Does the rank function do what you want? (With, perhaps, some minor
modification.) For the example you supplied,
sets X = {0,1,2,3} and Y = {1,2,2,3}.
Rank(X) = {1,2,3,4}, Rank(Y) = {1, 2.5, 2.5, 4}
(using the convention that for tied values you average the ranks
that would have been assigned if they weren't tied).
I believe these values can be adjusted to meet your criteria:
> 1. For the lowest number in A f(min)=0
Minimum rank = 1; so use Rank(X)-1.
> 2. For the average number in A f(av)=1
Average (rank-1) = (n-1)/2; so use [Rank(X)-1]*2/(n-1).
This leaves 0 at 0, while making the average = 1.
> 3. For the lowest number in A f(max)=an finite positive number.
Did you mean "For the highest number in A"?
Maximum [Rank-1]*2/(n-1) = (n-1)*2/(n-1) = 2, finite and positive.
It may also be convenient that this value does not depend on n.
> 4. For every a and b ( f(a) < f(b) ) <=> ( a < b ).
a < b <=> Rank(a) < Rank(b).
The more complicated function described above is also strictly
monotonic in Rank(a), and therefore in a.
> 5. Value of f(a) depends on:
> a.) the value of a (p. 4)
> b.) difference (imprecise term!) from a to the minimum of the A set,
> or the "difficulty" (dif) of getting an a value in A. By the
> "difficulty" I understand a value dependent on the number of values in
> A lower or equal a and on the number of values in A lower of a.
I'm not sure what you actually want here, so I'll just ask: does the
rank function meet this criterion? It certainly meets the inequalities
you specified for the function "dif":
difX(0) = difY(1) < difX(1) < difY(2) [!] < difX(2) < difX(3) = difY(3).
For the simple function Rank(X), we have
1 = 1 < 2 < 2.5 < 3 < 4 = 4.
For (rank-1) the same inequalities hold, obviously; as for
(rank-1)*2/(n-1).
On Fri, 13 Dec 2002, Ludwik G�rski wrote in part:
> I've a set A = {a: a in R}. I want to assign to every number being an
> element of the set A another number meeting these criteria:
> 1. For the lowest number in A f(min)=0
> 2. For the average number in A f(av)=1
> 3. For the lowest number in A f(max)=an finite positive number.
> 4. For every a and b ( f(a) < f(b) ) <=> ( a < b ).
> 5. Value of f(a) depends on:
> a.) the value of a (p. 4)
> b.) difference (imprecise term!) from a to the minimum of the A set,
> or the "difficulty" (dif) of getting an a value in A. By the
> "difficulty" I understand a value dependent on the number of values in
> A lower or equal a and on the number of values in A lower of a. So for
> sets X = {0,1,2,3} and Y = {1,2,2,3} difX(0) = difY(1) < difX(1) <
> difY(2) [!] < difX(2) < difX(3) = difY(3). Difference of value 2 in
> the Y set is lower because there is more 2s in Y than in X. So as if
> the difference was a relative result in a sports rally, while X and Y
> sets of results in these rallies.
> I thought of a formula like this:
> f(a) = a * ( a - min ) / ( average - min )
>
> but it doesn't comply with point 5b
I hope this has been helpful. Cheers! -- DFB.
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Donald F. Burrill [EMAIL PROTECTED]
56 Sebbins Pond Drive, Bedford, NH 03110 (603) 626-0816
[was: 184 Nashua Road, Bedford, NH 03110 (603) 471-7128]
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