You may have more luck with the triangle plot functions available in other R
libraries. In addition to triangle.plot in ade4, you have other options
(this may not be an exhaustive list):
plot.acomp in compositions
tri in cwhmisc.cwhtool
triax in plotrix
ternary in StatDA
ternaryplot in vcd
If you have your data organized in a matrix like the one printed at the
bottom of your jpg, barplot will produce the same graph:
set.seed(42)
a - matrix(round(runif(45)*100, 0), nrow=5, ncol=9)
a
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
[1,] 91 52 46 94 90 51 74 83
Pass a vector to list() instead of individual values:
env - list(c(0.8,0.9,1.0,1.1,1.2))
env
[[1]]
[1] 0.8 0.9 1.0 1.1 1.2
-
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
The OP indicated that the middle group should be closed on both ends, i.e.
[0.1, 0.6].
dat2 - rbind(dat, 0.1, 0.6)
dat2$group - factor(ifelse(dat2$Var.1, A, ifelse(dat2$Var.6, C,
B)))
dat2
Var group
1 0.0 A
2 0.2 B
3 0.5 B
4 1.0 C
5 4.0 C
6 6.0 C
7 0.1 B
8 0.6
Try again. I'm not having any trouble accessing those pages though the help
system or directly by pasting the urls into my browser.
-
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original
Another possibility:
IE - speciesTime1[,-1] - speciesTime2[,-1]
Imm - data.frame(addmargins(as.matrix(ifelse(IE0, -1, 0)*IE), 2))
Imm
Exti - data.frame(addmargins(as.matrix(ifelse(IE0, 1, 0)*IE), 2))
Exti
-
David L Carlson
Associate Professor of Anthropology
Probably because you have defined all of the variables as
supplemental. These will then be estimated from the other variables,
but there are no other variables.
---
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX
It may be easier if you convert the list you provided to a
data.frame:
dta.df - data.frame(place=dta$place, name=dta$name,
value=dta$value)
dta.df
Note that the last line is blank so you probably want to remove that
and remove the blank factor levels:
dta.df - dta.df[-nrow(dta.df),]
I believe you are looking for the offset= parameter. Consider the
following:
barplot(c(2, 3), ylim=c(1, 3))
barplot(c(2, 3), ylim=c(1, 3), offset=1)
-
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77840-4352
.
From your words,it seems that the quanti.sup and quali.sup. are not
parts of the original variables,but are estimated from the other
variables?
Many thanks.
At 2013-06-03 21:54:59,David Carlson dcarl...@tamu.edu
mailto:dcarl...@tamu.edu wrote:
Probably because you have defined all
at 7:55 AM, David Carlson dcarl...@tamu.edu
wrote:
I believe you are looking for the offset= parameter. Consider the
following:
barplot(c(2, 3), ylim=c(1, 3))
barplot(c(2, 3), ylim=c(1, 3), offset=1)
-
David L Carlson
Associate Professor of Anthropology
Texas AM
Something like this?
set.seed(42)
a - sample.int(10)
b - sample.int(5)
c - sample.int(3)
smpl - function(x) sample(x, ifelse(length(x)5, length(x), 5))
smpl(a)
[1] 4 8 1 10 2
smpl(b)
[1] 2 3 5 1 4
smpl(c)
[1] 2 1 3
-
David L Carlson
Associate
places ,
how can
I do it? So, by common I mean, it might be compared with 2, 3, 4
all
of them. All possible combinations
Many thanks
regards
Nico
On Mon, Jun 3, 2013 at 5:20 PM, David Carlson dcarl...@tamu.edu
wrote:
It may be easier if you convert the list you provided to a
data.frame
You didn't give us data, but this may give you enough to solve your problem:
set.seed(42)
nrows - 6
ncols - 5
mat1 - matrix(sample.int(100, 30), nrows, ncols)
mat1
[,1] [,2] [,3] [,4] [,5]
[1,] 92 70 83 397
[2,] 93 13 23 46 82
[3,] 29 61 40 73 98
[4,]
Works fine for me. Too bad you didn't include actual data:
set.seed(42)
DATA - data.frame(UnitName_1=factor(sample(c(lake, pond,
river),
+ 15, replace=TRUE)), Var=sample.int(100, 15))
DATA
UnitName_1 Var
1 river 95
2 river 97
3lake 12
4 river 47
5
Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77840-4352
Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of David Carlson
Sent: Thursday, June 13, 2013 9:00 AM
To: 'Shane Carey'; r-help@r-project.org
Density means that the AREAS of the bars add to 1, not the HEIGHTS
of the bars. You probably have intervals that are less than 1. Eg:
set.seed(42)
x - rpois(1000, 5)/100
info - hist(x, prob=TRUE)
info
$breaks
[1] 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11
0.12 0.13
$counts
Frequently Asked Questions 7.31 Why doesn't R think these numbers
are equal?
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think
-these-numbers-are-equal_003f
For more details read the First Circle of the R Inferno:
http://www.burns-stat.com/pages/Tutor/R_inferno.pdf
-Original Message-
From: Bert Gunter [mailto:gunter.ber...@gene.com]
Sent: Monday, June 17, 2013 12:35 PM
To: Elizabeth Beck
Cc: David Carlson; r-help@r-project.org
Subject: Re: [R] NMDS with missing data?
Just wanted to note that one does **not** use
prcomp() on the correlation matrix
: Elizabeth Beck [mailto:elizabethbe...@gmail.com]
Sent: Monday, June 17, 2013 1:43 PM
To: Bert Gunter
Cc: David Carlson; r-help@r-project.org
Subject: Re: [R] NMDS with missing data?
Hi Bert David -
I'm putting aside the issues with the missing data for the moment -
the NAs are due to not enough
It cannot be done because there is not enough time to create the
plot. In about a billion years the sun will be 10% brighter than
today and the oceans will start to boil away. You will still be
plotting genes when that happens.
:-(
-
David
-Original
Try this
set.seed(42)
Vec1 - sample(0:1, 15, replace=TRUE)
Vec2 - sample(0:1, 15, replace=TRUE)
Vec3 - sample(0:1, 15, replace=TRUE)
paste0(ifelse(Vec1,A,), ifelse(Vec2,B,),
ifelse(Vec3,C,))
[1] ABC ABC AC AB ABC A B ABC AC B
A
[13] AB C B
Perhaps you are talking about the variable.labels attribute returned
by read.spss in package foreign? If so, you should try the
attributes() or the attr() function:
?attributes
?attr
-
David L Carlson
Associate Professor of Anthropology
Texas AM University
Try this:
aggregate(Project~Country+Iso, df, length)
Country Iso Project
1 Burkina Faso BF 2
2 Ethiopia ET 1
3 Ghana GH 1
4 Kenya KE 2
5Madagascar MG 2
6 Mali ML 1
7Mozambique MZ 1
8 Nigeria NG
I think you have to assume there could be other coding errors as
well, such as misspellings or abbreviating Street as St. You
probably will need to use sub() to correct A2 and possibly A1 before
trying to merge. To figure where the problems are, you might try
something like this. The last command
Your rate variable is not a single value, but a vector of 200!.
I'm surprised you get anything at all. You have to pick one rate
value to use to create the values for the x-axis. For example run
rate - 1/4
And then run the qqmath() function.
Or, change the rate variable in the ggamma()
You should tell us where hdr.den() comes from as it is not one of
the default packages and there are thousands of packages. Some
searching suggests you are using package hdrcde.
Looking at the source code for hdr.den() it appears that the package
author used the default palette in selecting
Maybe something like this?
dataset - data.frame(x=c(y, y), y=c(x, z), g=rep(1:2, each=58))
dotplot(x~y, groups=g, data=dataset)
-
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original
You should read the help page
?kmeans
Especially the section labeled Value which tells you what kmeans
returns. You will see that the cluster membership is returned as a
vector of integers called cluster. If you don't know how to access
that from kmeans.results, you haven't read any of the basic
My_Vector[names(My_Vector)==a]
-
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Christofer
Your data is probably not arranged correctly. See if this works for
your data:
# Creating a reproducible example
set.seed(42)
dat - matrix(rnorm(100*15, 5, 5)+rep(1:15, each=100), 15, 100,
byrow=TRUE)
str(dat)
num [1:15, 1:100] 12.85 13 -2 8.98 16.67 ...
require(gplots)
# Rearranging dat
Another approach passing both D and delta:
Afct - function(D, delta) exp(-(outer(D, D, -)/delta)^2)
D - c(-1, -2/3, -1/3, 0, 1/3, 2/3, 1)
Delta - 2
Afct(D, delta)
-
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX
I don't see that you have set up mutually exclusive ranges. If we
modify your code so save the maxlogP and minlogP values:
niter = 1e5 # number of iterations is 10^5
CountLoss = rep(0,niter)
CountProf = rep (0,niter)
maxlogP - rep(0, ninter)
minlogP - rep(0, ninter)
set.seed(2009) # enables
You need to copy the lines you have typed in to R and send them
exactly. The only way I get an error message like the one you
indicate is with the following:
library(data)
Error in library(data) : there is no package called 'data'
There is no Data function in R so if you typed that you would
Refai [mailto:y_re...@hotmail.com]
Sent: Thursday, June 27, 2013 11:11 AM
To: David Carlson
Subject: Re: [R] Data Package Query
The same lines as what you typed below and I am getting the same
error as what you mentioned below. So what I can do to overcome that
problem?
Sent using BlackBerryR from
You can pass a matrix to by()
set.seed(42)
dat - data.frame(x=runif(50)*20, y=runif(50)*20,
g=rep(LETTERS[1:2], each=25))
as.vector(by(dat[,1:2], dat$g, function(x) cor(x)[1,2]))
[1] -0.05643063 0.16465040
-
David L Carlson
Associate Professor of
In addition to reading Frequently Asked Questions Section 7.31, you
should always run your own code before posting. Yours does not work.
Spend some time reading one or more of the R tutorials and learn
about vectorization. It will save you time and typing. There are
many to choose from at
Also
yrs - as.Date(c(2007-01-01, 2009-12-31))
day[day=yrs[1] day=yrs[2]]
[1] 2008-04-12 2008-04-12 2008-04-12
Should be fast if there are many days to process since it converts
the search criteria, not the data.
-
David L Carlson
Associate Professor of
I think this is what you are looking for
f - d[match(k, d[,1]), 2]
f
[1] 4 4 6 6 7 7 6
-
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org
It may be that single and efficient are opposing goals. Two steps
lets you create the subset and then just order each query.
Alternatively, if the data do not change often, create an ordered
version and query that.
David Carlson
Original Message-
From: r-help-boun...@r-project.org
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
First read the instructions for posting. You gave us no data, no
code, no example. Just an error message out of context.
?outer
e.g. output - outer(ap, am, func)
-
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of
Hard to say without a reproducible example, perhaps by using dput()
to include a subset of the data. You should indicate where
wtd.hist() comes from since it is not in base R. Perhaps the weights
package? It is not clear what weight=weight is unless you have
created a variable called weight. If
This should also work:
mask - randomized 0
mask
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] FALSE FALSE TRUE FALSE FALSE TRUE
[2,] TRUE TRUE FALSE TRUE TRUE TRUE
[3,] FALSE TRUE TRUE FALSE FALSE TRUE
[4,] TRUE TRUE FALSE TRUE FALSE TRUE
randomized[mask] - original[mask]
randomized
This may be close to what you want:
t.val - by(x, x$cat, function(y) if (min(table(y$name)1)) {
+ t.test(val~name, y)})
t.out - do.call(rbind, sapply(t.val, function(y) c(y$statistic,
+ p.value=y$p.value)))
t.out
t p.value
p178266580 -0.1156475 0.9144054453
That reduces the number of possibilities for the file structure to
ones that R can probably read, but we need to know more. Try opening
a sample file in a text editor to see if it is plain text. If so,
you can probably use read.table() or read.csv() to get the file into
R. If not, you will need
If you read ?dist, it says that:
binary:
(aka asymmetric binary): The vectors are regarded as
binary bits, so non-zero elements are ‘on’ and zero elements
are ‘off’. The distance is the proportion of bits in which
only one is on amongst those in which at least one is on.
The short answer is
The short answer is whichever version you want since they are
the same. Look at the t-test results, they are also different.
In one t is positive and in the other it is negative. That is
just a result of whether the smaller mean is subtracted from
the larger mean or vice versa. The same is
Go to Settings | Filters in Gmail and set filters to delete
the r-help messages you don't want or move them to a folder
for review. R-help messages are not tagged so you cannot
restrict them before they are sent.
-
David L Carlson
Associate Professor of
Actually the .0 on the first variable is not needed.
You could modify the reshape() call to search for the base
name of each variable so you would not need to change the code
if the number of replications changes:
reshape(df5, direction=long, v.names=c(dose, resp),
Benchmark is probably a subset from a larger dataframe. R does
not automatically remove empty levels but you can do it:
set.seed(42)
dataset - data.frame(Benchmark=factor(sample(LETTERS[1:26],
50,
replace=TRUE), levels=LETTERS[1:26]))
levels(dataset$Benchmark)
# [1] A B C D E F G H I J K L M
I may not understand, but if you are looking for a way to
extract the second item from each list element, you will need
to use lapply():
a - list(LETTERS[1:3], 10:12, LETTERS[26:24], 15:13)
b - list(LETTERS[4:6], 15:17, LETTERS[23:21], 10:8)
c - list(LETTERS[7:9], 20:22, LETTERS[20:18], 5:3)
abc
I'd start with the home page for R: http://www.r-project.org/
because you seem to have no idea what version is current
(3.0.1). You will find Google to be very helpful. As for
manuals, the official documentation is at
http://cran.r-project.org/manuals.html
and the user contributed manuals are at
Here's an approach that seems to work. I added an 11th case to
your data since you did not have a case where both Left and
Right had multiple values in the lookup table. This creates an
id value so that we can merge left and right separately and
then merge them back together:
# Create test data
When you use complete.cases(), it creates a logical vector
that selects cases with no missing values, but it does not
change the rownames in the data.frame and those are carried
through to the factor scores so you could link them up that
way. Alternatively, you could use na.exclude as the
Rui has shown you a much more efficient way to code your
function in R. To fix the code you posted, you need to add
brackets around the loop, test x[i] instead of i (which is
always = 1), and get the length of the loop from x not
pah1$P. Without the brackets only the first if() is included
in the
A triangular plot can represent three dimensions in a two
dimensional plane because the three dimensions are constrained
by the fact that they sum to 100% (e.g. sand/silt/clay
composition of a soil). That does not seem to apply to your
example. It sounds like you might need a 3d contour plot. You
Where did you find out about stat.table()? There is one in
package Epi, but who knows if it is the one you are looking
for? If you don't know about packages and the library()
function, you need to work through a basic tutorial on R. You
can't really expect to download a script file and run it
Or just
table(dat1$ID, dat1$Group1)
# 0 1
# 1 2 2
# 2 1 3
# 5 1 3
Or
xtabs(~ID+Group1, dat1)
#Group1
# ID 0 1
# 1 2 2
# 2 1 3
# 5 1 3
Or with labeling
dat1$Group1 - factor(dat1$Group1, labels=c(No, Yes))
xtabs(~ID+Group1, dat1)
#Group1
# ID No Yes
# 1 2 2
# 2
In your example all of the values are drawn from the same
distribution so there will not be substantial differences (row
means/variances and column means/variances will be
approximately the same).
set.seed(42)
d - matrix(rnorm(100),nrow=20)
# Start with your example and modify the row/col means
sapply(1:200, function(x) cor(mts1[,x], mts2[,x],
use=complete.obs, method=c(pearson)))
-
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org
It is hard to understand that your R code will not work with a
double matrix since double is just short for double precision
floating point matrix. Your only alternative would be integer.
From ?numeric
It is a historical anomaly that R has two names for its
floating-point vectors, double and
Read up on Gower's Distance measures (available in the ecodist
package) which can combine numeric and categorical data. You
didn't give us any information about how you numerically
transformed the categorical variables, but the usual approach
is to create indicator variables that code
Assuming you have created a data.frame from the text file you
sent that is called RP:
str(RP)
# 'data.frame': 28597 obs. of 8 variables:
# $ gene : chr XLOC_01 XLOC_02 XLOC_03
XLOC_04 ...
# $ ZPT.1: num 3516 342 2000 143 0 ...
# $ ZPT.0: num 626 82 361 30 0 0 0 0 0 1 ...
#
It depends on how fancy you want to get. A quick fix would be
pairs - paste0(colnames(matrix1), ., colnames(matrix2))
# lapply will be faster since you are returning a list
results - lapply(1:200, function(x)
summary(lm(formula=matrix1[,x]~matrix2[,x])))
names(results) - pairs
results
The
You cannot move or edit them interactively, but you can save
the locations of those points by using:
a - locator(type=o)
# e.g. click on three locations
a
$x
[1] 0.8559376 1.1778126 1.2340626
$y
[1] 1.2256976 1.2501162 0.8402324
You could now delete or modify the points and then use lines()
You might modify this function which computes Cramer's V using
the assocstats() function in package vcd:
catcor - function(x) {
require(vcd)
nc - ncol(x)
v - expand.grid(1:nc, 1:nc)
matrix(mapply(function(i1, i2) assocstats(table(x[,i1],
Not always true, but it is in this case:
?ifelse
David C
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Jeff Johnson
Sent: Tuesday, February 18, 2014 11:24 AM
To: R help
Subject: [R] Updating a data frame based on if condition
I
(ifelse(mydata$FNAME_LENGTH 35, TRUE, FALSE,
ifelse(regexpr(9, mydata$FNAME_PATTERN) 0, TRUE, FALSE)))
I have the R for Dummies book which covers it a bit, but I just
ordered the R Cookbook.
On Tue, Feb 18, 2014 at 10:16 AM, David Carlson
dcarl...@tamu.edu wrote:
Not always true
Both functions (not packages) are included in package stats
which is part of the standard R installation. Have you tried?
?prcomp
?princomp
-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original
Your code produces four line graphs, one above the other. What
do you mean by drawing a contour line for the entire panel? A
contour map requires three vectors, two for the horizontal
position and one for the elevation or height. You have four
vectors of y variables with the same x. Since you do
If the data are two week intervals, the frequency is 52/2 = 26
and that matches the amount of data 8*26=208:
vec.ts - ts(vec, start=c(2005, 1), frequency=26)
str(vec.ts)
Time-Series [1:208] from 2005 to 2013: 1 0 0 0 1 1 1 0 2 0 ...
print(vec.ts, calendar=TRUE)
p1 p2 p3 p4 p5 p6 p7 p8 p9
You have discovered two features of R with your example. Don
told you about the first. Data frames are considered to be lists
so if you provide only one index, you get the columns (the list
elements) when you type
str(leadership)
'data.frame': 5 obs. of 10 variables:
$ manager: num 1 2 3 4
The simplest way is to use clip():
clip(0, 33, -1, 0)
polygon(ff,gg,col=2)
clip(0, 33, 0, 1)
polygon(ff,gg,col=4)
-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
From:
One possibility would be to use the absolute value of the
differences. Then they would be all positive:
abs(Info$Attr1 - Info$Attr2)
[1] 0.46 3.21 0.01 0.99 1.07 10.08
-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX
Alternatively use type=c instead of l which will create line
breaks where the points will go.
plot(1:5,type=c, col=blue)
points(1:5,col=black)
David
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Jim Lemon
Sent: Wednesday, March
Not histograms, but here are two alternatives. The first gives
you kernel density plots for each value and the second uses
violin plots. Both plot points if there are fewer than 5.
set.seed(42)
y-rpois(500,2)
x-rnorm(500,y,1)
plot(x,y, type=n)
for (i in seq(min(y), max(y), by=1)) {
if
Does setting OutDec work for you?
options(OutDec=,)
x - rnorm(50, mean=2, sd=0.1)
y - rnorm(50, mean=2, sd=0.2)
plot(x, y)
# options(OutDec=.) to change it back if you want
-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX
If you read the manual page for read.csv (?read.csv), you will
see that the default for the header argument is TRUE, but your
example has no header. You do not need sep=, because that is
the default for read.csv. Also, the default is to convert
character strings to factors which you probably do
1. (alfa - X*Y) # Sends the result of the expression to
print()
2. Y - datos[,2, drop=FALSE] # see the bracket manual page:
?[
-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
From:
I think we're down to counting the number of characters in each
solution! Arun's 3 two-line versus your two-line solution (not
counting loading plyr). How about three short lines?
pop - rep(1:nrow(tab), tab$Freq)
ind - unlist(sapply(tab$Freq, seq_len))
tab2 - data.frame(pop, ind)
R-help strips attachments so your post has three problems:
1. No plot to look at
2. No data to replicate your code
3. Email is in html format instead of plain text
Try again using plain text mail, copy the results of
dput(your.data.frame) into your email and provide code that
generates the
Not exactly automatic, but if you are using Windows:
1. Open a blank spreadsheet in Excel
2. write.table(cor(nums2), file=clipboard-128, sep=\t) # In
R
3. Paste the clipboard into the spreadsheet
Package xlsx can write Excel format files, but does not put them
directly into Excel.
For the
78023, 43785, 69884, 12840, 54021 are listed as PersonID 3 in
rawData, but PersonID 4 in resultData.
Here is another way to get there:
# Split codes by PersonID creating a single vector for each
step1 - split(rawData$codes, rawData$PersonID)
# Figure out how many lines we need - here 3 lines
The digits option gives you the number of significant digits
(ignoring leading zeros after the decimal) and it operates on a
column by column basis. The first and second columns have values
with a leading zero after the decimal (-0.04074 in both columns)
so R prints five decimal places to get four
Look at the results of summary(body.pc). You will see that the
first component accounts for 70% of the variability. That is
very large and suggests that the variables are highly correlated
with one another. You could check with cor(bodysize). None of
the correlations is negative and the smallest
To add to Ranjan's reply, k-means can potentially find different
results with large nstart= numbers in a large data set. But you
are correct, with a large enough value, the results will be the
same unless there are two solutions that have exactly the same
between sum of squares (unlikely but not
That only requires two small changes in Sarah's first solution:
Finaldata[, !colnames(Finaldata) %in% V$v2[V$v1 == 1]]
Length Rate
1 0.53607323 0.01739951
2 0.15405615 0.11837908
3 0.04542388 0.53702702
4 0.15633703 0.68870041
5 0.35293973 0.38258981
Or tweaking Don's solution but using the original mu and sig
vectors:
set.seed(42)
x.matrix-matrix(0, nrow=27, ncol=5)
for(i in 1:27){
+ x.matrix[i, ] - rnorm(5, mu[i], sig[i])
+ }
set.seed(42)
xmat - matrix(rnorm(27*5, rep(mu, each=5), rep(sig, each=5)),
+ nrow=27, byrow=TRUE)
It may be possible to do this in a single step, but
x1 - aggregate(response~id+age, data, mean)
x2 - data[data$eye==l, c(id, response2)]
merge(x1, x2)
id age response response2
1 1 2 4.60 High
2 2 9 2.65 High
3 3 5 3.65 High
4 4 2 7.55 High
5
This is a bit more direct. It works by forcing R to treat test
as a matrix rather than a table:
tst2 - data.frame(ID=dimnames(test)$ID,
as.data.frame.matrix(test),
check.names=FALSE)
tst2
ID 1 2 3 4 5
10 10 10 20 30 40 50
11 11 0 0 60 70 0
12 12 0 0 0 80 0
rownames(tst2) -
Use an 3-dimensional array.
?array
And any basic introduction to R.
-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org
Something like (only 20 columns here):
x - matrix(rnorm(120*20), 120, 20)
xagg - aggregate(x, list(rep(1:12, each=10)), mean)
-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX 77840-4352
Original Message-
From:
You could randomly assign 1 to a single column in each row and
then use binomial draws on the remaining 0's:
set.seed(42)
dimMat - matrix(0, 1000, 4)
dimMat[cbind(1:1000, sample.int(4, 1000, replace=TRUE))] - 1
dimMat[dimMat1] - sample(0:1, 3000, replace=TRUE, prob=c(.6,
.4))
I was able to grab the logos by bringing the .svg file into
Inkscape (open source .svg editor). The three logos are grouped
so you can select them and then paste them into another
document. Then you'll have to ungroup the three logos and select
each logo separately.
This will work, as long as there are exactly 2 distinct modes:
runs - rle(sign(diff(d.rv$y)))
length(runs$lengths) # There should be 4 runs if there are
exactly 2 modes
[1] 4
mode1 - runs$lengths[1]+1
mode2 - length(d.rv$x)- runs$lengths[4]
d.rv$y[c(mode1, mode2)] # The 2 modes:
[1]
Look below at your question. R-help does not support html email
so you must use plain text if you want to get an answer.
-
David L Carlson
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Marta
...@gene.com]
Sent: Monday, April 21, 2014 4:17 PM
To: David Carlson
Cc: Luigi Marongiu; r-help@r-project.org
Subject: Re: [R] find maximum values on the density function of a bimodal
distribution
Well, yes and no, David.
Yours is one possible interpretation -- just work verbatim with the
finite
You can also use the legend() function which will draw the box
and arrange the text in a single column (or three columns if you
want a single row). It also has shortcuts for positioning the
box such as topleft or bottomright.
?legend
for more details.
-
David
This is a bit simpler. The function quantile() labels the
output whereas fivenum() does not:
aggregate(Age ~ Generation + Zygosity + Sex + Cohort +
ESstatus, data=x,
function(x) c(mean=mean(x), sd=sd(x), quantile(x)))
-
David L Carlson
Associate Professor
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