I want to rearrange the sentences so that each sentence is a prerequisites for the next sentence and here is what I have and it works fine:
const sentences = [
"he has eaten a pizza at the party last night",
"he has eaten a pizza",
"he has eaten a pizza at the party",
]
function sessionCardsSort(arr) {
arr = arr.map(part => part.split(' ')).sort((a, b) => a.length - b.length);
function sortBy(list, main) {
return list.sort((a, b) => {
let count1 = 0, count2 = 0;
a.forEach(part => { if(main.includes(part)) { count1++ } });
b.forEach(part => { if(main.includes(part)) { count2++ } });
return count1 - count2;
});
}
return sortBy(arr, arr[arr.length - 1]).map(part => part.join(' '));
}
console.log(sessionCardsSort(sentences));As you see I used a simple array named sentences as data structure to work with, but what if we want to use the same data of sentences array with new structure like this:
const sentences = [
{ reference: ['he has eaten a pizza at the party', 'last night'] }
{ reference: ['"he has', 'eaten a pizza'] },
{ reference: ['he', 'has', 'eaten a pizza at the party'] },
]
So the issue is I want to implement the exact same function this time for array of objects not just a simple array!
Here is desired result (only using the exact above function):
const sentences = [
{ reference: ['"he has', 'eaten a pizza'] },
{ reference: ['he', 'has', 'eaten a pizza at the party'] },
{ reference: ['he has eaten a pizza at the party', 'last night'] }
]
Note: In the second data structure joining elements inside each reference will give you the same sentence as the first example, for instance :
{ reference: ['he has eaten a pizza at the party', 'last night'] }
"he has eaten a pizza at the party last night"
Note: Please note that I want to implement the exact same function
Since the words of each sentence are either a strict subset of another sentence in the set, or the sentence is the largest one, it looks like it'd be a lot easier to just check the strings' overall length instead.
Give each sentence a length property (which is the length of the joined string), and then sort the resulting array of objects by that length, and finally remove the added length property.
const sentences = [
{ reference: ['"he has', 'eaten a pizza'] },
{ reference: ['he', 'has', 'eaten a pizza at the party'] },
{ reference: ['he has eaten a pizza at the party', 'last night'] }
];
function sessionCardsSort(sentences) {
const sentencesWithLength = sentences.map(
({ reference }) => ({ reference, length: reference.join(' ') })
);
return sentencesWithLength
.sort((a, b) => a.length - b.length)
.map(({ reference }) => ({ reference }));
}
console.log(sessionCardsSort(sentences));Or calculate the length each time inside the callback - more work (which could be a problem when dealing with huge datasets), but less code.
const sentences = [
{ reference: ['"he has', 'eaten a pizza'] },
{ reference: ['he', 'has', 'eaten a pizza at the party'] },
{ reference: ['he has eaten a pizza at the party', 'last night'] }
];
const toLength = arr => arr.reference.join(' ').length;
const sessionCardsSort = (sentences) => sentences
.sort((a, b) => toLength(a) - toLength(b));
console.log(sessionCardsSort(sentences));It doesn't look necessary given the input in the question, but if you really need to count words individually, then use a function with similar logic to your current sortBy to count up the number of included words.
const sentences = [
{ reference: ['"he has', 'eaten a pizza'] },
{ reference: ['he', 'has', 'eaten a pizza at the party'] },
{ reference: ['he has eaten a pizza at the party', 'last night'] }
];
const toLength = obj => obj.reference.join(' ');
const toWordsIncluded = (obj, mainWords) => obj.reference
.join(' ')
.split(' ')
.reduce((subtotal, word) => subtotal + mainWords.includes(word), 0)
const sessionCardsSort = (sentences) => {
const sorted = sentences
.sort((a, b) => toLength(a) - toLength(b));
const mainWords = sorted[sorted.length - 1].reference.join(' ').split(' ');
return sorted
.sort((a, b) => toWordsIncluded(a, mainWords) - toWordsIncluded(b, mainWords));
};
console.log(sessionCardsSort(sentences));