array = [['10','8'],['11','4'],['18','7'],['7','6'],['8',' 7'],['9','5']]
function sortNumber(a,b){return a[0]-b[0]}
let newArray = array.sort(sortNumber)

newArray should be the result you want

var arr = [['10','8'],['11','4'],['18','7'],['7','6'],['8','7'],['9','5']] arr.sort(function(a, b) { return a[0]-b[0] })

The result of the arrangement of your two-dimensional array seems to be compared with the first subscripted element in each element in the array (array). Then it would be bad if it were converted into a one-dimensional array and sorted. ? The one-dimensional array is sorted, and then the original two-dimensional array is sorted using index to correspond to the previous one-dimensional array.
Logical implementation idea:
var twoArray = [['10','8'],['11','4'],['18','7'],['7','6'], ['8','7'],['9','5']];

var oneArray = []; twoArray.map(function (item ,index) { oneArray.push({value: item[0], index: index}); }); console.log(oneArray); function sortNumber (a, b) { return a.value - b.value; } console.log(oneArray.sort(sortNumber)); // 此时的oneArray已排好序 var newTwoArray = []; oneArray.map(function (item) { newTwoArray.push(twoArray[item.index]); }); console.log(newTwoArray);// 即你要的排序

var groupNum = 3; var arr1 = [['10','8'],['11','4'],['18','7'],['7','6'],['8','7'],['9','5']]; var arr2 = []; for (var i = arr1.length; i > 0 ; i -= groupNum) { arr2.push(arr1.slice(i - groupNum, i)); } console.log(arr2) // [['7','6'],['8','7'],['9','5']，['10','8'],['11','4'],['18','7']]