如何将一组值转换为对它们进行计数的 HashMap? [英] How can I convert a collection of values into a HashMap that counts them?

问题描述

我有一个带有重复项的集合，我想创建一个 HashMap 将这些项作为键，以及它们在原始集合中出现的次数.>

我正在处理的具体示例是一个字符串，我想计算每个字符出现的次数.

我可以这样做:

fn into_character_map(word: &str) ->HashMap{让 mut 字符 = HashMap::new();for c in word.chars() {让 entry = characters.entry(c).or_insert(0);*条目 += 1;}人物}

但我想知道是否有更优雅的解决方案.我正在考虑使用 collect()，但它不维护项目之间的状态，因此似乎不支持我所需要的.

这是在我编写解决Anagram"问题时出现的.>

解决方案

是否更优雅值得怀疑，但fold 使用 HashMap 需要更少的行:

fn into_character_map(word: &str) ->HashMap{word.chars().fold(HashMap::new(), |mut acc, c| {*acc.entry(c).or_insert(0) += 1;acc})}

I have a collection of items, with repetitions, and I want to create a HashMap that has the items as keys, and the count of how many times they appear in the original collection.

The specific example I'm working on is a string where I want to count how many times each character appears.

I can do something like this:

fn into_character_map(word: &str) -> HashMap<char, i32> {
    let mut characters = HashMap::new();

    for c in word.chars() {
        let entry = characters.entry(c).or_insert(0);
        *entry += 1;
    }

    characters
}

But I was wondering if there's a more elegant solution. I was thinking of using collect(), but it doesn't maintain state between items, so doesn't seem to support what I need.

This came up as I was writing my solution to the 'Anagram' problem on Exercism.

解决方案

It's dubious if it's more elegant, but folding using a HashMap requires fewer lines:

fn into_character_map(word: &str) -> HashMap<char, i32> {
    word.chars().fold(HashMap::new(), |mut acc, c| {
        *acc.entry(c).or_insert(0) += 1;
        acc
    })
}

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