How to apply a function of several arguments to a list?

Question

I have a list of triplets:

list = {{1, 3, 5}, {4, 5, 1}, {4, 9, 2}}

How can I apply a function f to the last element of each triplet when f the last and second-to-last elements of each triplet as arguments? The result should look like:

{{1, 3, f[3, 5]}, {4, 5, f[5, 1]}, {4, 9, f[9, 2]}}

I tried:

list /. {a_, b_, c_} :> {a, b, f[b, c]}

But the result comes out as

{{1, 3, 5}, {4, 5, 1}, {f[4], f[9], f[2]}}

Answer 1

Here are two ways to do it.

data = {{1, 3, 5}, {4, 5, 1}, {4, 9, 2}};
Replace[data, {a_, b_, c_} :> {a, b, f[b, c]}, {1}]

{{1, 3, f[3, 5]}, {4, 5, f[5, 1]}, {4, 9, f[9, 2]}}

MapThread[{#1, #2, f[#2, #3]} &, data]

{{1, 3, f[3, 5]}, {4, 5, f[5, 1]}, {4, 9, f[9, 2]}}

A small variant of the last is

MapThread[{#1, #2, f[##2]} &, data]

Answer 2

When the sublists are short like in your example, I tend to use Apply:

 {#1, #2, f[#2, #3]}& @@@ {{1, 3, 5}, {4, 5, 1}, {4, 9, 2}}

Answer 3

{#[[1]], #[[2]], f[#[[2]], #[[3]]]} & /@ {{1, 3, 5}, {4, 5, 1}, {4, 9, 2}}

{{1, 3, f[3, 5]}, {4, 5, f[5, 1]}, {4, 9, f[9, 2]}}

Answer 4

# /. {a_, b_, c_} :> {a, b, f[b, c]} & /@ list