Find string in array of objects- javascript or jquery

You can use this code to put the data into the type of data structure that you asked for:

var response = {"COLUMNS":["SETTING_NAME","SETTING_VALUE","COLOR"],
"DATA":[["setting_1",100.0,"yellow"],["setting_2",150.0,"red"],
["setting_3",30.0,"green"],["setting_4",11.0,"blue"]]};

var data = response.DATA;
var columns = response.COLUMNS;
var hash = {}, item, name, i;
var cols = {};

// remember order of columns
for (i = 0; i < columns.length; i++) {
    cols[columns[i]] = i;
}
// fetch data from correct column
for (i = 0; i < data.length; i++) {
    item = data[i];
    name = item[cols["SETTING_NAME"]];
    hash[name + "_value"] = item[cols["SETTING_VALUE"]];
    hash[name + "_color"] = item[cols["COLOR"]];
}
hash.num = data.length;

As you requested, this gives you a data structure like this so you can directly read any value you want:

{
    "setting_1_value":100,
    "setting_1_color":"yellow",
    "setting_2_value":150,
    "setting_2_color":"red",
    "setting_3_value":30,
    "setting_3_color":"green",
    "setting_4_value":11,
    "setting_4_color":"blue",
    "num":4
}

jsFiddle here: http://jsfiddle.net/jfriend00/HZmYN/ that generated this result.

Personally, I would rather use this code to parse it into this type of data structure:

var response = {"COLUMNS":["SETTING_NAME","SETTING_VALUE","COLOR"],
"DATA":[["setting_1",100.0,"yellow"],["setting_2",150.0,"red"],
["setting_3",30.0,"green"],["setting_4",11.0,"blue"]]};

var data = response.DATA;
var columns = response.COLUMNS;
var newData = [], item, obj, i, num, match;

var cols = {};

// remember order of columns
for (i = 0; i < columns.length; i++) {
    cols[columns[i]] = i;
}

for (i = 0; i < data.length; i++) {
    item = data[i];
    obj = {};
    obj.value = item[cols["SETTING_VALUE"]];
    obj.color = item[cols["COLOR"]];
    obj.name = item[cols["SETTING_NAME"]];
    match = obj.name.match(/\d+$/);
    if (match && match.length > 0) {
        obj.settingNumber = parseInt(match[0], 10);
    }
    newData.push(obj);
}

// now sort the array by the number in the name setting
newData.sort(function(a, b) {
    return(a.settingNumber- b.settingNumber);
});

And generates this data structure:

[
  {"value":100,"color":"yellow","name":"setting_1","settingNumber":1},
  {"value":150,"color":"red","name":"setting_2","settingNumber":2},
  {"value":30,"color":"green","name":"setting_3","settingNumber":3},
  {"value":11,"color":"blue","name":"setting_4","settingNumber":4}
]

Illustrated in this jsFiddle: http://jsfiddle.net/jfriend00/A23Jd/.

The reason I prefer this structure, is you can more easily access the "n" settings as an array of objects:

newData[0].color
newData[0].value
newData[0].name
newData[1].color
....

And, it's easier to iterate through the various settings

You could simply reduce that list of DATA:

DATA.reduce(function (value, item) { if (item[0] === "setting_4") return item[2] })

You could wrap that whole thing into a function for easier use, passing in the "setting_4" part. E.g.

var getColour = function (id) {
  return DATA.reduce(function (value, item) {
    if (item[0] === id) return item[2]
  })
}

UPDATE: you could zip the two lists together, perhaps that would make access easier?

obj['DATA'].map(function (row) {
  return obj['COLUMNS'].reduce(function (memo, columnName, index) {
    memo[columnName] = row[index]
    return memo
  }, {})
})

This will return something like the following:

[{
   COLOR: "yellow",
   SETTING_NAME: "setting_1",
   SETTING_VALUE: 100
}]

You can do this with a simple for loop:

var obj = {"COLUMNS":["SETTING_NAME","SETTING_VALUE","COLOR"],
"DATA":[["setting_1",100.0,"yellow"],["setting_2",150.0,"red"],
["setting_3",30.0,"green"],["setting_4",11.0,"blue"]]};

for(var i = 0; i < obj.DATA.length; i++)
{
    var row = obj.DATA[i]
    if (row[0] == 'setting_4')
    {
        console.log(row[2]);
        break;
    }
}

Prints:

blue

Using $.grep will allow you to access the data without mapping them before:

var json={"COLUMNS":["SETTING_NAME","SETTING_VALUE","COLOR"],
          "DATA":[["setting_1",100.0,"yellow"],
                  ["setting_2",150.0,"red"],
                  ["setting_3",30.0,"green"],
                  ["setting_4",11.0,"blue"]]}

alert($.grep(json.DATA, function(item){return(item[0]=='setting_4');})[0][2])//returns 'blue'

//or using the labels provided by COLUMNS:

alert($.grep(json.DATA, 
             function(a){return(a[0]=='setting_4');})[0][$.inArray('COLOR',json.COLUMNS)])

A generic algorithm for translating the dataset into a more-easily-addressed structure.

var json = {
    "COLUMNS": [
        "SETTING_NAME",
        "SETTING_VALUE",
        "COLOR"],
    "DATA": [
        ["setting_1",100.0,"yellow"],
        ["setting_2",150.0,"red"],
        ["setting_3",30.0,"green"],
        ["setting_4",11.0,"blue"]
    ]
};

function translateJSON(json) {
    var oHash = {};

    var data = json['DATA'];
    var cols = json['COLUMNS'];

    for(var i = 0, imax = data.length; i < imax; i++) {
        var row = data[i]; // shorthand

        for(var j = 1, jmax = cols.length; j < jmax; j++) {
            var c = cols[j]; // shorthand
            oHash[(row[0] + '_' + c.replace(/[^_]+_/, '')).toLowerCase()] = row[j];
        }
    }

    return oHash;
}

var h = translateJSON(json);
console.log(h['setting_4_color']);

Edit: updated the code. translateJSON will transform the JSON into the data structure you described, for easier property access. If you anticipate needing to access more than one property from the same JSON payload, it will be much more efficient to do a one-time transform before data access than to use something like $.grep, and much less terse than doing the column-name cross-referencing by hand.

That said, I don't think the target data structure you asked for is necessarily the best one. Assuming you can't change the structure of the JSON payload, it would still probably be better to transform that into something like:

data = {
    'setting_1': { 'value': 100.0, 'color': 'yellow' },
    'setting_2': { 'value': 150.0, 'color': 'red' }
    ...
};