[R] Problem with apply list to function: numerical expression has 4 elements: only the first used

[Narua]

Dear kd,
thanks for your fast reply 
hm, I thougth, F is normally the abbrevation, but in R FALSE represents
"false"?
Anyway, your function works and I got the idea to work with "apply" 

I had actually a little more complicate functions (put I posted in the forum
first an easier example):
f<-function(a,b)((1/(sum(sqrt(1:a))))*sqrt(b))
F<-function(a,b)(sum(f(a,1:b)))
> F(5,3)
[1] 0.4946433
> F(5,5)
[1] 1
therefore, 0.4946433 is the distribution of b=3 with a=5 Elements.

FInv<-function(s,a,b) if(isTRUE(all.equal(F(a,b),s, tolerance=0.00001)))
cat("F(",s,",",a,",",b,") equals s; ") else cat("F(",a,",",b,") does not
equal to s; ")
FInv is the reverse function. you put a probability in the function and
recieve the distribution b of a special a.
I wanted to have a function FInv' to pass on s for all a and all b to
FInv(s,a,b) in order to find out, which probability fits to which a and b.

And I couln't pass on a and b so easily, I got the error: "In 1:i :
numerical expression has 4 elements: only the first used "
I think, the reason is, that R expects a number, but receives a list, and
that produces the error.

I tried 
>x<-c(1:10)
> FooInv<-function(s) foreach(i=1:10)%do%sapply(x,FInv,s=s,k=i) and that
> works:
> FooInv(0.4946433)
F(s = 0.4946433 ,m = 5 ,k = 3 ) equals s; [[1]]

Foreach is a for-loop, and sapply matches FInv for every element of the list
x.

But now another problem occured:
Obviously R gets anywhere a NULL-pointer, because the full output is:
F(s = 0.4946433 ,m = 5 ,k = 3 ) equals s; [[1]]
[[1]][[1]]
NULL

[[1]][[2]]
NULL

[[1]][[3]]
NULL

[[1]][[4]]
NULL

[[1]][[5]]
NULL
...
in every loop.

Do you or someone else know how  I can get rid of the NULL-outputs?

if this problem is solved, I have all I need in order to calculate the
distribution and its inverse function :-)

Thanks in advance,
Narua 



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