Dear R users,
I want to plot a one variable continuous function f(x) vs x, x=[0,1]. Say
for example: f(x)= x^2. Now, using the command plot(f~x) I will get a
curve where the range of x-axis is [0,1] with all equispaced label. But, I
need something else, and that is: my curve will be such that 80%
Dear R helpers,
today I found something interesting in R. 0^0 gives value 1 in R. But it
is undefined in mathematics. During debugging a R code, I found it and it
effects my program severely. So my question is why it is defined 1 in R?
Is there any particular reason or its a bug in the R software
Dear R-listers,
I am giving part of my R code :
###
n=15
m=1
library("partitions")
library("gregmisc")
library("combinat")
x = t(restrictedparts(n-m,m))
l = length(x[,1])
for(u in 1:l){
A= unique(matrix( unlist(permn(x[u,])), ncol=m,
Dear R-listers,
I am giving part of my R code :
###
n=15
m=1
library("partitions")
library("gregmisc")
library("combinat")
x = t(restrictedparts(n-m,m))
l = length(x[,1])
for(u in 1:l){
A= unique(matrix( unlist(permn(x[u,])), ncol=m,
Dear Sir/madam,
I'm getting a problem with a R-code which calculate Fisher Information
Matrix for Hybrid Censored Weibull Distribution. My problem is that:
when I take weibull(scale=1,shape=2) { i.e shape>1} I got my desired
result but when I take weibull(scale=1,shape=0.5) { i.e shape<1} it give
Dear Sir/Madam,
I'm getting a problem with a R-code which converts a data frame to a matrix.
It first generate a (m^(n-m) * m) matrix A and then regenerate another
matrix B having less dimension than A which satisfy some condition. Now I
wish to assign each row of B to a vector as individual.
My
Thanks David, your suggestion works fine.btw I have another
question..If I set (n,m) little bit large, say (n=20,m=10), R cannot
handle the large data frame generated through "expand.grid".Is there
any way to increase R-memory so that I can tackle large data.frame in R
?
regards
ri
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