Displaying 20 results from an estimated 7000 matches similar to: "Using of LME function in non-replicate data"
2006 Jul 08
2
String mathematical function to R-function
hello
I make a subroutine that give-me a (mathematical)
function in string format.
I would like transform this string into function ( R
function ).
thanks for any tips.
cleber
#e.g.
fun_String = "-100*x1 + 0*x2 + 100*x3"
fun <- function(x1,x2,x3){
return(
############
evaluation( fun_String )
############
)
True String mathematical function :-( :-(
> nomes
[1]
2006 Jan 27
1
about lm restrictions...
Hello all R-users
_question 1_
I need to make a statistical model and respective ANOVA table
but I get distinct results for
the T-test (in summary(lm.object) function) and
the F-test (in anova(lm.object) )
shouldn't this two approach give me the same result, i.e
to indicate the same significants terms in both tests???????
obs.
The system has two restrictions:
1) sum( x_i ) = 1
2) sum(
2009 Nov 09
3
How to transform the Matrix into the way I want it ???
Hi, R users,
I'm trying to transform a matrix A into B (see below). Anyone knows how to
do it in R? Thanks.
Matrix A (zone to zone travel time)
zone z1 z2 z3 z1 0 2.9 4.3 z2 2.9 0 2.5 z3 4.3 2.5 0
B:
from to time z1 z1 0 z1 z2 2.9 z1 z3 4.3 z2 z1 2.9 z2 z2 0 z2 z3 2.5 z3 z1
4.3 z3 z2 2.5 z3 z3 0
The real matrix I have is much larger, with more than 2000 zones. But I
think it should
2017 Jul 28
3
problem with "unique" function
I have the joint distribution of three discrete random variables z1, z2 and
z3 which is captured by "z"
and "prob" as described below.
For example, the probability for z1=0.46667, z2=-1 and z3=-1 is 2.752e-13.
Also, the probability adds up to 1.
> head(z) z1 z2 z3
[1,] -0.46667 -1.0000 -1.0000
[2,] -0.33333 -0.9333 -0.9333
[3,] -0.20000 -0.8667 -0.8667
2006 Jun 14
3
A question about stepwise procedures: step function
Dear all,
I tried to use "step" function to do model selection, but I got an error massage. What I don't understand is that data as data.frame worked well for my other programs, how come I cannot make it run this time. Could you please tell me how I can fix it?
***************************************************************************************************
2018 Feb 25
3
include
Thank you Jim,
I read the data as you suggested but I could not find K1 in col1.
rbind(preval,mydat) Col1 Col2 col3
1 <NA> <NA> <NA>
2 X1 <NA> <NA>
3 Y1 <NA> <NA>
4 K2 <NA> <NA>
5 W1 <NA> <NA>
6 Z1 K1 K2
7 Z2 <NA> <NA>
8 Z3 X1 <NA>
9 Z4 Y1 W1
On Sat, Feb 24, 2018 at 6:18 PM, Jim
2012 Mar 09
4
For loop and using its index
Dear All,
I have a data set with variables x1, x2, x3, ..., x20 and I want to
create z1, z2, z3, ..., z20 with the following formula:
z1 = 200 - x1
z2 = 200 - x2
z3 = 200 - x3
.
.
.
z20 = 200 - x20.
I tried using a for loop and its index as:
for (i in 1:20) {
z(i) = 200 - x(i)
}
But R gives the following error message: "Error: could not find function "x"".
Is there any
2010 Jan 29
1
use zoo package with multiple column data sets
Readers,
I am trying to use the zoo package with an array of data:
file1:
hh:mm:ss 1
hh:mm:ss 2
hh:mm:ss 3
hh:mm:ss 4
file2:
hh:mm:ss 11 55
hh:mm:ss 22 66
hh:mm:ss 33 77
hh:mm:ss 44 88
I wanted to merge these data set so I tried the following commands:
library(chron)
library(zoo)
z1<-read.zoo("path/to/file1.csv",header=TRUE,sep=",",FUN=times)
2009 Jul 02
1
lpSolve: how to allow variables to become negative
Dear all,
I am interested in solving a MIP problem with binary outcomes and
continuous variables, which ARE NOT RESTRICTED TO BE NEGATIVE. In
particular,
Max {z1,z2,z3,b1} z1 + z2 + z3
(s.t.)
# 7 z1 + 0 z2 + 0 z3 + b1 <= 5
# 0 z1 + 8 z2 + 0 z3 - b1 <= 5
# 0 z1 + 0 z2 + 6 z3 + b1 <= 7
# z1, z2, z3 BINARY {0,1}
# -5<= b1 <=5 (i.e. b1 <= 5; -b1 <= 5 )
Using
2018 Feb 25
2
include
HI Jim and all,
I want to put one more condition. Include col2 and col3 if they are not
in col1.
Here is the data
mydat <- read.table(textConnection("Col1 Col2 col3
K2 X1 NA
Z1 K1 K2
Z2 NA NA
Z3 X1 NA
Z4 Y1 W1"),header = TRUE,stringsAsFactors=FALSE)
The desired out put would be
Col1 Col2 col3
1 X1 0 0
2 K1 0 0
3 Y1 0 0
4 W1 0 0
6 K2 X1
2010 Aug 20
2
output values from within a function
Is it possible to get R to output the value of an expression, that is being calculated within a function? I've attached a very simple example but for more complicated ones would like to be able to debug by seeing what the value of specific expressions are each time it cycles through a loop that executes the expression. I'm relatively new to this so there may be much simpler more elegant
2018 Feb 25
0
include
Hi Val,
My fault - I assumed that the NA would be first in the result produced
by "unique":
mydat <- read.table(textConnection("Col1 Col2 col3
Z1 K1 K2
Z2 NA NA
Z3 X1 NA
Z4 Y1 W1"),header = TRUE,stringsAsFactors=FALSE)
val23<-unique(unlist(mydat[,c("Col2","col3")]))
napos<-which(is.na(val23))
preval<-data.frame(Col1=val23[-napos],
2017 Jun 25
2
Writing my 3D plot function
Hi all,I had a question last week on asking for a function that will help me draw three different circles on x,y,z axis based on polar coordinates (Each X,Y,Z circle are coming from three independent measurements of 1-360 polar coordinates). It turned out that there ?is no such function in R and thus I am trying to write my own piece of code that hopefully I will be able to share. I have spent
2017 Jul 28
0
problem with "unique" function
Most likely, previous computations have ended up giving slightly different values of say 0.13333. A pragmatic way out is to round to, say, 5 digits before applying unique. In this particular case, it seems like all numbers are multiples of 1/30, so another idea could be to multiply by 30, round, and divide by 30.
-pd
> On 28 Jul 2017, at 17:17 , li li <hannah.hlx at gmail.com> wrote:
2011 May 24
4
writing dates to a file
Hi,
I have attached the data files to this note. I use this code:
library(zoo)
z1 <- read.zoo("baltimorefludata.txt", format = "%m/%d/%Y", header = TRUE)
z2 <- read.zoo("baltimorew.txt", format = "%Y%m%d", header = TRUE)
z3<-merge(z1,z2)
write.table(z3, "fluweatherdata_baltimore2.txt", sep="\t")
R is writing the other data
2009 Nov 24
1
Titles in plots overlap
Hi,
I use fCopulae package to draw different graphs of univariate and bivariate skew t. But the plots titles overlap. I tried using cex.main, font.main to adjust the size but they still overlaps. Here is my code:
par(mfrow = c(3, 1))
mu = 0
Omega = 1
alpha1 = 0
alpha2 = 1.5
alpha3 = 2
alpha4 = 0.5
Z1 = matrix(dmvst(x, 1, mu, Omega, alpha1, df = Inf), length(x))
Z2 = matrix(dmvst(x, 1, mu,
2012 Aug 10
1
Solving binary integer optimization problem
Hi,
I am new to R for solving optimization problems, I have set of communication
channels with limited capacity with two types of costs, fixed and variable
cost. Each channel has expected gain for a single communication.
I want to determine optimal number of communications for each channel
maximizing ROI)return on investment) with overall budget as constraint.60000
is the budget allocated.
2018 Feb 25
2
include
Sorry , I hit the send key accidentally here is my complete message.
Thank you Jim and all, I got it.
I have one more question on the original question
What does this "[-1] " do?
preval<-data.frame(Col1=unique(unlist(mydat[,c("Col2","col3")]))[-1],
Col2=NA,col3=NA)
mydat <- read.table(textConnection("Col1 Col2 col3
Z1 K1 K2
Z2
2017 Jun 25
0
Writing my 3D plot function
Please look at what I see in your code below (run-on code mush) to understand part of why it is important for you to send your email as plain text as the Posting Guide indicates. You might find [1] helpful.
[1] https://wiki.openstack.org/wiki/MailingListEtiquette
--
Sent from my phone. Please excuse my brevity.
On June 25, 2017 2:42:26 PM EDT, Alaios via R-help <r-help at r-project.org>
2009 Sep 25
2
synchronisation of time series data using interpolation
Readers,
I have data with different time stamps that I wish to plot (for example):
data set 1
time(hh:mm:ss),datum
01:00:00,500
01:00:15,600
01:00:30,750
01:00:45,720
01:01:00,700
01:01:15,725
01:01:30,640
01:01:45,710
data set 2
time,datum
01:00:12,20
01:01:01,55
01:01:55,22
The time interval in data set 1 does not change, but the time interval
in data set 2 does change, such that for a