如何替換此數組中的倍數？

Question

array = ['blue', 'red', 'green', 'green', 'red', 'blue', 'black', 'blue'] 

...我試圖達到的輸出會導致

output = ['blue x3', 'red x2', 'green x2', 'black'] 

我有一個很難搞清楚這樣做的最有效途徑。

謝謝！

Answer 1

var array = ['blue', 'red', 'green', 'green', 'red', 'blue', 'black', 'blue'] 

var hash = {}; 
for(var i = 0; i < array.length; ++i){ 
    hash[array[i]] = (!hash.hasOwnProperty(array[i]) ? 1 : hash[array[i]]+1); 
} 

var output = []; 
for(var key in hash){ 
    output.push(key + (hash[key]>1 ? (" x"+hash[key]):"")); 
} 
console.log(output); //["blue x3", "red x2", "green x2", "black"] 

DEMO

Answer 2

var arr = ['blue', 'red', 'green', 'green', 'red', 'blue', 'black', 'blue']; 

var group = {}; 
for (var i = arr.length; --i >= 0;) { 
    var value = arr[i]; 
    group[value] = 1 - -(group[value] | 0); 
} 
var result = []; 
for (e in group) { 
    result.push(e + ' x' + group[e]); 
} 

Answer 3

這是我很快提出的一個俗氣的解決方案;

var array = ['blue', 'red', 'green', 'green', 'red', 'blue', 'black', 'blue'], 
    output = [], temp = {}, i; 

// loop through array and count the values 
array.forEach(function(a){ 
    temp[a] = temp[a] ? temp[a]+1 : 1; 
}); 

// loop though temp and add "xN" to the values 
for(i in temp){ 
    output.push(i + (temp[i] > 1 ? ' x'+temp[i] : '')); 
} 

console.log(output); 

注：.forEach不IE 8（或更低）工作。

Answer 4

// You can use an object to group strings 
var colourGroups = {}; 
var array = ['blue', 'red', 'green', 'green', 'red', 'blue', 'black', 'blue']; 

// loop and group each colour. 
array.forEach(function(colourName) { 

    if (colourGroups[colourName]) 
     // increment the count 
     colourGroups[colourName]++; 
    else 
     //initialize the property with 1 
     colourGroups[colourName] = 1; 

});  


// display results 
var results = []; 

for(var prop in colourGroups) { 
    var colourCountStats = prop + " x " + colourGroups[prop]; 
    results.push(colourCountStats); 
} 

console.log(results); 
document.write(results);​ 

JS小提琴例：http://jsfiddle.net/peterf/J7EUc/