hello all
I am writing a quite simple script to study dental wear patterns in humans
and I wrote this function
sqrt(var(Y1)+var(Y2))^2-4(var(Y1)*(var(Y2)-cov(Y1,Y2)^2)) but appear this
error message
Error: attempt to apply non-function
alternatively I wrote this
On 17-03-2013, at 16:47, Miguel Eduardo Delgado Burbano
mdelgadoburb...@gmail.com wrote:
hello all
I am writing a quite simple script to study dental wear patterns in humans
and I wrote this function
sqrt(var(Y1)+var(Y2))^2-4(var(Y1)*(var(Y2)-cov(Y1,Y2)^2)) but appear this
error
Hi,
Y1- 1:4
Y2- 5:8
sqrt(var(Y1)+var(Y2)^2)-4*((var(Y1)*(var(Y2)-cov(Y1,Y2)^2)))
#[1] 9.515593
A.K.
- Original Message -
From: Miguel Eduardo Delgado Burbano mdelgadoburb...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Sunday, March 17, 2013 11:47 AM
Subject: [R] help with simple
Hello,
You are missing a '*' in your first try.
(As for the second, in R parenthesis are round, not [])
sqrt(var(Y1) + var(Y2))^2 - 4*(var(Y1)*(var(Y2) - cov(Y1, Y2)^2))
This written as a function becomes
fun - function(Y1, Y2)
sqrt(var(Y1) + var(Y2))^2 - 4*(var(Y1)*(var(Y2) - cov(Y1,
I'm trying to build on Jim's approach to change the parameters in the
function, with new rules:
1. if (x[i]==0) NA
2. if (x[i]0) log(x[i]/(number of consecutive zeros preceding it +1))
x-c(1,0,1,0,0,1,0,0,0,1,0,0,0,0,1)
# i.e. output desired = c(0, NA, -0.69, NA, NA, -1.098, NA, NA, NA, -1.38,
I'm trying to build on Jim's approach to change the parameters in the
function, with new rules:
1. if (x[i]0) log(x[i]/(number of consecutive zeros immediately preceding
it +1))
2. if (x[i]==0) NA
x-c(1,0,1,0,0,1,0,0,0,1,0,0,0,0,1)
# i.e. output desired = c(0, NA, -0.69, NA, NA, -1.098, NA,
I'm trying to build on Jim's approach to change the parameters in the
function, with new rules:
1. if (x[i]==0) NA
2. if (x[i]0) log(x[i]/(number of consecutive zeros preceding it +1))
x-c(1,0,1,0,0,1,0,0,0,1,0,0,0,0,1)
# i.e. output desired = c(0, NA, -0.69, NA, NA, -1.098, NA, NA, NA,
I'm trying to build on Jim's approach to change the parameters in the
function, with new rules:
1. if (x[i]==0) NA
2. if (x[i]0) log(x[i]/(number of consecutive zeros immediately preceding
it +1))
x-c(1,0,1,0,0,1,0,0,0,1,0,0,0,0,1)
# i.e. output desired = c(0, NA, -0.69, NA, NA, -1.098,
Here's an added caveat, with subsequently a more detailed explanation of the
output desired:
The data this will apply to includes a variety of whole numbers not limited
to 1 0, a number of which may appear consecutively and not separated by
zeros!
e.g. x-c(3,2,0,1,0,2,0,0,1,0,0,0,0,4,1)
Here's an added caveat, with subsequently a more detailed explanation of the
output desired:
The data this will apply to includes a variety of whole numbers not limited
to 1 0, a number of which may appear consecutively and not separated by
zeros!
e.g. x-c(3,2,0,1,0,2,0,0,1,0,0,0,0,4,1)
I have a matrix of frequency counts from 0-160.
x-as.matrix(c(0,1,0,0,1,0,0,0,1,0,0,0,0,1))
I would like to apply a function creating a new column (x[,2])containing
values equal to:
a) log(x[m,1]) if x[m,1] 0; and
b) for all x[m,1]= 0, log(next x[m,1] 0 / count of preceding zero values
+1)
On Tue, 27 May 2008, T.D.Rudolph wrote:
I have a matrix of frequency counts from 0-160.
x-as.matrix(c(0,1,0,0,1,0,0,0,1,0,0,0,0,1))
I would like to apply a function creating a new column (x[,2])containing
values equal to:
a) log(x[m,1]) if x[m,1] 0; and
b) for all x[m,1]= 0, log(next x[m,1]
In fact x[4,2] should = log(x[5,1]/2]
whereas x[3,2] = log(x[5,1/3])
i.e. The denominator in the log function equals the number of rows between
m==0 and m0 (inclusive, hence the +1)
Hope this helps!...
Charles C. Berry wrote:
On Tue, 27 May 2008, T.D.Rudolph wrote:
I have a matrix
Does this do what you want:
x-c(0,1,0,0,1,0,0,0,1,0,0,0,0,1)
y - rle(x)
result - lapply(seq_along(y$lengths), function(.indx){
+ if (y$values[.indx] == 0)
log(y$values[.indx+1]/seq(y$lengths[.indx]+1, by=-1,
length=y$lengths[.indx]))
+ else rep(log(y$values[.indx]), y$lengths[.indx])
+
In fact x[4,2] should = log(x[5,1]/2]
whereas x[3,2] = log(x[5,1/3])
i.e. The denominator in the log function equals the number of rows between
m==0 and m0 (inclusive, hence the +1)
Hope this helps!
Charles C. Berry wrote:
On Tue, 27 May 2008, T.D.Rudolph wrote:
I have a matrix of
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