Displaying 20 results from an estimated 1524 matches for "indded".
2012 Nov 17
2
problem with "any" function
Dear R users,
I have the "any" function of R
any(ind.c, ind.r, ind.sgn)
all are logical factors
it works fine when any of three is true
but when they are combined it doesnt work.
i tried
this
any(ind.c, ind.r, ind.sgn,((ind.c==TRUE) & (ind.r==TRUE)),
((ind.c==TRUE) & (ind.sgn==TRUE)),((ind.r==TRUE) & (ind.sgn==TRUE)),
((ind.c==TRUE) & (ind.r==TRUE) &
2008 Mar 28
1
Beginner help with retrieving frequency and transforming a matrix
Hi All,
Just hoping some one can give me a hand with a problem...
I have a dataframe (DF) with about 5 million entries that looks something
like the following:
>DF
ID Cl Co Brd Ind A AB AB
1 S-3 IND A BR_F BR_F01 1 0 0
2 S-3 IND A BR_F BR_F01 1 0 0
3 S-3 IND A BR_F BR_F01 1 0 0
4 S-3 IND A BR_F BR_F01 1 0 0
5 S-3 IND A BR_F BR_F01 1 0 0
6 S-3 IND A BR_F
2011 Jan 11
0
modified FAST Script from package SensoMineR for the R community - Reg
###Dear R users
###I have been using SensoMineR package from CRAN for most of my work in
sensory data analysis and from my usage experience, I encountered some
areas for improvement and considered ###modifying the function in
SensoMineR package for my personal use. I felt that it could be useful to
share this to the community for enabling adoption by other users where
they might require a
2001 Nov 28
1
Reading output from "debugfs -R stat <8>"
Hi,
I'm trying to understand if my journal is fragmented. Here's some output
from `debugfs -R "stat <8>" /dev/sda3`:
Inode: 8 Type: regular Mode: 0600 Flags: 0x0 Generation: 0
User: 0 Group: 0 Size: 104857600
File ACL: 0 Directory ACL: 0
Links: 1 Blockcount: 205016
Fragment: Address: 0 Number: 0 Size: 0
ctime: 0x3c0442fd -- Tue Nov 27
2004 Jun 18
2
Barplots and error indicators: Some R-Code
I' ve seen that several people are looking for a function that creates a
barplot with an error indicators (I was one of them myself). Maybe you will
find the following code helpful (There are some examples how to use it at
the end):
# Creates a barplot.
#bar.plot() needs a datavector for the height of bars and a error
#indicator for the interval
#many of the usual R parameters can be set:
2007 Jul 10
1
Get journal position
Hi,
There is any way to figure out where physically is the journal on a ext3
fs and it's size?
Thanks!
Jordi
2011 Apr 29
3
why doesn't ifelse work ?
I have the following lines of code:
ind <- rollapply(GSPC, 200, mean)
signal <- ifelse(diff(ind, 5) > 0 , 1 , -1)
signal[is.na(signal)] <- 0
I never get a value of -1 for signal even though I know diff(ind , 5) is
less than zero frequently. It looks like when diff(ind , 5) is less than
zero, signal gets set to 0 instead of - 1. Any ideas why ? Here's some
information on ind and
2006 Jun 03
2
Busy Signals after hangup
I've not seen an answer to this in any forum.
I make a call through Asterisk, with a VOIP phone, doesn't matter which.
The call gets made, I leave a voicemail, or complete the call in some
manner, and the other side hangs up. I hear a busy signal on the phone
on my end.
If I have an extension that looks like this, after the hangup() is
executed, my phone gives busy signals until I
2009 Aug 27
5
Help on efficiency/vectorization
Dear R users,
I am trying to extract the rownames of a data set for which each columns
meet a certain criteria. (condition - elements of each column to be equal
1)
I have the correct result, however I am seeking for more efficient (desire
vectorization) way in implementing such problem as it can get quite messy if
there are hundreds of columns.
Arbitrary data set and codes are shown below for
2006 May 26
2
combinatorial programming problem
Hola!
I am programming a class (S3) "symarray" for
storing the results of functions symmetric in its
k arguments. Intended use is for association indices
for more than two variables, for instance coresistivity
against antibiotics.
There is one programming problem I haven't solved, making an inverse
of the index function indx() --- se code below. It could for instance
return the
2010 Feb 02
1
ggplot/time series with indicators question
Hello, I am trying to plot time-series data with certain weeks highlighted using symbols.
require(ggplot2)
#plotting time series data
timescale <- seq(as.Date("01/01/09","%m/%d/%y"), length.out=12, by=7)
data.all <- data.frame(
id = c(rep('111',12),rep('222',12),rep('333',12)),
week=c(timescale,timescale,timescale),
2012 Aug 03
5
replacement has length zero. In addition: Warning message: In max(i) : no non-missing arguments to max; returning -Inf
Hi,
Here is my data, the first 10 rows
> u=regCond_all[1:10,]
> dput(u)
structure(c(999, 999, 999, 999, 999, 999, 999, 999, 999, 999,
99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99, 99,
99, 99, 99, 99, 1.9, 2, 1.97, 1.99, 1.83, 1.78, 1.6, 1.52, 1.52,
1.36, 10.53, 9.88, 9.88, 10.53, 10.53, 10.53, 5.26, 9.88, 10.53,
10.53, 5.4, 5.57, 5.46, 5.34, 5.5, 5.59, 5.62, 5.76, 6.23, 6.19,
2011 Feb 25
1
help please ..simple question regarding output the p-value inside a function and lm
Dear R community members and R experts
I am stuck at a point and I tried with my colleagues and did not get it out.
Sorry, I need your help.
Here my data (just created to show the example):
# generating a dataset just to show how my dataset look like, here I have x
variables
# x1 .........to X1000 plus ind and y
ind <- c(1:100)
y <- rnorm(100, 10,2)
set.seed(201)
P <-
2005 Apr 19
2
indexing an array using an index-array, but one entry being ', '
Hi
I have the following array:
test <- array(c(1:16), dim = c(3,4,3))
test
## I call some enries using an index array
test.ind <- array(rbind(c(1,2,1), c(3,3,2)), dim = c(2,3))
test[test.ind]
## suppose I want all values in the 2nd row and 4th col over
## all three 3rd dimensions
test[2,4,]
how to specify a test.ind array with the last index left with ',' i.e
test.ind should be
2009 Feb 23
1
Slow indexing access for Matrix
Consider the following little "benchmark"
> require(Matrix)
> tmp <- Matrix(c(rep(1,1000),rep(0,9000)),ncol=1)
> ind <- sample(1:10000,10000)
> system.time(tmp[ind,])
user system elapsed
0.004 0.001 0.005
> ind <- sample(1:1000,10000,replace=TRUE)
> system.time(tmp[ind,])
user system elapsed
0.654 0.006 0.703
>
2009 Jun 17
2
Re gression by groups questions
I have a large dataset grouped by a factor and I want to perform a regression
on each data subset based on this factor. There are many ways to do this,
posted here and elsewhere. I have tried several. However I found one method
posted on the R wiki which works exactly as I want, and I like the elegance
and simplicity of the solution, but I don't understand how it works. Its
all in the formula
2007 Jul 13
2
Suggestion to extend aggregate() to return multiple and/or named values
Hi all,
This is my first post to the developers list. As I understand it,
aggregate() currently repeats a function across cells in a dataframe
but is only able to handle functions with single value returns.
Aggregate() also lacks the ability to retain the names given to the
returned value. I've created an agg() function (pasted below) that is
apparently backwards compatible (i.e.
2018 Feb 28
1
Missed opportunity in the midend, unsigned comparison
Hi everybody, I see a missed optimization opportunity in LLVM that GCC
catches and I'd love to hear community's input.
Here's the original C code:
1 char arr[2];
2 char *get(unsigned ind) {
3 if (ind >= 1) {
4 return 0;
5 }
6 return &(arr[ind]);
7 }
The variable `ind` is unsigned so, based on the comparison, if it is not
greater or equals to one, than it is
2011 Mar 27
6
Asking Favor For the Script of Median Filter
Hello,everybody. My name is Chuan Zun Liang. I come from Malaysia. I am just
a beginner for R. Kindly to ask favor about median filter. The problem I
facing as below:
> x<-matrix(sample(1:30,25),5,5)
> x
[,1] [,2] [,3] [,4] [,5]
[1,] 7 8 30 29 13
[2,] 4 6 12 5 9
[3,] 25 3 22 14 24
[4,] 2 15 26 23 19
[5,] 28 18 10 11 20
This
2006 Jul 03
2
help a newbie with a loop
Hi,
I am new in R and stumbled on a problem my (more experienced) friends
can not help with with. Why isnt this code working?
The function is working, also with the loop and the graph appears,
only when I build another loop around it (for different values of p) ,
R stays in a loop?
Can't it take more then 2 loops in one program?
powerb<-function(x,sp2,a,b,b1,m)
{