JavaScript函数自动计算字符串中的连续字母

Question

I am attempting (unsuccessfully) to write JavaScript function LetterCount to count the consecutive letters in a string (and not the total number).

Ideally: LetterCount("eeeeeoooohhoooee") = [["e", 5],["o",3],["h",2],["o",3],["e",2]]

The following code attempts to count the number of consecutive letters in a string only when I already know what they are:

function LetterCount(str) {
for (var i=0; i<str.length;i++) {
    var arr1=[]; arr2=[]; arr3=[]; arr4=[]; arr5=[];
    var e=0; o=0; h=0; o2=0; e2=0;
    if(str[i]="e") {
        arr1 += "e";
        e++;
    }
    arr1.push(e);
    if(str[i]="o") {
        arr2 += "o";
        o++;
    }
    arr2.push(o);
    if(str[i]="h") {
        arr3 += "h";
        h++;
    }
    arr3.push(h);
    if(str[i]="o") {
        arr4 += "o";
        o2++;
    }
    arr4.push(o2);
    if(str[i]="e") {
        arr5 += "e";
        e2++;
    }
    arr5.push(e2);
}
return arr1.concat(arr2).concat(arr3).concat(arr4).concat(arr5);
}

In the code above, I need to first know what the letters in the string are, and how many of them are present, in what order.

INSTEAD: How do you write a function that will automatically recognize the letter themselves, and then return the count of consecutive letters. Would also be great if the answer is in the following format:

 LetterCount("eeeeeoooohhoooee") = [["e", 5],["o",3],["h",2],["o",3],["e",2]]

Any help is much appreciated!

Answer 1

You can use a regular expression to match any letter followed by zero or more instances of the same letter.

rx=/([a-zA-Z])\\1*/g;

Your example matches ["eeeee","oooo","hh","ooo","ee"].

Using map, return the initial letter and the number of occurences in a new array for each index.

function letterCount(str){
    var s= str.match(/([a-zA-Z])\1*/g)||[];
    return s.map(function(itm){
        return [itm.charAt(0), itm.length];
    });
}

letterCount("eeeeeoooohhoooee")

returned value: (Array)

[["e",5],["o",4],["h",2],["o",3],["e",2]]

NOTES:

1. var s= str.match(/([a-zA-Z])\\1*/g)||[];

returns an array of matches (repeated letters) or an empty array([]). Otherwise, if the string does not contain any letters an error would be thrown (from calling map on null).

2. \\1* is used to allow matching instances of a single letter with any or no sequential repetition. '\\1+' would not match a single unrepeated letter.

3. Array map expects a function and passes three arguments- the value at each index, the index number, and a reference to the entire array. In this case, only the value of each index is used, so we can ignore the other arguments.

Answer 2

This is my answer:

function LetterCount(str) {
    var current, i = 0, l = str.length;
    var outputArr = [];
    while(i<l) {
        current = str.charAt(i);
        if(!i++ || outputArr[outputArr.length-1][0] !== current)
            outputArr[outputArr.length] = [current, 1];
        else outputArr[outputArr.length-1][1]++;
        }
    return outputArr;
    }

As a modification to kennebec's (awesome) answer, so that the anonymous function isn't declared each time the parent function is called. This is only to reference a better programming practice in comparison to pure simplicity (this is probably the most efficient method):

var letterCount = (function(){
    var r = /([A-z])\1*/g,
        f = function(itm){
        return [itm.charAt(0), itm.length];
        };
    return function letterCount(str){
        return str.match(r).map(f);
        };
    }());

Answer 3

Actually "fixed" ["o",3] to ["o",4] ;)

// node v0.10.31
// assert v1.3.0

var assert = require('assert');

function letterCount(str) {
    var i = 0,
        seq = 0,
        results = [];

    while (i < str.length) {
        var current = str[i],
            next = str[i + 1];

        if (typeof results[seq] === 'undefined') {
            results[seq] = [current, 0];
        }

        results[seq][1]++;

        if (current !== next) {
            seq++;
        }

        i++;
    }

    return results;
}

var actual = letterCount('eeeeeoooohhoooee');
var expected = [["e", 5],["o",4],["h",2],["o",3],["e",2]];

assert.deepEqual(actual, expected);

Answer 4

I'd use a map keyed on the character to store the count of consecutive chars and then build the output structure at the end. I'm not sure I understand exactly what you mean by consecutive based on your example but you can tweak the logic to identify a consecutive number accordingly.

function LetterCount(str) {
  var counts = {};
  for (var i = 0, prevChar = null; i < str.length; i++) {
    var char = str.charAt(i);
    if(counts.hasOwnProperty(char) && char === prevChar) {
      counts[char] = counts[char] + 1;  
    } else if (!counts.hasOwnProperty(char)) {
      counts[char] = 0;
    }
    prevChar = char;
  }
  var res = [];
  for (var char in counts) {
    if (counts.hasOwnProperty(char)) {
      res.push([char,counts[char]);
    }
  }
  return res;
}

Answer 5

function LetterCount(text){
    arr = [];
    letter = text[0];
    counter = 0;
    for (c in text+' '){
        if (text[c] != letter){
            newArr = [letter, counter];
            arr.push(newArr);
            letter = text[c];
            counter = 0;
        }
        counter += 1;
    };
    return arr;
}