javascript 使用javascript计算字符串中字符的频率
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counting frequency of characters in a string using javascript
提问by Samrat
I need to write some kind of loop that can count the frequency of each letter in a string.
我需要编写某种循环来计算字符串中每个字母的频率。
For example: "aabsssd"
例如：“aabsssd”
output: a:2, b:1, s:3, d:1
输出：a:2, b:1, s:3, d:1
Also want to map same character as property name in object. Any good idea how to do this?
还想映射与对象中的属性名称相同的字符。任何好主意如何做到这一点？
I am not sure how to do it.
我不知道该怎么做。
This is where I am so far:
这是我到目前为止的地方：
var arr = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"];
function counter(x) {
var count=0, temp = [];
x = x.split('');
console.log(x);
for(var i=0, len = x.length; i < len; i++) {
if(x[i] == "a") {
count++;
}
}
return count;
}
var a = "aabbddd";
console.log(counter(a));
回答by Jonathan Crowe
Here you go:
干得好：
function getFrequency(string) {
var freq = {};
for (var i=0; i<string.length;i++) {
var character = string.charAt(i);
if (freq[character]) {
freq[character]++;
} else {
freq[character] = 1;
}
}
return freq;
};
回答by russiansummer
some ES6 syntax with reduce:
一些带有 reduce 的 ES6 语法：
let counter = str => {
return str.split('').reduce((total, letter) => {
total[letter] ? total[letter]++ : total[letter] = 1;
return total;
}, {});
};
counter("aabsssd"); // => { a: 2, b: 1, s: 3, d: 1 }
回答by sarunast
Another solution:
另一种解决方案：
function count (string) {
var count = {};
string.split('').forEach(function(s) {
count[s] ? count[s]++ : count[s] = 1;
});
return count;
}
回答by frederick99
With some ES6 features and short-circuiting:
使用一些 ES6 特性和短路：
const counter = s => [...s].reduce((a,c) => (a[c] = a[c]+1 || 1) && a, {})
counter("hello") // {h: 1, e: 1, l: 2, o: 1}
回答by vitkon
More declarative way to get a word histogram will be to utilise reduce to iterate through letters and come up with a new object that contains letters as keys and frequencies as values.
获得单词直方图的更明确的方法是利用 reduce 遍历字母并提出一个包含字母作为键和频率作为值的新对象。
function getFrequency(str) {
return str.split('').reduce( (prev, curr) => {
prev[curr] = prev[curr] ? prev[curr] + 1 : 1;
return prev;
}, {});
};
console.log(getFrequency('test')); // => {t: 2, e: 1, s: 1}回答by Yoni Rabinovitch
Here's another way:
这是另一种方式：
const freqMap = s => [...s].reduce((freq,c) => {freq[c] = -~freq[c]; return freq} ,{})
Or, if you prefer a "for" loop:
或者，如果您更喜欢“for”循环：
function freqMap(s) {
freq={};
for (let c of s)
freq[c]=-~freq[c];
return freq;
}
e.g. freqMap("MaMaMia") returns Object{M : 3, a : 3, i : 1}
例如 freqMap("MaMaMia") 返回 Object{M : 3, a : 3, i : 1}
This method leverages the fact that in javascript, bitwise not on "undefined" gives -1, (whereas "undefined+1" gives NaN). So, -~undefined is 1, -~1 is 2, -~2 is 3 etc.
这种方法利用了这样一个事实，即在 javascript 中，按位不在“未定义”上给出 -1，（而“未定义+1”给出 NaN）。所以，-~undefined 是 1，-~1 是 2，-~2 是 3，等等。
We can thus iterate over the characters of the string, and simply increment freq[c] without any "if". The first time we encounter a character c, freq[c] will be undefined, so we set it to -~freq[c] which is 1. If we subsequently encounter c again, we again set freq[c] to -~freq[c], which will now be 2, etc.
因此，我们可以遍历字符串的字符，并简单地增加 freq[c] 而没有任何“if”。第一次遇到字符 c，freq[c] 将是未定义的，因此我们将它设置为 -~freq[c]，即 1。如果我们随后再次遇到 c，我们再次将 freq[c] 设置为 -~freq [c]，现在是 2，依此类推。
Simple, elegant, concise.
简洁，优雅，简洁。
回答by leerssej
a leaner, functional solution:
一个更精简、更实用的解决方案：
using ES6 Arrows && Logical Operators:
使用 ES6 箭头 && 逻辑运算符：
const buildFreqDict = string =>
string.split('').reduce((freqDict, char) => {
freqDict[char] = (freqDict[char] || 0) + 1;
return freqDict;
}, {})
console.log(buildFreqDict("banana"))Explained
解释
- splitstringinto array of characters.
- and then feed it into a reducemethod (using method.chaining()).
- if charis already logged in countDictthen add 1 to it.
- orif character not found in countDictthen set it to 1.
- return new values back up to reduce's accumulator object
- NB: don't forget about including the third argument of .reduce(): in this case it is a {}(object literal) that serves to initialize the freqDictobject.
- 将字符串拆分为字符数组。
- 然后将其输入到一个reduce方法中（使用method.chaining()）。
- 如果char已经登录countDict则加 1。
- 或者如果在countDict 中找不到字符，则将其设置为 1。
- 将新值返回到reduce的累加器对象
- 注意：不要忘记包含.reduce()的第三个参数：在这种情况下，它是一个{}（对象文字），用于初始化freqDict对象。
for more info see Counting instances of values in an objecthalf way down the page here: MDN Reduce
and for more info about using logical operatorsplease see here: MDN Logical Operators
有关更多信息，请参阅在页面下方的对象中计数值的实例：MDN Reduce
以及有关使用逻辑运算符的更多信息，请参阅此处：MDN 逻辑运算符
回答by Sergei Volynkin
// Count frequency of characters in a string
// input: 'Hello, I'm Paul!'
// result: {
// H: 1,
// E: 1,
// L: 3,
// ... and so on ...
// }
const countChars = (string) => {
let charStats = {};
string = string.replace(' ', '').toUpperCase().split('');
string.forEach((char) => {
if (charStats[char]) {
charStats[char]++;
} else {
charStats[char] = 1;
}
});
return charStats;
};
回答by bajran
Another Solution
另一种解决方案
function maxChar(str) {
const charMap = {};
let max = 0;
let maxChar = '';
for(let char of str){
if(charMap[char]){
charMap[char]++;
}else{
charMap[char] = 1;
}
}
for(let char in charMap){
if(charMap[char] > max){
max = charMap[char];
maxChar = char;
}
}
return maxChar;
}
===>
maxChar('355385') "5"
===>
maxChar('355385') "5"
回答by Aaron
var str = 'abcccdddd';
function maxCharCount(target) {
const chars = {};
let maxChar = '';
let maxValue = 1;
for (let char of target) {
chars[char] = chars[char] + 1 || 1;
}
return chars;
}
console.log(maxCharCount(str));
