Use optimize() to find the minimum and feed that value into uniroot().
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, Apr 16, 2015 at 7:47 AM, li li wrote:
> Hi Jeff,
> Thanks for the reply. I am aware that the sign needs to be different at
> the ends of the starting interval.
>
>An
Look more at help(rep) and help(seq):
> n <- 7
> rep(seq_len(n), each=4)
[1] 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Sun, Apr 19, 2015 at 6:44 AM, John Sorkin
wrote:
> Windows 7 64-bit
> R 3.1.3
> RStudio 0.98.1103
>
>
>
The hyphen without a following digit confuses tidyr::extract_numeric().
E.g.,
> extract_numeric("23 ft-lbs")
Warning message:
In extract_numeric("23 ft-lbs") : NAs introduced by coercion
[1] NA
> extract_numeric("23 ft*lbs")
[1] 23
Contact the BugReports address for the package
Use the str() function to see the internal structure of most objects. In
your case it would show something like:
> Data <- data.frame(theData=round(sin(1:38),1))
> x <- ts(Data[[1]], frequency=12) # or Data[,1]
> y <- ts(Data, frequency=12)
> str(x)
Time-Series [1:38] from 1 to 4.08: 0.8 0.9 0.1
> Interesting that a 2D matrix of size Nx1 is treated as a different
> animal from a length N vector.
I think we can call this a bug in stl().
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Apr 21, 2015 at 6:39 PM, Paul wrote:
> William Dunlap tibco.com> writes:
>
Are there missing values in your data? If so, try adding
the argument
na.action = na.exclude
to your original call to glm or lm. It is like the default
na.omit except that it records which rows were omitted
(because they contained missing values) and fills in
the corresponding entries in the p
You could do something tricky like
> do.call(cbind, lapply(big.char, as.name))
dog cat tree
[1,] 1 25
[2,] 2 36
[3,] 3 47
but you are usually better off creating these things as part of a list
and passing that to do.call(cbind, list).
There is a slight danger
Your function ln() does not return a scalar.
> ln(theta=c(1,2))
[1] 48.5342640972 48.5342640972 48.5342640972 48.5342640972 48.5342640972
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Apr 28, 2015 at 6:40 PM, Hanze Zhang wrote:
> I still cannot solve the problem: 'invalid functio
# invoke user defined function
myFunc ( newdata=nmpr_nmproid,
oridata=xanloid_set,
cohort=xanloid_set$cohort_type,
value= "NMPR_Cohort",
xdate_to_int_time=anl_to_int_time,
xflag=xanloid_set$anlflag,
idate=xanloid_set$intdate,
You can sort the eigenvalues in the order you want with
o <- order(Re(e$values), decreasing = TRUE)
or
o <- order(abs(Re(e$values)), decreasing = TRUE)
followed by
e$values[o]
where 'e' is the object that eigen returns.
The main argument to order() is what you want to sort by and the s
It looks like a problem in the Matrix package. I made the file KE.rda
containing the Matrix objects K and E constructed in calc.diffusion.kernel
by adding a save() call just before where R dies in the original example:
K = lam$values[1] * I - M
E = I - matrix(1, ncol = ncol(I), nr
Your nLL function returns 1e+308 in near-boundary cases. Since 1e+308 is so
close to machine infinity, it is easy to get into Inf-Inf (=NaN) or Inf/Inf
(=NaN)
situations when working with it. Try making that limiting value something
smaller,
like 1e+30, and you may have better luck.
Bill Dunlap
Software
wdunlap tibco.com
On Thu, May 7, 2015 at 4:46 PM, Jean Marchal
wrote:
> Yes, indeed! Problem solved!
>
> Thanks a lot!
>
> Jean
>
> 2015-05-07 14:06 GMT-07:00 William Dunlap :
> > Your nLL function returns 1e+308 in near-boundary cases. Since 1e+308
> is so
txt <- c("A B C D E", "A 1232 0.565", "B 2323 0.5656 0.5656 0.5656",
"C 2323 0.5656", "D 2323 0.5656", "E 2323 0.5656", "F 2323 0.5656",
"G 2323 0.5656", "G 2323 0.5656 0.5656 0.5656")
z <- read.table(text=txt, fill=TRUE, header=TRUE)
str(z)
#'data.frame': 8 obs. of 5 variables:
# $ A: Factor
You can do the timing yourself on a dataset which you feel is typical of
your usage.
E.g., define a function the implements each algorithm
> f1 <- function(foo) lapply(foo, function(x) { if (x[1] == 1) x[2] <- 0
; x })
> f2 <- function(foo) { for(i in seq_along(foo)) if (foo[[i]][1] == 1)
foo[[
You could use a 'replacement function' named 'bar<-', whose last argument
is called 'value', and use bar(variable) <- newValue where you currently
use foo(variable, newValue).
bar <- function(x) {
x + 3
}
`bar<-` <- function(x, value) {
bar(value)
}
a <- NA
bar(a) <- 4
a
# [1] 7
b <- list
Note that when using double precision arithmetic you cannot tell
if that number is different from other numbers within 10 billion of
it -- they all have the same representation.
> 1.677721599849e+29 == 1.677721599849e+29 + 1e10
[1] TRUE
You need to use other, time consuming, methods (e.g.
eval( parse(text=txt), parent.frame() )
>
> in `foo()` will do the trick.
>
> Bests
> Sören
>
> > On 15.05.2015, at 00:02, William Dunlap wrote:
> >
> > You could use a 'replacement function' named 'bar<-', whose last argument
>
You can automate the adding of the names to the list with the following
function, so you
can replace the
dflist<-list(df1,df2,df3)
names(dflist)<-c("df1","df2","df3")
with
dflist <- namedList(df1, df2, df3)
If you supply names, such in
dflist <- namedList(df1, Second=df2, log(df3))
it will
Here is a self-contained example of what you might be trying to do.
You would get better answers if you supplied this yourself.
dataset1 <-
data.frame(Date=as.POSIXct(c("2015-04-01","2015-04-01","2015-04-07",
"2015-04-19")), Weight=11:14)
datums <- as.POSIXct(c("2015-04-01", "2015-04-08", "2015-04
Can you show a small self-contained example of you data and expected
results?
I tried to make one and your expression returned a single number in a 1 by
1 matrix.
library(doBy)
Generation<-list(
data.frame(Wgt=c(1,2,4), SPCLORatingValue=c(10,11,12)),
data.frame(Wgt=c(8,16), SPCLORatingValue=
If you add a print statement to trace the evaluation of the input argument
you can see how the lazy evaluation works:
> update <- function (newtime) {
cat("# update() is changing global ginput's time from", ginput$time,
"to", newtime, "\n")
ginput <<- list(time = newtime)
}
> server <-
I renamed your 'c' to be 'toyData' and your 'e' to be 'desiredResult'. Do
you
want the following, which uses only base R code?
> vapply(toyData,
FUN=function(V)with(V, sum(Wgt[SPCLORatingValue>16])),
FUN.VALUE=0)
V5 V8 V10 V44
One way to use variable names in functions like Predict() that
do not evaluate their arguments in the standard way is to use
do.call() along with as.name(). E.g.,
varName<-"age"
do.call("Predict", list(fit, as.name(varName), np=4))})
gives the same result as
Predict(fit, age, np=4)
Bill D
match() will do what you want. E.g., run your data through
the following function.
f <- function (data)
{
uniqStrings <- unique(c(data[, 2], data[, 3], data[, 4]))
uniqStrings <- setdiff(uniqStrings, "0")
for (j in 2:4) {
data[[j]] <- match(data[[j]], uniqStrings, nomatch = 0L
x) where 'x' contains the strings from all the columns
> that you want to replace:
>
> m <- matrix(c(
> "X0001", "BYX859",0,0, 2, 1, "BYX859",
> "X0001", "BYX894",0,0, 1, 1, &
I don't know for sure what result you want, but it may be that
using a length(dim(array))-column matrix as your subscript will
do what you want. E.g.,
> a <- array(101:124, dim=4:2)
> subscriptMatrix <- cbind(c(1,3), c(2,2), c(2,1))
> subscriptMatrix
[,1] [,2] [,3]
[1,]12
Their 'srcref' attributes differ. srcref describes the text that
the parser was given when creating the function.
> identical(f1 <- function(x)1, f2 <- function(x)1)
[1] FALSE
> str(f1)
function (x)
- attr(*, "srcref")=Class 'srcref' atomic [1:8] 1 17 1 28 17 28 1 1
.. ..- attr(*, "srcfile")=
Points later in the input vectors may be obscuring earlier
points. If that is the problem then use pch="." or cex=.2
(or some other small number) to make the plot symbols
smaller so they don't overlap as much. Sometimes using
transparency helps also - try using adjustcolor(Color(n), alpha.f=0.5)
You can use
labels=as.expression(lapply(1:9, function(i)bquote(hat(lambda)[.(i)])))
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Wed, Jun 3, 2015 at 12:39 PM, Frank S. wrote:
> Hi everyone, I have allocated greek letters with subscript numbers on
> x-axis of my R plot this way:
> plot(10
Does the following do what you want?
> d <- list() # empty list
> for(i in 1:8) if (i %% 2 == 0) {
newElement <- structure(list(i), names=LETTERS[i])
d <- c(d, newElement)
}
> str(d)
List of 4
$ B: int 2
$ D: int 4
$ F: int 6
$ H: int 8
Bill Dunlap
TIBCO Software
wdunlap tibco.c
Your f1() has an unneeded for loop in it.
f1a <- function(mat) mat > 0.1, 1, 0)
would do the same thing in a bit less time.
However, I think that a simple
mat > 0.1
would be preferable. The resulting TRUEs and FALSEs
are easier to interpret than the 1s and 0s that f1a()
produces and arithme
> mynames
[1] "x.y" "x.y" "x.y" "x.y"
> mynames <- paste(mynames, seq_along(mynames), sep="_")
In addition, if there were a variety of names in mynames and you
wanted to number each unique name separately you could use ave():
> origNames <- c("X", "Y", "Y", "X", "Z", "X")
> ave(origNames
Use is.element(elements,set), or its equivalent, elements %in% set:
df <- data.frame(dd = c(1, 2, 3),
rows = c("A1", "A2", "A3"),
columns = c("B1", "B2", "B3"),
numbers = c(400, 500, 600))
test_rows <-c("A1","A3")
df[ is.element(df$rows, test_rows
participant.files <- list.files("/Users/cdanyluck/Documents/Studies/MIG -
Dissertation/Data & Syntax/MIG_RAW DATA & TXT Files/Plain Text Files")
Try adding the argument full.names=TRUE to that call to list.files().
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, Jun 8, 2015 at 7:15 PM, C
> combnWithRepetition <- function(n, k) combn(n+k-1, k) - seq(from=0, len=k)
> combnWithRepetition(2, 2)
[,1] [,2] [,3]
[1,]112
[2,]122
> combnWithRepetition(3, 2)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]111223
[2,]123233
.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Jun 9, 2015 at 8:56 AM, William Dunlap wrote:
> > combnWithRepetition <- function(n, k) combn(n+k-1, k) - seq(from=0,
> len=k)
> > combnWithRepetition(2, 2)
> [,1] [,2] [,3]
> [1,]11
I am not sure what you mean by "a block", but you can probably use
scan() instead of read.table():
> txt <- "1 21 41 61 81"
> d <- data.frame(Numbers=scan(text=txt, what="numeric", quiet=TRUE))
> d
Numbers
1 1
2 21
3 41
4 61
5 81
Bill Dunlap
TIBCO Sof
Usually, along with that error message, you get a warning
that gives a few details about the problem, like the name
of the file and the reason it could be opened. E.g.,
> filename <- "no\\such\\file.blah.blah"
> file(filename, "r", encoding="UTF-8")
Error in file(filename, "r", encoding = "U
> The CRAN guidelines should be rewritten so that they say what they *mean*.
> If a complete sentence is not actually required --- and it seems
abundantly clear
> that it is not --- then guidelines should not say so. Rather they should
say,
> clearly and comprehensibly, what actually *is* required
for (i in 1:(ncol(MD_dist) - 1)){
...
}
Even better, replace
1:(n-1)
with
seq_len(n-1)
The latter does what you want (and empty integer vector) when n is 1;
the former would give c(1,0).
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Jul 7, 2015 at 2:11 PM, David Barron wr
You can use an environment instead of a list using the same [[ syntax. It
is like 'get0(..., inherit=FALSE)' on the left side of the <- and like
'assign(...)' on the right side. E.g.,
myData <- new.env()
varName <- "v1"
myData[[varName]] <- 1:10
myData[[varName]][4] <- myData[[varNam
Look at your data.frame with str() and see if the variables you think
are numeric are actually factors. is.finite(factor()) reports TRUE but
lots of functions expecting numeric data will abort when given factor
data.
E.g.
library(ggplot2)
> d <- data.frame(X=factor(round(sin(1:1000),1)),
Y=factor
't assign() do the trick? Most
> similar threads seem to include this approach (among others, indeed).
>
> Regards,
> Bastien
>
>
> On Wednesday, July 8, 2015 at 7:30:04 PM UTC+2, William Dunlap wrote:
> > You can use an environment instead of a list using the same [[
Try doing
rownames(x) <- x["Species"]
before running dbFD. The function is probably using the row and column
names of its inputs as species names, not a certain row or column of the
data.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Wed, Jul 15, 2015 at 6:47 AM, lauraRIG wrote:
> Dear R
Read about the 'makepredictcall' generic function. There is a method,
makepredictcall.poly(), for poly() that attaches the polynomial coefficients
used during the fitting procedure to the call to poly() that predict()
makes.
You ought to supply a similar method for your xpoly(), and xpoly() needs
~ log(u) + OPoly(u,1), data = clotting, family = Gamma)
>
> > predict(g3,dc)Error in poly(4 * (rep(x, weight) -
> > mean(range(x)))/diff(range(x)), degree) :
> missing values are not allowed in 'poly'
>
> I thought it might be due to the /diff(range(x) in the func
fit)[c(3,8)]
predict(fit, newdata=data.frame(X=d$X[c(3,8)])) # same result
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, Jul 16, 2015 at 4:39 PM, William Dunlap wrote:
> OPoly<-function(x,degree=1,weight=1){
> weight=round(weight,0)# weight need to be integer
> if(leng
You could do something like the following
> rowsToShiftLeft <- c(2,4,5) # 4, not the 3 that was in the original post
> mat <- as.matrix(df_start)
> mat[rowsToShiftLeft, 1:3] <- mat[rowsToShiftLeft, 2:4]
> result <- data.frame(mat[, 1:3], stringsAsFactors=FALSE)
> str(result)
'data.frame
You can get a more informative error message from system("bash -c 'cd
yourDirectory'"), although it will not take R to that directory if there
are no problems. E.g., I did in a shell
% mkdir -p dir/subdir
% chmod -x dir
to make an untraversable directory 'dir' and a subdirectory of it. Then
This will happen if you have redefined 'T':
> T <- 101:104
> write.table(df, file="junk.txt", sep=",", append=T, quote=F,
row.names=F, col.names=F)
Error in file(file, ifelse(append, "a", "w")) : invalid 'open' argument
Use 'TRUE' and 'FALSE' instead of 'T' and 'F' this sort of problem.
Bil
Missing words:
Use 'TRUE' and 'FALSE' instead of 'T' and 'F' >to avoid< this sort of
problem
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, Jul 27, 2015 at 1:50 PM, William Dunlap wrote:
> This will happen if you have redefined &
If you return just the row that the foreach procedure produces instead of
the entire matrix containing that row and use .combine=rbind then you will
end up with the matrix of interest. E.g.,
Simpar3a <- function (n1)
{
L2distance <- matrix(NA, ncol = n1, nrow = n1)
data <- rnorm(n1)
d
> During installation EDC_HOME was set to /home/robert/EDC
> and the directory definitely exists.
Are you sure that EDC_HOME is set now? What do you get from the following
command?
Sys.getenv("EDC_HOME")
If that is set to something other than "", what do you get from
getwd()
setwd(Sys.ge
# your data, from dump("xxx", file=stdout())
xxx <-
structure(list(Registered = c(1327, 2129, 10, 433, 5, 166, 1784,
17, 787), GA_Total = c(127, 150, 0, 32, 1, 76, 153, 7, 31), PCT = c("1120",
"1121", "1121", "1122", "1124", "1125", "1125", "1125", "1126"
)), .Names = c("Registered", "GA_Total", "P
help(distcritmulti) says that its 2nd argument is
"clustering: vector of integers indicating the clustering"
and you supplied the output of hclust, which is not a
vector of group identifiers. You can make a group
identifier vector from hc with cutree(hc, k=4), where k is
the number of groups.
The || operator will always return a result of type 'logical' and length 1.
You gave it two operands of length 0, so it returned the logical value NA,
meaning it had no idea what the result should be. If you give it operands
of length > 1, it will use the only the first elements of them. (S and S
In the R GUI the output is
> path.expand("~")
[1] "~"
Did you set the environment variable R_USER to something odd like "~"?
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Aug 11, 2015 at 7:41 AM, Thierry Onkelinx
wrote:
> Dear all,
>
> I'm puzzled by the behaviour of path.ex
You can do this in 2 steps - have cut() make a factor with a different
level for each time period
then use levels<-() to merge some of the levels.
> z <- cut(.5:3.5, breaks=c(0,1,2,3,4), labels=c("0-1", "1-2", "2-3",
"3-4"))
> levels(z)
[1] "0-1" "1-2" "2-3" "3-4"
> levels(z) <- c("betw
You could try using R.utils::withTimeout(expr, timeout=10) to cause an
error when evaluating the expression expr takes longer than 10 seconds.
Wrap that in tryCatch or try to catch the error and examine the outputs of
lapply for ones of class "TimeoutException" to find the ones that took too
long.
Does the following do what you want?
> raw <- c("A/B", " /B", "A/", "/ ")
> tmp <- sub("^ */", "./", raw)
> cleaned <- sub("/ *$", "/.", tmp)
> cleaned
[1] "A/B" "./B" "A/." "./."
(The " *" is to allow optional spaces before or after the slash.)
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On F
"\\" is stored as a single backslash, just as "\n" is a single newline
character. It is printed with an extra backslash.
> nchar("\\")
[1] 1
> cat(paste0("\\", "\n"))
\
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Sat, Dec 6, 2014 at 11:00 AM, Prof J C Nash (U30A)
wrote:
> This i
strsplit(split=":") does almost what you want, but it omits the colons from
the output. You can use perl zero-length look-ahead and look-behind
operators in the split argument to get the colons as well:
> strsplit(c(":sad", "happy:", "happy:sad"), split="(?<=:)|(?=:)",
perl=TRUE)
[[1]]
[1] ":"
":"
[[3]]
[1] "happy" ":" "sad"
[[4]]
[1] ":" "happy" ":" "sad" ":" "subdued" ":"
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, Dec 8, 2014 at 1:1
Have you looked at the merge() function?
Here is an example. I don't know if it resembles your problem.
> M1 <- data.frame(V1=letters[1:3], V2=LETTERS[26:24], N1=101:103)
> M2 <- data.frame(V1=letters[c(3,1,2,3,2)],
V2=LETTERS[c(23,26,22,24,24)], N2=c(1003,1001,1002,1003,1002))
> merge(M
Here is a reproducible example
> d <- read.csv(text="Name,Age\nBob,2\nXavier,25\nAdam,1")
> str(d)
'data.frame': 3 obs. of 2 variables:
$ Name: Factor w/ 3 levels "Adam","Bob","Xavier": 2 3 1
$ Age : int 2 25 1
Do you get something similar? If not, show us what you have (you
could
s is fundamentally
> sound and something I will need to get my head around. For now though, I
> think I'll stick to exploring ggplot2 so that I can visualise this data set
> more easily.
>
> Thanks again.
>
> Best
>
> Sun
>
>
> On 11/12/14 16:06, William Dunla
Your 'x' has length 2, so x[[3]] cannot be calculated ('subscript out of
bounds' is what I get). You can check for this with length(x)<3.
In general, you want to be more precise: 'does not have a value', 'is
NULL', and 'is empty' are not synonymous. I'm not sure what 'does not have
a value' mean
Another approach to to make a table and extract your summaries from the
table:
> tbl <- with(Test, table(ID, IsZero=X==0))
> tbl
IsZero
ID FALSE TRUE
1 31
2 02
3 03
> sum(tbl[,"FALSE"] == 0)
[1] 2
> sum(tbl[,"FALSE"] == 0 & rowSums(tbl)>=3)
[
A while ago I wrote for a questioner on this list a function, 'f1', below,
that would give the start and stop times of runs of data that started when
then the data went above a threshold and stopped when it first dropped
below a different (lower) threshold). It used no loops and was pretty
quick.
> ave( as.character(1:5), gl(2,2,5), FUN=length )
[1] "3" "3" "2" "2" "3"The output has character type, but it is supposed
to be a collection of vector lengths.
ave() uses its first argument, 'x', to set the length of its output and to
make
an initial guess at the type of its output. The re
as.vector(x) will return x without any attributes and
structure(x, attrA=NULL, attrB=NULL) will return x
without the named attributes.
> z <- f(1:3, 4)
> z
[1] 14
attr(,"gradient")
[1] -6 -4 -2
> as.vector(z)
[1] 14
> structure(z, gradient=NULL)
[1] 14
as.vector is a gen
> But this heterogeneity comes even with only supposed numeric data.frame
> (gene expression). here an example
>
> ibrary(cgdsr)
> GeneList <- c("DDR2", "HPGDS", "MS4A2","SSUH2","MLH1" ,"MSH2", "ATM"
> ,"ATR", "MDC1" ,"PARP1")
> cgds<-CGDS("http://www.cbioportal.org/public-portal/";)
>
> str(getPr
You can use options(scipen=n) to control when numbers are printed
using scientific notation (the default value is 0). Compare
options(scipen=0); plot( 1e-17*(1:10), 1e5*(1:10))
options(scipen=10); plot( 1e-17*(1:10), 1e5*(1:10))
options(scipen=20); plot( 1e-17*(1:10), 1e5*(1:10))
Bill Dunlap
f <- function (x) {
isState <- is.element(tolower(x), tolower(state.name))
w <- which(isState)
data.frame(State = x[rep(w, diff(c(w, length(x) + 1)) - 1L)],
City = x[!isState])
}
E.g.,
V1 <-c("alabama", "bates", "tuscaloosa", "smith", "arkansas", "fayette",
"little rock", "alas
Supply a vector of colors. E.g.,
barplot(table(mtcars$cyl), col=c("red","yellow","green"))
boxplot(with(mtcars, split(wt, cyl)), col=c("red","yellow","green"))
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, Jan 8, 2015 at 12:06 PM, varin sacha wrote:
> Hi R-experts,
> Basic question
Change that column to be a factor with the levels in the order that you
wish.
dataFrame$column <- factor(dataFrame$column, levels=c(1:150, "S", "Z"))
Then it will sort in that order. E.g.,
> d <- data.frame(One=c(5,21,10,"Z","S",9), Two=2^(1:6))
> d[order(d$One),]
One Two
3 10 8
I think it would be nice if predict methods returned NA in appropriate
spots instead of aborting when a categorical predictor contains levels not
found in the training set. It should not be that hard to implement, as the
'xlevels' component of the model is already being used to put factor levels
i
There are lots of ways to do this. You have to decide on how you want to
organize the results.
Here are two ways that use only core R packages. Many people like the plyr
package for this
split-data/analyze-parts/combine-results sort of thing.
> df <- data.frame(x=1:27,response=log2(1:27),
Are you sure the factors of T are in the order you think they are? (Are you
sure you are using the expected version of T.) Use print(levels(T)) to
make
sure.
I tried
timeCats <- c("Presurgery", "Day 30", "Day 60", "Day 180", "Day 365")
d <- data.frame(T = factor(rep(timeCats, 11:15), lev
The following is one way to parse your file using R (using R-3.1.2 on
Windows
in a US English locale). I downloaded it from Google Docs in tab-separated
format.
I could not get read.table() to do the job, but I don't completely
understand
the encoding/fileEncoding business there.
> file <- "examp
You did not show what answer you expected, but does the following do what
you want?
> match(sc$TERM, TermList)
[1] 11 20 30
Making a factor whose levels are TermList may also be useful. (The
exclude=NULL
is to factor doesn't drop NA from the levels).
> sc$fTERM <- factor(sc$fTERM, levels=T
Here is one way. Sort the data.frame, first by Name then break ties with
CheckInDate.
Then choose the rows that are the last in a run of identical Name values.
> txt <- "NameCheckInDate Temp
+ John 1/3/2014 97
+ Mary 1/3/2014 98.1
+ Sam 1/4/2014
last (as ordered in the original file).
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Sun, Jan 25, 2015 at 1:01 AM, Göran Broström
wrote:
> On 2015-01-24 01:14, William Dunlap wrote:
>
>> Here is one way. Sort the data.frame, first by Name then break ties with
>> CheckIn
> with(dat1, ave(integer(length(Date)), Date, FUN=seq_along))
[1] 1 1 2 1 1 2 1 2 1 2 1
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Wed, Jan 28, 2015 at 4:54 PM, Morway, Eric wrote:
> The two datasets below are excerpts from much larger datasets. Note that
> there are duplicate dates in
Does
help(curve)
talk about its 'xname' argument?
Try
curve(10*foofoo, from=0, to=17, xname="foofoo")
You will have to modify your function, since curve() will
call it once with a long vector for the independent variable
and func(rnorm(10), rnorm(10), mu=seq(0,5,len=501)) won't
work right.
You successfully used an 'expression' object in your first example
plot(1,1,xlab=expression(paste("D50 [", mu,"m]")))
so use a vector of expressions in your second example
nams<-c(expression(P[205]~"[%]"), expression(paste("D50 [", mu, "m]")))
par(mfrow=c(2,1))
for (i in 1:2) plot(1, 1,
For class 'ts' the 'window' function in the base package will do it.
> x <- ts(101:117, start=2001.75, frequency=4)
> x
Qtr1 Qtr2 Qtr3 Qtr4
2001 101
2002 102 103 104 105
2003 106 107 108 109
2004 110 111 112 113
2005 114 115 116 117
> window(x, start=2002.5)
Does the following work for you. The only trick is working around the fact
that matrix subscripting does not allow out-of-bounds subscripts but vector
subscripting does. We do the subscripting in two steps, relying on the
drop=TRUE default in the matrix subscripting operator that converts the
sin
Try adding the line
#define R_NO_REMAP
to the top of your file (before any #include's) so the R
include files don't define length(x).
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, Feb 2, 2015 at 1:11 PM, wrote:
> Hi,
>
> when writing a simple cpp function to be run in R I obtain a comp
I don't think that worked as OP would like it too - all columns of the
output
are factors.
>
data.frame(rbind(as.matrix(data.frame(a=1:3,b=letters[1:3])),as.matrix(data.frame(x=1:5,b=LETTERS[1:5]
a b
1 1 a
2 2 b
3 3 c
4 1 A
5 2 B
6 3 C
7 4 D
8 5 E
> str(.Last.value)
'data.frame': 8 obs. o
A useful technique when it is easy to compute a vector from an ordered
data.frame but you need to do it for an unordered one is to compute the
order
vector 'ord', compute the vector from df[ord,], and use df[ord,...] <-
vector
to reorder the vector. In your case you could do:
> dat_2<-data.frame
A potential problem with
ave(dat_2$D, dat_2$S, FUN=order)
is that it will silently give the wrong answer
or give an error if dat_2$D is not numeric.
E.g., if D is a Date vector we get
> dat_3 <- dat_2[,1:2]
> dat_3$D <- as.Date(paste0("2015-02-", dat_2$D))
> with(dat_3, ave(D, S, FUN=orde
> ff <- reformulate(termlabels=c("time","as.factor(gvhd)"), response=yname,
intercept=TRUE)
If the right hand side of the formula were more complicated than an
additive list of terms,
say '~ time * as.factor(gvhd)' or the left side were more than a name, say
'log(yname)' you
could use bquote() ins
You can let lapply() do the preallocation and the looping for you with
ASL <- lapply(1:5, function(j) lapply(1:5, function(i) i^j))
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, Feb 16, 2015 at 9:46 AM, Jeff Newmiller
wrote:
> You have two named objects when your goal is to have one
You did not put the altered columns back into the data.frame,
so glm() never saw them. Does the following work?
func <- function(x,y,z) {
#x is a data frame
#y is a formula for the regression
#z vector of names of columns of x to convert to factors
for (name in z) {
x[[name]] <- facto
The elNamed(x, name) function can simplify this code a bit. The following
gives the same
result as David W's get_shas() for the sample dataset provided:
get_shas2 <- function (input) {
lapply(input, function(el) elNamed(elNamed(el, "content")[[1]],
"sha")[1])
}
Bill Dunlap
TIBCO Sof
Using lapply() where Jim used sapply() would keep the types
right and be a fair bit faster than a solution based on repeatedly
appending to a list (like your getFirst).
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Fri, Feb 20, 2015 at 1:52 PM, JS Huang wrote:
> Hi,
>
> Jim's answer is neat
> Even though I was looking in several r-books
> I could not find a suitable function to this problem
Which R books did you look through?
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, Feb 26, 2015 at 4:02 AM, David Studer wrote:
> Hello everybody!
>
> I have a (probabely very easy) prob
You could define functions like
is.true <- function(x) !is.na(x) & x
is.false <- function(x) !is.na(x) & !x
and use them in your selections. E.g.,
> x <- data.frame(a=1:10,b=2:11,c=c(1,NA,3,NA,5,NA,7,NA,NA,10))
> x[is.true(x$c >= 6), ]
a b c
7 7 8 7
10 10 11 10
Bill Dun
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