Subsection of an Array in JS/JQuery - javascript

I'm looking to extract a subsection of a javascript array of objects based on a certain condition, e.g:
object.property == 2
I know that I could look through and build a new array from the ones that match, but I was wondering whether there was a shorthand for this.

You can use grep().
var arr = [ 1, 2, 3 ];
var subset = $.grep(arr,function(n,i){ return n >= 2 });
// subset = [2, 3]

Use http://api.jquery.com/filter/
e.g.
$(yourCollection).filter(function(){
return (this.property == 2);
});
EDIT:
jsFiddle with both approaches benchmarked: http://jsfiddle.net/StuperUser/6AfQj/

Related

Sum of two multi dimensional array

I have a array in JavaScript like this.
var arr=
[
['A'],[1,2,3,4],
['A'],[4,3,2,1],
['B'],[10,12,3,1],
['B'],[1,2,3,4],
.
.
.
.
['AZ'],[1,2,3,4]
]
and I want the output to summarize the array like -
var output=
[
['A'],[5,5,5,5],
['B'],[11,14,6,5],
['AZ'],[1,2,3,4]
]
Thanks.
Script
You can use the following script to achieve what you want to do.
const arr = [
["A"],
[1, 2, 3, 4],
["A"],
[4, 3, 2, 1],
["B"],
[10, 12, 3, 1],
["B"],
[1, 2, 3, 4],
["AZ"],
[1, 2, 3, 4],
];
/**
* Generator to return key-value pairs with array[i] being the key and array[i+1] being the value
* #param {Array<any>} array
*/
function* keyValue(array) {
// make sure we can build pairs (other ways of handling this are also possible)
if (array.length % 2 !== 0)
throw new RangeError(
"Array length must be dividable by 2 without remainder!"
);
for (let i = 0; i < array.length; i += 2) {
yield [array[i], array[i + 1]];
}
}
// here the created key-value pairs
console.log("Key-value pairs created by keyValue() generator function:");
console.log([...keyValue(arr)]);
// loop over key value pairs and sum up all the individul arrays based on the letter assigned to them
const result = [...keyValue(arr)].reduce((all, [[key], array]) => {
// if we don't have values for this letter, assing copy of the array to that letter
if (!all[key]) all[key] = [...array];
// we have some values for that letter already, sum up each value
else all[key] = all[key].map((prev, idx) => prev + array[idx]);
return all;
}, {});
// this would be a "better" result to my mind as there is no point wrapping single string values in arrays
// When using objects the values can easily be accessed in O(1)
console.log(result);
// now transform JS object to array of arrays
console.log("Result:");
const transformed = Object.entries(result).flatMap(([key, value]) => [[key], value]);
console.log(transformed);
.as-console-wrapper { max-height: 100% !important; top: 0; }
Please note: This implementation assumes that the arrays for a given letter have the same length (as is the case in your example).
Explanation
First of all, I use a generator function keyValue() to always group two consecutive values in the array (a key and a value) together. One could also do this differently but once you understand how generators work that's an easy and elegant approach, I think. For this demo I just throw an error if the array is not dividable by 2 without remainder, but one could also handle this more gracefully.
Then, using reduce(), I iterate over the array created by using keyValue() and for each element in the array I check if I've encountered that value before. If I have not, I create a copy of the array (for immutablility) and assign it to the key i.e. a letter. If I have encountered a certain letter before I add up the values that I have previously saved assigned to that letter with the ones I am currently processing. After iteration all sums are calculated and I have a JavaScript object containing the results.
To my mind, this would be a good output because your output is a bit odd, since there is no point storing single letters in an array or even arrays of arrays. Using a JavaScript object is much more convenient and faster for lookups.
Nevertheless, you can easily deduct your result from the created object using flatMap().

Finding unique values in multiple arrays

I'm trying to solve a freeCodeCamp exercise with this goal:
Write a function that takes two or more arrays and returns a new array
of unique values in the order of the original provided arrays.
In other words, all values present from all arrays should be included
in their original order, but with no duplicates in the final array.
The unique numbers should be sorted by their original order, but the
final array should not be sorted in numerical order.
So what I do is concatenate all the arguments into a single array called everything. I then search the array for duplicates, then search the arguments for these duplicates and .splice() them out.
So far everything works as expected, but the last number of the last argument does not get removed and I can't really figure out why.
Can anybody please point out what I'm doing wrong? Please keep in mind that I'm trying to learn, so obvious things probably won't be obvious to me and need to be pointed out. Thanks in advance.
function unite(arr1, arr2, arr3) {
var everything = [];
//concat all arrays except the first one
for(var x = 0; x < arguments.length; x++) {
for(var y = 0; y < arguments[x].length; y++) {
everything.push(arguments[x][y]);
}
}
//function that returns duplicates
function returnUnique(arr) {
return arr.reduce(function(dupes, val, i) {
if (arr.indexOf(val) !== i && dupes.indexOf(val) === -1) {
dupes.push(val);
}
return dupes;
}, []);
}
//return duplicates
var dupes = returnUnique(everything);
//remove duplicates from all arguments except the first one
for(var n = 1; n < arguments.length; n++) {
for(var m = 0; m < dupes.length; m++) {
if(arguments[n].hasOwnProperty(dupes[m])) {
arguments[n].splice(arguments[n].indexOf(dupes[m]), 1);
}
}
}
//return concatenation of the reduced arguments
return arr1.concat(arr2).concat(arr3);
}
//this returns [1, 3, 2, 5, 4, 2]
unite([1, 3, 2], [5, 2, 1, 4], [2, 1]);
Looks like you overcomplicated it a bit ;)
function unite() {
return [].concat.apply([], arguments).filter(function(elem, index, self) {
return self.indexOf(elem) === index;
});
}
res = unite([1, 2, 3], [5, 2, 1, 4], [2, 1], [6, 7, 8]);
document.write('<pre>'+JSON.stringify(res));
Explanations
We split the problem into two steps:
combine arguments into one big array
remove non-unique elements from this big array
This part handles the first step:
[].concat.apply([], arguments)
The built-in method someArray.concat(array1, array2 etc) appends given arrays to the target. For example,
[1,2,3].concat([4,5],[6],[7,8]) == [1,2,3,4,5,6,7,8]
If our function had fixed arguments, we could call concat directly:
function unite(array1, array2, array3) {
var combined = [].concat(array1, array2, array3);
// or
var combined = array1.concat(array2, array3);
but as we don't know how many args we're going to receive, we have to use apply.
someFunction.apply(thisObject, [arg1, arg2, etc])
is the same as
thisObject.someFunction(arg1, arg2, etc)
so the above line
var combined = [].concat(array1, array2, array3);
can be written as
var combined = concat.apply([], [array1, array2, array3]);
or simply
var combined = concat.apply([], arguments);
where arguments is a special array-like object that contains all function arguments (actual parameters).
Actually, last two lines are not going to work, because concat isn't a plain function, it's a method of Array objects and therefore a member of Array.prototype structure. We have to tell the JS engine where to find concat. We can use Array.prototype directly:
var combined = Array.prototype.concat.apply([], arguments);
or create a new, unrelated, array object and pull concat from there:
var combined = [].concat.apply([], arguments);
This prototype method is slightly more efficient (since we're not creating a dummy object), but also more verbose.
Anyways, the first step is now complete. To eliminate duplicates, we use the following method:
combined.filter(function(elem, index) {
return combined.indexOf(elem) === index;
})
For explanations and alternatives see this post.
Finally, we get rid of the temporary variable (combined) and chain "combine" and "dedupe" calls together:
return [].concat.apply([], arguments).filter(function(elem, index, self) {
return self.indexOf(elem) === index;
});
using the 3rd argument ("this array") of filter because we don't have a variable anymore.
Simple, isn't it? ;) Let us know if you have questions.
Finally, a small exercise if you're interested:
Write combine and dedupe as separate functions. Create a function compose that takes two functions a and b and returns a new function that runs these functions in reverse order, so that compose(a,b)(argument) will be the same as b(a(argument)). Replace the above definition of unite with unite = compose(combine, dedupe) and make sure it works exactly the same.
You can also try this :
var Data = [[1, 2, 3], [5, 2, 1, 4], [2, 1], [6, 7, 8]]
var UniqueValues = []
for (var i = 0; i < Data.length; i++) {
 UniqueValues = [...new Set(UniqueValues.concat(Data[i]))]
}
console.log(UniqueValues)

Javascript: Why array variable assignment is behaving differently inside and outside this function?

For the life of me, I just can't figure out what I'm doing wrong here.
I'm trying to use both the reduce and concat array methods to take all of the values of a 2d array and combine them into a single value (basically condense them into a single array and then sum them up).
The problem that I keep running into is that when I try to make a for/loop to concat each array element, the argument that I'm passing into the function is not being recognized as an array, thus my call to .concat() is failing. I've placed a console.log() at the beginning of the function to see if the element is being recognized as the first array element in the 2d array, and it's coming up as "1"(?).
I tried another test outside of the function, and it logs as the actual array element. What am I doing wrong here? code below:
var arrays = [[1, 2, 3], [4, 5], [6]];
var myArray = arrays[0]; // Test
console.log(myArray); // Test
var flatArray = arrays.reduce(function(arrays)
{
console.log(arrays[0]); // Test
for (var i = 0; i < arrays.length - 1; i++)
{
arrays[0].concat(arrays[i+1]);
}
return arrays;
});
console.log(flatArray);
This is the output that I keep getting:
Array [ 1, 2, 3 ]
1
TypeError: arrays[0].concat is not a function
It's almost seems like array is being converted to a number-type when inside the function...?
You have an error in your code here:
var flatArray = arrays.reduce(function(param) {})
that param will be an element of your arrays vector.
Check this https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/Array/reduce
You are using .reduce() incorrectly and you don't even need to use it to flatten an array. You can just do this:
var flatArray = [].concat.apply([],arrays);
Working demo: http://jsfiddle.net/jfriend00/wfjyfp42/
To understand .reduce(), the callback you pass it gets four arguments (see MDN reference). The first two arguments are important in using .reduce() correctly:
callback(previousValue, currentValue, index, array)
The previousValue is the accumulated value so far in the reduction. The currentValue is the next element of the array that is being iterated. The last two arguments do not need to be used if not needed.
Your code is only using the previousValue so it is never looking at the next item in the array as passed in by .reduce().
You could make a solution work using .reduce() like this:
var flatArray = arrays.reduce(function(previousValue, currentValue) {
return previousValue.concat(currentValue);
}, []);
Working demo: http://jsfiddle.net/jfriend00/2doohfc5/
Reduce performs an operation on two elements.
var sum = [[1, 2, 3], [4, 5], [6]].reduce(function(a, b) {
return a.concat(b);
}).reduce(function(a, b) {
return a + b;
});

How to split an specified number of elements from an array?

I have an array below:
var arr = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20];
and the length of the array is 20 and i am spliting it with 5
When i specify 1, it should return [1,2,3,4,5] and if i specify 2, it should return [6,7,8,9,10]
How can i get it through scripting?
Thanks for the help
You can use the .slice() method to return a segment from an array.
MDN documentation
For your case, you'd do something like this.
arr.slice(5 * n, 5 * n + 5)

Multiply a number to an array

From the following thread (Multiplying an array with a single value by a number?) I got the information that it is not possible to multiply a number to all elements within an array by executing [1, 2]*3. But since I have to do this I was wondering if there is an smart way of doing this?
I know I could write a function which iterates through all elements and multiply a number to each element manually. But I was wondering if there is a smarter way out there?
Maybe there are some libraries out there which implements some math functions where I can multiply a number to all elements of an array?
Or do you think the map function of Array can be used for this purpose?
You can indeed use the map function:
function multiply(input) {
return input * 3;
}
var myArray = [1, 2];
var myNewArray = myArray.map(multiply);
What this does is perform a function you provide on each element in the array, and return a new array with the results.
If you're already using jQuery, you can make it a little more concise with the each function:
$.each(myArray, function(index, value) {
myArray[index] = value * 3;
});
This changes the existing array. Although, using the plain-JS approach, you could do myArray = myArray.map(multiply); if you don't want a new variable.
I made a jsFiddle as an example.
This is easy with a library like math.js, which comes with matrix support. You could do something like:
var result = math.multiply([1, 2], 3);
or using the expression parser:
var result = math.eval('[1, 2] * 3');
Similar to Jos de Jong's answer, if you are using numericjs you can do
var result = numeric.mul([1, 2], 3);
To give the result [3, 6].
You have not defined your array as an array variable: Just do as following:
function times2(theArray){
var newArray = [];
for (i = 0; i < theArray.length; i++) {
newArray.push(theArray[i] * 2);
}
return newArray;
}
document.write("Array double = ",times2([1,2,3,4,5]).toString() , "<br>");
Now it does not matter how many elements you have in your array, it multiplies them by 2. You can choose any number instead of 2.

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