I wanted to do an F-test to compare model L2 with L3 assuming
heteroskedasticity. The variable KONK2 only has three levels and are for
that reason replaced by dummyvariables KONKD1 and KONKD2 in L3. Hence making
L4 equivalent to L2.
L2-lm(PRMRES ~ factor(NYPR) + KONK2 + OMS + factor(NYPR)*OMS )
Is there a way to find all roots of a polynomial equation?
Lets say
x^5+a*x^4+b*x^3+c*x^2+d*x+e=0
how to find its all roots?
The package rootSolve might proove interesting
__
R-help@ mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
Is there anyway I can convert a vectors objectname to a string to be used in
fx:
Monkey-c(0,0,0,1,1,1)
Wax-c(1,0,1,0,1,0)
f-function(x,y){ table(x,y) }
f(Monkey,Wax)
so that the printout is not
y
x 0 1
0 1 2
1 2 1
but
Wax
Monkey 0 1
0 1 2
1 2 1
--
View
Or try:
datbig-numeric(1000)
for (i in 1:1000){
datbig[i] - mean(sample(1:6, 1000, prob = c(1, 1, 2, 3, 2, 1)/10, replace
= TRUE))
hist(datbig,breaks=FD)
--
View this message in context:
http://r.789695.n4.nabble.com/ifelse-command-tp2329538p2329711.html
Sent from the R help mailing list
sorry missed an }
datbig-numeric(1000)
for (i in 1:1000){
datbig[i] - mean(sample(1:6, 1000, prob = c(1, 1, 2, 3, 2, 1)/10, replace
= TRUE))
hist(datbig,breaks=FD)
}
--
View this message in context:
http://r.789695.n4.nabble.com/ifelse-command-tp2329538p2329717.html
Sent from the R help
How can I control label on y-axis? Specifically how high relative to the
y-axis it is plotted?
At the moment I am using mtext appearing in the function quotet below.
The function is applied to a dataframe and plots histograms for all
numerical variables of the frame generating several
how to plot 4 graphs simultaneously?
Depends on what graphs were talking about: A Side-by-side boxplot containing
fx 4 boxplots can be generated by a single plot() command. However often you
use a primary command that activivates
the plot window and plots one graph, then afterwards you use
You need {}
for (i in 1:500){ ...commands...}
Like:
x-rep(c(0.8,1.2),250)
for (i in 1:500) {
if (x[i]1) x[i]=1
}
BR
JEsper
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-perform-a-substitution-in-a-loop-tp2325048p2325073.html
Sent from the R help mailing
Example is spot on - sr for not providing one myself.
The results you calculate are what I'm looking for.
Would like a function F where I could type:
F(weight ~ Time, data = ChickWeight, SOME ARGUMENT = Diet))
Resulting in
for (i in 1:4){
print(
lm(weight ~ Time, data = ChickWeight, subset =
If you're trying to follow the youtube video you have a typing mistake here:
InsectSprays.aov -(test01$count ~ test01$spray)
I think this should be:
InsectSprays.aov -aov(test01$count ~ test01$spray)
youre missing the functioncall aov on the right hand side of the
assignment operator '-'.
I think it would be helpful if you could clarify youre question - do you want
distinct sets - maybe use
unique()
but why (5,20) when its (5,10) in the row in youre example? What criteria do
you want the function to select the sets by and what kind of output do you
need?
Maybe it's just me
I have a simple dataset of a numerical dependent Y, a numerical independent X
and a categorial variable Z with three levels. I want to do linear
regression Y~X for each level of Z. How can I do this in a single command
that is without using lm() applied three isolated times?
--
View this message
You can use following scriptI think
#create a vector of random numbers on which to test script
v-sample(1:3,size=90,replace=TRUE)
#creates two matrixes out of vector v which can be assigned to M to test
script
M2-matrix(v,ncol=2)
M3-matrix(v,ncol=3)
M- #Assign you're matrix or a
13 matches
Mail list logo
