Split array based on element contents?

Question

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')

@NinaScholz if I'm honest, I got mixed up and tried to use array.split('goose') which is so nonsensical it wasn't worth posting — AncientSwordRage
– AncientSwordRage, Commented Mar 31, 2022 at 15:06

Nina Scholz · Accepted Answer · 2022-03-31 15:18:20Z

2

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; }

answered Mar 31, 2022 at 15:18

Nina Scholz

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Comments

p4avinash · Accepted Answer · 2022-03-31 12:18:58Z

1

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)

edited Mar 31, 2022 at 12:18

answered Mar 31, 2022 at 12:12

p4avinash

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Comments

ahsan · Accepted Answer · 2022-03-31 12:15:06Z

0

Here's the first solution that came up in my mind.

We keep pushing elements to temp array until we get a string containing the word goose.

const words = [
 'duck foo bar',
 'duck',
 'duck bing ',
 'bing',
 'Bloop#12 goose',
 'duck 12',
 'duck goose',
];

const splits = [];
let temp = [];

words.forEach(word => {
 if(word.includes('goose')) {
    temp.push(word);
  splits.push(temp);
  temp = [];
 } else {
    temp.push(word);
 }
})

console.log(splits)

answered Mar 31, 2022 at 12:15

ahsan

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Comments

Amila Senadheera · Accepted Answer · 2022-03-31 12:17:41Z

0

This would also work.

const words = [
  "duck foo bar",
  "duck",
  "duck bing ",
  "bing",
  "Bloop#12 goose",
  "duck 12",
  "duck goose",
];

const output = [];

let chunk = [];
words.forEach((item) => {
  const splits = item.split(" ");
  const last =  splits[splits.length - 1];
  if(last === "goose") {
    chunk.push(item)
    output.push(chunk)
    chunk = [];
  } else {
    chunk.push(item)
  }
})

console.log(output);

answered Mar 31, 2022 at 12:17

Amila Senadheera

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Comments

AncientSwordRage · Accepted Answer · 2022-03-31 16:31:58Z

0

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)

edited Mar 31, 2022 at 16:31

AncientSwordRage

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answered Mar 31, 2022 at 12:11

Maik Lowrey

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1 Comment

Maik Lowrey Over a year ago

@AncientSwordRage Thank you for editing. My english is horrible. Hope my code helps a little bit?

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