javascript 基于变量javascript对数组项进行分组
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Group array items based on variable javascript
提问by Tobias
I have an array that is created dynamic from an xml document looking something like this:
我有一个从 xml 文档动态创建的数组，如下所示：
myArray[0] = [1,The Melting Pot,A]
myArray[1] = [5,Mama's MexicanKitchen,C]
myArray[2] = [6,Wingdome,D]
myArray[3] = [7,Piroshky Piroshky,D]
myArray[4] = [4,Crab Pot,F]
myArray[5] = [2,Ipanema Grill,G]
myArray[6] = [0,Pan Africa Market,Z]
This array is created within a for loop and could contain whatever based on the xml document
这个数组是在 for 循环中创建的，可以包含基于 xml 文档的任何内容
What I need to accomplish is grouping the items from this array based on the letters so that all array objects that have the letter A in them get stored in another array as this
我需要完成的是根据字母对该数组中的项目进行分组，以便所有包含字母 A 的数组对象都存储在另一个数组中，如下所示
other['A'] = ['item 1', 'item 2', 'item 3'];
other['B'] = ['item 4', 'item 5'];
other['C'] = ['item 6'];
To clarify I need to sort out items based on variables from within the array, in this case the letters so that all array objects containing the letter A goes under the new array by letter
为了澄清，我需要根据数组中的变量对项目进行排序，在这种情况下是字母，以便包含字母 A 的所有数组对象都在新数组下一个字母
Thanks for any help!
谢谢你的帮助！
采纳答案by nnnnnn
You shouldn't use arrays with non-integer indexes. Your othervariable should be a plain object rather than an array. (It does work with arrays, but it's not the best option.)
您不应该使用具有非整数索引的数组。您的other变量应该是一个普通对象而不是数组。（它确实适用于数组，但它不是最佳选择。）
// assume myArray is already declared and populated as per the question
var other = {},
letter,
i;
for (i=0; i < myArray.length; i++) {
letter = myArray[i][2];
// if other doesn't already have a property for the current letter
// create it and assign it to a new empty array
if (!(letter in other))
other[letter] = [];
other[letter].push(myArray[i]);
}
Given an item in myArray[1,"The Melting Pot","A"], your example doesn't make it clear whether you want to store that whole thing in otheror just the string field in the second array position - your example output only has strings but they don't match your strings in myArray. My code originally stored just the string part by saying other[letter].push(myArray[i][1]);, but some anonymous person has edited my post to change it to other[letter].push(myArray[i]);which stores all of [1,"The Melting Pot","A"]. Up to you to figure out what you want to do there, I've given you the basic code you need.
给定myArray[1,"The Melting Pot","A"] 中的一个项目，您的示例并未说明您是要将整个内容存储other在第二个数组位置中还是仅存储在第二个数组位置的字符串字段中 - 您的示例仅输出有字符串，但它们与您在myArray. 我的代码最初只存储了字符串部分other[letter].push(myArray[i][1]);，但是一些匿名的人已经编辑了我的帖子，将其更改为other[letter].push(myArray[i]);存储所有 [1,"The Melting Pot","A"] 的内容。由你决定你想在那里做什么，我已经给了你你需要的基本代码。
回答by slv007
Try groupBy function offered by http://underscorejs.org/#groupBy
试试http://underscorejs.org/#groupBy提供的 groupBy 函数
_.groupBy([1.3, 2.1, 2.4], function(num){ return Math.floor(num); });
Result => {1: [1.3], 2: [2.1, 2.4]}
回答by Jose Faeti
You have to create an empty JavaScript object and assign an array to it for each letter.
您必须创建一个空的 JavaScript 对象并为每个字母分配一个数组。
var object = {};
for ( var x = 0; x < myArray.length; x++ )
{
var letter = myArray[x][2];
// create array for this letter if it doesn't exist
if ( ! object[letter] )
{
object[letter] = [];
}
object[ myArray[x][2] ].push[ myArray[x] ];
}
Demo fiddle here.
演示小提琴在这里。
回答by kzh
Good ol' ES5 Array Extras are great.
Good ol' ES5 Array Extras 很棒。
var other = {};
myArray.forEach(function(n, i, ary){
other[n[2]] = n.slice(0,2);
});
回答by user278064
This codewill work for your example.
此代码适用于您的example.
var other = Object.create(null), // you can safely use in opeator.
letter,
item,
max,
i;
for (i = 0, max = myArray.length; i < max; i += 1) {
item = myArray[i];
letter = myArray[2];
// If the letter does not exist in the other dict,
// create its items list
other[letter] = other[letter] || [];
other.push(item);
}
回答by Jayendra
Try -
尝试 -
var myArray = new Array();
myArray[0] = [1,"The Melting Pot,A,3,Sake House","B"];
myArray[1] = [5,"Mama's MexicanKitchen","C"];
myArray[2] = [6,"Wingdome","D"];
myArray[3] = [7,"Piroshky Piroshky","D"];
myArray[4] = [4,"Crab Pot","F"];
myArray[5] = [2,"Ipanema Grill","G"];
myArray[6] = [0,"Pan Africa Market","Z"];
var map = new Object();
for(i =0 ; i < myArray.length; i++){
var key = myArray[i][2];
if(!map[key]){
var array = new Array();
map[key] = array;
}
map[key].push(myArray[i]);
}
