I have an array of words like the one below:
arr = ["id1 abc test", "id#2 XX car house", "id-3 abc home"]
I would like to sort it, but ignoring the first word. For example, the output would be:
arr = ["id-3 abc home", "id1 abc test", "id#2 XX car house"]
Thanks
You can use array.sort where you searchb the index of the space from the string and take for the comparison the original-string just from one position after the space.
Remark: For implementing the sort-function you have pay attention that there are 3 different return values for (a,b):
< 0 => a<b= 0 => a=b< 0 => a>bThe result is now ok after I consider this (I had made here a mistake which is now corrected):
"id#2 XX car house", "id-3 abc home", "id1 abc test"
Note: XX stands before abc because uppercases are smaller than lowercases e.g. ASCII from "A" is 65 and from "a" is 97.
console.log('A'.charCodeAt(0));
console.log('a'.charCodeAt(0));Here is the code-sample in extended version because it's not 100% clearified if for sorting the upper/lowercase-problematic is to be considered or not. So I added an extra parameter, so that now both variants are possible:
let arr = ["id1 abc test", "id#2 XX car house", "id-3 abc home"];
function specialSort(ignoreUpperCase) {
arr.sort( (a,b) => {
if (ignoreUpperCase) {
a = a.toUpperCase();
b = b.toUpperCase()
}
a = a.substr(a.indexOf(' ')+1);
b= b.substr(b.indexOf(' ')+1);
return (a<b) ? -1 : (a>b) ? 1 : 0;
});
}
specialSort(true);
console.log('Ignore UpperCases:',arr);
specialSort(false);
console.log('With UpperCases:',arr);
id-3 abc home should come before id1 abc test
"id#2 XX car house", "id-3 abc home", "id1 abc test" because X<a. For explanation look at my answer above.function getStringWithoutFirstWord(str) {
return str.split(/\s+/).slice(1).join(' ');
}
function compareLoacallyWhilstIgnoringFirstWord(a, b) {
a = getStringWithoutFirstWord(a); // yes, this is a ...
b = getStringWithoutFirstWord(b); // ... highly repetitive task.
return a.localeCompare(b);
}
function compareStronglyAscendingWhilstIgnoringFirstWord(a, b) {
a = getStringWithoutFirstWord(a); // yes, this is a ...
b = getStringWithoutFirstWord(b); // ... highly repetitive task.
return ((a < b && -1) || (a > b && 1) || 0);
}
const arr = ["id1 abc test", "id#2 XX car house", "id-3 abc home"];
console.log(
arr.sort(compareLoacallyWhilstIgnoringFirstWord)
);
console.log(
arr.sort(compareStronglyAscendingWhilstIgnoringFirstWord)
);.as-console-wrapper { min-height: 100%!important; top: 0; }... a working implementation of the already by Hassan Imam mentioned decorate-sort-undecorate approach ...
function getStringWithoutFirstWord(str) {
return str.split(/\s+/).slice(1).join(' ');
}
function restoreStringFromDecoratedIgnoredFirstWordData(data) {
return data.original;
}
function createDecoratedIgnoredFirstWordData(str) {
return {
original: str,
sortable: getStringWithoutFirstWord(str)
}
}
function compareSortablesLoacally(a, b) {
return a.sortable.localeCompare(b.sortable);
}
function compareSortablesStronglyAscending(a, b) {
a = a.sortable;
b = b.sortable;
return ((a < b && -1) || (a > b && 1) || 0);
}
const arr = ["id1 abc test", "id#2 XX car house", "id-3 abc home"];
console.log(arr
.map(createDecoratedIgnoredFirstWordData)
.sort(compareSortablesLoacally)
.map(restoreStringFromDecoratedIgnoredFirstWordData)
);
console.log(arr
.map(createDecoratedIgnoredFirstWordData)
.sort(compareSortablesStronglyAscending)
.map(restoreStringFromDecoratedIgnoredFirstWordData)
);.as-console-wrapper { min-height: 100%!important; top: 0; }This can be done with regular Array.prototype.sort and a small regex that ignores first word:
const arr = ["id1 abc test", "id#2 XX car house", "id-3 abc home"]
const result = arr.sort((a,b) => {
const [aStr, bStr] = [a,b].map(l => l.replace(/^\S+/g,''));
return aStr.localeCompare(bStr)
})
console.log(result);
id1 abc testis beforeid-3 abc home?id-3 abc homeshould be before