Sorting other arrays in order of a specific array?

Question

I have a set of arrays in this form:

var myRows = [ [{idx: 0, val: 90}, {idx: 1, val: 75}, {idx: 2, val: 35}], [{idx: 0, val: 50}, {idx: 1, val: 17}, {idx: 2, val: 95}], [{idx: 0, val: 10}, {idx: 1, val: 24}, {idx: 2, val: 80}] // ... ]; 

Suppose I would like to sort the first line going back to val , so it becomes:

 [{idx: 2, val: 35}, {idx: 1, val: 75}, {idx: 0, val: 90}] 

Is there an easy way to sort the remaining arrays so that their order matches the idx -order of the sorted first row?

 myArrays = [ [{idx: 2, val: 35}, {idx: 1, val: 75}, {idx: 0, val: 90}] , [{idx: 2, val: 95}, {idx: 1, val: 17}, {idx: 0, val: 50}] , [{idx: 2, val: 80}, {idx: 1, val: 24}, {idx: 0, val: 10}] // ... ]; 

Perhaps this is possible even without the idx property?

Answer 1

You can use sorting with a map and apply a mapping to all elements.

This clause stores indexes, arranges an array, and applies order to all other arrays.

 // the array to be sorted var list = [[{ idx: 0, val: 90 }, { idx: 1, val: 75 }, { idx: 2, val: 35 }], [{ idx: 0, val: 50 }, { idx: 1, val: 17 }, { idx: 2, val: 95 }], [{ idx: 0, val: 10 }, { idx: 1, val: 24 }, { idx: 2, val: 80 }]]; // temporary array holds objects with position and sort-value var mapped = list[0].map(function (el, i) { return { index: i, value: el.val }; }) // sorting the mapped array containing the reduced values mapped.sort(function (a, b) { return a.value - b.value; }); // rearrange all items in list list.forEach(function (a, i, aa) { aa[i] = mapped.map(function (el) { return a[el.index]; }); }); console.log(list); 

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Answer 2

When you drop the idx property, you can just use an array:

 // Function copied from here: http://stackoverflow.com/a/36164530/5710637 var transpose = m => m[0].map((x,i) => m.map(x => x[i])) var sortByRow = 0 var myRows = [ [90, 75, 35], [50, 17, 95], [10, 24, 80] ] var myCols = transpose(myRows) myCols.sort((x, y) => x[sortByRow] - y[sortByRow]) myRows = transpose(myCols) console.log(myRows) 

Answer 3

You can do something like this.

 var order = myRows[0].map(function(e) { return e.idx }) myRows.forEach(function(row) { row.sort(function(a,b) { return order.indexOf(a.idx) - order.indexOf(b.idx); }); }); 

This is a very simple code to demonstrate the idea. It will probably be slow for very large arrays.

Answer 4

You can do the following by doing

• Sort the first row of array ¹ and store their idx es in temporary array ²
• Assign the remaining array using the temp property according to the first idx ³
• Sort the remaining array based on their temp ⁴ property (which is based on the first array)
• Remove the temp ⁵ property

eg.

 var filteredRows = []; var myRows = [ [{idx: 0, val: 90}, {idx: 1, val: 75}, {idx: 2, val: 35}], [{idx: 0, val: 50}, {idx: 1, val: 17}, {idx: 2, val: 95}], [{idx: 0, val: 10}, {idx: 1, val: 24}, {idx: 2, val: 80}] ]; /* 1. Sort the first row */ myRows[0].sort(function(a, b) { return a.val - b.val; }); filteredRows.push(myRows[0]); /* 2. Get indexes */ var idxs = []; for (var obj of myRows[0]) { idxs.push(obj.idx); } /* Handle the remaining array */ myRows.slice(1).map(function (val) { /* 3. Assign temp value */ val.map(function (obj, i) { obj.temp = idxs[i]; }); /* 4. Sort them */ val.sort(function (a, b) { return a.temp - b.temp; }); /* 5. Remove temp value */ val.map(function (obj, i) { delete obj.temp; }); }); console.log(JSON.stringify(myRows)); 

Answer 5

Use hash table to create sorting criteria based on the first row - see the demo below:

 var myRows=[[{idx:0,val:90},{idx:1,val:75},{idx:2,val:35}],[{idx:0,val:50},{idx:1,val:17},{idx:2,val:95}],[{idx:0,val:10},{idx:1,val:24},{idx:2,val:80}]]; // sort the first row (as desired) myRows[0].sort((a,b) => a.val - b.val); myRows.forEach(function(c,i){ if(i === 0){ // create order criteria based on first row c.forEach(function(e, k){ this[e.idx] = k; }); } else { c.sort(function(a,b) { return this[a.idx] - this[b.idx]; }); } }, Object.create(null)); console.log(myRows); 

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