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How to split an array into multple arrays each with a unique name

I have an array = [A,1,0,1,0,1,B,1,0,0,1,A,1]
I need to split this array into multiple arrays. The split will occur at the "A" or "B" position as seen in the new arrays below. The names of the new arrays use the string "group" plus an incremented number starting with 1 or 0.
The end result should look like:
group1 = [A,1,0,1,0,1]
group2 = [B,1,0,0,1]
group3 = [A,1]
I can get the section of the array I need by creating an array (arrTemp), so I can store the positions (indexes) and later use slice() to get the sections I want (A,1,0,1,0,1), (A,1,0,0,1), and (A,1). But I don't know how to store the results of my slice()'s in arrays with unique names incremented by 1.
This is what I have tried so far:
var arr = [A,1,0,1,0,1,B,1,0,0,1,A,1];
arr.forEach(myFunction)
function myFunction(item, index) {
if ((item=="A") || (item=="B")) {
arrTemp.push(index);
arrTemp=arrTemp; //not sure I need this. I did this so it array would be global
}
}
for (var i = 0; i < arr.length; i++){
sectArray = arr.slice(arrTemp[i]+1,arrTemp[i + 1])
'group' + [i] = [arrTemp[i],sectArray]; //here is my problem.
}

It seems like you're trying to dynamically create variables. That seems tricky and probably won't work. What you should probably have is some collection of results. Probably a parent array that holds all of them.
For example:
var containerArray = [];
Then:
for (var i = 0; i < arr.length; i++){
sectArray = arr.slice(arrTemp[i]+1,arrTemp[i + 1])
containerArray[i] = [arrTemp[i],sectArray];
}
Now containerArray will have all of your stuff. You can also do this with an object:
var containerObject = {};
And the same thing after.

you only need one loop here, keep an empty temp array, iterate over arr and keep pushing elements in temp each time you see 'A' or 'B' push temp to final array, and at last push temp once more into final array because last section will be left.
var arr = ['A',1,0,1,0,1,'B',1,0,0,1,'A',1];
var temp = [];
var sectArray = [];
arr.forEach(myFunction)
function myFunction(item, index) {
if (((item=="A") || (item=="B")) && temp.length) {
sectArray.push(temp);
temp = [item];
}else{
temp.push(item);
}
}
sectArray.push(temp);
console.log(sectArray);

Check this solution that use a combination of string and array methods:
var data = ['A',1,0,1,0,1,'B',1,0,0,1,'A',1];
var results = data.toString().split(/(?=[a-zA-Z]+)/)
.map(function(value){
return value.split(',').filter(function (item) {
return item.length ? true: false;
})
})
.map(function(item) {
return item.map(function (value) {
return isNaN(parseInt(value)) ? value : parseInt(value);
})
});
console.log(results);
// results = [["A", 1, 0, 1, 0, 1], ["B", 1, 0, 0, 1], ["A", 1]]

Another solution using Array#reduce function.
var x = ["A", 1, 0, 1, 0, 1, "B", 1, 0, 0, 1, "A", 1];
function reformat(arr) {
var smallArrCounter = 0;
return arr.reduce(function (acc, item) {
if (item === "A" || item === "B") {
acc["group" + (++smallArrCounter)] = [item];
} else {
acc["group" + smallArrCounter].push(item);
}
return acc;
}, {});
}
var result = reformat(x);
console.log(result.group1); // ["A", 1, 0, 1, 0, 1]
console.log(result.group2); // ["B", 1, 0, 0, 1]
console.log(result.group3); // ["A", 1]

There may be a more performant approach that doesn't require two iterations of the array, but my thought is:
Determine the indices of the group delimiters (characters)
Slice the array into groups based on those delimiters, using either the next index as the end, or arr.length if slicing the last group
This has the assumption that the array delimiters may not be known in advance.
const charIndices = [];
const groups = [];
const arr = ['A',1,0,1,0,1,'B',1,0,0,1,'A',1];
// get the indices of the characters
arr.forEach((v, i) => ('' + v).match(/[A-Z]+/) ? charIndices.push(i) : undefined);
// use the found indices to split into groups
charIndices.reduce((a, b, i) => {
a.push(arr.slice(b, charIndices[i+1] ? charIndices[i+1]-1 : arr.length));
return a;
}, groups);
console.log(groups);

Split nested array into multiple arrays

I have a string that looks like this:
str = {1|2|3|4|5}{a|b|c|d|e}
I want to split it into multiple arrays. One containing all the first elements in each {}, one containing the second element, etc. Like this:
arr_0 = [1,a]
arr_1 = [2,b]
arr_2 = [3,c]
.....
The best I can come up with is:
var str_array = str.split(/}{/);
for(var i = 0; i < str_array.length; i++){
var str_row = str_array[i];
var str_row_array = str_row.split('|');
arr_0.push(str_row_array[0]);
arr_1.push(str_row_array[1]);
arr_2.push(str_row_array[2]);
arr_3.push(str_row_array[3]);
arr_4.push(str_row_array[4]);
}
Is there a better way to accomplish this?

Try the following:
var zip = function(xs, ys) {
var out = []
for (var i = 0; i < xs.length; i++) {
out[i] = [xs[i], ys[i]]
}
return out
}
var res = str
.split(/\{|\}/) // ['', '1|2|3|4|5', '', 'a|b|c|d|e', '']
.filter(Boolean) // ['1|2|3|4|5', 'a|b|c|d|e']
.map(function(x){return x.split('|')}) // [['1','2','3','4','5'], ['a','b','c','d','e']]
.reduce(zip)
/*^
[['1','a'],
['2','b'],
['3','c'],
['4','d'],
['5','e']]
*/

Solution
var str = '{1|2|3|4|5}{a|b|c|d|e}'.match(/[^{}]+/g).map(function(a) {
return a.match(/[^|]+/g);
}),
i,
result = {};
for (i = 0; i < str[0].length; i += 1) {
result["arr_" + i] = [+str[0][i], str[1][i]];
}
How it works
The first part, takes the string, and splits it into the two halves. The map will return an array after splitting them after the |. So str is left equal to:
[
[1,2,3,4,5],
['a', 'b', 'c', 'd', 'e']
]
The for loop will iterate over the [1,2,3,4,5] array and make the array with the appropriate values. The array's are stored in a object. The object we are using is called result. If you don't wish for it to be kept in result, read Other
Other
Because you can't make variable names from another variable, feel free to change result to window or maybe even this (I don't know if that'll work) You can also make this an array
Alternate
var str = '{1|2|3|4|5}{a|b|c|d|e}'.match(/[^{}]+/g).map(function(a) { return a.match(/[^|]+/g); }),
result = [];
for (var i = 0; i < str[0].length; i += 1) {
result[i] = [+str[0][i], str[1][i]];
}
This is very similar except will generate an Array containing arrays like the other answers,

Finding nested duplicate arrays in JavaScript. (Nested Array uniq in lodash/underscore)

I am trying to determine if an array of JavaScript arrays contains duplicates. Is this possible? I am first trying to see if I can strip the duplicates out and then do an equality check but I cannot get past the first part. Here is what underscore returns:
var arr1 = [[1,2], [2,3], [1,2]];
var arr2 = _.uniq(arr1);
var arraysAreEqual = _.isEqual(arr1, arr2);
console.log(arraysAreEqual, arr1, arr2);
// true
Jsbin: http://jsbin.com/vogumo/1/edit?js,console
Anyone know of a way to determine if the array contains duplicate arrays?

It's a little sloppy, but (possible)
var arr2 = _.uniq(arr1, function(item) {
return JSON.stringify(item);
});
will give you a correct result

Try This:
var numArray = [1, 7, 3, 0, 9, 7, 8, 6, 2, 3];
var duplicates = [];
var sortednumArray = numArray.sort();
for (var i = 0; i < sortednumArray.length; i++) {
//console.log(sortednumArray[i]);
if (sortednumArray[i] == sortednumArray[i + 1]) {
duplicates.push(sortednumArray[i]);
}
}
if (duplicates.length == 0) {
console.log("Soted Array:");
for(var i = 0; i < sortednumArray.length; i++) {
console.log(sortednumArray[i]);
}
} else {
console.log("Duplicates:");
for(var i = 0; i < duplicates.length; i++){
console.log(duplicates[i]);
}
}
Program pushes all duplicates to an array called 'duplicates' then displays it, but if none are present, it displays the sorted version of numArray

From the underscore.js documentation:
uniq _.uniq(array, [isSorted], [iteratee]) Alias: unique
Produces a
duplicate-free version of the array, using === to test object
equality. If you know in advance that the array is sorted, passing
true for isSorted will run a much faster algorithm. If you want to
compute unique items based on a transformation, pass an iteratee
function.
But arrays can't be strictly compared in JavaScript.
Therefore, you can use a transformation function to enable comparison with uniq. For example:
console.log([1,2] === [1,2]) // false, can't strict compare arrays
console.log([1,2].toString()) // "1,2" - string representation
console.log([1,2].toString() === [1,2].toString()) // true, strings can be compared
var valueToString = function(v) {return v.toString()}; // transform array to string
var arr1 = [[1,2], [2,3], [1,2]];
var arr2 = _.uniq(arr1, false, valueToString); // compare based on transformation
var arraysAreEqual = _.isEqual(arr1, arr2);
console.log("arraysAreEqual:", arraysAreEqual, arr1, arr2);
// false
// [[1, 2], [2, 3], [1, 2]]
// [[1, 2], [2, 3]]
Note that transforming to string is "hacky": you would be better off comparing each value of the array, as discussed in this StackOverflow question.
By using the proposed equals implementation in that question, you would need to implement your own version of uniq that uses equals instead of ===.
The implementation of uniq in Underscore is very straight-forward - it creates a new result array and loops through the given array. If the current value is not already in result, insert it.
console.log("Using array comparison:");
arrayEquals = function (array1, array2) {
// if any array is a falsy value, return
if (!array1 || !array2)
return false;
// compare lengths - can save a lot of time
if (array1.length != array2.length)
return false;
for (var i = 0, l=array1.length; i < l; i++) {
// Check if we have nested arrays
if (array1[i] instanceof Array && array2[i] instanceof Array) {
// recurse into the nested arrays
if (!arrayEquals(array1[i],array2[i]))
return false;
}
else if (array1[i] !== array2[i]) {
return false;
}
}
return true;
};
_.uniqArrays = function(array) {
if (array == null) return [];
var result = [];
for (var i = 0, length = array.length; i < length; i++) {
var value = array[i];
var arrayEqualsToValue = arrayEquals.bind(this, value); // arrayEquals with first argument set to value
var existing = _.find(result, arrayEqualsToValue); // did we already find this?
if (!existing) {
result.push(value);
}
}
return result;
};
var arr3 = _.uniqArrays(arr1);
arraysAreEqual = _.isEqual(arr1, arr3);
console.log("arraysAreEqual:", arraysAreEqual, arr1, arr3); // false
I made a jsbin with all the code, if you want to play around.

In the latest lodash (4.6.1) you could do something like this:
if (_.uniqWith(arr, _.isEqual).length < arr.length) {
// then there were duplicates
}

Hashing array of strings in javascript

Just wondering if there is some other way than this.
var hashStringArray = function(array) {
array.sort();
return array.join('|');
};
I don't like sorting much and using that delimiter is not safe either if it's contained in one of the strings. In overall I need to produce same hash no matter the order of strings. It will be rather short arrays (up to 10 items), but it will be required very often so it shouldn't be too slow.
I intend to use it with ES6 Map object and I need to easily find same array collection.
Updated example of use
var theMap = new Map();
var lookup = function(arr) {
var item = null;
var hashed = hashStringArray(arr);
if (item = theMap.get( hashed )) {
return item;
}
theMap.set( hashed, itemBasedOnInput );
return itemBasedOnInput;
}
var arr1 = ['alpha','beta','gama'];
var arr2 = ['beta','alpha','gama'];
lookup(arr1) === lookup(arr2)
Performance tests
http://jsperf.com/hashing-array-of-strings/5

Two things occurred to me as the basis of a solution:
summing doesn't depend on order, which is actually a flaw in simple checksums (they don't catch changes in block order within a word), and
we can convert strings to summable numbers using their charcodes
Here's a function to do (2) :
charsum = function(s) {
var i, sum = 0;
for (i = 0; i < s.length; i++) {
sum += (s.charCodeAt(i) * (i+1));
}
return sum
}
Here's a version of (1) that computes an array hash by summing the charsum values:
array_hash = function(a) {
var i, sum = 0
for (i = 0; i < a.length; i++) {
var cs = charsum(a[i])
sum = sum + (65027 / cs)
}
return ("" + sum).slice(0,16)
}
Fiddle here: http://jsfiddle.net/WS9dC/11/
If we did a straight sum of the charsum values, then the array ["a", "d"] would have the same hash as the array ["b", "c"] - leading to undesired collisions. So based on using non-UTF strings, where charcodes go up to 255, and allowing for 255 characters in each string, then the max return value of charsum is 255 * 255 = 65025. So I picked the next prime number up, 65027, and used (65027 / cs) to compute the hash. I am not 100% convinced this removes collisions... perhaps more thought needed... but it certainly fixes the [a, d] versus [b, c] case.
Testing:
var arr1 = ['alpha','beta','gama'];
var arr2 = ['beta','alpha','gama'];
console.log(array_hash(arr1))
console.log(array_hash(arr2))
console.log(array_hash(arr1) == array_hash(arr2))
Outputs:
443.5322979371356
443.5322979371356
true
And testing a case that shows different hashes:
var arr3 = ['a', 'd'];
var arr4 = ['b', 'c'];
console.log(array_hash(arr3))
console.log(array_hash(arr4))
console.log(array_hash(arr3) == array_hash(arr4))
outputs:
1320.651443298969
1320.3792001649144
false
Edit:
Here's a revised version, which ignore duplicates from the arrays as it goes, and return the hash based on unique items only:
http://jsfiddle.net/WS9dC/7/
array_hash = function(a) {
var i, sum = 0, product = 1
for (i = 0; i < a.length; i++) {
var cs = charsum(a[i])
if (product % cs > 0) {
product = product * cs
sum = sum + (65027 / cs)
}
}
return ("" + sum).slice(0, 16)
}
testing:
var arr1 = ['alpha', 'beta', 'gama', 'delta', 'theta', 'alpha', 'gama'];
var arr2 = ["beta", "gama", "alpha", "theta", "delta", "beta"];
console.log(array_hash(arr1))
console.log(array_hash(arr2))
console.log(array_hash(arr1) === array_hash(arr2))
returns:
689.878503111701
689.878503111701
true
Edit
I've revised the answer above to account for arrays of words that have the same letters. We need these to return different hashes, which they now do:
var arr1 = ['alpha', 'beta']
var arr2 = ['alhpa', 'ateb']
The fix was to add a multiplier to the charsum func based on the char index:
sum += (s.charCodeAt(i) * (i+1));

If you calculate a numeric hash code for each string, then you can combine them with an operator where the order doesn't matter, like the ^ XOR operator, then you don't need to sort the array:
function hashStringArray(array) {
var code = 0;
for (var i = 0; i < array.length; i++) {
var n = 0;
for (var j = 0; j < array[i].length; j++) {
n = n * 251 ^ array[i].charCodeAt(j);
}
code ^= n;
}
return code
};

You can do this:
var hashStringArray = function(array) {
return array.sort().join('\u200b');
};
The \u200b character is an unicode character that also means null, but is not the same as the \0 character, which is most widely used.
'\u200b' == '\0'
> false

An idea to have very fast hash if your set of possible string is less than 32 items long : hash the string with a built-in hash function that will return power-of two as hash :
function getStringHash(aString) {
var currentPO2 = 0;
var hashSet = [];
getStringHash = function ( aString) {
var aHash = hashSet[aString];
if (aHash) return aHash;
aHash = 1 << currentPO2++;
hashSet[aString] = aHash;
return aHash;
}
return getStringHash(aString);
}
Then use this hash on your string array, ORing the hashes ( | ) :
function getStringArrayHash( aStringArray) {
var aHash = 0;
for (var i=0; i<aStringArray.length; i++) {
aHash |= getStringHash(aStringArray[i]);
}
return aHash;
}
So to test a bit :
console.log(getStringHash('alpha')); // 1
console.log(getStringHash('beta')); // 2
console.log(getStringHash('gamma')); // 4
console.log(getStringHash('alpha')); // 1 again
var arr1 = ['alpha','beta','gama'];
var arr2 = ['beta','alpha','gama'];
var arr3 = ['alpha', 'teta'];
console.log(getStringArrayHash(arr1)); // 11
console.log(getStringArrayHash(arr2)); // 11 also, like for arr1
var arr3 = ['alpha', 'teta'];
console.log(getStringArrayHash(arr3)); // 17 : a different array has != hashset
jsbin is here : http://jsbin.com/rozanufa/1/edit?js,console
RQ !!! with this method, arrays are considered as set, meaning that a repeated item won't change the hash of an array !!!
This HAS to be faster since it uses only 1) function call 2) lookup 3) integer arithmetic.
So no sort, no (long) string, no concat.
jsperf confirms that :
http://jsperf.com/hashing-array-of-strings/4
EDIT :
version with prime numbers, here : http://jsbin.com/rozanufa/3/edit?js,console
// return the unique prime associated with the string.
function getPrimeStringHash(aString) {
var hashSet = [];
var currentPrimeIndex = 0;
var primes = [ 2, 3, 5, 7, 11, 13, 17 ];
getPrimeStringHash = function ( aString) {
var aPrime = hashSet[aString];
if (aPrime) return aPrime;
if (currentPrimeIndex == primes.length) aPrime = getNextPrime();
else aPrime = primes[currentPrimeIndex];
currentPrimeIndex++
hashSet[aString] = aPrime;
return aPrime;
};
return getPrimeStringHash(aString);
// compute next prime number, store it and returns it.
function getNextPrime() {
var pr = primes[primes.length-1];
do {
pr+=2;
var divides = false;
// discard the number if it divides by one earlier prime.
for (var i=0; i<primes.length; i++) {
if ( ( pr % primes[i] ) == 0 ) {
divides = true;
break;
}
}
} while (divides == true)
primes.push(pr);
return pr;
}
}
function getStringPrimeArrayHash( aStringArray) {
var primeMul = 1;
for (var i=0; i<aStringArray.length; i++) {
primeMul *= getPrimeStringHash(aStringArray[i]);
}
return primeMul;
}
function compareByPrimeHash( aStringArray, anotherStringArray) {
var mul1 = getStringPrimeArrayHash ( aStringArray ) ;
var mul2 = getStringPrimeArrayHash ( anotherStringArray ) ;
return ( mul1 > mul2 ) ?
! ( mul1 % mul2 )
: ! ( mul2 % mul1 );
// Rq : just test for mul1 == mul2 if you are sure there's no duplicates
}
Tests :
console.log(getPrimeStringHash('alpha')); // 2
console.log(getPrimeStringHash('beta')); // 3
console.log(getPrimeStringHash('gamma')); // 5
console.log(getPrimeStringHash('alpha')); // 2 again
console.log(getPrimeStringHash('a1')); // 7
console.log(getPrimeStringHash('a2')); // 11
var arr1 = ['alpha','beta','gamma'];
var arr2 = ['beta','alpha','gamma'];
var arr3 = ['alpha', 'teta'];
var arr4 = ['alpha','beta','gamma', 'alpha']; // == arr1 + duplicate 'alpha'
console.log(getStringPrimeArrayHash(arr1)); // 30
console.log(getStringPrimeArrayHash(arr2)); // 30 also, like for arr1
var arr3 = ['alpha', 'teta'];
console.log(getStringPrimeArrayHash(arr3)); // 26 : a different array has != hashset
console.log(compareByPrimeHash(arr1, arr2) ); // true
console.log(compareByPrimeHash(arr1, arr3) ); // false
console.log(compareByPrimeHash(arr1, arr4) ); // true despite duplicate

Javascript merge 2 arrays and sum same key values

I have 2 array:
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
Want to get 1 merged array with the sum of corresponding keys;
var array1 = [[1,10],[2,10],[3,10],[4,10],[5,50],[6,50],[7,10],[8,10],[9,10]];
Both arrays have unique keys, but the corresponding keys needs to be summed.
I tried loops, concat, etc but can't get the result i need.
anybody done this before?

You can use .reduce() to pass along an object that tracks the found sets, and does the addition.
DEMO: http://jsfiddle.net/aUXLV/
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
var result =
array1.concat(array2)
.reduce(function(ob, ar) {
if (!(ar[0] in ob.nums)) {
ob.nums[ar[0]] = ar
ob.result.push(ar)
} else
ob.nums[ar[0]][1] += ar[1]
return ob
}, {nums:{}, result:[]}).result
If you need the result to be sorted, then add this to the end:
.sort(function(a,b) {
return a[0] - b[0];
})

This is one way to do it:
var sums = {}; // will keep a map of number => sum
// for each input array (insert as many as you like)
[array1, array2].forEach(function(array) {
//for each pair in that array
array.forEach(function(pair) {
// increase the appropriate sum
sums[pair[0]] = pair[1] + (sums[pair[0]] || 0);
});
});
// now transform the object sums back into an array of pairs
var results = [];
for(var key in sums) {
results.push([key, sums[key]]);
}
See it in action.

a short routine can be coded using [].map()
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
array1=array2.concat(array1).map(function(a){
var v=this[a[0]]=this[a[0]]||[a[0]];
v[1]=(v[1]||0)+a[1];
return this;
},[])[0].slice(1);
alert(JSON.stringify(array1));
//shows: [[1,10],[2,10],[3,10],[4,10],[5,50],[6,50],[7,10],[8,10],[9,10]]
i like how it's just 3 line of code, doesn't need any internal function calls like push() or sort() or even an if() statement.

Try this:
var array1 = [[5,10],[6,10],[7,10],[8,10],[9,10]];
var array2 = [[1,10],[2,10],[3,10],[4,10],[5,40],[6,40]];
var res = [];
someReasonableName(array1, res);
someReasonableName(array2, res);
function someReasonableName(arr, res) {
var arrLen = arr.length
, i = 0
;
for(i; i < arrLen; i++) {
var ar = arr[i]
, index = ar[0]
, value = ar[1]
;
if(!res[index]) {
res[index] = [index, 0];
}
res[index][1] += value;
}
}
console.log(JSON.stringify(res, null, 2));
So, the result may have holes. Just like 0th index. Use the below function if you want to ensure there are no holes.
function compact(arr) {
var i = 0
, arrLen = arr.length
, res = []
;
for(i; i < arrLen; i++) {
var v = arr[i]
;
if(v) {
res[res.length] = v;
}
}
return res;
}
So, you can do:
var holesRemoved = compact(res);
And finally if you don't want the 0th elem of res. Do res.shift();
Disclaimer: I am not good with giving reasonable names.

The simple solution is like this.
function sumArrays(...arrays) {
const n = arrays.reduce((max, xs) => Math.max(max, xs.length), 0);
const result = Array.from({ length: n });
return result.map((_, i) => arrays.map(xs => xs[i] || 0).reduce((sum, x) => sum + x, 0));
}
console.log(...sumArrays([0, 1, 2], [1, 2, 3, 4], [1, 2])); // 2 5 5 4