'How does one create a HashMap with a default value in Rust?
Being fairly new to Rust, I was wondering on how to create a HashMap
with a default value for a key? For example, having a default value 0
for any key inserted in the HashMap
.
In Rust, I know this creates an empty HashMap:
let mut mymap: HashMap<char, usize> = HashMap::new();
I am looking to maintain a counter for a set of keys, for which one way to go about it seems to be:
for ch in "AABCCDDD".chars() {
mymap.insert(ch, 0)
}
Is there a way to do it in a much better way in Rust, maybe something equivalent to what Ruby provides:
mymap = Hash.new(0)
mymap["b"] = 1
mymap["a"] # 0
Solution 1:[1]
Answering the problem you have...
I am looking to maintain a counter for a set of keys.
Then you want to look at How to lookup from and insert into a HashMap efficiently?. Hint: *map.entry(key).or_insert(0) += 1
Answering the question you asked...
How does one create a HashMap with a default value in Rust?
No, HashMap
s do not have a place to store a default. Doing so would cause every user of that data structure to allocate space to store it, which would be a waste. You'd also have to handle the case where there is no appropriate default, or when a default cannot be easily created.
Instead, you can look up a value using HashMap::get
and provide a default if it's missing using Option::unwrap_or
:
use std::collections::HashMap;
fn main() {
let mut map: HashMap<char, usize> = HashMap::new();
map.insert('a', 42);
let a = map.get(&'a').cloned().unwrap_or(0);
let b = map.get(&'b').cloned().unwrap_or(0);
println!("{}, {}", a, b); // 42, 0
}
If unwrap_or
doesn't work for your case, there are several similar functions that might:
Of course, you are welcome to wrap this in a function or a data structure to provide a nicer API.
ArtemGr brings up an interesting point:
in C++ there's a notion of a map inserting a default value when a key is accessed. That always seemed a bit leaky though: what if the type doesn't have a default? Rust is less demanding on the mapped types and more explicit about the presence (or absence) of a key.
Rust adds an additional wrinkle to this. Actually inserting a value would require that simply getting a value can also change the HashMap
. This would invalidate any existing references to values in the HashMap
, as a reallocation might be required. Thus you'd no longer be able to get references to two values at the same time! That would be very restrictive.
Solution 2:[2]
What about using entry
to get an element from the HashMap, and then modify it.
From the docs:
fn entry(&mut self, key: K) -> Entry<K, V>
Gets the given key's corresponding entry in the map for in-place manipulation.
use std::collections::HashMap;
let mut letters = HashMap::new();
for ch in "a short treatise on fungi".chars() {
let counter = letters.entry(ch).or_insert(0);
*counter += 1;
}
assert_eq!(letters[&'s'], 2);
assert_eq!(letters[&'t'], 3);
assert_eq!(letters[&'u'], 1);
assert_eq!(letters.get(&'y'), None);
Solution 3:[3]
.or_insert() and .or_insert_with()
Adding to the existing example for .entry().or_insert()
, I wanted to mention that if the default value passed to .or_insert()
is dynamically generated, it's better to use .or_insert_with()
.
Using .or_insert_with()
as below, the default value is not generated if the key already exists. It only gets created when necessary.
for v in 0..s.len() {
components.entry(unions.get_root(v))
.or_insert_with(|| vec![]) // vec only created if needed.
.push(v);
}
In the snipped below, the default vector passed to .or_insert()
is generated on every call. If the key exists, a vector is being created and then disposed of, which can be wasteful.
components.entry(unions.get_root(v))
.or_insert(vec![]) // vec always created.
.push(v);
So for fixed values that don't have much creation overhead, use .or_insert()
, and for values that have appreciable creation overhead, use .or_insert_with()
.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | |
Solution 2 | Akavall |
Solution 3 |