I'm trying to duplicate each element in an array, but using functional style.
I have this currently:
["a", "b", "c"]
And I'm getting this:
["a","a","b","b","c","c"]
So far I have tried the following, mapping each element to an array, then using flat() to get a 1d array. Is there a cleaner way because it feels like I'm abusing map and flat.
["a", "b", "c"].map(item => [item, item]).flat();
Is there a better way to do this?
I was trying to provide a example as simple as possible but left some details out. The real input is not sorted because elements are not comparable.
It's something like:
[
{
a:"a"
b:"b"
},
{
c: 1
d: 2
},
{
apple: {},
sellers: ["me", "her"]
}
]
The duplicated result should be something like this, where duplicated elements are next to each other:
[
{
a:"a"
b:"b"
},
{
a:"a"
b:"b"
},
{
c: 1
d: 2
},
{
c: 1
d: 2
},
{
apple: {},
sellers: ["me", "her"]
},
{
apple: {},
sellers: ["me", "her"]
}
]
Array.reduce is semantically the appropriate method here: take an object (in this case an array) and return an object of a different type, or with a different length or shape (note: edited to use Array.push for faster performance per #slider suggestion):
EDIT: I've edited my answer to reflect OP's updated input data. Note also, that this solution is cross-browser and NodeJS compatible without requiring transpilation.
let data = [
{
a:"a",
b:"b",
},
{
c: 1,
d: 2
},
{
apple: {},
sellers: ["me", "her"]
}
];
let result = data
.reduce((acc, el) => {
acc.push(el, el);
return acc;
}, []);
console.log(JSON.stringify(result, null, 2));
Otherwise you could map each element, duplicating it, then combine them:
let data = [
{
a:"a",
b:"b",
},
{
c: 1,
d: 2
},
{
apple: {},
sellers: ["me", "her"]
}
];
let result = data.map(item => [item, item]).reduce((acc, arr) => acc.concat(arr));
console.log(JSON.stringify(result, null, 2));
As mentioned in other answers here, either of these approaches have the advantage of not requiring the original array to have been sorted.
You can use the function reduce and concatenate the same object on each iteration.
let array = ["a", "b", "c"],
result = array.reduce((a, c) => a.concat(c, c), []);
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }
I would recommend Array.prototype.flatMap -
const twice = x =>
[ x, x ]
console .log
( [ 'a', 'b', 'c' ] .flatMap (twice) // [ 'a', 'a', 'b', 'b', 'c', 'c' ]
, [ 1, 2, 3, 4, 5 ] .flatMap (twice) // [ 1, 1, 2, 2, 3, 3, 4, 4, 5, 5 ]
)
flatMap is useful for all kinds of things -
const tree =
[ 0, [ 1 ], [ 2, [ 3 ], [ 4, [ 5 ] ] ] ]
const all = ([ value, ...children ]) =>
[ value ] .concat (children .flatMap (all))
console .log (all (tree))
// [ 0, 1, 2, 3, 4, 5 ]
really cool things -
const ranks =
[ 'J', 'Q', 'K', 'A' ]
const suits =
[ '♡', '♢', '♤', '♧' ]
console .log
( ranks .flatMap (r =>
suits .flatMap (s =>
[ [ r, s ] ]
)
)
)
// [ ['J','♡'], ['J','♢'], ['J','♤'], ['J','♧']
// , ['Q','♡'], ['Q','♢'], ['Q','♤'], ['Q','♧']
// , ['K','♡'], ['K','♢'], ['K','♤'], ['K','♧']
// , ['A','♡'], ['A','♢'], ['A','♤'], ['A','♧']
// ]
flatMap is just a specialised Array.prototype.reduce and is easy to implement in environments where Array.prototype.flatMap is not already supported -
const identity = x =>
x
const flatMap = (xs = [], f = identity) =>
xs .reduce ((r, x) => r . concat (f (x)), [])
const ranks =
[ 'J', 'Q', 'K', 'A' ]
const suits =
[ '♡', '♢', '♤', '♧' ]
console.log
( flatMap (ranks, r =>
flatMap (suits, s =>
[ [ r, s ] ]
)
)
)
// [ ['J','♡'], ['J','♢'], ['J','♤'], ['J','♧']
// , ['Q','♡'], ['Q','♢'], ['Q','♤'], ['Q','♧']
// , ['K','♡'], ['K','♢'], ['K','♤'], ['K','♧']
// , ['A','♡'], ['A','♢'], ['A','♤'], ['A','♧']
// ]
You could just do this:
var arr = ["a", "b", "c"];
arr = arr.concat(arr).sort();
This is one of the simplest methods to do what you are asking to do.
The simplest solution is to use flatMap():
const source = ["a", "b", "c"];
const result = source.flatMap(item => [item, item]);
[ 'a', 'a', 'b', 'b', 'c', 'c' ]
A little bit of classic:
let source = ["a", "b", "c"];
const originalLength = source.length;
for(let i = 0; i <= originalLength + (originalLength - 2); i++) {
source.splice(i, 0, source[i++]);
}
[ 'a', 'a', 'b', 'b', 'c', 'c' ]
Related
I have the following data array:
const data = [
{
value: [
'a',
'b',
'a',
'a'
]
},
{
value: [
'c',
'c',
'd',
'c'
]
}
];
So there's is 4 combination here based on index:
combination 1 : a - c (index 0 in each value arrays)
combination 2 : b - c (index 1 in each value arrays)
combination 3 : a - d (index 2 in each value arrays)
combination 4 : a - c (index 3 in each value arrays)
As you can see the first and the last combinations are the same, so i want to remove the second occurrence from each array, the result should be:
[
{
value: [
'a',
'b',
'a'
]
},
{
value: [
'c',
'c',
'd'
]
}
]
You can zip the values arrays from both objects to form an array which looks like:
["a-c", "b-c", ...]
As these are now strings, you can turn this array into a Set using new Set(), which will remove all duplicate occurrences. You can then turn this set back into an array which you can then use .reduce() on to build you array of objects from. For each value you can obtain the list of values by using .split() on the '-', and from that, populate your reduced array.
See example below:
const data = [{ value: [ 'a', 'b', 'a', 'a' ] }, { value: [ 'c', 'c', 'd', 'c' ] } ];
const unq = [...new Set(
data[0].value.map((_,c)=> data.map(({value})=>value[c]).join('-'))
)];
const res = unq.reduce((acc, str) => {
const values = str.split('-');
values.forEach((value, i) => acc[i].value.push(value));
return acc;
}, Array.from({length: data.length}, _ => ({value: []})))
console.log(res);
Limitations of the above method assume that you won't have a - character as your string value. If this is an issue, you can consider using a different delimiter, or find unique values within your array using .filter() instead of a Set.
You could save a lookup object for unique pairs of value based with index
Given your input is, below solution could help you
const data = [
{
value: ["a", "b", "a", "a"],
},
{
value: ["c", "c", "d", "c"],
},
]
const lookup = {}
data[0].value.forEach((_, index) => {
lookup[`${data[0].value[index]}-${data[1].value[index]}`] = true
})
const res = Object.keys(lookup).reduce(
(acc, key) => {
const [val1, val2] = key.split("-")
acc[0].value.push(val1)
acc[1].value.push(val2)
return acc
},
[{ value: [] }, { value: [] }]
)
console.log(res)
Below is a two step solution with a generator function and a single pass.
const data = [ { value: [ 'a', 'b', 'a', 'a' ] }, { value: [ 'c', 'c', 'd', 'c', ] } ];
const zipDataValues = function* (data) {
const iterators = data.map(item => item.value[Symbol.iterator]())
let iterations = iterators.map(iter => iter.next())
while (iterations.some(iteration => !iteration.done)) {
yield iterations.map(iteration => iteration.value)
iterations = iterators.map(iter => iter.next())
}
}
const filterOutDuplicateCombos = function (values) {
const combosSet = new Set(),
resultData = [{ value: [] }, { value: [] }]
for (const [valueA, valueB] of values) {
const setKey = [valueA, valueB].join('')
if (combosSet.has(setKey)) {
continue
}
combosSet.add(setKey)
resultData[0].value.push(valueA)
resultData[1].value.push(valueB)
}
return resultData
}
console.log(
filterOutDuplicateCombos(zipDataValues(data))
) // [ { value: [ 'a', 'b', 'a' ] }, { value: [ 'c', 'c', 'd' ] } ]
Here is a reference on generators and iterators
Filter combinations + sorting by the first occurrence:
const data = [{
value: ['a', 'b', 'a', 'a']
},{
value: ['c', 'c', 'd', 'c']
}];
var res = {}, i, t;
for (i = 0; i < data[0].value.length; ++i) {
res[data[0].value[i]] = res[data[0].value[i]] || {};
res[data[0].value[i]][data[1].value[i]] = true;
}
data[0].value = [];
data[1].value = [];
for (i in res) {
for (t in res[i]) {
data[0].value[data[0].value.length] = i;
data[1].value[data[1].value.length] = t;
}
}
console.log(data);
I have an array that looks like:
[
["A","B","C","D"],
["E","F","G","H"],
[6,43,2,4]
]
I want to sort this array in descending order of the items in the third row, such that it looks like:
[
["B","A","D","C"],
["F","E","H","G"],
[43,6,4,2]
]
So far this is what I wrote:
var arr = [
["A","B","C","D"],
["E","F","G","H"],
[6,43,2,4]
]
arr = arr.sort((a,b) => {
return b[2] - a[2]
});
console.log(arr);
But the array does not get sorted. Where did I go wrong?
You could take the indices of the array for sorting, sort this array by the values of the original array and map all reordered items.
var array = [["A", "B", "C", "D"], ["E", "F", "G", "H"], [6, 43, 2, 4]],
indices = [...array[2].keys()].sort((a, b) => array[2][b] - array[2][a]);
array = array.map(a => indices.map(i => a[i]));
console.log(array.map(a => a.join(' ')));
You want to sort columns. It is easier to sort rows. So, you could transpose then sort then retranspose:
const arr = [
["A","B","C","D"],
["E","F","G","H"],
[6,43,2,4]
];
const transpose = array => array[0].map((col, i) => array.map(row => row[i]));
const t = transpose(arr);
t.sort((rowi,rowj) => (rowj[rowj.length - 1] - rowi[rowi.length-1]));
console.log(transpose(t));
Which has the intended output of:
[ [ 'B', 'A', 'D', 'C' ],
[ 'F', 'E', 'H', 'G' ],
[ 43, 6, 4, 2 ] ]
This is clearly not as efficient as the other answers, but the fact that you are asking the question suggests that your current way of storing the data isn't convenient for your purposes. Perhaps you might want to transpose then skip the retranspose part.
You need to first sort the number store indexes and then sort the other arrays.
var arr = [
["A","B","C","D"],
["E","F","G","H"],
[6,43,2,4]
]
var tmp = arr[2].map((value, i) => ({value, i})).sort((a, b) => b.value - a.value);
arr = arr.map(function(arr) {
return tmp.map(({value, i}) => arr[i]);
});
console.log(arr);
It would be easier if you could depict the data differently. The Array.sort function will not help you with your current example.
var arr = [{
a: "A", b: "E", c: 6
},{
a: "B", b: "F", c: 43
},{
a: "C", b: "G", c: 2
},{
a: "D", b: "H", c: 4
}]
arr.sort((a, b) => {
return b.c - a.c;
})
console.log(arr);
If you can't, I could provide a different answer; but the answer would be very inefficient.
I want to create an object with the given structure:
const questions = [
{
id: '1',
instruction: 'Question1',
options: {
'a': 'SomeText1',
'b': 'SomeText2',
'c': 'SomeText3'
},
correct: ['c']
},
{
id: '2',
instruction: 'Question2',
options: {
'a': 'SomeText1',
'b': 'SomeText2',
'c': 'SomeText3',
'd': 'SomeText4'
},
correct: ['a']
},
{
id: '3',
instruction: 'Question3,
options: {
'a': 'SomeText1',
'b': 'SomeText2'
},
correct: ['b']
}
]
I have an arrays containing necessary information to fill this object and create it using .map() .
const questions =
[ 'Question1',
'Question2',
'Question3'
]
const answers =
[
'Answer1',
'Answer2',
'Answer3'
]
const options = [
[ 'Option1', 'Option2', 'Option3' ],
[ 'Option4', 'Option5', 'Option6' ],
[ 'Option7', 'Option8', 'Option9' ]
]
function toJson()
const alphabet = ['a', 'b', 'c', 'd', 'e'];
const json = questions.map((question, index) => (
{
id: index + 1,
instruction: question,
options: Object.assign({}, options[index]),
correct: answers[index]
}
))
}
I have only problem with options key. As You see I want to have a letters as keys, depending on how many answers question has.
This function gives me numbers as keys when I use Object.assign(), and I don't know how to replace them with letters from alphabet array.
EDIT:
So the solution for the options key in desired object is:
options: Object.assign(
{},
...options[index].map((a, i) => ({ [alphabet[i]]: a }))
),
Now I'm able to create an object with consecutive alphabet letters with assigned answer.
options[index] returns an array. It contains values by index. By passing it to Object.assign, you add all values by their array index as a string: "0", "1", etc.
If we map the array in to a list of { "a": option } first, and spread the result in to the Object.assign call, we can change those indexes to the letters you want:
const questions =
[ 'Question1',
'Question2',
'Question3'
]
const answers =
[
'Answer1',
'Answer2',
'Answer3'
]
const options = [
[ 'Option1', 'Option2', 'Option3' ],
[ 'Option4', 'Option5', 'Option6' ],
[ 'Option7', 'Option8', 'Option9' ]
]
function toJson() {
const alphabet = ['a', 'b', 'c', 'd', 'e'];
const json = questions.map((question, index) => (
{
id: index + 1,
instruction: question,
options: Object.assign(
{},
...options[index].map((a, i) => ({ [alphabet[i]]: a }))
),
correct: answers[index]
}
));
return json;
}
console.log(toJson());
I suggest using something like zip and objectFromPairs (both snippets from 30secondsofcode, a project/website I am a maintainer of). From the website:
zip
Creates an array of elements, grouped based on the position in the original arrays.
Use Math.max.apply() to get the longest array in the arguments. Creates an array with that length as return value and use Array.from() with a map-function to create an array of grouped elements. If lengths of the argument-arrays vary, undefined is used where no value could be found.
objectFromPairs
Creates an object from the given key-value pairs.
Use Array.reduce() to create and combine key-value pairs.
The only extra step I took was to trim each zipped array to the length of options[index].
const questions = ['Question1',
'Question2',
'Question3'
]
const answers = [
'Answer1',
'Answer2',
'Answer3'
]
const options = [
['Option1', 'Option2', 'Option3'],
['Option4', 'Option5', 'Option6'],
['Option7', 'Option8', 'Option9']
]
const zip = (...arrays) => {
const maxLength = Math.max(...arrays.map(x => x.length));
return Array.from({
length: maxLength
}).map((_, i) => {
return Array.from({
length: arrays.length
}, (_, k) => arrays[k][i]);
});
};
const objectFromPairs = arr => arr.reduce((a, [key, val]) => ((a[key] = val), a), {});
function toJson() {
const alphabet = ['a', 'b', 'c', 'd', 'e'];
const json = questions.map((question, index) => ({
id: index + 1,
instruction: question,
options: objectFromPairs(zip(alphabet, options[index]).slice(0, options[index].length)),
correct: answers[index]
}))
console.log(json);
}
toJson();
the below should work (i believe)
options: alphabet.reduce((acc, letter, i) => {
let option = options[index][i] || 'DefaultText' + i;
acc[letter] = option;
return acc;
}, {})
Edit: Corrected typos
EDIT:
So the solution for the options key in desired object is:
options: Object.assign(
{},
...options[index].map((a, i) => ({ [alphabet[i]]: a }))
),
I'm new to node js/express. I'm having the problem cause I need to insert bulk in MySQL. I use body-parser, but to simplify my code this is the analogy.
I have two objects from req.body:
Numbers = { 1, 2, 3 }
Letters = { a, b, c }
Then, I need it to be like this,
Object = [ { '1', 'a' }, { '2', 'b' }, { '3', 'c' } ]
What can I use to do this?
const Numbers = [1, 2, 3]
const Letters = ['a', 'b', 'c']
const result = []
Numbers.forEach((el, i) => {
result.push({[el]: Letters[i]})
})
console.log(result)
or
const Numbers = [1, 2, 3]
const Letters = ['a', 'b', 'c']
const result = Numbers.map((el, i) => ({[el]: Letters[i]}))
console.log(result)
This question already has answers here:
Merge/flatten an array of arrays
(84 answers)
Closed 6 years ago.
I have array like this
var arr = [
[ {"c": 1},{"d": 2} ],
[ {"c": 2},{"d": 3} ]
]
I want to change into (by javascript only)
var newArr = [{"c": 1},{"d": 2},{"c": 2},{"d": 3}]
I have tried this by map but not getting expected output .
You can use reduce() and spread syntax.
var arr = [
[ {"c": 1},{"d": 2} ],
[ {"c": 2},{"d": 3} ]
]
var result = arr.reduce((r, e) => (r.push(...e), r), [])
console.log(result)
You could use Array#reduce with Array#concat.
var arr = [[{ c: 1 }, { d: 2 }], [{ c: 2 }, { d: 3 }]],
result = arr.reduce((r, a) => r.concat(a), []);
console.log(result)
ES5
var arr = [[{ c: 1 }, { d: 2 }], [{ c: 2 }, { d: 3 }]],
result = arr.reduce(function (r, a) { return r.concat(a); }, []);
console.log(result)
In plain JavaScript you can do this using Array.portotype.forEach.
var arr = [
[ {"c": 1},{"d": 2} ],
[ {"c": 2},{"d": 3} ]
];
var newArray = [];
arr.forEach(function(e){
e.forEach(function(e1){
newArray.push(e1);
});
})
console.log(newArray);
Use this..