Re: [R] A coding question involving variable assignments in ifelse()
Is this an ad hominem comment or a comment of brevity? Unless my eyes
are playing tricks on me, I can't seem to find any language in the
Posting Guide on what is considered a reasonable vs. unreasonable
request from an anonymous poster. Kindly point me to it if it exists.
In any case, thanks for your time and suggestion.
On 4/26/07, Duncan Murdoch [EMAIL PROTECTED] wrote:
On 4/26/2007 5:21 PM, xpRt.wannabe wrote:
I made a few slight modifications to the original model in an effort
to see the inner workings of the code:
deductible - 1
coverage.limit - 2
insurance.threshold - deductible + coverage.limit
snip
set.seed(123)
loss - abs(rnorm(rpois(1, 5), 1, 3))
n - length(loss)
accept - runif(n) 0.8
payout - runif(n) 0.999
sum(ifelse(accept payout, ifelse(loss insurance.threshold,
loss - coverage.limit, pmin(loss, deductible)), 0))
[1] 6.188817
snip
To tease out the data as well as to see the effect of 'accept
payout', I did the following:
loss
[1] 3.401663 4.570620 4.068667 4.718488
accept
[1] TRUE FALSE TRUE TRUE # The second loss claim is NOT accepted
by the insurance company.
payout
[1] TRUE TRUE TRUE TRUE
accept payout
[1] TRUE FALSE TRUE TRUE # The second entry is FALSE because of
the second entry in 'accept.'
Based on the inner ifelse() expression, the original loss numbers
become : 1.401663, 2.570620, 2.068667, 2.718488, respectively (which
is fine and what I wanted).
Because the second entry in 'accept payout' is FALSE, the second
altered loss number (2.570620) becomes 0, making sum(...) equal
6.188817. Unfortunately this is _not_ what I want, and I apologize
for not being clear in the first place. What I want is: for any FALSE
entry, the original loss number is unaltered, as opposed to become 0.
So in the example above, the four numbers that should have been added
are: 1.401663, 4.570620, 2.068667, 2.718488, yielding 10.759438
instead of 6.188817.
Any further suggestions would be greatly appreciated.
I'm sorry, but from an anonymous poster that's not a reasonable request.
Just work it out yourself.
Duncan Murdoch
On 4/26/07, Duncan Murdoch [EMAIL PROTECTED] wrote:
On 4/26/2007 2:31 PM, xpRt.wannabe wrote:
Just to be sure, is what I have below the right intepretation of your
suggestion:
Yes, that's what I suggested.
Duncan Murdoch
deductible - 15
coverage.limit - 75
insurance.threshold - deductible + coverage.limit
tmpf - function() {
loss - rlnorm(rpois(1, 3), 2, 5)
n - length(loss)
accept - runif(n) 0.8
payout - runif(n) 0.999
sum(ifelse(accept payout, ifelse(loss insurance.threshold, loss -
coverage.limit, pmin(loss, deductible)), 0))
}
net - replicate(100, tmpf())
On 4/26/07, Duncan Murdoch [EMAIL PROTECTED] wrote:
On 4/26/2007 12:48 PM, xpRt.wannabe wrote:
Dear List,
Below is a simple, standard loss model that takes into account the
terms of an insurance policy:
deductible - 15
coverage.limit - 75
insurance.threshold - deductible + coverage.limit
tmpf - function() {
loss - rlnorm(rpois(1, 3), 2, 5)
sum(ifelse(loss insurance.threshold, loss - coverage.limit,
pmin(loss, deductible)))
}
net - replicate(100, tmpf())
Now, I would like to enhance the model by incorporating the following
two probabilities:
1. Probability of claim being accepted by the insurance company, say,
0.8
2. Probability of payout by the insurance company, say, 0.999
Could anyone suggest how one might do this?
A general way to generate events with probability p is runif(n) p. So
I'd add
n - length(loss)
accept - runif(n) 0.8
payout - runif(n) 0.999
and then require accept payout before any payment at all, e.g.
sum(ifelse(accept payout, [ your old ifelse expression ], 0))
There are a lot of implicit independence assumptions here; they may not
be very realistic.
Duncan Murdoch
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] A coding question involving variable assignments in ifelse()
Dear List,
Below is a simple, standard loss model that takes into account the
terms of an insurance policy:
deductible - 15
coverage.limit - 75
insurance.threshold - deductible + coverage.limit
tmpf - function() {
loss - rlnorm(rpois(1, 3), 2, 5)
sum(ifelse(loss insurance.threshold, loss - coverage.limit,
pmin(loss, deductible)))
}
net - replicate(100, tmpf())
Now, I would like to enhance the model by incorporating the following
two probabilities:
1. Probability of claim being accepted by the insurance company, say, 0.8
2. Probability of payout by the insurance company, say, 0.999
Could anyone suggest how one might do this?
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
Thanks,
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] A coding question involving variable assignments in ifelse()
Just to be sure, is what I have below the right intepretation of your
suggestion:
deductible - 15
coverage.limit - 75
insurance.threshold - deductible + coverage.limit
tmpf - function() {
loss - rlnorm(rpois(1, 3), 2, 5)
n - length(loss)
accept - runif(n) 0.8
payout - runif(n) 0.999
sum(ifelse(accept payout, ifelse(loss insurance.threshold, loss -
coverage.limit, pmin(loss, deductible)), 0))
}
net - replicate(100, tmpf())
On 4/26/07, Duncan Murdoch [EMAIL PROTECTED] wrote:
On 4/26/2007 12:48 PM, xpRt.wannabe wrote:
Dear List,
Below is a simple, standard loss model that takes into account the
terms of an insurance policy:
deductible - 15
coverage.limit - 75
insurance.threshold - deductible + coverage.limit
tmpf - function() {
loss - rlnorm(rpois(1, 3), 2, 5)
sum(ifelse(loss insurance.threshold, loss - coverage.limit,
pmin(loss, deductible)))
}
net - replicate(100, tmpf())
Now, I would like to enhance the model by incorporating the following
two probabilities:
1. Probability of claim being accepted by the insurance company, say, 0.8
2. Probability of payout by the insurance company, say, 0.999
Could anyone suggest how one might do this?
A general way to generate events with probability p is runif(n) p. So
I'd add
n - length(loss)
accept - runif(n) 0.8
payout - runif(n) 0.999
and then require accept payout before any payment at all, e.g.
sum(ifelse(accept payout, [ your old ifelse expression ], 0))
There are a lot of implicit independence assumptions here; they may not
be very realistic.
Duncan Murdoch
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] A coding question involving variable assignments in ifelse()
I made a few slight modifications to the original model in an effort
to see the inner workings of the code:
deductible - 1
coverage.limit - 2
insurance.threshold - deductible + coverage.limit
snip
set.seed(123)
loss - abs(rnorm(rpois(1, 5), 1, 3))
n - length(loss)
accept - runif(n) 0.8
payout - runif(n) 0.999
sum(ifelse(accept payout, ifelse(loss insurance.threshold,
loss - coverage.limit, pmin(loss, deductible)), 0))
[1] 6.188817
snip
To tease out the data as well as to see the effect of 'accept
payout', I did the following:
loss
[1] 3.401663 4.570620 4.068667 4.718488
accept
[1] TRUE FALSE TRUE TRUE # The second loss claim is NOT accepted
by the insurance company.
payout
[1] TRUE TRUE TRUE TRUE
accept payout
[1] TRUE FALSE TRUE TRUE # The second entry is FALSE because of
the second entry in 'accept.'
Based on the inner ifelse() expression, the original loss numbers
become : 1.401663, 2.570620, 2.068667, 2.718488, respectively (which
is fine and what I wanted).
Because the second entry in 'accept payout' is FALSE, the second
altered loss number (2.570620) becomes 0, making sum(...) equal
6.188817. Unfortunately this is _not_ what I want, and I apologize
for not being clear in the first place. What I want is: for any FALSE
entry, the original loss number is unaltered, as opposed to become 0.
So in the example above, the four numbers that should have been added
are: 1.401663, 4.570620, 2.068667, 2.718488, yielding 10.759438
instead of 6.188817.
Any further suggestions would be greatly appreciated.
On 4/26/07, Duncan Murdoch [EMAIL PROTECTED] wrote:
On 4/26/2007 2:31 PM, xpRt.wannabe wrote:
Just to be sure, is what I have below the right intepretation of your
suggestion:
Yes, that's what I suggested.
Duncan Murdoch
deductible - 15
coverage.limit - 75
insurance.threshold - deductible + coverage.limit
tmpf - function() {
loss - rlnorm(rpois(1, 3), 2, 5)
n - length(loss)
accept - runif(n) 0.8
payout - runif(n) 0.999
sum(ifelse(accept payout, ifelse(loss insurance.threshold, loss -
coverage.limit, pmin(loss, deductible)), 0))
}
net - replicate(100, tmpf())
On 4/26/07, Duncan Murdoch [EMAIL PROTECTED] wrote:
On 4/26/2007 12:48 PM, xpRt.wannabe wrote:
Dear List,
Below is a simple, standard loss model that takes into account the
terms of an insurance policy:
deductible - 15
coverage.limit - 75
insurance.threshold - deductible + coverage.limit
tmpf - function() {
loss - rlnorm(rpois(1, 3), 2, 5)
sum(ifelse(loss insurance.threshold, loss - coverage.limit,
pmin(loss, deductible)))
}
net - replicate(100, tmpf())
Now, I would like to enhance the model by incorporating the following
two probabilities:
1. Probability of claim being accepted by the insurance company, say, 0.8
2. Probability of payout by the insurance company, say, 0.999
Could anyone suggest how one might do this?
A general way to generate events with probability p is runif(n) p. So
I'd add
n - length(loss)
accept - runif(n) 0.8
payout - runif(n) 0.999
and then require accept payout before any payment at all, e.g.
sum(ifelse(accept payout, [ your old ifelse expression ], 0))
There are a lot of implicit independence assumptions here; they may not
be very realistic.
Duncan Murdoch
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to iteratively extract elements out of a list
Thank you!
On 8/26/06, Patrick Burns [EMAIL PROTECTED] wrote:
sub.m - lapply(m, function(x)x[x2])
sub.m
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
sub.m[unlist(lapply(sub.m, function(x) length(x) == 2))]
[[1]]
[1] 3 4
[[2]]
[1] 4 5
sub4.m - lapply(m, function(x)x[x4])
sub4.m[unlist(lapply(sub4.m, function(x) length(x) 0))]
[[1]]
[1] 5
Patrick Burns
[EMAIL PROTECTED]
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and A Guide for the Unwilling S User)
xpRt.wannabe wrote:
Yet another question:
Let's say I do the following:
set.seed(123)
tmpf - function(){
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n, tmpf())
m
sub.m - lapply(m, function(x)x[x2])
'sub.m' gives me:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
The question is: What do I need to do such that I can extract
componets of length 2 in 'sub.m' into another sublist, which would
look like this:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
I think that's all the questions I can think of -- for now.
Many, many thanks!!!
On 8/25/06, xpRt. wannabe [EMAIL PROTECTED] wrote:
Jim and Patrick,
Both of you made the same suggestion, which works great!
A follow-up question: Suppose I change the condition 'x2' in 'lapply'
to 'x4', as follows:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
sub.m - lapply(m, function(x)x[x4]) # was x2
As a result, I'd get:
sub.m
[[1]]
numeric(0)
[[2]]
[1] 5
[[3]]
numeric(0)
However, what would I need to do such that 'sub.m' contains only the
non-zero length component; namely, the 'sub.m[[2]]'? In essence, I'd
like to drop all the components of zero length such that 'sub.m'
results in:
[[1]]
[1] 5
My best effort was to use 'lapply' again:
lapply(sub.m, function(x)x[length(x)0])
which still gives me:
[[1]]
numeric(0)
[[2]]
[1] 5
[[3]]
numeric(0)
Again, any help would be greately appreciated.
p.s. Sorry to bug you again. I should have thought through a little
more prior to composing an example that would represent all possible
scenarios.
On 8/25/06, jim holtman [EMAIL PROTECTED] wrote:
try this:
set.seed(123)
tmpf - function() {
+ x - rpois(rpois(1,4),2)
+ }
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
lapply(m, function(x)x[x2])
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
On 8/25/06, xpRt.wannabe [EMAIL PROTECTED] wrote:
Dear List,
The following code produces a list, which is what I what:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
Now I need something that would to extract iteratively (or as many
times as
the size of 'n') the values that are greater than 2 in each component
of
'm' into another list such that the sub-list would be:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
Below is what I tried:
for(i in 1:3)
sub.list - lapply(m,subset,m[[i]]2)
sub.list
which gives me something different from what I want:
[[1]]
[1] 4
[[2]]
[1] 4
[[3]]
[1] 4
Any help would be appreciated.
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] How to iteratively extract elements out of a list
Dear List,
The following code produces a list, which is what I what:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
Now I need something that would to extract iteratively (or as many
times as
the size of 'n') the values that are greater than 2 in each component
of
'm' into another list such that the sub-list would be:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
Below is what I tried:
for(i in 1:3)
sub.list - lapply(m,subset,m[[i]]2)
sub.list
which gives me something different from what I want:
[[1]]
[1] 4
[[2]]
[1] 4
[[3]]
[1] 4
Any help would be appreciated.
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to iteratively extract elements out of a list
Jim and Patrick,
Both of you made the same suggestion, which works great!
A follow-up question: Suppose I change the condition 'x2' in 'lapply'
to 'x4', as follows:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
sub.m - lapply(m, function(x)x[x4]) # was x2
As a result, I'd get:
sub.m
[[1]]
numeric(0)
[[2]]
[1] 5
[[3]]
numeric(0)
However, what would I need to do such that 'sub.m' contains only the
non-zero length component; namely, the 'sub.m[[2]]'? In essence, I'd
like to drop all the components of zero length such that 'sub.m'
results in:
[[1]]
[1] 5
My best effort was to use 'lapply' again:
lapply(sub.m, function(x)x[length(x)0])
which still gives me:
[[1]]
numeric(0)
[[2]]
[1] 5
[[3]]
numeric(0)
Again, any help would be greately appreciated.
p.s. Sorry to bug you again. I should have thought through a little
more prior to composing an example that would represent all possible
scenarios.
On 8/25/06, jim holtman [EMAIL PROTECTED] wrote:
try this:
set.seed(123)
tmpf - function() {
+ x - rpois(rpois(1,4),2)
+ }
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
lapply(m, function(x)x[x2])
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
On 8/25/06, xpRt.wannabe [EMAIL PROTECTED] wrote:
Dear List,
The following code produces a list, which is what I what:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
Now I need something that would to extract iteratively (or as many
times as
the size of 'n') the values that are greater than 2 in each component
of
'm' into another list such that the sub-list would be:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
Below is what I tried:
for(i in 1:3)
sub.list - lapply(m,subset,m[[i]]2)
sub.list
which gives me something different from what I want:
[[1]]
[1] 4
[[2]]
[1] 4
[[3]]
[1] 4
Any help would be appreciated.
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to iteratively extract elements out of a list
Yet another question:
Let's say I do the following:
set.seed(123)
tmpf - function(){
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n, tmpf())
m
sub.m - lapply(m, function(x)x[x2])
'sub.m' gives me:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
The question is: What do I need to do such that I can extract
componets of length 2 in 'sub.m' into another sublist, which would
look like this:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
I think that's all the questions I can think of -- for now.
Many, many thanks!!!
On 8/25/06, xpRt. wannabe [EMAIL PROTECTED] wrote:
Jim and Patrick,
Both of you made the same suggestion, which works great!
A follow-up question: Suppose I change the condition 'x2' in 'lapply'
to 'x4', as follows:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
sub.m - lapply(m, function(x)x[x4]) # was x2
As a result, I'd get:
sub.m
[[1]]
numeric(0)
[[2]]
[1] 5
[[3]]
numeric(0)
However, what would I need to do such that 'sub.m' contains only the
non-zero length component; namely, the 'sub.m[[2]]'? In essence, I'd
like to drop all the components of zero length such that 'sub.m'
results in:
[[1]]
[1] 5
My best effort was to use 'lapply' again:
lapply(sub.m, function(x)x[length(x)0])
which still gives me:
[[1]]
numeric(0)
[[2]]
[1] 5
[[3]]
numeric(0)
Again, any help would be greately appreciated.
p.s. Sorry to bug you again. I should have thought through a little
more prior to composing an example that would represent all possible
scenarios.
On 8/25/06, jim holtman [EMAIL PROTECTED] wrote:
try this:
set.seed(123)
tmpf - function() {
+ x - rpois(rpois(1,4),2)
+ }
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
lapply(m, function(x)x[x2])
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
On 8/25/06, xpRt.wannabe [EMAIL PROTECTED] wrote:
Dear List,
The following code produces a list, which is what I what:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
Now I need something that would to extract iteratively (or as many
times as
the size of 'n') the values that are greater than 2 in each component
of
'm' into another list such that the sub-list would be:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
Below is what I tried:
for(i in 1:3)
sub.list - lapply(m,subset,m[[i]]2)
sub.list
which gives me something different from what I want:
[[1]]
[1] 4
[[2]]
[1] 4
[[3]]
[1] 4
Any help would be appreciated.
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Fitting Truncated Lognormal to a truncated data set (was: fitting truncated normal distribution)
Dear List,
I am trying to fit Truncated Lognormal to a data set that is
'truncated' from above a certain value, say, 0.01. Below is what I
was able to come up with. I would appreciate it if you could review
and make any necessary changes.
# This is modified off the code for 'dtnorm' of library(msm).
dtlnorm - function (n, mean = 0, sd = 1, lower = -Inf, upper = Inf)
{
ret - numeric()
if (length(n) 1)
n - length(n)
while (length(ret) n) {
y - rlnorm(n - length(ret), mean, sd)
y - y[y = lower y = upper]
ret - c(ret, y)
}
stopifnot(length(ret) == n)
ret
}
# This is modified off the code for 'rtnorm' of the library(msm).
rtlnorm - function (n, mean = 0, sd = 1, lower = -Inf, upper = Inf)
{
ret - numeric()
if (length(n) 1)
n - length(n)
while (length(ret) n) {
y - rlnorm(n - length(ret), mean, sd)
y - y[y = lower y = upper]
ret - c(ret, y)
}
stopifnot(length(ret) == n)
ret
}
x - rtlnorm(100, mean=-11.64857, sd = 3.422795, 0.01)
fitting truncated normal distribution on 8/16/2006.
dtlnorm0 - function(x, mean = 0, sd = 1)
{
dtlnorm(x, mean, sd, 0.01, Inf)
}
fitdistr(x, dtlnorm0, start = list(mean = -11, sd = 1))
Thank you.
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] A coding question
Uwe and Ben,
Thank you both for your help.
To me, both sets of code seem to do the job and should produce the
same results. However, as a test I inserted set.seed( ) as follows.
Unless I put set.seed( ) in the wrong lines, the results produced by
both sets of code turn out to be different. I am baffled.
y - replicate(10, {
set.seed(123)
rp - rpois(8, 5)
mysum - sapply(rp, function(x) {
set.seed(123)
x - rnorm(x)
x - x - max(0, x-15) + max(0, x-90)
sum(x)
})
})
##
tmpf - function() {
set.seed(123)
x - rnorm(rpois(1,5))
sum(ifelse(x90,x-75,pmin(x,15)))
}
replicate(10,replicate(8,tmpf()))
On 6/2/06, Uwe Ligges [EMAIL PROTECTED] wrote:
xpRt.wannabe wrote:
Dear List:
I have the follow code:
y - replicate(10,replicate(8,sum(rnorm(rpois(1,5)
Now I need to apply the following condition to _every_ randomly generated
Normal number in the code above:
x - max(0,x-15) + max(0,x-90), where x represents the individual Normal
numbers.
In other words, the said condition needs to be applied before
replicate(...(replicate(...(sum(...))) takes place.
Any help would be greatly appreciated.
y - replicate(10, {
rp - rpois(8, 5)
mysum - sapply(rp, function(x) {
x - rnorm(x)
x - x - max(0, x-15) + max(0, x-90)
sum(x)
})
})
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] A coding question
Dear List: I have the follow code: y - replicate(10,replicate(8,sum(rnorm(rpois(1,5) Now I need to apply the following condition to _every_ randomly generated Normal number in the code above: x - max(0,x-15) + max(0,x-90), where x represents the individual Normal numbers. In other words, the said condition needs to be applied before replicate(...(replicate(...(sum(...))) takes place. Any help would be greatly appreciated. platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 2.1 year 2005 month 12 day 20 svn rev 36812 language R __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Is there a way to view S-PLUS script files in R
Dear List, I have some S-PLUS script files (.ssc). Does there exist an R function/command that can read such files? I simply want to view the code and practice in R to help me learn the subject matter. Any help would be greatly appreciated. platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major2 minor2.1 year 2005 month12 day 20 svn rev 36812 language R __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] How to show the intermediate tick marks without the values in a different color
Is there a way to show also the intermediate tick marks without the values? Example: plot(cars) What would one do to show the tick marks in, say, gray color at the increment of 1 without showing the actual values in-between the default x values: 5, 10, 15, 20, 25? platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major2 minor2.1 year 2005 month12 day 20 svn rev 36812 language R Thanks, __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Another question related to box-percentile plot, bpplot(): How to color the box or middle half of the data
Dear List, A follow-up question related to bpplot() of the Hmisc package: How does one color the box , or the middle half of the data, of a box-percentile plot? bpplot(... , col=lightgray) apparently does not work, though boxplot(... , col=lightgray) works. With much appreciation, platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major2 minor1.0 year 2005 month04 day 18 language R __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] A two-part question about box-percentile plots, bpplot(): (1) yaxt=n doesn't seem to work (2) how to display mean values
Dear List, I have a two-part question related to bpplot(), a box-percentile plot function in the Hmisc package. Take the example given in the Help for bpplot(), for instance. (1) How does one set but not draw the y-axis? What I did was, bpplot(... , yaxt=n), but that apparently does not work (though it works for boxplot()). (2) How does one display the mean value of each variable inside each respective box-percentile box? The following is what I did but to no avail: bpplot(x1, x2, x3) points(1:1, mean(x1), pch = 1) points(1:1, mean(x2), pch = 2) points(1:1, mean(x3), pch = 3) Your help is much appreciated. platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major2 minor1.0 year 2005 month04 day 18 language R __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Add lines to density plot
Dear List, I am using R to learn bootstrapping concepts. Not to be oblivious to the contributed packages of boot and bootstrap, I have opted to do the following as a way to hone my R skills. One example I am trying to work through is the following: x - 1:20 Observed - sample(x,20) Resamples - replicate(1000,sample(Observed,replace=TRUE)) Boot.Means - apply(Resamples,2,mean) plot(density(Boot.Means)) For the life of me, I can't seem to figure out how to add two lines to the density plot: 1. A line that represents the mean of Boot.Means 2. A line that represents the mean of Observed Any help/pointer would be greatly appreciated. I have tried lines() but to no avail. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] A coding question involving variable assignments
Dear List, I have the following code that does what I want: x - replicate(5,replicate(10,sum(rnorm(rpois(1,10) How might one change it such that the maximum value generated by rnorm(rpois(1,10)) can be retrieved for later use? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] A coding question involving variable assignments
Jim and List, Thank you for the prompt reply. Perhaps I should have been more specific in the way I phrased the question. The code you gave would return the max value just one time. I was interested in getting as many max values generated by rnorm(rpois(1,10)) as specified by: replicate(5,replicate(10 ... ) In the end, I expect to get 10 x 5 max values. In that context, how might the code be changed? On 8/11/05, jim holtman [EMAIL PROTECTED] wrote: temp - rnorm(rpois(1,10)) x - replicate(5,replicate(10,sum(temp))) temp - max(temp) On 8/11/05, xpRt.wannabe [EMAIL PROTECTED] wrote: Dear List, I have the following code that does what I want: x - replicate(5,replicate(10,sum(rnorm(rpois(1,10) How might one change it such that the maximum value generated by rnorm(rpois(1,10)) can be retrieved for later use? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Jim Holtman Convergys +1 513 723 2929 What the problem you are trying to solve? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] A coding question involving variable assignments
Ted and List, In your code that produced 'mx', you dropped sum() from my original code though. As a result, the 10 x 5 max's are of the same value. Unfortunately, that's not what I need. On 8/11/05, ecatchpole [EMAIL PROTECTED] wrote: On 12/08/05 05:55, xpRt.wannabe wrote,: Dear List, I have the following code that does what I want: x - replicate(5,replicate(10,sum(rnorm(rpois(1,10) How might one change it such that the maximum value generated by rnorm(rpois(1,10)) can be retrieved for later use? set.seed(99) x - replicate(5,replicate(10,sum(rnorm(rpois(1,10) set.seed(99) mx - replicate(5,replicate(10,max(rnorm(rpois(1,10) should work? Ted. -- Dr E.A. Catchpole Visiting Fellow Univ of New South Wales at ADFA, Canberra, Australia and University of Kent, Canterbury, England - www.ma.adfa.edu.au/~eac - fax: +61 2 6268 8786 - ph: +61 2 6268 8895 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] A coding question involving variable assignments
Ted and List, What I need is I need to know what max of rnorm(rpois(1,10)) is before R does sum(), replicate(10, ...) and replicate(5, ...). The fact that you have set.seed(99) twice, does that mean, say, entry [1,1] 0.4896243 in 'mx' is one of the z number of values generated by rnorm(rpois(1,10)) that add up to [1,1] -2.0071674 in 'x'? Another way to ask the question, I guess, is, by doing set.seed(99) twice are the values generated by rnorm(rpois(1,10)) for 'x' same as those for 'mx'? On 8/11/05, ecatchpole [EMAIL PROTECTED] wrote: Sorry, I don't follow. What's wrong with this? Ted. set.seed(99) x - replicate(5,replicate(10,sum(rnorm(rpois(1,10) set.seed(99) mx - replicate(5,replicate(10,max(rnorm(rpois(1,10) x [,1][,2][,3] [,4] [,5] [1,] -2.0071674 -7.0883335 -0.03766649 2.6859415 -5.4685172 [2,] -4.7211799 2.4177121 -0.92575948 -7.2201952 -2.5969177 [3,] 0.2325584 -0.1790635 -3.17988580 -2.0249829 1.6994276 [4,] -1.7738725 1.5836438 5.06193854 -5.4798269 -2.4363479 [5,] -3.0394562 2.8859440 2.67993750 5.0534413 0.6560877 [6,] 6.2436591 -4.0226431 1.97545757 -1.5641548 4.0443831 [7,] 0.4641453 -10.4417831 1.08048629 2.4675178 -5.5114109 [8,] 1.1570728 -3.6081361 1.37858782 0.3534015 1.8282236 [9,] 4.3988625 3.0692562 -0.69898483 -1.6882952 -1.1548913 [10,] -3.7288105 -1.4455309 5.80146323 -6.1962790 -1.3698381 mx [,1] [,2] [,3] [,4] [,5] [1,] 0.4896243 1.4110132 0.7329387 3.2700493 1.861891 [2,] 1.0989215 1.6215407 1.5980779 2.2921963 1.417614 [3,] 1.4000518 1.4867612 1.0130372 0.5686142 1.442630 [4,] 0.5981696 2.0916016 2.0894395 0.9760749 1.419316 [5,] 1.0066032 1.8084703 1.6556502 2.1431458 2.393037 [6,] 2.7329641 1.8689793 1.1494738 1.2899945 1.702919 [7,] 0.5851713 0.6224785 2.4466643 1.1955567 0.951106 [8,] 1.3850466 0.9305735 1.4003689 1.5209779 1.864211 [9,] 1.8645760 1.1958389 0.9270208 0.4971312 1.576020 [10,] 1.5889265 1.0253490 4.6865908 0.9852816 1.032410 On 12/08/05 12:17, xpRt.wannabe wrote,: Ted and List, In your code that produced 'mx', you dropped sum() from my original code though. As a result, the 10 x 5 max's are of the same value. Unfortunately, that's not what I need. On 8/11/05, ecatchpole [EMAIL PROTECTED] wrote: On 12/08/05 05:55, xpRt.wannabe wrote,: Dear List, I have the following code that does what I want: x - replicate(5,replicate(10,sum(rnorm(rpois(1,10) How might one change it such that the maximum value generated by rnorm(rpois(1,10)) can be retrieved for later use? set.seed(99) x - replicate(5,replicate(10,sum(rnorm(rpois(1,10) set.seed(99) mx - replicate(5,replicate(10,max(rnorm(rpois(1,10) should work? Ted. -- Dr E.A. Catchpole Visiting Fellow Univ of New South Wales at ADFA, Canberra, Australia and University of Kent, Canterbury, England - www.ma.adfa.edu.au/~eac - fax: +61 2 6268 8786 - ph: +61 2 6268 8895 -- Dr E.A. Catchpole Visiting Fellow Univ of New South Wales at ADFA, Canberra, Australia and University of Kent, Canterbury, England - www.ma.adfa.edu.au/~eac - fax: +61 2 6268 8786 - ph: +61 2 6268 8895 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] A coding question involving variable assignments
Thank you! That was helpful. Another thing I learned was that I would need to do set.seed(99) not once but twice in this context. On 8/11/05, ecatchpole [EMAIL PROTECTED] wrote: On 12/08/05 13:27, xpRt.wannabe wrote,: Ted and List, What I need is I need to know what max of rnorm(rpois(1,10)) is before R does sum(), replicate(10, ...) and replicate(5, ...). The fact that you have set.seed(99) twice, does that mean, say, entry [1,1] 0.4896243 in 'mx' is one of the z number of values generated by rnorm(rpois(1,10)) that add up to [1,1] -2.0071674 in 'x'? Another way to ask the question, I guess, is, by doing set.seed(99) twice are the values generated by rnorm(rpois(1,10)) for 'x' same as those for 'mx'? Yes, that's the reason for having a set.seed() function. Ted. -- Dr E.A. Catchpole Visiting Fellow Univ of New South Wales at ADFA, Canberra, Australia and University of Kent, Canterbury, England - www.ma.adfa.edu.au/~eac - fax: +61 2 6268 8786 - ph: +61 2 6268 8895 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
