Split vector by a field, preserving order - vector

Question

In Rust is there a simple way to split a vector like this:
let original_vec = vec![
(14, "Foo", 62.3),
(17, "Foo", 77.8),
(25, "Bar", 33.7),
(27, "Bar", 99.2),
(61, "Foo", 17.4),
(64, "Bar", 55.5),
(77, "Bar", 31.2),
];
into this:
let res = vec![
vec![
(14, "Foo", 62.3),
(17, "Foo", 77.8),
],
vec![
(25, "Bar", 33.7),
(27, "Bar", 99.2),
],
vec![
(61, "Foo", 17.4), // "Foo" again
],
vec![
(61, "Bar", 55.5), // A separate group of "Bars"
(77, "Bar", 31.2),
],
]
Just to clarify: I want to split my input vector by one of the fields in its elements. Only contiguous elements with the same field value must be put in a sub-vector. The "flattened" result vector must have the same order as the original vector.

Answer 1

I found exactly what I was looking for:
original_vec.group_by(|a, b| a.1 == b.1)
However, the group_by method is only available in Nightly at the moment of writing this and requires adding:
#![feature(slice_group_by)]
As an alternative, as Sebastian suggested, group_by() from the itertools crate could be used.

Answer 2

Basically I imagine the following solution
let original_vec = vec![
(14, "Foo", 62.3),
(17, "Foo", 77.8),
(25, "Bar", 33.7),
(27, "Bar", 99.2),
(61, "Foo", 17.4),
(64, "Bar", 55.5),
(77, "Bar", 31.2),
];
type DATA = (i32, &'static str, f32);
let mut t: Vec<Vec<DATA>> = Vec::new();
let mut q: Vec<DATA> = Vec::new();
let mut p: DATA = (0, "", 0.0);
for i in original_vec
{
if i.1 != p.1 && p.1 != "" {
t.push(q);
q = Vec::new();
q.push(i);
}
q.push(i);
p = i;
}
println!("{:?}", t);