2],
procTime[3] - Stack[Level, 1],
procTime[3],
memory.size()))
if ((oper == pop) (Level 1)) Level - Level - 1
else if (oper == reset) Level - 1
invisible(flush.console()) # force a write to the console
}
})
It produces output like this:
my.stats('start')
start (1) - Rgui : 14:53:16 39.2 597822.1 597822.1 : 1213.8MB
system.time(for(i in 1:col) ym[, i] - cumsum(x[,i]))
user system elapsed
1.770.011.80
my.stats('done')
done (1) - Rgui : 14:53:23 41.0 597828.6 597828.6 : 1213.8MB
This says that between 'start' and 'done', 1.8 CPU seconds were used
(41.0 - 39.2) which is what syste.time was reporting.
On Sat, Nov 3, 2012 at 11:52 AM, mrzung mrzun...@gmail.com wrote:
Hi all;
I want to print system.time whenever I execute any command.
It takes too much time to type system.time() function to all command.
is there any solution on it?
And,
apply(matrix,1,cumsum) command is too slow to some large matrix.
is there any function like rowCumSums ?
thank u!
--
View this message in context:
http://r.789695.n4.nabble.com/to-print-system-time-always-tp4648314.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] mergeing a large number of large .csvs
A faster way would be to use something like 'per', 'awk' or 'sed'.
You can strip off the header line of each CSV (if it has one) and then
concatenate the files together. This is very efficient use of memory
since you are just reading one file at a time and then writing it out.
Will probably be a lot faster since no conversions have to be done.
Once you have the one large file, then you can play with it (load it
if you have enough memory, or load it into a database).
On Sat, Nov 3, 2012 at 11:37 AM, Jeff Newmiller
jdnew...@dcn.davis.ca.us wrote:
On the absence of any data examples from you per the posting guidelines, I
will refer you to the help files for the melt function in the reshape2
package. Note that there can be various mixtures of wide versus long... such
as a wide file with one date column and columns representing all stock prices
and all trade volumes. The longest format would be what melt gives (date,
column name, and value) but an in-between format would have one distinct
column each for dollar values and volume values with a column indicating
ticker label and of course another for date.
If your csv files can be grouped according to those with similar column
types, then as you read them in you can use cbind( csvlabel=somelabel,
csvdf) to distinguish it and then rbind those data frames together to create
an intermediate-width data frame. When dealing with large amounts of data you
will want to minimize the amount of reshaping you do, but it would require
knowledge of your data and algorithms to say any more.
---
Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
Benjamin Caldwell btcaldw...@berkeley.edu wrote:
Jeff,
If you're willing to educate, I'd be happy to learn what wide vs long
format means. I'll give rbind a shot in the meantime.
Ben
On Nov 2, 2012 4:31 PM, Jeff Newmiller jdnew...@dcn.davis.ca.us
wrote:
I would first confirm that you need the data in wide format... many
algorithms are more efficient in long format anyway, and rbind is way
more
efficient than merge.
If you feel this is not negotiable, you may want to consider sqldf.
Yes,
you need to learn a bit of SQL, but it is very well integrated into
R.
---
Jeff NewmillerThe . . Go
Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live
Go...
Live: OO#.. Dead: OO#..
Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#.
rocks...1k
---
Sent from my phone. Please excuse my brevity.
Benjamin Caldwell btcaldw...@berkeley.edu wrote:
Dear R help;
I'm currently trying to combine a large number (about 30 x 30) of
large
.csvs together (each at least 1 records). They are organized by
plots,
hence 30 X 30, with each group of csvs in a folder which corresponds
to
the
plot. The unmerged csvs all have the same number of columns (5). The
fifth
column has a different name for each csv. The number of rows is
different.
The combined csvs are of course quite large, and the code I'm
running
is
quite slow - I'm currently running it on a computer with 10 GB ram,
ssd,
and quad core 2.3 ghz processor; it's taken 8 hours and it's only
75%
of
the way through (it's hung up on one of the largest data groupings
now
for
an hour, and using 3.5 gigs of RAM.
I know that R isn't the most efficient way of doing this, but I'm
not
familiar with sql or C. I wonder if anyone has suggestions for a
different
way to do this in the R environment. For instance, the key function
now
is
merge, but I haven't tried join from the plyr package or rbind from
base.
I'm willing to provide a dropbox link to a couple of these files if
you'd
like to see the data. My code is as follows:
#multmerge is based on code by Tony cookson,
http://www.r-bloggers.com/merging-multiple-data-files-into-one-data-frame/
;
The function takes a path. This path should be the name of a folder
that
contains all of the files you would like to read and merge together
and
only those files you would like to merge.
multmerge = function(mypath){
filenames=list.files(path=mypath, full.names=TRUE)
datalist = try(lapply(filenames,
function(x){read.csv(file=x,header=T)}))
[R] Violin plot of categorical/binned data
Hi, I'm trying to create a plot showing the density distribution of some shipping data. I like the look of violin plots, but my data is not continuous but rather binned and I want to make sure its binned nature (not smooth) is apparent in the final plot. So for example, I have the number of individuals per vessel, but rather than having the actual number of individuals I have data in the format of: 7 values of zero, 11 values between 1-10, 6 values between 10-100, 13 values between 100-1000, etc. To plot this data I generated a new dataset with the first 7 values being 0, representing the 7 values of 0, the next 11 values being 5.5, representing the 11 values between 1-10, etc. Sample data below. I can make a violin plot (code below) using a log y-axis, which looks alright (though I do have to deal with the zeros still), but in its default format it hides the fact that these are binned data, which seems a bit misleading. Is it possible to make a violin plot that looks a bit more angular (more corners, less smoothing) or in someway shows the distribution, but also clearly shows the true nature of these data? I've tried playing with the bandwidth adjustment and the kernel but haven't been able to get a figure that seems to work. Anyone have some thoughts on this? Thanks, Nate library(ggplot2) library(scales) p=ggplot(data2,(aes(vessel,values))) p+geom_violin()+ scale_y_log10(breaks = trans_breaks(log10, function(x) 10^x),labels = trans_format(log10, math_format(10^.x))) data2-read.table(textConnection( vessel values rec 0.0e+00 rec 0.0e+00 rec 0.0e+00 rec 0.0e+00 rec 0.0e+00 rec 0.0e+00 rec 0.0e+00 rec 5.5e+00 rec 5.5e+00 rec 5.5e+00 rec 5.5e+00 rec 5.5e+00 rec 5.5e+00 rec 5.5e+00 rec 5.5e+00 rec 5.5e+00 rec 5.5e+00 rec 5.5e+00 rec 5.5e+01 rec 5.5e+01 rec 5.5e+01 rec 5.5e+01 rec 5.5e+01 rec 5.5e+01 rec 5.5e+02 rec 5.5e+02 rec 5.5e+02 rec 5.5e+02 rec 5.5e+02 rec 5.5e+02 rec 5.5e+02 rec 5.5e+02 rec 5.5e+02 rec 5.5e+02 rec 5.5e+02 rec 5.5e+02 rec 5.5e+02 rec 5.5e+03 rec 5.5e+03 rec 5.5e+03 rec 5.5e+03 rec 5.5e+03 rec 5.5e+03 rec 5.5e+03 rec 5.5e+04 rec 5.5e+04 rec 5.5e+04 rec 5.5e+05 rec 5.5e+05,header=T) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] some help
On Nov 3, 2012, at 9:07 AM, dattel_palme wrote: Hi People! I have following concern consisting of some steps to do in R: I have an ascii file (table) consisting of many columns and rows. 1. I would like to order all values of the columns one under each other. It will begin with column 1, then column 2 under column 1, column 3 under column 2 etc. until at the end there is only 1 column. How do I do it? something along the lines of dat - read.table(filename, sep=separator, header=TRUE) stacked - do.call(rbind, dat) 2. Second problem is to make a scatterplot of two variables (I think after further explanation scatter plot itself will not be needed). I have two columns of two different variables (that I produces before), Did you now? But from the data produced above you only have one column. Is this another data-object where these column have names? column 1 with variable 1 and column 2 with variable 2. I would like to order them by one variable and 0,01 interval (the varibale values will range between 0 and 1). From each 0,01 interval (100 intervals) i want to pick the maximum and minimum value of variable 2. That is incoherent to this native speaker of English who is sometimes confused by presentations of problems without concrete references. An example would help greatly. 3. From the obtained max and min of values of each interval i would like to make a linear least square regression. Definitely need an example. Please read the Posting Guide. -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] some help
Hello, Without data it's not easy to answer to your questions, but 1. Use ?unlist. If the data is in a file, read it with ?read.table and the unlist the result. All columns will be stacked. dat - read.table(filename, ...) unlist(dat) 2. At best confusing. But to divide a vector into groups use ?cut or ?findInterval and then, to find the maximum and minimum of each group, ?tapply or ?ave. 3. Regress what on what? Provide a data example using dput for better answers: dput( head(mydata, 30) ) # paste the output of this in a post Hope this helps, Rui Barradas Em 03-11-2012 16:07, dattel_palme escreveu: Hi People! I have following concern consisting of some steps to do in R: I have an ascii file (table) consisting of many columns and rows. 1. I would like to order all values of the columns one under each other. It will begin with column 1, then column 2 under column 1, column 3 under column 2 etc. until at the end there is only 1 column. How do I do it? 2. Second problem is to make a scatterplot of two variables (I think after further explanation scatter plot itself will not be needed). I have two columns of two different variables (that I produces before), column 1 with variable 1 and column 2 with variable 2. I would like to order them by one variable and 0,01 interval (the varibale values will range between 0 and 1). From each 0,01 interval (100 intervals) i want to pick the maximum and minimum value of variable 2. 3. From the obtained max and min of values of each interval i would like to make a linear least square regression. I hope someone can help me out! Thanks Stefan -- View this message in context: http://r.789695.n4.nabble.com/some-help-tp4648316.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] mergeing a large number of large .csvs
Jim, Where can I find documentation of the commands you mention? Thanks On Sat, Nov 3, 2012 at 12:15 PM, jim holtman jholt...@gmail.com wrote: A faster way would be to use something like 'per', 'awk' or 'sed'. You can strip off the header line of each CSV (if it has one) and then concatenate the files together. This is very efficient use of memory since you are just reading one file at a time and then writing it out. Will probably be a lot faster since no conversions have to be done. Once you have the one large file, then you can play with it (load it if you have enough memory, or load it into a database). On Sat, Nov 3, 2012 at 11:37 AM, Jeff Newmiller jdnew...@dcn.davis.ca.us wrote: On the absence of any data examples from you per the posting guidelines, I will refer you to the help files for the melt function in the reshape2 package. Note that there can be various mixtures of wide versus long... such as a wide file with one date column and columns representing all stock prices and all trade volumes. The longest format would be what melt gives (date, column name, and value) but an in-between format would have one distinct column each for dollar values and volume values with a column indicating ticker label and of course another for date. If your csv files can be grouped according to those with similar column types, then as you read them in you can use cbind( csvlabel=somelabel, csvdf) to distinguish it and then rbind those data frames together to create an intermediate-width data frame. When dealing with large amounts of data you will want to minimize the amount of reshaping you do, but it would require knowledge of your data and algorithms to say any more. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. Benjamin Caldwell btcaldw...@berkeley.edu wrote: Jeff, If you're willing to educate, I'd be happy to learn what wide vs long format means. I'll give rbind a shot in the meantime. Ben On Nov 2, 2012 4:31 PM, Jeff Newmiller jdnew...@dcn.davis.ca.us wrote: I would first confirm that you need the data in wide format... many algorithms are more efficient in long format anyway, and rbind is way more efficient than merge. If you feel this is not negotiable, you may want to consider sqldf. Yes, you need to learn a bit of SQL, but it is very well integrated into R. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. Benjamin Caldwell btcaldw...@berkeley.edu wrote: Dear R help; I'm currently trying to combine a large number (about 30 x 30) of large .csvs together (each at least 1 records). They are organized by plots, hence 30 X 30, with each group of csvs in a folder which corresponds to the plot. The unmerged csvs all have the same number of columns (5). The fifth column has a different name for each csv. The number of rows is different. The combined csvs are of course quite large, and the code I'm running is quite slow - I'm currently running it on a computer with 10 GB ram, ssd, and quad core 2.3 ghz processor; it's taken 8 hours and it's only 75% of the way through (it's hung up on one of the largest data groupings now for an hour, and using 3.5 gigs of RAM. I know that R isn't the most efficient way of doing this, but I'm not familiar with sql or C. I wonder if anyone has suggestions for a different way to do this in the R environment. For instance, the key function now is merge, but I haven't tried join from the plyr package or rbind from base. I'm willing to provide a dropbox link to a couple of these files if you'd like to see the data. My code is as follows: #multmerge is based on code by Tony cookson, http://www.r-bloggers.com/merging-multiple-data-files-into-one-data-frame/ ; The function takes a path. This path should be the name of a folder that contains all of the files
Re: [R] backreferences in gregexpr
On 11/2/2012 5:14 PM, Gabor Grothendieck wrote: On Fri, Nov 2, 2012 at 6:02 PM, Alexander Shenkin ashen...@ufl.edu wrote: Hi Folks, I'm trying to extract just the backreferences from a regex. temp = abcd1234abcd1234 regmatches(temp, gregexpr((?:abcd)(1234), temp)) [[1]] [1] abcd1234 abcd1234 What I would like is: [1] 1234 1234 Note: I know I can just match 1234 here, but the actual example is complicated enough that I have to match a larger string, and just want to pass out the backreferenced portion. Any help greatly appreciated! Try this: library(gsubfn) strapplyc(temp, abcd(1234)) [[1]] [1] 1234 1234 Thanks Gabor. Didn't find strapplyc in package gsubfun, but did find strapply, and that worked well. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] mergeing a large number of large .csvs
These are not commands, but programs you can use. Here is a file copy
program in perl (I spelt it wrong in the email); This will copy all
the files that have daily in their names. It also skips the first
line of each file assuming that it is the header.
perl can be found on most systems. www.activestate.com has a
version that runs under Windows and that is what I am using.
chdir /temp/csv; # my directory with files
@files = glob daily*csv; # get files to copy (daily..csv)
open OUTPUT, combined.csv; # output file
# loop for each file
foreach $file (@files) {
print $file, \n; # print file being processed
open INPUT, . $file;
# assume that the first line is a header, so skip it
$header = INPUT;
@all = INPUT; # read rest of the file
close INPUT;
print OUTPUT @all; # append to the output
}
close OUTPUT;
Here is what was printed on the console:
C:\Users\Ownerperl copyFiles.pl
daily.BO.csv
daily.C.csv
daily.CL.csv
daily.CT.csv
daily.GC.csv
daily.HO.csv
daily.KC.csv
daily.LA.csv
daily.LN.csv
daily.LP.csv
daily.LX.csv
daily.NG.csv
daily.S.csv
daily.SB.csv
daily.SI.csv
daily.SM.csv
Which was a list of all the files copied.
On Sat, Nov 3, 2012 at 4:08 PM, Benjamin Caldwell
btcaldw...@berkeley.edu wrote:
Jim,
Where can I find documentation of the commands you mention?
Thanks
On Sat, Nov 3, 2012 at 12:15 PM, jim holtman jholt...@gmail.com wrote:
A faster way would be to use something like 'per', 'awk' or 'sed'.
You can strip off the header line of each CSV (if it has one) and then
concatenate the files together. This is very efficient use of memory
since you are just reading one file at a time and then writing it out.
Will probably be a lot faster since no conversions have to be done.
Once you have the one large file, then you can play with it (load it
if you have enough memory, or load it into a database).
On Sat, Nov 3, 2012 at 11:37 AM, Jeff Newmiller
jdnew...@dcn.davis.ca.us wrote:
On the absence of any data examples from you per the posting guidelines,
I will refer you to the help files for the melt function in the reshape2
package. Note that there can be various mixtures of wide versus long...
such as a wide file with one date column and columns representing all stock
prices and all trade volumes. The longest format would be what melt gives
(date, column name, and value) but an in-between format would have one
distinct column each for dollar values and volume values with a column
indicating ticker label and of course another for date.
If your csv files can be grouped according to those with similar column
types, then as you read them in you can use cbind( csvlabel=somelabel,
csvdf) to distinguish it and then rbind those data frames together to
create
an intermediate-width data frame. When dealing with large amounts of data
you will want to minimize the amount of reshaping you do, but it would
require knowledge of your data and algorithms to say any more.
---
Jeff NewmillerThe . . Go
Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live
Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#.
rocks...1k
---
Sent from my phone. Please excuse my brevity.
Benjamin Caldwell btcaldw...@berkeley.edu wrote:
Jeff,
If you're willing to educate, I'd be happy to learn what wide vs long
format means. I'll give rbind a shot in the meantime.
Ben
On Nov 2, 2012 4:31 PM, Jeff Newmiller jdnew...@dcn.davis.ca.us
wrote:
I would first confirm that you need the data in wide format... many
algorithms are more efficient in long format anyway, and rbind is way
more
efficient than merge.
If you feel this is not negotiable, you may want to consider sqldf.
Yes,
you need to learn a bit of SQL, but it is very well integrated into
R.
---
Jeff NewmillerThe . . Go
Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live
Go...
Live: OO#.. Dead: OO#..
Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#.
rocks...1k
---
Sent from my phone. Please excuse my brevity.
Benjamin Caldwell btcaldw...@berkeley.edu wrote:
Dear R help;
I'm currently trying to combine a large number (about 30 x 30) of
large
.csvs together (each at least 1 records). They are organized
Re: [R] mergeing a large number of large .csvs
It easier than that. I forgot I can do it entirely within R:
setwd(/temp/csv)
files - Sys.glob(daily*csv)
output - file('Rcombined.csv', 'w')
for (i in files){
cat(i, '\n') # write out file processing
input - readLines(i)
input - input[-1L] # delete header
writeLines(input, output)
}
close(output)
On Sat, Nov 3, 2012 at 4:56 PM, jim holtman jholt...@gmail.com wrote:
These are not commands, but programs you can use. Here is a file copy
program in perl (I spelt it wrong in the email); This will copy all
the files that have daily in their names. It also skips the first
line of each file assuming that it is the header.
perl can be found on most systems. www.activestate.com has a
version that runs under Windows and that is what I am using.
chdir /temp/csv; # my directory with files
@files = glob daily*csv; # get files to copy (daily..csv)
open OUTPUT, combined.csv; # output file
# loop for each file
foreach $file (@files) {
print $file, \n; # print file being processed
open INPUT, . $file;
# assume that the first line is a header, so skip it
$header = INPUT;
@all = INPUT; # read rest of the file
close INPUT;
print OUTPUT @all; # append to the output
}
close OUTPUT;
Here is what was printed on the console:
C:\Users\Ownerperl copyFiles.pl
daily.BO.csv
daily.C.csv
daily.CL.csv
daily.CT.csv
daily.GC.csv
daily.HO.csv
daily.KC.csv
daily.LA.csv
daily.LN.csv
daily.LP.csv
daily.LX.csv
daily.NG.csv
daily.S.csv
daily.SB.csv
daily.SI.csv
daily.SM.csv
Which was a list of all the files copied.
On Sat, Nov 3, 2012 at 4:08 PM, Benjamin Caldwell
btcaldw...@berkeley.edu wrote:
Jim,
Where can I find documentation of the commands you mention?
Thanks
On Sat, Nov 3, 2012 at 12:15 PM, jim holtman jholt...@gmail.com wrote:
A faster way would be to use something like 'per', 'awk' or 'sed'.
You can strip off the header line of each CSV (if it has one) and then
concatenate the files together. This is very efficient use of memory
since you are just reading one file at a time and then writing it out.
Will probably be a lot faster since no conversions have to be done.
Once you have the one large file, then you can play with it (load it
if you have enough memory, or load it into a database).
On Sat, Nov 3, 2012 at 11:37 AM, Jeff Newmiller
jdnew...@dcn.davis.ca.us wrote:
On the absence of any data examples from you per the posting guidelines,
I will refer you to the help files for the melt function in the reshape2
package. Note that there can be various mixtures of wide versus long...
such as a wide file with one date column and columns representing all
stock
prices and all trade volumes. The longest format would be what melt gives
(date, column name, and value) but an in-between format would have one
distinct column each for dollar values and volume values with a column
indicating ticker label and of course another for date.
If your csv files can be grouped according to those with similar column
types, then as you read them in you can use cbind( csvlabel=somelabel,
csvdf) to distinguish it and then rbind those data frames together to
create
an intermediate-width data frame. When dealing with large amounts of data
you will want to minimize the amount of reshaping you do, but it would
require knowledge of your data and algorithms to say any more.
---
Jeff NewmillerThe . . Go
Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live
Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#.
rocks...1k
---
Sent from my phone. Please excuse my brevity.
Benjamin Caldwell btcaldw...@berkeley.edu wrote:
Jeff,
If you're willing to educate, I'd be happy to learn what wide vs long
format means. I'll give rbind a shot in the meantime.
Ben
On Nov 2, 2012 4:31 PM, Jeff Newmiller jdnew...@dcn.davis.ca.us
wrote:
I would first confirm that you need the data in wide format... many
algorithms are more efficient in long format anyway, and rbind is way
more
efficient than merge.
If you feel this is not negotiable, you may want to consider sqldf.
Yes,
you need to learn a bit of SQL, but it is very well integrated into
R.
---
Jeff NewmillerThe . . Go
Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live
Go...
Live: OO#.. Dead: OO#..
Playing
Research Engineer (Solar/BatteriesO.O#.
[R] Can you turn a string into a (working) symbol?
Dear folks--
Suppose I have an expression that evaluates to a string, and that that
string, were it not a character vector, would be a symbol. I would like a
function, call it doppel(), that will take that expression as an argument
and produce something that functions exactly like the symbol would have if I
typed it in the place of the function of the expression. It should go as
far along the path to evaluation as the symbol would have, and then stop,
and be available for subsequent manipulation. For example, if
aa - 3.1416
bb - function(x) {x^2}
r - 2
xx - c(aa, bb)
out - doppel(xx[1])*doppel(xx[2])(r)
Then out should be 13.3664
Or similarly, after
doppel(paste(a, a, sep='')) - 3
aa
typing aa should return 3.
Is there such a function? Can there be?
I thought as.symbol would do this, but it does not.
as.symbol (xx[1])*as.symbol (xx[2])(r)
Error: attempt to apply non-function
Looking forward to hearing from y'all.--andrewH
--
View this message in context:
http://r.789695.n4.nabble.com/Can-you-turn-a-string-into-a-working-symbol-tp4648343.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] backreferences in gregexpr
On Sat, Nov 3, 2012 at 4:08 PM, Alexander Shenkin ashen...@ufl.edu wrote: On 11/2/2012 5:14 PM, Gabor Grothendieck wrote: On Fri, Nov 2, 2012 at 6:02 PM, Alexander Shenkin ashen...@ufl.edu wrote: Hi Folks, I'm trying to extract just the backreferences from a regex. temp = abcd1234abcd1234 regmatches(temp, gregexpr((?:abcd)(1234), temp)) [[1]] [1] abcd1234 abcd1234 What I would like is: [1] 1234 1234 Note: I know I can just match 1234 here, but the actual example is complicated enough that I have to match a larger string, and just want to pass out the backreferenced portion. Any help greatly appreciated! Try this: library(gsubfn) strapplyc(temp, abcd(1234)) [[1]] [1] 1234 1234 Thanks Gabor. Didn't find strapplyc in package gsubfun, but did find strapply, and that worked well. You must have an old version of the package. Time to upgrade. -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Logical vector-based extraction
Hello, The most part of the program works except that the following logical variable does not get created although the second logical variable-based extraction works. I don't understand what I am doing wrong here. state_pflt200 - df$p_fatal 200 df[state_pflt200, c(state.name,p_fatal)] I would appreciate receiving your help. Thanks, Pradip Muhuri # Below is the code that includes the reproducible example. df - data.frame (state.name= c(Alabama,Alaska,Arizona,Arkansas,California,Colorado,Connecticut, Delaware,DC, Florida,Georgia,Hawaii,Idaho,Illinois,Indiana, Iowa,Kansas,Kentucky, Louisiana,Maine,Maryland,Massachusetts,Michigan, Minnesota,Mississippi,Missouri,Montana,Nebraska,Nevada,New Hampshire, New Jersey,New Mexico,New York,North Carolina,North Dakota,Ohio,Oklahoma, Oregon,Pennsylvania,Rhode Island,South Carolina,South Dakota,Tennessee,Texas, Utah, Vermont,Virginia,Washington,West Virginia,Wisconsin,Wyoming), p_fatal = sample(200:500,51,replace=TRUE), t_safety_score = sample(1:10,51,replace=TRUE) ) options (width=120) # The following logical variable does not get created - Don't understand what I am doing wrong state_pflt200 - df$p_fatal 200 df[state_pflt200, c(state.name,p_fatal)] # The following works state_sslt5 - df$t_safety_score 5 df[state_sslt5,c(state.name, t_safety_score)] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Logical vector-based extraction
works fine for me creating: state_pflt200 - df$p_fatal 200 df[state_pflt200, c(state.name,p_fatal)] [1] state.name p_fatal 0 rows (or 0-length row.names) considering that there were no values less than 200 in your data, the result is correct. So what is the problem? On Sat, Nov 3, 2012 at 5:41 PM, Muhuri, Pradip (SAMHSA/CBHSQ) pradip.muh...@samhsa.hhs.gov wrote: Hello, The most part of the program works except that the following logical variable does not get created although the second logical variable-based extraction works. I don't understand what I am doing wrong here. state_pflt200 - df$p_fatal 200 df[state_pflt200, c(state.name,p_fatal)] I would appreciate receiving your help. Thanks, Pradip Muhuri # Below is the code that includes the reproducible example. df - data.frame (state.name= c(Alabama,Alaska,Arizona,Arkansas,California,Colorado,Connecticut, Delaware,DC, Florida,Georgia,Hawaii,Idaho,Illinois,Indiana, Iowa,Kansas,Kentucky, Louisiana,Maine,Maryland,Massachusetts,Michigan, Minnesota,Mississippi,Missouri,Montana,Nebraska,Nevada,New Hampshire, New Jersey,New Mexico,New York,North Carolina,North Dakota,Ohio,Oklahoma, Oregon,Pennsylvania,Rhode Island,South Carolina,South Dakota,Tennessee,Texas, Utah, Vermont,Virginia,Washington,West Virginia,Wisconsin,Wyoming), p_fatal = sample(200:500,51,replace=TRUE), t_safety_score = sample(1:10,51,replace=TRUE) ) options (width=120) # The following logical variable does not get created - Don't understand what I am doing wrong state_pflt200 - df$p_fatal 200 df[state_pflt200, c(state.name,p_fatal)] # The following works state_sslt5 - df$t_safety_score 5 df[state_sslt5,c(state.name, t_safety_score)] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Can you turn a string into a (working) symbol?
Is this what you want (the answer you wanted is not correct):
aa - 3.1416
bb - function(x) {x^2}
r - 2
xx - c(aa, bb)
doppel - function(x) get(x)
out - doppel(xx[1])*doppel(xx[2])(r)
out
[1] 12.5664
On Sat, Nov 3, 2012 at 5:31 PM, andrewH ahoer...@rprogress.org wrote:
Dear folks--
Suppose I have an expression that evaluates to a string, and that that
string, were it not a character vector, would be a symbol. I would like a
function, call it doppel(), that will take that expression as an argument
and produce something that functions exactly like the symbol would have if I
typed it in the place of the function of the expression. It should go as
far along the path to evaluation as the symbol would have, and then stop,
and be available for subsequent manipulation. For example, if
aa - 3.1416
bb - function(x) {x^2}
r - 2
xx - c(aa, bb)
out - doppel(xx[1])*doppel(xx[2])(r)
Then out should be 13.3664
Or similarly, after
doppel(paste(a, a, sep='')) - 3
aa
typing aa should return 3.
Is there such a function? Can there be?
I thought as.symbol would do this, but it does not.
as.symbol (xx[1])*as.symbol (xx[2])(r)
Error: attempt to apply non-function
Looking forward to hearing from y'all.--andrewH
--
View this message in context:
http://r.789695.n4.nabble.com/Can-you-turn-a-string-into-a-working-symbol-tp4648343.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Can you turn a string into a (working) symbol?
for the second part use 'assign'
assign(paste0('a', 'a'), 3)
aa
[1] 3
On Sat, Nov 3, 2012 at 5:31 PM, andrewH ahoer...@rprogress.org wrote:
Dear folks--
Suppose I have an expression that evaluates to a string, and that that
string, were it not a character vector, would be a symbol. I would like a
function, call it doppel(), that will take that expression as an argument
and produce something that functions exactly like the symbol would have if I
typed it in the place of the function of the expression. It should go as
far along the path to evaluation as the symbol would have, and then stop,
and be available for subsequent manipulation. For example, if
aa - 3.1416
bb - function(x) {x^2}
r - 2
xx - c(aa, bb)
out - doppel(xx[1])*doppel(xx[2])(r)
Then out should be 13.3664
Or similarly, after
doppel(paste(a, a, sep='')) - 3
aa
typing aa should return 3.
Is there such a function? Can there be?
I thought as.symbol would do this, but it does not.
as.symbol (xx[1])*as.symbol (xx[2])(r)
Error: attempt to apply non-function
Looking forward to hearing from y'all.--andrewH
--
View this message in context:
http://r.789695.n4.nabble.com/Can-you-turn-a-string-into-a-working-symbol-tp4648343.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Replacing NAs in long format
Hi Jim,
Thank you so much. That does exactly what I want.
Chris
On Sat, Nov 3, 2012 at 1:30 PM, jim holtman jholt...@gmail.com wrote:
x - read.table(text = idr schyear year
+ 1 80
+ 1 91
+ 1 10 NA
+ 2 4 NA
+ 2 5 -1
+ 2 60
+ 2 71
+ 2 82
+ 2 93
+ 2 104
+ 2 11 NA
+ 2 126
+ 3 4 NA
+ 3 5 -2
+ 3 6 -1
+ 3 70
+ 3 81
+ 3 92
+ 3 103
+ 3 11 NA, header = TRUE)
# you did not specify if there might be multiple contiguous NAs,
# so there are a lot of checks to be made
x.l - lapply(split(x, x$idr), function(.idr){
+ # check for all NAs -- just return indeterminate state
+ if (sum(is.na(.idr$year)) == nrow(.idr)) return(.idr)
+ # repeat until all NAs have been fixed; takes care of contiguous ones
+ while (any(is.na(.idr$year))){
+ # find all the NAs
+ for (i in which(is.na(.idr$year))){
+ if ((i == 1L) (!is.na(.idr$year[i + 1L]))){
+ .idr$year[i] - .idr$year[i + 1L] - 1
+ } else if ((i 1L) (!is.na(.idr$year[i - 1L]))){
+ .idr$year[i] - .idr$year[i - 1L] + 1
+ } else if ((i nrow(.idr)) (!is.na(.idr$year[i + 1L]))){
+ .idr$year[i] - .idr$year[i + 1L] -1
+ }
+ }
+ }
+ return(.idr)
+ })
do.call(rbind, x.l)
idr schyear year
1.11 80
1.21 91
1.31 102
2.42 4 -2
2.52 5 -1
2.62 60
2.72 71
2.82 82
2.92 93
2.10 2 104
2.11 2 115
2.12 2 126
3.13 3 4 -3
3.14 3 5 -2
3.15 3 6 -1
3.16 3 70
3.17 3 81
3.18 3 92
3.19 3 103
3.20 3 114
On Sat, Nov 3, 2012 at 1:14 PM, Christopher Desjardins
cddesjard...@gmail.com wrote:
Hi,
I have the following data:
data[1:20,c(1,2,20)]
idr schyear year
1 80
1 91
1 10 NA
2 4 NA
2 5 -1
2 60
2 71
2 82
2 93
2 104
2 11 NA
2 126
3 4 NA
3 5 -2
3 6 -1
3 70
3 81
3 92
3 103
3 11 NA
What I want to do is replace the NAs in the year variable with the
following:
idr schyear year
1 80
1 91
1 10 2
2 4 -2
2 5 -1
2 60
2 71
2 82
2 93
2 104
2 11 5
2 126
3 4 -3
3 5 -2
3 6 -1
3 70
3 81
3 92
3 103
3 11 4
I have no idea how to do this. What it needs to do is make sure that for
each subject (idr) that it either adds a 1 if it is preceded by a value
in
year or subtracts a 1 if it comes before a year value.
Does that make sense? I could do this in Excel but I am at a loss for how
to do this in R. Please reply to me as well as the list if you respond.
Thanks!
Chris
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Replacing NAs in long format
I have a similar sort of follow up and I bet I could reuse some of this
code but I'm not sure how.
Let's say I want to create a flag that will be equal to 1 if schyear = 5
and year = 0 for a given idr. For example
dat
idr schyear year
1 4 -1
1 50
1 61
1 72
2 90
2101
211 2
How could I make the data look like this?
idr schyear year flag
1 4 -1 1
1 50 1
1 61 1
1 72 1
2 90 0
21010
211 2 0
I am not sure how to end up not getting both 0s and 1s for the 'flag'
variable for an idr. For example,
dat$flag = ifelse(schyear = 5 year ==0, 1, 0)
Does not work because it will create:
idr schyear year flag
1 4 -1 0
1 50 1
1 61 0
1 72 0
2 90 0
21010
211 2 0
And thus flag changes for an idr. Which it shouldn't.
Thanks,
Chris
On Sat, Nov 3, 2012 at 5:50 PM, Christopher Desjardins
cddesjard...@gmail.com wrote:
Hi Jim,
Thank you so much. That does exactly what I want.
Chris
On Sat, Nov 3, 2012 at 1:30 PM, jim holtman jholt...@gmail.com wrote:
x - read.table(text = idr schyear year
+ 1 80
+ 1 91
+ 1 10 NA
+ 2 4 NA
+ 2 5 -1
+ 2 60
+ 2 71
+ 2 82
+ 2 93
+ 2 104
+ 2 11 NA
+ 2 126
+ 3 4 NA
+ 3 5 -2
+ 3 6 -1
+ 3 70
+ 3 81
+ 3 92
+ 3 103
+ 3 11 NA, header = TRUE)
# you did not specify if there might be multiple contiguous NAs,
# so there are a lot of checks to be made
x.l - lapply(split(x, x$idr), function(.idr){
+ # check for all NAs -- just return indeterminate state
+ if (sum(is.na(.idr$year)) == nrow(.idr)) return(.idr)
+ # repeat until all NAs have been fixed; takes care of contiguous
ones
+ while (any(is.na(.idr$year))){
+ # find all the NAs
+ for (i in which(is.na(.idr$year))){
+ if ((i == 1L) (!is.na(.idr$year[i + 1L]))){
+ .idr$year[i] - .idr$year[i + 1L] - 1
+ } else if ((i 1L) (!is.na(.idr$year[i - 1L]))){
+ .idr$year[i] - .idr$year[i - 1L] + 1
+ } else if ((i nrow(.idr)) (!is.na(.idr$year[i +
1L]))){
+ .idr$year[i] - .idr$year[i + 1L] -1
+ }
+ }
+ }
+ return(.idr)
+ })
do.call(rbind, x.l)
idr schyear year
1.11 80
1.21 91
1.31 102
2.42 4 -2
2.52 5 -1
2.62 60
2.72 71
2.82 82
2.92 93
2.10 2 104
2.11 2 115
2.12 2 126
3.13 3 4 -3
3.14 3 5 -2
3.15 3 6 -1
3.16 3 70
3.17 3 81
3.18 3 92
3.19 3 103
3.20 3 114
On Sat, Nov 3, 2012 at 1:14 PM, Christopher Desjardins
cddesjard...@gmail.com wrote:
Hi,
I have the following data:
data[1:20,c(1,2,20)]
idr schyear year
1 80
1 91
1 10 NA
2 4 NA
2 5 -1
2 60
2 71
2 82
2 93
2 104
2 11 NA
2 126
3 4 NA
3 5 -2
3 6 -1
3 70
3 81
3 92
3 103
3 11 NA
What I want to do is replace the NAs in the year variable with the
following:
idr schyear year
1 80
1 91
1 10 2
2 4 -2
2 5 -1
2 60
2 71
2 82
2 93
2 104
2 11 5
2 126
3 4 -3
3 5 -2
3 6 -1
3 70
3 81
3 92
3 103
3 11 4
I have no idea how to do this. What it needs to do is make sure that for
each subject (idr) that it either adds a 1 if it is preceded by a value
in
year or subtracts a 1 if it comes before a year value.
Does that make sense? I could do this in Excel but I am at a loss for
how
to do this in R. Please reply to me as well as the list if you respond.
Thanks!
Chris
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Data
Re: [R] Replacing NAs in long format
Hi,
May be this helps:
dat2-read.table(text=
idr schyear year
1 4 -1
1 5 0
1 6 1
1 7 2
2 9 0
2 10 1
2 11 2
,sep=,header=TRUE)
dat2$flag-unlist(lapply(split(dat2,dat2$idr),function(x)
rep(ifelse(any(apply(x,1,function(x) x[2]=5
x[3]==0)),1,0),nrow(x))),use.names=FALSE)
dat2
# idr schyear year flag
#1 1 4 -1 1
#2 1 5 0 1
#3 1 6 1 1
#4 1 7 2 1
#5 2 9 0 0
#6 2 10 1 0
#7 2 11 2 0
A.K.
- Original Message -
From: Christopher Desjardins cddesjard...@gmail.com
To: jim holtman jholt...@gmail.com
Cc: r-help@r-project.org
Sent: Saturday, November 3, 2012 7:09 PM
Subject: Re: [R] Replacing NAs in long format
I have a similar sort of follow up and I bet I could reuse some of this
code but I'm not sure how.
Let's say I want to create a flag that will be equal to 1 if schyear = 5
and year = 0 for a given idr. For example
dat
idr schyear year
1 4 -1
1 5 0
1 6 1
1 7 2
2 9 0
2 10 1
2 11 2
How could I make the data look like this?
idr schyear year flag
1 4 -1 1
1 5 0 1
1 6 1 1
1 7 2 1
2 9 0 0
2 10 1 0
2 11 2 0
I am not sure how to end up not getting both 0s and 1s for the 'flag'
variable for an idr. For example,
dat$flag = ifelse(schyear = 5 year ==0, 1, 0)
Does not work because it will create:
idr schyear year flag
1 4 -1 0
1 5 0 1
1 6 1 0
1 7 2 0
2 9 0 0
2 10 1 0
2 11 2 0
And thus flag changes for an idr. Which it shouldn't.
Thanks,
Chris
On Sat, Nov 3, 2012 at 5:50 PM, Christopher Desjardins
cddesjard...@gmail.com wrote:
Hi Jim,
Thank you so much. That does exactly what I want.
Chris
On Sat, Nov 3, 2012 at 1:30 PM, jim holtman jholt...@gmail.com wrote:
x - read.table(text = idr schyear year
+ 1 8 0
+ 1 9 1
+ 1 10 NA
+ 2 4 NA
+ 2 5 -1
+ 2 6 0
+ 2 7 1
+ 2 8 2
+ 2 9 3
+ 2 10 4
+ 2 11 NA
+ 2 12 6
+ 3 4 NA
+ 3 5 -2
+ 3 6 -1
+ 3 7 0
+ 3 8 1
+ 3 9 2
+ 3 10 3
+ 3 11 NA, header = TRUE)
# you did not specify if there might be multiple contiguous NAs,
# so there are a lot of checks to be made
x.l - lapply(split(x, x$idr), function(.idr){
+ # check for all NAs -- just return indeterminate state
+ if (sum(is.na(.idr$year)) == nrow(.idr)) return(.idr)
+ # repeat until all NAs have been fixed; takes care of contiguous
ones
+ while (any(is.na(.idr$year))){
+ # find all the NAs
+ for (i in which(is.na(.idr$year))){
+ if ((i == 1L) (!is.na(.idr$year[i + 1L]))){
+ .idr$year[i] - .idr$year[i + 1L] - 1
+ } else if ((i 1L) (!is.na(.idr$year[i - 1L]))){
+ .idr$year[i] - .idr$year[i - 1L] + 1
+ } else if ((i nrow(.idr)) (!is.na(.idr$year[i +
1L]))){
+ .idr$year[i] - .idr$year[i + 1L] -1
+ }
+ }
+ }
+ return(.idr)
+ })
do.call(rbind, x.l)
idr schyear year
1.1 1 8 0
1.2 1 9 1
1.3 1 10 2
2.4 2 4 -2
2.5 2 5 -1
2.6 2 6 0
2.7 2 7 1
2.8 2 8 2
2.9 2 9 3
2.10 2 10 4
2.11 2 11 5
2.12 2 12 6
3.13 3 4 -3
3.14 3 5 -2
3.15 3 6 -1
3.16 3 7 0
3.17 3 8 1
3.18 3 9 2
3.19 3 10 3
3.20 3 11 4
On Sat, Nov 3, 2012 at 1:14 PM, Christopher Desjardins
cddesjard...@gmail.com wrote:
Hi,
I have the following data:
data[1:20,c(1,2,20)]
idr schyear year
1 8 0
1 9 1
1 10 NA
2 4 NA
2 5 -1
2 6 0
2 7 1
2 8 2
2 9 3
2 10 4
2 11 NA
2 12 6
3 4 NA
3 5 -2
3 6 -1
3 7 0
3 8 1
3 9 2
3 10 3
3 11 NA
What I want to do is replace the NAs in the year variable with the
following:
idr schyear year
1 8 0
1 9 1
1 10 2
2 4 -2
2 5 -1
2 6 0
2 7 1
2 8 2
2 9 3
2 10 4
2 11 5
2 12 6
3 4 -3
3 5 -2
3
Re: [R] finding global variables in a function containing formulae
-Original Message-
From: William Dunlap
Sent: Saturday, November 03, 2012 11:23 AM
To: 'Hafen, Ryan P'; Bert Gunter
Cc: r-help@r-project.org
Subject: RE: [R] finding global variables in a function containing formulae
findGlobals must be explicitly ignoring calls to the ~ function.
You could poke through the source code of codetools and find
where this is happening.
I looked through some old notes and found you could
disable the special handler for ~ by removing it from
the environment codetools:::collectUsageHandlers:
findGlobals(function(y)lm(y~x)) # doesn't note 'x' as a global reference
[1] ~ lm
tildeHandler - codetools:::collectUsageHandlers[[~]]
remove(~, envir=codetools:::collectUsageHandlers)
findGlobals(function(y)lm(y~x)) # notes 'x'
[1] ~ lm x
# reinstall ~ handler to get original behavior
# or detach(package:codetools, unload=TRUE) and reattach
assign(~, tildeHandler, envir=codetools:::collectUsageHandlers)
findGlobals(function(y)lm(y~x)) # does not note 'x'
[1] ~ lm
You still have the false alarm problem.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
Or, if you have the source code for the package you are investigating,
use sed to change all ~ to %TILDE% and then use findGlobals on
the resulting source code. The messages will be a bit garbled but
should give you a start. E.g., compare the following two, in which y
is defined in the function but x is not:
findGlobals(function(y)lm(y~x))
[1] ~ lm
findGlobals(function(y)lm(y %TILDE% x))
[1] lm %TILDE% x
You will get false alarms, since in a call like lm(y~x+z, data=dat)
findGlobals
cannot know if dat includes columns called 'x', 'y', and 'z' and the above
approach errs on the side of reporting the potential problem.
You could use code in codetools to analyze S code instead of source code
to globally replace all calls to ~ with calls to %TILDE% but that is more
work than using sed on the source code.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of Hafen, Ryan P
Sent: Friday, November 02, 2012 4:28 PM
To: Bert Gunter
Cc: r-help@r-project.org
Subject: Re: [R] finding global variables in a function containing formulae
Thanks. That works if I a have the formula expression handy. But suppose
I want a function, findGlobalVars() that takes a function as an argument
and finds globals in it, where I have absolutely no idea what is in the
supplied function:
findGlobalVars - function(f) {
require(codetools)
findGlobals(f, merge=FALSE)$variables
}
findGlobalVars(plotFn1)
I would like findGlobalVars() to be able to find variables in formulae
that might be present in f.
On 11/1/12 1:19 PM, Bert Gunter gunter.ber...@gene.com wrote:
Does
?all.vars
##as in
all.vars(y~x)
[1] y x
help?
-- Bert
On Thu, Nov 1, 2012 at 11:04 AM, Hafen, Ryan P ryan.ha...@pnnl.gov
wrote:
I need to find all global variables being used in a function and
findGlobals() in the codetools package works quite nicely. However, I
am not able to find variables that are used in formulae. Simply
avoiding formulae in functions is not an option because I do not have
control over what functions this will be applied to.
Here is an example to illustrate:
library(codetools)
xGlobal - rnorm(10)
yGlobal - rnorm(10)
plotFn1 - function() {
plot(yGlobal ~ xGlobal)
}
plotFn2 - function() {
y - yGlobal
x - xGlobal
plot(y ~ x)
}
plotFn3 - function() {
plot(xGlobal, yGlobal)
}
findGlobals(plotFn1, merge=FALSE)$variables
# character(0)
findGlobals(plotFn2, merge=FALSE)$variables
# [1] xGlobal yGlobal
findGlobals(plotFn3, merge=FALSE)$variables
# [1] xGlobal yGlobal
I would like to find that plotFn1 also uses globals xGlobal and
yGlobal. Any suggestions on how I might do this?
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Genentech Nonclinical Biostatistics
Internal Contact Info:
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Re: [R] Replacing NAs in long format
ave() or split-() can make that easier to write, although it
may take some time to internalize the idiom. E.g.,
flag - rep(NA, nrow(dat2)) # add as.integer if you prefer 1,0 over
TRUE,FALSE
split(flag, dat2$idr) - lapply(split(dat2, dat2$idr), function(d)with(d,
any(schyear=5 year==0)))
data.frame(dat2, flag)
idr schyear year flag
1 1 4 -1 TRUE
2 1 50 TRUE
3 1 61 TRUE
4 1 72 TRUE
5 2 90 FALSE
6 2 101 FALSE
7 2 112 FALSE
or
ave(seq_len(nrow(dat2)), dat2$idr, FUN=function(i)with(dat2[i,],
any(schyear=5 year==0)))
[1] 1 1 1 1 0 0 0
flag - ave(seq_len(nrow(dat2)), dat2$idr, FUN=function(i)with(dat2[i,],
any(schyear=5 year==0)))
data.frame(dat2, flag)
idr schyear year flag
1 1 4 -11
2 1 501
3 1 611
4 1 721
5 2 900
6 2 1010
7 2 1120
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of arun
Sent: Saturday, November 03, 2012 5:01 PM
To: Christopher Desjardins
Cc: R help
Subject: Re: [R] Replacing NAs in long format
Hi,
May be this helps:
dat2-read.table(text=
idr schyear year
1 4 -1
1 5 0
1 6 1
1 7 2
2 9 0
2 10 1
2 11 2
,sep=,header=TRUE)
dat2$flag-unlist(lapply(split(dat2,dat2$idr),function(x)
rep(ifelse(any(apply(x,1,function(x) x[2]=5
x[3]==0)),1,0),nrow(x))),use.names=FALSE)
dat2
# idr schyear year flag
#1 1 4 -1 1
#2 1 5 0 1
#3 1 6 1 1
#4 1 7 2 1
#5 2 9 0 0
#6 2 10 1 0
#7 2 11 2 0
A.K.
- Original Message -
From: Christopher Desjardins cddesjard...@gmail.com
To: jim holtman jholt...@gmail.com
Cc: r-help@r-project.org
Sent: Saturday, November 3, 2012 7:09 PM
Subject: Re: [R] Replacing NAs in long format
I have a similar sort of follow up and I bet I could reuse some of this
code but I'm not sure how.
Let's say I want to create a flag that will be equal to 1 if schyear = 5
and year = 0 for a given idr. For example
dat
idr schyear year
1 4 -1
1 5 0
1 6 1
1 7 2
2 9 0
2 10 1
2 11 2
How could I make the data look like this?
idr schyear year flag
1 4 -1 1
1 5 0 1
1 6 1 1
1 7 2 1
2 9 0 0
2 10 1 0
2 11 2 0
I am not sure how to end up not getting both 0s and 1s for the 'flag'
variable for an idr. For example,
dat$flag = ifelse(schyear = 5 year ==0, 1, 0)
Does not work because it will create:
idr schyear year flag
1 4 -1 0
1 5 0 1
1 6 1 0
1 7 2 0
2 9 0 0
2 10 1 0
2 11 2 0
And thus flag changes for an idr. Which it shouldn't.
Thanks,
Chris
On Sat, Nov 3, 2012 at 5:50 PM, Christopher Desjardins
cddesjard...@gmail.com wrote:
Hi Jim,
Thank you so much. That does exactly what I want.
Chris
On Sat, Nov 3, 2012 at 1:30 PM, jim holtman jholt...@gmail.com wrote:
x - read.table(text = idr schyear year
+ 1 8 0
+ 1 9 1
+ 1 10 NA
+ 2 4 NA
+ 2 5 -1
+ 2 6 0
+ 2 7 1
+ 2 8 2
+ 2 9 3
+ 2 10 4
+ 2 11 NA
+ 2 12 6
+ 3 4 NA
+ 3 5 -2
+ 3 6 -1
+ 3 7 0
+ 3 8 1
+ 3 9 2
+ 3 10 3
+ 3 11 NA, header = TRUE)
# you did not specify if there might be multiple contiguous NAs,
# so there are a lot of checks to be made
x.l - lapply(split(x, x$idr), function(.idr){
+ # check for all NAs -- just return indeterminate state
+ if (sum(is.na(.idr$year)) == nrow(.idr)) return(.idr)
+ # repeat until all NAs have been fixed; takes care of contiguous
ones
+ while (any(is.na(.idr$year))){
+ # find all the NAs
+ for (i in which(is.na(.idr$year))){
+ if ((i == 1L) (!is.na(.idr$year[i + 1L]))){
+ .idr$year[i] - .idr$year[i + 1L] - 1
+ } else if ((i 1L) (!is.na(.idr$year[i - 1L]))){
+ .idr$year[i] - .idr$year[i - 1L] + 1
+ } else if ((i nrow(.idr)) (!is.na(.idr$year[i +
1L]))){
+ .idr$year[i] -
Re: [R] Violin plot of categorical/binned data
On 11/04/2012 06:27 AM, Nathan Miller wrote: Hi, I'm trying to create a plot showing the density distribution of some shipping data. I like the look of violin plots, but my data is not continuous but rather binned and I want to make sure its binned nature (not smooth) is apparent in the final plot. So for example, I have the number of individuals per vessel, but rather than having the actual number of individuals I have data in the format of: 7 values of zero, 11 values between 1-10, 6 values between 10-100, 13 values between 100-1000, etc. To plot this data I generated a new dataset with the first 7 values being 0, representing the 7 values of 0, the next 11 values being 5.5, representing the 11 values between 1-10, etc. Sample data below. I can make a violin plot (code below) using a log y-axis, which looks alright (though I do have to deal with the zeros still), but in its default format it hides the fact that these are binned data, which seems a bit misleading. Is it possible to make a violin plot that looks a bit more angular (more corners, less smoothing) or in someway shows the distribution, but also clearly shows the true nature of these data? I've tried playing with the bandwidth adjustment and the kernel but haven't been able to get a figure that seems to work. Anyone have some thoughts on this? Hi Nate, I'm not exactly sure what you are doing in the data transformation, but you can display this type of information as a single polygon for each instance (kiteChart) or separate rectangles (battleship.plot). library(plotrix) vessels-matrix(c(zero=sample(1:10,5),one2ten=sample(5:20,5), ten2hundred=sample(15:36,5),hundred2thousand=sample(10:16,5)), ncol=4) battleship.plot(vessels,xlab=Number of passengers, yaxlab=c(Barnacle,Maelstrom,Poopdeck,Seasick,Wallower), xaxlab=c(0,1-10,10-100,100-1000)) kiteChart(vessels,xlab=Number of passengers,ylab=Vessel, varlabels=c(Barnacle,Maelstrom,Poopdeck,Seasick,Wallower), timelabels=c(0,1-10,10-100,100-1000)) Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Replacing NAs in long format
Or, even simpler, flag - with(dat2, ave(schyear=5 year==0, idr, FUN=any)) data.frame(dat2, flag) idr schyear year flag 1 1 4 -1 TRUE 2 1 50 TRUE 3 1 61 TRUE 4 1 72 TRUE 5 2 90 FALSE 6 2 101 FALSE 7 2 112 FALSE Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of William Dunlap Sent: Saturday, November 03, 2012 5:38 PM To: arun; Christopher Desjardins Cc: R help Subject: Re: [R] Replacing NAs in long format ave() or split-() can make that easier to write, although it may take some time to internalize the idiom. E.g., flag - rep(NA, nrow(dat2)) # add as.integer if you prefer 1,0 over TRUE,FALSE split(flag, dat2$idr) - lapply(split(dat2, dat2$idr), function(d)with(d, any(schyear=5 year==0))) data.frame(dat2, flag) idr schyear year flag 1 1 4 -1 TRUE 2 1 50 TRUE 3 1 61 TRUE 4 1 72 TRUE 5 2 90 FALSE 6 2 101 FALSE 7 2 112 FALSE or ave(seq_len(nrow(dat2)), dat2$idr, FUN=function(i)with(dat2[i,], any(schyear=5 year==0))) [1] 1 1 1 1 0 0 0 flag - ave(seq_len(nrow(dat2)), dat2$idr, FUN=function(i)with(dat2[i,], any(schyear=5 year==0))) data.frame(dat2, flag) idr schyear year flag 1 1 4 -11 2 1 501 3 1 611 4 1 721 5 2 900 6 2 1010 7 2 1120 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of arun Sent: Saturday, November 03, 2012 5:01 PM To: Christopher Desjardins Cc: R help Subject: Re: [R] Replacing NAs in long format Hi, May be this helps: dat2-read.table(text= idr schyear year 1 4 -1 1 5 0 1 6 1 1 7 2 2 9 0 2 10 1 2 11 2 ,sep=,header=TRUE) dat2$flag-unlist(lapply(split(dat2,dat2$idr),function(x) rep(ifelse(any(apply(x,1,function(x) x[2]=5 x[3]==0)),1,0),nrow(x))),use.names=FALSE) dat2 # idr schyear year flag #1 1 4 -1 1 #2 1 5 0 1 #3 1 6 1 1 #4 1 7 2 1 #5 2 9 0 0 #6 2 10 1 0 #7 2 11 2 0 A.K. - Original Message - From: Christopher Desjardins cddesjard...@gmail.com To: jim holtman jholt...@gmail.com Cc: r-help@r-project.org Sent: Saturday, November 3, 2012 7:09 PM Subject: Re: [R] Replacing NAs in long format I have a similar sort of follow up and I bet I could reuse some of this code but I'm not sure how. Let's say I want to create a flag that will be equal to 1 if schyear = 5 and year = 0 for a given idr. For example dat idr schyear year 1 4 -1 1 5 0 1 6 1 1 7 2 2 9 0 2 10 1 2 11 2 How could I make the data look like this? idr schyear year flag 1 4 -1 1 1 5 0 1 1 6 1 1 1 7 2 1 2 9 0 0 2 10 1 0 2 11 2 0 I am not sure how to end up not getting both 0s and 1s for the 'flag' variable for an idr. For example, dat$flag = ifelse(schyear = 5 year ==0, 1, 0) Does not work because it will create: idr schyear year flag 1 4 -1 0 1 5 0 1 1 6 1 0 1 7 2 0 2 9 0 0 2 10 1 0 2 11 2 0 And thus flag changes for an idr. Which it shouldn't. Thanks, Chris On Sat, Nov 3, 2012 at 5:50 PM, Christopher Desjardins cddesjard...@gmail.com wrote: Hi Jim, Thank you so much. That does exactly what I want. Chris On Sat, Nov 3, 2012 at 1:30 PM, jim holtman jholt...@gmail.com wrote: x - read.table(text = idr schyear year + 1 8 0 + 1 9 1 + 1 10 NA + 2 4 NA + 2 5 -1 + 2 6 0 + 2 7 1 + 2 8 2 + 2 9 3 + 2 10 4 + 2 11 NA + 2 12 6 + 3 4 NA + 3 5 -2 + 3 6 -1 + 3 7 0 + 3 8 1 + 3 9 2 + 3 10 3 + 3 11 NA, header = TRUE) # you did not specify if there might be multiple contiguous NAs, #
Re: [R] Replacing NAs in long format
HI Bill,
It is much simpler.
# with aggregate() and merge()
res1-with(dat2,aggregate(seq_len(nrow(dat2)),by=list(idr=idr),FUN=function(i)
with(dat2[i,], any(schyear=5 year ==0
res2-merge(dat2,res1,by=idr)
colnames(res2)[4]-flag
within(res2,{flag-as.integer(flag)})
#idr schyear year flag
#1 1 4 -1 1
#2 1 5 0 1
#3 1 6 1 1
#4 1 7 2 1
#5 2 9 0 0
#6 2 10 1 0
#7 2 11 2 0
A.K.
- Original Message -
From: William Dunlap wdun...@tibco.com
To: arun smartpink...@yahoo.com; Christopher Desjardins
cddesjard...@gmail.com
Cc: R help r-help@r-project.org
Sent: Saturday, November 3, 2012 9:21 PM
Subject: RE: [R] Replacing NAs in long format
Or, even simpler,
flag - with(dat2, ave(schyear=5 year==0, idr, FUN=any))
data.frame(dat2, flag)
idr schyear year flag
1 1 4 -1 TRUE
2 1 5 0 TRUE
3 1 6 1 TRUE
4 1 7 2 TRUE
5 2 9 0 FALSE
6 2 10 1 FALSE
7 2 11 2 FALSE
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of William Dunlap
Sent: Saturday, November 03, 2012 5:38 PM
To: arun; Christopher Desjardins
Cc: R help
Subject: Re: [R] Replacing NAs in long format
ave() or split-() can make that easier to write, although it
may take some time to internalize the idiom. E.g.,
flag - rep(NA, nrow(dat2)) # add as.integer if you prefer 1,0 over
TRUE,FALSE
split(flag, dat2$idr) - lapply(split(dat2, dat2$idr), function(d)with(d,
any(schyear=5
year==0)))
data.frame(dat2, flag)
idr schyear year flag
1 1 4 -1 TRUE
2 1 5 0 TRUE
3 1 6 1 TRUE
4 1 7 2 TRUE
5 2 9 0 FALSE
6 2 10 1 FALSE
7 2 11 2 FALSE
or
ave(seq_len(nrow(dat2)), dat2$idr, FUN=function(i)with(dat2[i,],
any(schyear=5
year==0)))
[1] 1 1 1 1 0 0 0
flag - ave(seq_len(nrow(dat2)), dat2$idr, FUN=function(i)with(dat2[i,],
any(schyear=5 year==0)))
data.frame(dat2, flag)
idr schyear year flag
1 1 4 -1 1
2 1 5 0 1
3 1 6 1 1
4 1 7 2 1
5 2 9 0 0
6 2 10 1 0
7 2 11 2 0
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of arun
Sent: Saturday, November 03, 2012 5:01 PM
To: Christopher Desjardins
Cc: R help
Subject: Re: [R] Replacing NAs in long format
Hi,
May be this helps:
dat2-read.table(text=
idr schyear year
1 4 -1
1 5 0
1 6 1
1 7 2
2 9 0
2 10 1
2 11 2
,sep=,header=TRUE)
dat2$flag-unlist(lapply(split(dat2,dat2$idr),function(x)
rep(ifelse(any(apply(x,1,function(x) x[2]=5
x[3]==0)),1,0),nrow(x))),use.names=FALSE)
dat2
# idr schyear year flag
#1 1 4 -1 1
#2 1 5 0 1
#3 1 6 1 1
#4 1 7 2 1
#5 2 9 0 0
#6 2 10 1 0
#7 2 11 2 0
A.K.
- Original Message -
From: Christopher Desjardins cddesjard...@gmail.com
To: jim holtman jholt...@gmail.com
Cc: r-help@r-project.org
Sent: Saturday, November 3, 2012 7:09 PM
Subject: Re: [R] Replacing NAs in long format
I have a similar sort of follow up and I bet I could reuse some of this
code but I'm not sure how.
Let's say I want to create a flag that will be equal to 1 if schyear = 5
and year = 0 for a given idr. For example
dat
idr schyear year
1 4 -1
1 5 0
1 6 1
1 7 2
2 9 0
2 10 1
2 11 2
How could I make the data look like this?
idr schyear year flag
1 4 -1 1
1 5 0 1
1 6 1 1
1 7 2 1
2 9 0 0
2 10 1 0
2 11 2 0
I am not sure how to end up not getting both 0s and 1s for the 'flag'
variable for an idr. For example,
dat$flag = ifelse(schyear = 5 year ==0, 1, 0)
Does not work because it will create:
idr schyear year flag
1 4 -1 0
1 5 0 1
1 6 1 0
1 7 2 0
2 9 0 0
2 10 1 0
2 11 2 0
And thus flag changes for an idr. Which it shouldn't.
Thanks,
Chris
On Sat, Nov 3, 2012 at 5:50 PM, Christopher
[R] sqldf Date problem
Dear R-help readers, i've created a database for quotes data (for 4 years; 2007 -- 2010) with the sqldf package. This database contains a column Date in the format mm/dd/. The table in the database is called main.data and the database itself Honda. I tried to get the Data just for certain period, say from 01/01/2007 until 01/10/2007 with the following code: sqldf(select * from main.data where Date='01/10/2007' and Date='01/01/2007'), dbname=Honda) I get the data for this period for every year(2007,2008,2009,2010) not only for 2007. It seems that the year is overlooked and just looked for the fitting days and months. Because I haven't really much experience with sql I decide to send my problem to the list. Many thanks in advance. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Date format conversion from 2012-09-20 to 2012:09:20
Hi,
thanks A.K
try this not working
#*
# Load historical data
#**
library('quantmod')
endDate =Sys.Date()
startDate = as.Date(endDate-10, order=ymd)
dataspy = getSymbols(SPY, from = startDate, to=endDate, auto.assign =
FALSE)
myStDt- startDate
while (myStDt = endDate){
startEndDate-paste(startDate,myStDt,sep=::)
print(dataspy)
dataspy=Cl(dataspy[startEndDate])
#display the subseted data
print(dataspy)
myStDt=myStDt+1
}
--
View this message in context:
http://r.789695.n4.nabble.com/Date-format-conversion-from-2012-09-20-to-2012-09-20-tp4643710p4648327.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] lmPerm p-values and multiple testing
Even you used perm=Exact, the maximum observations allowed is only 10. If data exceeds this, perm=Prob is used instead of Exact. So, the p-values are always changed. The Porb method will approximate the permutation distribution by randomly exchanging pairs of Y elements. -- View this message in context: http://r.789695.n4.nabble.com/lmPerm-p-values-and-multiple-testing-tp4643219p4648350.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] importing jpeg 2000
Mike, thanks for your answer. I thought GDAL is a library itself. You suggest underlying libraries. Do you know where can I find them? The other option: to convert the jpeg2000 files outside R: is it possible in R to execute an extern program? Btw I'm working with OSX Lion and still in the steep learning curve of R :) Best Fred -- View this message in context: http://r.789695.n4.nabble.com/importing-jpeg-2000-tp4648242p4648329.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] sqldf Date problem
Dear R-help readers, i've created a database for quotes data (for 4 years; 2007 -- 2010) with the sqldf package. This database contains a column Date in the format mm/dd/. The table in the database is called main.data and the database itself Honda. I tried to get the Data just for certain period, say from 01/01/2007 until 01/10/2007 with the following code: sqldf(select * from main.data where Date='01/10/2007' and Date='01/01/2007'), dbname=Honda) I get the data for this period for every year(2007,2008,2009,2010) not only for 2007. It seems that the year is overlooked and just looked for the fitting days and months. Because I haven't really much experience with sql I decide to send my problem to the list. Many thanks in advance. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Changing Date Variables as Continuous Variables
I am very new to R, so I apologize if this question is trivial. I have a row in my data of dates in the format mm/dd/; about 3500 rows. I am using this variable in a logistic regression model, and need to treat it as continuous, not a factor as r has decided it is. I tried the as.numeric function but it resulted in all NA's and the message: NAs introduced by coercion If anyone knows a solution, I would greatly appreciate it. Cheers, Jake -- View this message in context: http://r.789695.n4.nabble.com/Changing-Date-Variables-as-Continuous-Variables-tp4648354.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Changing Date Variables as Continuous Variables
Here is how to convert your column of factors into Dates: x - read.table(text = '2/10/2011 + 2/20/2011 + 3/4/2011') # read in as factors str(x) 'data.frame': 3 obs. of 1 variable: $ V1: Factor w/ 3 levels 2/10/2011,2/20/2011,..: 1 2 3 # convert to Date x$date - as.Date(as.character(x$V1), format = %m/%d/%Y) x V1 date 1 2/10/2011 2011-02-10 2 2/20/2011 2011-02-20 3 3/4/2011 2011-03-04 str(x) 'data.frame': 3 obs. of 2 variables: $ V1 : Factor w/ 3 levels 2/10/2011,2/20/2011,..: 1 2 3 $ date: Date, format: 2011-02-10 2011-02-20 2011-03-04 On Sat, Nov 3, 2012 at 8:09 PM, hoguejm hogu...@gmail.com wrote: I am very new to R, so I apologize if this question is trivial. I have a row in my data of dates in the format mm/dd/; about 3500 rows. I am using this variable in a logistic regression model, and need to treat it as continuous, not a factor as r has decided it is. I tried the as.numeric function but it resulted in all NA's and the message: NAs introduced by coercion If anyone knows a solution, I would greatly appreciate it. Cheers, Jake -- View this message in context: http://r.789695.n4.nabble.com/Changing-Date-Variables-as-Continuous-Variables-tp4648354.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] sqldf Date problem
Most likely your Date is either a character or a factor (you need to
provide an 'str' of the dataframe). You are therefore most likely
doing a character compare and that is the reason for your problem.
You need to convert to a character string of the format -MM-DD to
do the correct character comparison.
##
x - data.frame(Date = paste0('1/', 1:31, '/2011'))
str(x)
'data.frame': 31 obs. of 1 variable:
$ Date: Factor w/ 31 levels 1/1/2011,1/10/2011,..: 1 12 23 26 27
28 29 30 31 2 ...
x
Date
1 1/1/2011
2 1/2/2011
3 1/3/2011
4 1/4/2011
5 1/5/2011
6 1/6/2011
7 1/7/2011
8 1/8/2011
9 1/9/2011
10 1/10/2011
11 1/11/2011
12 1/12/2011
13 1/13/2011
14 1/14/2011
15 1/15/2011
16 1/16/2011
17 1/17/2011
18 1/18/2011
19 1/19/2011
20 1/20/2011
21 1/21/2011
22 1/22/2011
23 1/23/2011
24 1/24/2011
25 1/25/2011
26 1/26/2011
27 1/27/2011
28 1/28/2011
29 1/29/2011
30 1/30/2011
31 1/31/2011
require(sqldf)
# not correct because of character compares
sqldf('select * from x where Date 1/13/2011 and Date 1/25/2011')
Date
1 1/2/2011
2 1/14/2011
3 1/15/2011
4 1/16/2011
5 1/17/2011
6 1/18/2011
7 1/19/2011
8 1/20/2011
9 1/21/2011
10 1/22/2011
11 1/23/2011
12 1/24/2011
# convert the date to /MM/DD for character compares
x$newDate - as.character(as.Date(as.character(x$Date), format = %m/%d/%Y))
# now do the select
sqldf('select * from x where newDate between 2011-01-13 and 2011-01-25')
DatenewDate
1 1/13/2011 2011-01-13
2 1/14/2011 2011-01-14
3 1/15/2011 2011-01-15
4 1/16/2011 2011-01-16
5 1/17/2011 2011-01-17
6 1/18/2011 2011-01-18
7 1/19/2011 2011-01-19
8 1/20/2011 2011-01-20
9 1/21/2011 2011-01-21
10 1/22/2011 2011-01-22
11 1/23/2011 2011-01-23
12 1/24/2011 2011-01-24
13 1/25/2011 2011-01-25
On Sat, Nov 3, 2012 at 4:22 PM, Andreas Recktenwald
a.recktenw...@mx.uni-saarland.de wrote:
Dear R-help readers,
i've created a database for quotes data (for 4 years; 2007 -- 2010) with the
sqldf package. This database contains a column Date in the format
mm/dd/.
The table in the database is called main.data and the database itself
Honda. I tried to get the Data just for certain period, say from
01/01/2007 until 01/10/2007 with the following code:
sqldf(select * from main.data where Date='01/10/2007' and
Date='01/01/2007'),
dbname=Honda)
I get the data for this period for every year(2007,2008,2009,2010) not only
for 2007. It seems that the year is overlooked and just looked for the
fitting days and months.
Because I haven't really much experience with sql I decide to send my
problem to the list.
Many thanks in advance.
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Re: [R] Changing Date Variables as Continuous Variables
There is a phenomenon that occurs here which, it seems to me, merits some emphasis. The glm() function appears to be perfectly willing to take a variable of class Date and treat it as a continuous variable. Apparently what it does (on the basis of one little experiment that I did) is convert the vector of dates to a vector of Julian date values, taking the origin to be the minimum value of the Date vector. I would have thought that the user would have to effect this conversion her/himself. But not so; the software is so cleverly written that it's all handled for the user. The designers of R have thought of just about *everything*! We think it 'mazing! :-) cheers, Rolf Turner On 04/11/12 14:58, jim holtman wrote: Here is how to convert your column of factors into Dates: x - read.table(text = '2/10/2011 + 2/20/2011 + 3/4/2011') # read in as factors str(x) 'data.frame': 3 obs. of 1 variable: $ V1: Factor w/ 3 levels 2/10/2011,2/20/2011,..: 1 2 3 # convert to Date x$date - as.Date(as.character(x$V1), format = %m/%d/%Y) x V1 date 1 2/10/2011 2011-02-10 2 2/20/2011 2011-02-20 3 3/4/2011 2011-03-04 str(x) 'data.frame': 3 obs. of 2 variables: $ V1 : Factor w/ 3 levels 2/10/2011,2/20/2011,..: 1 2 3 $ date: Date, format: 2011-02-10 2011-02-20 2011-03-04 On Sat, Nov 3, 2012 at 8:09 PM, hoguejm hogu...@gmail.com wrote: I am very new to R, so I apologize if this question is trivial. I have a row in my data of dates in the format mm/dd/; about 3500 rows. I am using this variable in a logistic regression model, and need to treat it as continuous, not a factor as r has decided it is. I tried the as.numeric function but it resulted in all NA's and the message: NAs introduced by coercion If anyone knows a solution, I would greatly appreciate it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Can you turn a string into a (working) symbol?
Ah! Excellent! That will be most useful. And sorry about the typo.
I found another function in a different discussion that also seems to work,
at least in most cases I have tried. I do not at all understand the
difference between the two.
doppel - function(x) {eval(parse(text=x))
However, neither one seems to work on the left hand side of a -, a -,
or an =.
Again, my thanks.--andrewH
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Re: [R] Can you turn a string into a (working) symbol?
Yes, the assign command goes a little way toward what what I was hoping for. But it requires a different syntax, and it does not in general let you use quoted expressions that you could use with other assignment operators. For instance, DD - 1:3 assign(DD[2], 5) DD [1] 1 2 3 So I am still looking for a function that produces an output that is fully equivalent to the string without quotation marks. Or for a definite statement that no such function can exist. Thanks so much for your attention to this problem. andrewH -- View this message in context: http://r.789695.n4.nabble.com/Can-you-turn-a-string-into-a-working-symbol-tp4648343p4648366.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
