Let's say my elements look like this:
const words = [
'duck foo bar',
'duck',
'duck bing ',
'bing',
'Bloop#12 goose',
'duck 12',
'duck goose',
...
]
What I'd like is to split this into chunks where 'goose' is the final element in a chunk:
const words = [
[
'duck foo bar',
'duck',
'duck bing',
'bing',
'Bloop#12 goose',
],
[
'duck 12',
'duck goose',
],
[
...
],
];
There's no regularity to how many elements precede a 'goose', nor what is in each element except that 1) goose is always the last part of an element, and 2) goose never appears in any other element besides the one I want a chunk to end on (i.e. I never get 'goose foo', but I might get 'duck goose')
You could reduce the array and have a look to the previous string and add an array to the result set for a new group.
const
words = ['duck foo bar', 'duck', 'duck bing ', 'bing', 'Bloop#12 goose', 'duck 12', 'duck goose'],
separator = 'goose',
groups = words.reduce((r, s, i, a) => {
if (!i || a[i - 1].includes(separator)) r.push([]);
r[r.length - 1].push(s);
return r;
}, []);
console.log(groups);.as-console-wrapper { max-height: 100% !important; top: 0; }Try this. This should do what you're looking for.
const words = [
'duck foo bar',
'duck',
'duck bing ',
'bing',
'Bloop#12 goose',
'duck 12',
'duck goose'
]
const answer = []
let temp = []
for(let i = 0; i < words.length; i++){
if(words[i].includes('goose')){
temp.push(words[i])
answer.push(temp)
temp = []
} else{
temp.push(words[i])
}
}
console.log(answer)You can achieve this with a simple forEach loop. In my example below I store the values first temporally.
Then I push the chunk into the new array if the word contains goose.
const words = [
'duck foo bar',
'duck',
'duck bing ',
'bing',
'Bloop#12 goose',
'duck 12',
'duck goose',
'abc'
]
const n = []
let tmp = [];
words.forEach((w) => {
if (! w.includes('goose')) {
tmp.push(w);
} else {
tmp.push(w);
n.push(tmp);
tmp = [];
}
});
console.log(n)