solving an equation for two lists of values sequentially

Question

I'm very new to this. My goal is to run a function for 10 different pairs of values. The function looks like this:

f[groupsize_, solvedforq_] = 
Sum[Binomial[groupsize, t] Chop[solvedforq]^t (1 - Chop[solvedforq])^(groupsize - t), {t, 3, groupsize}]

The objects groupsize and solvedforq are lists, each containing 10 values. I would like to create a List or Table of outputs, where the first output is the solution for taking the first element of each list (i.e., the pair of values that I get when taking the first element of groupsize and the first element of solvedforq), the second output is the solution for taking the second element of each list, etc...

So I will ideally have a List or Table with 10 output values. I am ware for the command Map[], but I wouldn't know how to use it in this more complex case. Thanks for all help!

Answer 1

One way is with MapThread:

MapThread[f, {group, solved}]

This will give f[group[[1]], solved[[1]]], f[group[[2]], solved[[2]]], etc. Note that your function f should probably be defined with := instead of =.

Bob Hanlon suggest the use of Transpose, and this brings up a good point. There is a tight relationship between here: Thread is a synonym for Transpose (at least for 2D matix-style lists), so this could also be written:

Map[f, Thread[{group, solved}]]

or using the shortcut

f /@ Thread[{group, solved}]

Here it is easy to see the relationship between Map and Thread, and MapThread. This also demonstrates how similar the functioning is between Map /@ and Apply @@@:

f @@@ Thread[{group, solved}]

or the more pedantic

Apply[f, Thread[{group, solved}], 2]

which also give the same output.

Answer 2

Here is another variation, with Table.

Table[f[group[[i]], solved[[i]]], {i, 1, 4}]

I assume

group = {1, 2, 3, 4};
solved = {5, 6, 7, 8};
f[groupsize_, solvedforq_] := etc

For these sets, the output is the same as in the previous case. {0, 0, 343, -10240}