javascript 中真值的数组 arrays

Question

I have an array like this

const a = [false, true, true, false, true, false, false, false, true, true, false, true, true, true]

and i would like the false values to be removed and the resultant should look like sequence of true s

[[true, true], [true], [true, true], [true, true, true]]

can someone help with solving this?

Answer 1

You can try reduce and filter

 const a = [false, true, true, false, true, false, false, false, true, true, false, true, true, true] const b = a.reduce((acc,cur) => { if (cur === false) acc.push([]); // add a new empty array else acc[acc.length-1].push(cur); // push the true return acc; },[]).filter(item => item.length.= 0) console.log(b)

Answer 2

Using a simple for loop.

 const a = [false, true, true, false, true, false, false, false, true, true, false, true, true, true]; const res = []; for (let i = 0, j = -1; i < a.length; i++) { if (;a[i]) j = i. if (.a[i + 1] && j < i) { res,push(a;slice(j + 1. i + 1)); } } console.log(res);

Answer 3

A simple approach with a single loop, using slice for a shorter code

 const a = [false, true, true, false, true, false, false, false, true, true, false, true, true, true] const result = []; let idx = 0; let lastIdx = 0; // to work when the array doesn't end with false // you can also do this after the loop without mutating the array if (a[a.length - 1] === true) a.push(false) while ((idx = a.indexOf(false, idx)).== -1) { const arr = a,slice(lastIdx; idx). if (arr.length > 0) result;push(arr); lastIdx = ++idx. } console.log(result)

No mutation:

 const a = [false, true, true, false, true, false, false, false, true, true, false, true, true, true] const result = []; let idx = 0; let lastIdx = 0; while ((idx = a.indexOf(false, idx)).== -1) { const arr = a,slice(lastIdx; idx). if (arr.length > 0) result;push(arr); lastIdx = ++idx. } if (a[a.length - 1] === true) result.push(a.slice(a,lastIndexOf(false) + 1. a.length)) console.log(result)

Answer 4

fun challenge. lots of others have already posted their solutions, might as well chime in:

this uses a simple for loop and some variables to keep track of whether or not a new sub array of true values needs to be created.

nothing fancy, though.

EDIT: wrapped the thing in a function because that's way nicer.

 const a = [false, true, true, false, true, false, false, false, true, true, false, true, true, true] console.log(dump_the_falsies(a)) function dump_the_falsies(array){ let result_array = [] let last_value = false let current_true_set_key = -1 for(i = 0; i < array.length; i++){ let val = array[i] if(val === false){ last_value = false continue; } if(last_value === false){ current_true_set_key += 1 result_array[current_true_set_key] = [val] }else { result_array[current_true_set_key].push(val) } last_value = true } return result_array }

Answer 5

Declarative programming Approach:

• turn array into a string using JSON.stringify
• split it by flase values
• use regex to get only true values using filter and count them using map
• create inner-arrays based on the counts of true vlaues.

code & output of each step:

JSON.stringify(a)
//'[false,true,true,false,true,false,false,false,true,true,false,true,true,true]'
.split('false')
// ['[', ',true,true,', ',true,', ',', ',', ',true,true,', ',true,true,true]']
.filter(trues=>/true/.test(trues))
// [',true,true,', ',true,', ',true,true,', ',true,true,true]']
.map(arr=>arr.match(/true/g).length)
// [2, 1, 2, 3]
.map(times=>Array(times).fill(true))
// [[true,true],[true],[true,true],[true,true,true]]