algorithm - sort outer array based on values in inner array, javascript - Stack Overflow

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I have an array with arrays in it, where I want to sort the outer arrays based on values in a specific column in the inner.

I bet that sounded more than a bit confusing, so I'll skip straight to an example.

Initial data:

var data = [
  [
    "row_1-col1",
    "2-row_1-col2",
    "c-row_1-coln"
  ],
  [
    "row_2-col1",
    "1-row_2-col2",
    "b-row_2-coln"
  ],
  [
    "row_m-col1",
    "3-row_m-col2",
    "a-row_m-coln"
  ]
];

Sort data, based on column with index 1

data.sortFuncOfSomeKind(1);

where the object then would look like this;

var data = [
  [
    "row_2-col1",
    "1-row_2-col2",
    "b-row_2-coln"
  ],
  [
    "row_1-col1",
    "2-row_1-col2",
    "c-row_1-coln"
  ],
  [
    "row_m-col1",
    "3-row_m-col2",
    "a-row_m-coln"
  ]
];

Sort data, based on column with index 2

data.sortFuncOfSomeKind(2);

where the object then would look like this;

var data = [
  [
    "row_m-col1",
    "3-row_m-col2",
    "a-row_m-coln"
  ],
  [
    "row_2-col1",
    "1-row_2-col2",
    "b-row_2-coln"
  ],
  [
    "row_1-col1",
    "2-row_1-col2",
    "c-row_1-coln"
  ]
];

The big Q

Is there an existing solution to this that you know of, or would I have to write one myself? If so, which would be the easiest sort algorithm to use? QuickSort?

_L

I have an array with arrays in it, where I want to sort the outer arrays based on values in a specific column in the inner.

I bet that sounded more than a bit confusing, so I'll skip straight to an example.

Initial data:

var data = [
  [
    "row_1-col1",
    "2-row_1-col2",
    "c-row_1-coln"
  ],
  [
    "row_2-col1",
    "1-row_2-col2",
    "b-row_2-coln"
  ],
  [
    "row_m-col1",
    "3-row_m-col2",
    "a-row_m-coln"
  ]
];

Sort data, based on column with index 1

data.sortFuncOfSomeKind(1);

where the object then would look like this;

var data = [
  [
    "row_2-col1",
    "1-row_2-col2",
    "b-row_2-coln"
  ],
  [
    "row_1-col1",
    "2-row_1-col2",
    "c-row_1-coln"
  ],
  [
    "row_m-col1",
    "3-row_m-col2",
    "a-row_m-coln"
  ]
];

Sort data, based on column with index 2

data.sortFuncOfSomeKind(2);

where the object then would look like this;

var data = [
  [
    "row_m-col1",
    "3-row_m-col2",
    "a-row_m-coln"
  ],
  [
    "row_2-col1",
    "1-row_2-col2",
    "b-row_2-coln"
  ],
  [
    "row_1-col1",
    "2-row_1-col2",
    "c-row_1-coln"
  ]
];

The big Q

Is there an existing solution to this that you know of, or would I have to write one myself? If so, which would be the easiest sort algorithm to use? QuickSort?

_L

• javascript
• algorithm
• sorting

Share Improve this question Follow asked May 8, 2010 at 11:19 ptrnptrn 5,20266 gold badges3030 silver badges3030 bronze badges 1

• 2 possible duplicate of Sorting objects in an array by a field value in JavaScript, Javascript: How to sort an array of records by values in one of the fields? – outis Commented May 8, 2010 at 11:33

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3 Answers 3

Sorted by: Reset to default 16

Array#sort (see section 15.4.4.11 of the spec, or MDC) accepts an optional function parameter which will be used to compare two entries for sorting purposes. The function should return -1 if the first argument is "less than" the second, 0 if they're equal, or 1 if the first is "greater than" the second. So:

outerArray.sort(function(a, b) {
    var valueA, valueB;

    valueA = a[1]; // Where 1 is your index, from your example
    valueB = b[1];
    if (valueA < valueB) {
        return -1;
    }
    else if (valueA > valueB) {
        return 1;
    }
    return 0;
});

(You can obviously compress that code a bit; I've kept it verbose for clarity.)

Here is a solution not needing a separate variable to contain the index

var arr = [.....]
arr.sort((function(index){
    return function(a, b){
        return (a[index] === b[index] ? 0 : (a[index] < b[index] ? -1 : 1));
    };
})(2)); // 2 is the index

This sorts on index 2

Here used to be a sort implementation that returned the result of a simple x<y comparison. This solution is disencouraged and this post is left only to preserve the ensuing discussion.

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