Related
I have an array like so: [1, 2, 3, 4, 5, 6, 7, 9, 10]. I need to chunk it into different size chunks, yet with a simple pattern of: 4, 3, 3, 3, 4, 3, 3, 3 like so:
[
[ // four
1,
2,
3,
4
],
[ // three (1/3)
5,
6,
7
],
[ // three (2/3)
8,
9,
10
],
[ // three (3/3)
11,
12,
13
],
[ // four
14,
15,
16,
17
],
[ // three (1/3)
18,
19,
20
], // and so on..
]
I have tried with this code I have customized:
const arr; // my array of values
const chuncked = arr.reduce((acc, product, i) => {
if (i % 3) {
return acc;
} else if (!didFourWayReduce) {
didFourWayReduce = true;
fourWayReduces++;
if ((fourWayReduces - 1) % 2) { // only make every second a "4 row"
return [...acc, arr.slice(i, i + 3)];
} else {
return [...acc, arr.slice(i, i + 4)];
}
} else {
didFourWayReduce = false;
return [...acc, arr.slice(i, i + 3)];
}
}, []);
And it works, almost, expect that the first chunk of threes (1/3) have the last element of the chunk with 4. So 1 key is repeated every first chunk of three. Like so:
[
[
1,
2,
3,
4
],
[
4, // this one is repeated, and it shouldn't be
5,
6
]
]
You could take two indices, one for the data array and one for sizes. Then slice the array with a given length and push the chunk to the chunks array.
Proceed until end of data.
var data = Array.from({ length: 26 }, (_, i) => i + 1),
sizes = [4, 3, 3, 3],
i = 0,
j = 0,
chunks = [];
while (i < data.length) chunks.push(data.slice(i, i += sizes[j++ % sizes.length]));
console.log(chunks);
.as-console-wrapper { max-height: 100% !important; top: 0; }
const arr = Array.from({ length: 100 }, (_, i) => i);
const copy = [...arr];
const sizes = [4, 3, 3, 3];
const result = [];
let i = 0;
while (i <= arr.length && copy.length) {
result.push(copy.splice(0, sizes[i % sizes.length]));
i++;
}
console.log(result);
A recursive approach is fairly elegant:
const chunks = (xs, [s, ...ss]) =>
xs.length ? [xs .slice (0, s), ... chunks (xs .slice (s), [...ss, s])] : []
const data = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
const sizes = [4, 3, 3, 3]
console .log (chunks (data, sizes))
.as-console-wrapper { max-height: 100% !important; top: 0; }
By replacing [s, ...ss] with [...ss, s], we pass a cycled version of the sizes array, so that for instance, [4, 3, 3, 3] becomes [3, 3, 3, 4]. This makes it easy to parse step-by-step.
Mod operator to check if it should be 4 or 3. Use two arrays just to make it easier (can be done with one)
const groupIt = arr => arr.reduce(({
group,
out
}, v, i) => {
var max = out.length % 4 === 0 ? 4 : 3
group.push(v)
if (group.length === max || i === arr.length - 1) {
out.push(group)
group = []
}
return {
group,
out
}
}, {
group: [],
out: []
}).out
console.log(groupIt([1, 2, 3, 4, 5, 6, 7, 8]))
var test = (new Array(30)).fill(0).map((x,i) => i + 1)
console.log(groupIt(test))
with just one:
const groupIt = arr => arr.reduce((out, v, i) => {
var max = (out.length - 1) % 4 === 0 ? 4 : 3
out[out.length - 1].push(v)
if (out[out.length - 1].length === max) {
out.push([])
}
return out
}, [[]])
console.log(groupIt([1, 2, 3, 4, 5, 6, 7, 8]))
var test = (new Array(30)).fill(0).map((x, i) => i + 1)
console.log(groupIt(test))
This answer is similar to that of Nina Scholz, but uses a for loop, which I personally find more clear.
const arr = Array.from({length: 100}, (_, i) => i + 1);
const sizes = [4, 3, 3, 3];
const result = [];
for (let i = 0, j = 0; i < arr.length; i += sizes[j], j = (j + 1) % sizes.length) {
result.push(arr.slice(i, i + sizes[j]));
}
console.log(result);
I'm trying to find the smallest common multiple from one array by comparing it to the values in the other by using % num === 0. The answer should be 6, but because each number in the first array is true at least once, they all get returned. How do I find the value that is only true and never false?
let arr = [1, 2, 3]
function test(arr) {
let x = [1, 2, 3, 4, 5, 6, 7, 8]
for (let n of arr) {
return x.filter(k => k % n === 0)
}
}
console.log(test(arr))
You need loop over the x array and return the first element that gets divided by every value in arr.
let arr = [1, 2, 3];
function test(arr) {
let x = [1, 2, 3, 4, 5, 6, 7, 8];
for (let n of x) {
if (arr.every((a) => n % a === 0)) {
return n;
}
}
}
console.log(test(arr));
You can also simply the solution using Array.prototype.find.
const
arr = [1, 2, 3],
test = (arr) =>
[1, 2, 3, 4, 5, 6, 7, 8].find((n) => arr.every((a) => n % a === 0));
console.log(test(arr));
Note: If x is not sorted, then you will have to sort it first.
const arr = [1, 2, 3],
test = (arr) =>
[8, 7, 6, 5, 4, 3, 2, 1]
.sort((a, b) => a - b)
.find((n) => arr.every((a) => n % a === 0));
console.log(test(arr));
Update based on OP's comment
You can use Array.prototype.filter and Array.prototype.some.
const arr = [1, 2, 3],
test = (arr) =>
[1, 2, 3, 4, 5, 6, 7, 8].filter((n) => arr.some((a) => n / a === 2));
console.log(test(arr));
If x is sorted can use find and every
let arr = [1, 2, 3]
let x = [1, 2, 3, 4, 5, 6, 7, 8, 12]
let res = x.find(n => arr.every(a => n%a === 0))
console.log(res)
if unsorted x
let arr = [1, 2, 3]
let x = [1, 12, 6, 4, 2, 7, 8]
let res = [...x].sort((a,b)=> a-b).find(n => arr.every(a => n%a === 0))
console.log(res)
Filter and intersect
let arr = [1, 2, 3]
let x = [1, 2, 3, 4, 5, 6, 7, 8]
function test(arr) {
let common = []
arr.forEach(n => common.push(x.filter(k => k % n === 0)))
return common.reduce((acc, cur) => acc.filter(e => cur.includes(e)));
}
console.log(test(arr))
Late to the party answer
You could solved this with one line of code using Array.reduce. Here we initialize it with an empty array and then use a ternary operator to append values that pass the remainder test. Note that we don't need to check for zero (c % n === 0) because we treat the result as a boolean.
x.reduce((p,c) => c % n ? p : p.concat([c]), [])
// TEST
const x = [1, 2, 3, 4, 5, 6, 7, 8];
[1, 2, 3].forEach(n => {
console.log(
"n =" + n,
x.reduce((p,c) => c % n ? p : p.concat([c]), [])
);
});
My input is an array like so:
[7, 7, 7, 7, 4, 4, 5, 5, 5, 1, 9, 2, 7, 7]
I want to group together the numbers and add them, but by neighbors, not by total in the array. So the output would be:
['7:4', '4:2', '5:3', 1, 9, 2, '7:2']
I've attempted a few different methods using reduce, and gotten close but using the built-in Javascript methods I end up counting ALL in the array, not by neighbors.
const firstArray = [7, 7, 7, 7, 4, 4, 5, 5, 5, 1, 9, 2, 7, 7];
const masterArray = [];
const unique = new Set (numberArray); // Set {7, 4, 5, 1, 9, 2, 7}
unique.forEach(u => {
masterArray.push(numberArray.filter(e => e === u));
});
console.log(masterArray);
Set is obviously wrong to use here because that gets the unique values and counts them, but I want to do it by neighbor only. So then I think I should be using a reduce but I run into the same problem.
var test = [7, 7, 7, 7, 4, 4, 5, 5, 5, 1, 9, 2, 7, 7];
console.log(
test.reduce((acc, element) => {
if (acc.length && acc[acc.length - 1].key == element) {
acc[acc.length - 1].count++;
} else {
acc.push({ key: element, count: 1 });
}
return acc;
}, []).map(element => `${element.key}:${element.count}`)
);
So the logic first reduces the number to an array, tracking the last key and count. So long as the key is the same, the count is incremented. Once the key changes, a new object is pushed to start the next run. After the reduce is done, a map is performed to convert the key and counts into the desired strings.
You'll need to keep track of the last number iterated over in a persistent variable, as well as the number of occurrences of the last number which gets incremented:
const arr = [7, 7, 7, 7, 4, 4, 5, 5, 5, 1, 9, 2, 7, 7];
let lastNum = arr[0];
let count = 0;
const results = [];
const doPush = () => {
results.push(
count === 1
? lastNum
: `${lastNum}:${count}`
);
};
for (const num of arr) {
if (num !== lastNum) {
doPush();
lastNum = num;
count = 1;
} else count++;
}
doPush();
console.log(results);
You could reduce the array by having a look to the last value a the index before the actual value.
const
array = [7, 7, 7, 7, 4, 4, 5, 5, 5, 1, 9, 2, 7, 7],
result = array.reduce((r, value, i, { [i - 1]: last }) => {
let count = 0;
if (last === value) count = (+r.pop().toString().split(':')[1] || 0) + 1;
return [...r, count ? `${value}:${count}`: value];
}, []);
console.log(result);
Here is a solution that uses a recursive function to group the neighbors, and then Array.prototype.map() to format with a colon:
const arr = [7, 7, 7, 7, 4, 4, 5, 5, 5, 1, 9, 2, 7, 7],
output = (function groupNeighbors([first, ...rest], output = [], last) {
last === first ? output[output.length - 1].push(first) : output.push([first]);
return rest.length ? groupNeighbors(rest, output, first) : output;
})(arr).map(({ [0]: num, length }) => length > 1 ? [num, length].join(':') : num);
console.log(output);
As it uses recursion, it is limited in terms of input array length, you can find the stack size limits per browser here: Browser Javascript Stack size limit
Variant with Array.prototype.reduce() (slightly shortest, no recursion, unlimited input length):
const arr = [7, 7, 7, 7, 4, 4, 5, 5, 5, 1, 9, 2, 7, 7],
output = arr
.reduce(
(acc, cur, i, { [i - 1]: last }) =>
(cur === last ? acc[acc.length - 1].push(cur) : acc.push([cur])) && acc,
[]
)
.map(({ [0]: num, length }) => length > 1 ? [num, length].join(':') : num);
console.log(output);
Yes, the proper choice here is reduce:
const countDuplicatesQuantity = (arr) => {
const duplicatesObj = arr.reduce((duplicates, item) => {
duplicates[item] = duplicates[item] ? duplicates[item] + 1 : 1;
return duplicates;
}, {});
return Object.keys(duplicatesObj).map(
(key) => `${key}:${duplicatesObj[key]}`,
);
};
You wanted reduce, how about twice? :) (I don't know if I've done something stupid here.)
First reduce finds where the values change in the array, second uses that to build new array.
const firstArray = [7, 7, 7, 7, 4, 4, 5, 5, 5, 1, 9, 2, 7, 7];
const masterArray = firstArray
.reduce((acc, v, i, all) => all[i + 1] !== v ? acc.concat(i) : acc, [])
.reduce(
(acc, v, i, all) => {
const range = v - (i === 0 ? -1 : all[i - 1]);
return acc.concat(range > 1 ? firstArray[v] + ':' + range : firstArray[v]);
}, []
);
console.log(masterArray);
Using array.reduce:
const firstArray = [7, 7, 7, 7, 4, 4, 5, 5, 5, 1, 9, 2, 7, 7];
var newArray = [];
var count = 0;
firstArray.reduce((acc, cur, index) => {
if (acc == cur) {
count++;
}
if (acc != cur || index+1 == firstArray.length) {
newArray.push(acc + ":" + count);
count=1
}
return cur;
})
console.log(newArray);
my way ... I felt that there could be "simpler"
const firstArray = [7, 7, 7, 7, 4, 4, 5, 5, 5, 1, 9, 2, 7, 7];
const result = firstArray.reduceRight((r,v)=>
{
let [n,c] = !r.length ? [-v,0] : isNaN(r[0]) ? r[0].split(':') : [r[0],1]
if (n==v) r[0] = `${n}:${++c}`
else r.unshift(v)
return r
},[])
console.log( JSON.stringify(result) )
// [ "7:4", "4:2", "5:3", 1, 9, 2, "7:2" ]
.as-console-wrapper { max-height: 100% !important; top: 0; }
I have an array like so: [1, 2, 3, 4, 5, 6, 7, 9, 10]. I need to chunk it into different size chunks, yet with a simple pattern of: 4, 3, 3, 3, 4, 3, 3, 3 like so:
[
[ // four
1,
2,
3,
4
],
[ // three (1/3)
5,
6,
7
],
[ // three (2/3)
8,
9,
10
],
[ // three (3/3)
11,
12,
13
],
[ // four
14,
15,
16,
17
],
[ // three (1/3)
18,
19,
20
], // and so on..
]
I have tried with this code I have customized:
const arr; // my array of values
const chuncked = arr.reduce((acc, product, i) => {
if (i % 3) {
return acc;
} else if (!didFourWayReduce) {
didFourWayReduce = true;
fourWayReduces++;
if ((fourWayReduces - 1) % 2) { // only make every second a "4 row"
return [...acc, arr.slice(i, i + 3)];
} else {
return [...acc, arr.slice(i, i + 4)];
}
} else {
didFourWayReduce = false;
return [...acc, arr.slice(i, i + 3)];
}
}, []);
And it works, almost, expect that the first chunk of threes (1/3) have the last element of the chunk with 4. So 1 key is repeated every first chunk of three. Like so:
[
[
1,
2,
3,
4
],
[
4, // this one is repeated, and it shouldn't be
5,
6
]
]
You could take two indices, one for the data array and one for sizes. Then slice the array with a given length and push the chunk to the chunks array.
Proceed until end of data.
var data = Array.from({ length: 26 }, (_, i) => i + 1),
sizes = [4, 3, 3, 3],
i = 0,
j = 0,
chunks = [];
while (i < data.length) chunks.push(data.slice(i, i += sizes[j++ % sizes.length]));
console.log(chunks);
.as-console-wrapper { max-height: 100% !important; top: 0; }
const arr = Array.from({ length: 100 }, (_, i) => i);
const copy = [...arr];
const sizes = [4, 3, 3, 3];
const result = [];
let i = 0;
while (i <= arr.length && copy.length) {
result.push(copy.splice(0, sizes[i % sizes.length]));
i++;
}
console.log(result);
A recursive approach is fairly elegant:
const chunks = (xs, [s, ...ss]) =>
xs.length ? [xs .slice (0, s), ... chunks (xs .slice (s), [...ss, s])] : []
const data = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
const sizes = [4, 3, 3, 3]
console .log (chunks (data, sizes))
.as-console-wrapper { max-height: 100% !important; top: 0; }
By replacing [s, ...ss] with [...ss, s], we pass a cycled version of the sizes array, so that for instance, [4, 3, 3, 3] becomes [3, 3, 3, 4]. This makes it easy to parse step-by-step.
Mod operator to check if it should be 4 or 3. Use two arrays just to make it easier (can be done with one)
const groupIt = arr => arr.reduce(({
group,
out
}, v, i) => {
var max = out.length % 4 === 0 ? 4 : 3
group.push(v)
if (group.length === max || i === arr.length - 1) {
out.push(group)
group = []
}
return {
group,
out
}
}, {
group: [],
out: []
}).out
console.log(groupIt([1, 2, 3, 4, 5, 6, 7, 8]))
var test = (new Array(30)).fill(0).map((x,i) => i + 1)
console.log(groupIt(test))
with just one:
const groupIt = arr => arr.reduce((out, v, i) => {
var max = (out.length - 1) % 4 === 0 ? 4 : 3
out[out.length - 1].push(v)
if (out[out.length - 1].length === max) {
out.push([])
}
return out
}, [[]])
console.log(groupIt([1, 2, 3, 4, 5, 6, 7, 8]))
var test = (new Array(30)).fill(0).map((x, i) => i + 1)
console.log(groupIt(test))
This answer is similar to that of Nina Scholz, but uses a for loop, which I personally find more clear.
const arr = Array.from({length: 100}, (_, i) => i + 1);
const sizes = [4, 3, 3, 3];
const result = [];
for (let i = 0, j = 0; i < arr.length; i += sizes[j], j = (j + 1) % sizes.length) {
result.push(arr.slice(i, i + sizes[j]));
}
console.log(result);
I'm stuck with this problem for 3 days now... Someone please help me.
Challenge 5
Construct a function intersection that compares input arrays and returns a new array with elements found in all of the inputs.
function intersection(arrayOfArrays) {
}
console.log(intersection([[5, 10, 15, 20], [15, 88, 1, 5, 7], [1, 10, 15, 5, 20]]));
// should log: [5, 15]
Reduce the arrays to a Map of counts, with the value as key. Spread the Map to entries. Use Array.filter() on the Map's entries to remove all entries, which value is not equal to the arrayOfArrays lenth. Extract the original number from the entries using Array.map():
function intersection(arrayOfArrays) {
return [...arrayOfArrays.reduce((r, s) => {
s.forEach((n) => r.set(n, (r.get(n) || 0) + 1));
return r;
}, new Map())]
.filter(([k, v]) => v === arrayOfArrays.length)
.map(([k]) => k);
}
console.log(intersection([[5, 10, 15, 20], [15, 88, 1, 5, 7], [1, 10, 15, 5, 20]]));
You could reduce the array by filtering with just checking if the other array contains the value.
This works for arrays with unique values.
Array#reduce:
If no initialValue is provided, then accumulator will be equal to the first value in the array, and currentValue will be equal to the second.
The callback
a.filter(v => b.includes(v))
filters array a. If the array b includes the value of a, then this value v is included in the accumulator for the next iteration or as final result.
accumulator currentValue new accumulator
a b result
-------------------- -------------------- --------------------
[ 5, 10, 15, 20] [15, 88, 1, 5, 7] [ 5, 15]
[ 5, 15] [ 1, 10, 15, 5, 20] [ 5, 15]
function intersection(arrayOfArrays) {
return arrayOfArrays.reduce((a, b) => a.filter(v => b.includes(v)));
}
console.log(intersection([[5, 10, 15, 20], [15, 88, 1, 5, 7], [1, 10, 15, 5, 20]]));
First try to find out the intersection of two arrays which is the base problem. Then try to build up for variable number of arrays passed as arguments for intersection. You can use reduce() for doing that.
function intersectionOfTwoArrays(arr1, arr2)
{
return arr1.filter(x => arr2.some(y => y === x));
}
function intersection(...arrayOfArrays)
{
return arrayOfArrays
.reduce((a, b) => intersectionOfTwoArrays(a, b));
}
intersection(
[5, 10, 15, 20],
[15, 88, 1, 5, 7],
[1, 10, 15, 5, 20]
);
You can go through the first array in the array of arrays and check which of its value is present in all the other arrays.
Here is an example:
function intersection(input) {
let firstArray = input[0];
let restOfArrays = input.splice(1);
return firstArray.filter(v => restOfArrays.every(arr => arr.includes(v)));
}
const input = [[5, 10, 15, 20], [15, 88, 1, 5, 7], [1, 10, 15, 5, 20]];
const result = intersection(input);
console.log(result);
Works with even if there is duplicate in same array.. like in my example added 5 twice in arrayEle[1];
var arrayEle = [[5, 10, 15, 20], [15, 88, 1, 5, 5], [1, 10, 15, 5, 20]]
var startIndex = 1;
var newArray = [];
for (var x = 0; x < arrayEle[0].length; x++) {
var temVal = 1;
var value;
for (var y = 1; y < arrayEle.length; y++) {
for (var z = 0; z < arrayEle[y].length; z++) {
if (arrayEle[y][z] == arrayEle[0][x]) {
temVal++;
value = arrayEle[y][z];
break;
}
}
}
if (temVal == arrayEle.length) {
newArray.push(value);
console.log(value);
}
}
console.log(newArray);
//log: [5, 15]
I think you want the common elements. Let me show you how:
var Array1 = [5, 10, 15, 20]
var Array2 = [15, 88, 1, 5, 7]
var Array3 = [1, 10, 15, 5, 20]
var found = []
var Final = []
var c = 1;e = 1;
for (i = 1;i<=Array1.length;i++){
for (k = 1;k<=Array2.length;i++){
if (Array1[i] == Array2[k]){
Found[c] = Array[i];
c++;
}
}
}
for (n = 1;n <= Found.length ; n++){
for (m = 1;m <= Array3.length ; n++){
if (Found[n] == Array3[m]){
Final[e] = Found[n]
e++;
}
}
}
//the Array Final Contains 5 , 15