On Dec 26, 2009, at 4:56 PM, James Rome wrote:
Thanks David.
I wanted to calculate the Poisson distribution from my histograms to
see
how closely they match it. So the formula is something like
pprob=((lambda**cnts)/factorial(cnts))*exp(lambda)
(Please don't reply privately.)
You have two vectors of different lengths, the lambdas and the counts,
and you need to create the 2-way combinations in a form that can be
supplied to a function.
And I was hoping to do it for the whole table at once.
Something like this?
> lambdas <-scan() # scan can read a sequence of numbers separated
by white space.
1: 0.199190283 0.013765182 0.006477733 0.017813765 0.093117409
0.160323887
7: 0.401619433
8:
Read 7 items
> ldf <- expand.grid(lambdas, 1:15)
> str(ldf)
'data.frame': 105 obs. of 2 variables:
$ Var1: num 0.19919 0.01377 0.00648 0.01781 0.09312 ...
$ Var2: int 1 1 1 1 1 1 1 2 2 2 ...
- attr(*, "out.attrs")=List of 2
..$ dim : int 7 15
..$ dimnames:List of 2
.. ..$ Var1: chr "Var1=0.199190283" "Var1=0.013765182"
"Var1=0.006477733" "Var1=0.017813765" ...
.. ..$ Var2: chr "Var2= 1" "Var2= 2" "Var2= 3" "Var2= 4" ...
> ldf$countprob <- (ldf$Var1**ldf$Var2)/ factorial(ldf$Var2)*exp(ldf
$Var1)
> ldf
Var1 Var2 countprob
1 0.199190283 1 2.430946e-01
2 0.013765182 1 1.395597e-02
3 0.006477733 1 6.519830e-03
4 0.017813765 1 1.813394e-02
snipped remaining 100 lines...
--
David
Jim Rome
On 12/26/09 2:28 PM, David Winsemius wrote:
On Dec 26, 2009, at 2:02 PM, James Rome wrote:
I am sorry to be bothering the list so much.
I made a table of counts of flight arrivals by hour:
No, you made a list of tables, which is different.
cnts=tapply(Arrival4,list(Hour),table). There are up to 15
arrivals in a
bin.
Why not work with something more regular, like (this is all guesswork
and untested in absence of test data objects):
table(Arrival4, factor(Hour, levels=1:15) ) ?
# or use whatever your max(arrivals) might be.
xx <- factor(sample(1:10, 4))
xx
[1] 9 4 7 1
Levels: 1 4 7 9
table(factor(xx, levels=1:10) )
1 2 3 4 5 6 7 8 9 10
1 0 0 1 0 0 1 0 1 0
cnts
$`0`
1 2 3 4 5 6 7 8 9 10 13
1 2 5 9 2 7 5 4 2 4 1
$`1`
1 2 3 4
3 2 2 1
$`2`
1 3
2 2
. . .
My first problem is how to get this table filled in with the 0
slots.
E.g., I want
$`0`
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
1 2 5 9 2 7 5 4 2 4 0 0 1 0 0
for all 24 hours. The elements of the tables are lists, but I do not
seem to be able to extract the names of each list, which I think
would
allow this manipulation.
My second problem is that I want to compute probability
distributions. I
have
lambda
0 1 2 4 5
6 7
0.199190283 0.013765182 0.006477733 0.017813765 0.093117409
0.160323887
0.401619433
8 9 10 11 12
13 14
0.191093117 0.177327935 0.318218623 0.404858300 0.463157895
0.495546559
0.435627530
15 16 17 18 19
20 21
0.418623482 0.307692308 0.405668016 0.484210526 0.580566802
0.585425101
0.519028340
22 23
0.556275304 0.503643725
and I need to calculate lambda**cnts for each bin, and each hour.
I am
also unsure of how to do this.
Can you give a justification for this procedure? (I'm not saying it
is
nonsense, only that it does not make make sense to me. So perhaps
there is some mathematical property about which I need education.)
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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