the code below is set to perform the sum of even valued terms in a Fibonacci sequence. The first part creates the terms so the first 2 ones and the addition from that. The second part creates an array that stores all these terms. The third part seeks out only the even terms and adds them and a condition is added to only do such for those less than 5 million.
//The below creates Fibonacci sequence as we know it
var memo = [0, 1];
function fib (n) {
if(memo.length-1 < n) {
memo[n] = fib(n-1)+fib(n-2);
}
return memo[n];
}
// creates array with elements from function that takes the index as argument while a given condition holds
function takeWhile (fromFunc, cond, arr=[]) {
var n = arr.length;
var val = fromFunc(n);
if(cond(val)) {
arr.push(val);
return takeWhile(fromFunc, cond, arr);
}
return arr;
}
//adds all the numbers in the sequence below 5 million
var sum =
takeWhile(fib, n => n < 5000000)
//the below line makes sure that only the even numbers in the sequence are added
.filter(n => n%2==0)
.reduce((acc, c) => acc+c, 0);
console.log(sum);
You need to take the index, not the value for filtering.
.filter((_, i) => i % 2 == 0)
function fib(n) {
if (memo.length - 1 < n) memo[n] = fib(n - 1) + fib(n - 2);
return memo[n];
}
function takeWhile(fromFunc, cond, arr = []) {
var val = fromFunc(arr.length);
if (!cond(val)) return arr;
arr.push(val);
return takeWhile(fromFunc, cond, arr);
}
var memo = [0, 1],
sum = takeWhile(fib, n => n < 5)
.filter((_, i) => i % 2 == 0)
.reduce((acc, c) => acc + c, 0);
console.log(sum);
console.log(...memo);
function getFrequency(str){
var sorted = str.split("").sort();
var data = [];
for(var i = 0; i <sorted.length; i++){
var last = data[data.length - 1]
if(last && last.character === sorted[i]) last.count++;
else data.push({character : sorted[i], count: 1});
}
return data;
}
it returns an array that consists of characters and their count. I want to arrange these in descending order of count. What to do?
You could sort the data array and use the property count for taking the delta as return value.
data.sort((a, b) => b.count - a.count);
function getFrequency(str){
var sorted = str.split("").sort(),
data = [],
i,
last;
for (i = 0; i < sorted.length; i++){
last = data[data.length - 1];
if (last && last.character === sorted[i]) {
last.count++;
} else {
data.push({ character: sorted[i], count: 1 });
}
}
return data.sort((a, b) => b.count - a.count);
}
console.log(getFrequency('zzzwwwwwabcdabaa'))
.as-console-wrapper { max-height: 100% !important; top: 0; }
Sort data based on count prop.
function getFrequency(str){
var sorted = str.split("").sort();
var data = [];
for(var i = 0; i <sorted.length; i++){
var last = data[data.length - 1]
if(last && last.character === sorted[i]) last.count++;
else data.push({character : sorted[i], count: 1});
}
data = data.sort(function(a,b){
return b.count - a.count
});
return data;
}
console.log( getFrequency('aaaabddyww') )
You could use reduce() to create an object for each character and add it to an array.
function getFrequency(str) {
var chars = str.split("");
return chars.reduce((a, c) => {
var f = a.find(e => e.character === c);
if (f) f.count++;
else a.push({
character: c,
count: 1
});
return a;
}, []);
}
console.log(getFrequency('zzzwwwwwabcdabaa'))
Reduce the string into an object of characters and counts. Convert to an array using Object.values(), then sort by b.count - a.count:
function getFrequency(str) {
return Object.values(str.split('')
.reduce(function(r, c) {
r[c] = r[c] || { character: c, count: 0 };
r[c].count += 1;
return r;
}, Object.create(null))
).sort(function(a, b) {
return b.count - a.count;
});
}
console.log(getFrequency('who let the dogs out who who'));
I've been trying to calculate median but still I've got some mathematical issues I guess as I couldn't get the correct median value and couldn't figure out why. Here's the code;
class StatsCollector {
constructor() {
this.inputNumber = 0;
this.average = 0;
this.timeout = 19000;
this.frequencies = new Map();
for (let i of Array(this.timeout).keys()) {
this.frequencies.set(i, 0);
}
}
pushValue(responseTimeMs) {
let req = responseTimeMs;
if (req > this.timeout) {
req = this.timeout;
}
this.average = (this.average * this.inputNumber + req) / (this.inputNumber + 1);
console.log(responseTimeMs / 1000)
let groupIndex = Math.floor(responseTimeMs / 1000);
this.frequencies.set(groupIndex, this.frequencies.get(groupIndex) + 1);
this.inputNumber += 1;
}
getMedian() {
let medianElement = 0;
if (this.inputNumber <= 0) {
return 0;
}
if (this.inputNumber == 1) {
return this.average
}
if (this.inputNumber == 2) {
return this.average
}
if (this.inputNumber > 2) {
medianElement = this.inputNumber / 2;
}
let minCumulativeFreq = 0;
let maxCumulativeFreq = 0;
let cumulativeFreq = 0;
let freqGroup = 0;
for (let i of Array(20).keys()) {
if (medianElement <= cumulativeFreq + this.frequencies.get(i)) {
minCumulativeFreq = cumulativeFreq;
maxCumulativeFreq = cumulativeFreq + this.frequencies.get(i);
freqGroup = i;
break;
}
cumulativeFreq += this.frequencies.get(i);
}
return (((medianElement - minCumulativeFreq) / (maxCumulativeFreq - minCumulativeFreq)) + (freqGroup)) * 1000;
}
getAverage() {
return this.average;
}
}
Here's the snapshot of the results when I enter the values of
342,654,987,1093,2234,6243,7087,20123
The correct result should be;
Median: 1663.5
Change your median method to this:
function median(values){
if(values.length ===0) throw new Error("No inputs");
values.sort(function(a,b){
return a-b;
});
var half = Math.floor(values.length / 2);
if (values.length % 2)
return values[half];
return (values[half - 1] + values[half]) / 2.0;
}
fiddle
Here's another solution:
function median(numbers) {
const sorted = Array.from(numbers).sort((a, b) => a - b);
const middle = Math.floor(sorted.length / 2);
if (sorted.length % 2 === 0) {
return (sorted[middle - 1] + sorted[middle]) / 2;
}
return sorted[middle];
}
console.log(median([4, 5, 7, 1, 33]));
The solutions above - sort then find middle - are fine, but slow on large data sets. Sorting the data first has a complexity of n x log(n).
There is a faster median algorithm, which consists in segregating the array in two according to a pivot, then looking for the median in the larger set. Here is some javascript code, but here is a more detailed explanation
// Trying some array
alert(quickselect_median([7,3,5])); // 2300,5,4,0,123,2,76,768,28]));
function quickselect_median(arr) {
const L = arr.length, halfL = L/2;
if (L % 2 == 1)
return quickselect(arr, halfL);
else
return 0.5 * (quickselect(arr, halfL - 1) + quickselect(arr, halfL));
}
function quickselect(arr, k) {
// Select the kth element in arr
// arr: List of numerics
// k: Index
// return: The kth element (in numerical order) of arr
if (arr.length == 1)
return arr[0];
else {
const pivot = arr[0];
const lows = arr.filter((e)=>(e<pivot));
const highs = arr.filter((e)=>(e>pivot));
const pivots = arr.filter((e)=>(e==pivot));
if (k < lows.length) // the pivot is too high
return quickselect(lows, k);
else if (k < lows.length + pivots.length)// We got lucky and guessed the median
return pivot;
else // the pivot is too low
return quickselect(highs, k - lows.length - pivots.length);
}
}
Astute readers will notice a few things:
I simply transliterated Russel Cohen's Python solution into JS,
so all kudos to him.
There are several small optimisations worth
doing, but there's parallelisation worth doing, and the code as is
is easier to change in either a quicker single-threaded, or quicker
multi-threaded, version.
This is the average linear time
algorithm, there is more efficient a deterministic linear time version, see Russel's
post for details, including performance data.
ADDITION 19 Sept. 2019:
One comment asks whether this is worth doing in javascript. I ran the code in JSPerf and it gives interesting results.
if the array has an odd number of elements (one figure to find), sorting is 20% slower that this "fast median" proposition.
if there is an even number of elements, the "fast" algorithm is 40% slower, because it filters through the data twice, to find elements number k and k+1 to average. It is possible to write a version of fast median that doesn't do this.
The test used rather small arrays (29 elements in the jsperf test). The effect appears to be more pronounced as arrays get larger. A more general point to make is: it shows these kinds of optimisations are worth doing in Javascript. An awful lot of computation is done in JS, including with large amounts of data (think of dashboards, spreadsheets, data visualisations), and in systems with limited resources (think of mobile and embedded computing).
var arr = {
max: function(array) {
return Math.max.apply(null, array);
},
min: function(array) {
return Math.min.apply(null, array);
},
range: function(array) {
return arr.max(array) - arr.min(array);
},
midrange: function(array) {
return arr.range(array) / 2;
},
sum: function(array) {
var num = 0;
for (var i = 0, l = array.length; i < l; i++) num += array[i];
return num;
},
mean: function(array) {
return arr.sum(array) / array.length;
},
median: function(array) {
array.sort(function(a, b) {
return a - b;
});
var mid = array.length / 2;
return mid % 1 ? array[mid - 0.5] : (array[mid - 1] + array[mid]) / 2;
},
modes: function(array) {
if (!array.length) return [];
var modeMap = {},
maxCount = 1,
modes = [array[0]];
array.forEach(function(val) {
if (!modeMap[val]) modeMap[val] = 1;
else modeMap[val]++;
if (modeMap[val] > maxCount) {
modes = [val];
maxCount = modeMap[val];
}
else if (modeMap[val] === maxCount) {
modes.push(val);
maxCount = modeMap[val];
}
});
return modes;
},
variance: function(array) {
var mean = arr.mean(array);
return arr.mean(array.map(function(num) {
return Math.pow(num - mean, 2);
}));
},
standardDeviation: function(array) {
return Math.sqrt(arr.variance(array));
},
meanAbsoluteDeviation: function(array) {
var mean = arr.mean(array);
return arr.mean(array.map(function(num) {
return Math.abs(num - mean);
}));
},
zScores: function(array) {
var mean = arr.mean(array);
var standardDeviation = arr.standardDeviation(array);
return array.map(function(num) {
return (num - mean) / standardDeviation;
});
}
};
2022 TypeScript Approach
const median = (arr: number[]): number | undefined => {
if (!arr.length) return undefined;
const s = [...arr].sort((a, b) => a - b);
const mid = Math.floor(s.length / 2);
return s.length % 2 === 0 ? ((s[mid - 1] + s[mid]) / 2) : s[mid];
};
Notes:
The type in the function signature (number[]) ensures only an array of numbers can be passed to the function. It could possibly be empty though.
if (!arr.length) return undefined; checks for the possible empty array, which would not have a median.
[...arr] creates a copy of the passed-in array to ensure we don't overwrite the original.
.sort((a, b) => a - b) sorts the array of numbers in ascending order.
Math.floor(s.length / 2) finds the index of the middle element if the array has odd length, or the element just to the right of the middle if the array has even length.
s.length % 2 === 0 determines whether the array has an even length.
(s[mid - 1] + s[mid]) / 2 averages the two middle items of the array if the array's length is even.
s[mid] is the middle item of an odd-length array.
TypeScript Answer 2020:
// Calculate Median
const calculateMedian = (array: Array<number>) => {
// Check If Data Exists
if (array.length >= 1) {
// Sort Array
array = array.sort((a: number, b: number) => {
return a - b;
});
// Array Length: Even
if (array.length % 2 === 0) {
// Average Of Two Middle Numbers
return (array[(array.length / 2) - 1] + array[array.length / 2]) / 2;
}
// Array Length: Odd
else {
// Middle Number
return array[(array.length - 1) / 2];
}
}
else {
// Error
console.error('Error: Empty Array (calculateMedian)');
}
};
const median = (arr) => {
return arr.slice().sort((a, b) => a - b)[Math.floor(arr.length / 2)];
};
Short and sweet.
Array.prototype.median = function () {
return this.slice().sort((a, b) => a - b)[Math.floor(this.length / 2)];
};
Usage
[4, 5, 7, 1, 33].median()
Works with strings as well
["a","a","b","b","c","d","e"].median()
For better performance in terms of time complexity, use MaxHeap - MinHeap to find the median of stream of array.
Simpler & more efficient
const median = dataSet => {
if (dataSet.length === 1) return dataSet[0]
const sorted = ([ ...dataSet ]).sort()
const ceil = Math.ceil(sorted.length / 2)
const floor = Math.floor(sorted.length / 2)
if (ceil === floor) return sorted[floor]
return ((sorted[ceil] + sorted[floor]) / 2)
}
Simple solution:
function calcMedian(array) {
const {
length
} = array;
if (length < 1)
return 0;
//sort array asc
array.sort((a, b) => a - b);
if (length % 2) {
//length of array is odd
return array[(length + 1) / 2 - 1];
} else {
//length of array is even
return 0.5 * [(array[length / 2 - 1] + array[length / 2])];
}
}
console.log(2, calcMedian([1, 2, 2, 5, 6]));
console.log(3.5, calcMedian([1, 2, 2, 5, 6, 7]));
console.log(9, calcMedian([13, 9, 8, 15, 7]));
console.log(3.5, calcMedian([1, 4, 6, 3]));
console.log(5, calcMedian([5, 1, 11, 2, 8]));
Simpler, more efficient, and easy to read
cloned the data to avoid alterations to the original data.
sort the list of values.
get the middle point.
get the median from the list.
return the median.
function getMedian(data) {
const values = [...data];
const v = values.sort( (a, b) => a - b);
const mid = Math.floor( v.length / 2);
const median = (v.length % 2 !== 0) ? v[mid] : (v[mid - 1] + v[mid]) / 2;
return median;
}
const medianArr = (x) => {
let sortedx = x.sort((a,b)=> a-b);
let halfIndex = Math.floor(sortedx.length/2);
return (sortedx.length%2) ? (sortedx[Math.floor(sortedx.length/2)]) : ((sortedx[halfIndex-1]+sortedx[halfIndex])/2)
}
console.log(medianArr([1,2,3,4,5]));
console.log(medianArr([1,2,3,4,5,6]));
function Median(arr){
let len = arr.length;
arr = arr.sort();
let result = 0;
let mid = Math.floor(len/2);
if(len % 2 !== 0){
result += arr[mid];
}
if(len % 2 === 0){
result += (arr[mid] + arr[mid+1])/2
}
return result;
}
console.log(`The median is ${Median([0,1,2,3,4,5,6])}`)
function median(arr) {
let n = arr.length;
let med = Math.floor(n/2);
if(n % 2 != 0){
return arr[med];
} else{
return (arr[med -1] + arr[med])/ 2.0
}
}
console.log(median[1,2,3,4,5,6]);
The arr.sort() method sorts the elements of an array in place and returns the array. By default, it sorts the elements alphabetically, so if the array contains numbers, they will not be sorted in numerical order.
On the other hand, the arr.sort((a, b) => a - b) method uses a callback function to specify how the array should be sorted. The callback function compares the two elements a and b and returns a negative number if a should be sorted before b, a positive number if b should be sorted before a, and zero if the elements are equal. In this case, the callback function subtracts b from a, which results in a sorting order that is numerical in ascending order.
So, if you want to sort an array of numbers in ascending order, you should use arr.sort((a, b) => a - b), whereas if you want to sort an array of strings alphabetically, you can use arr.sort():
function median(numbers) {
const sorted = Array.from(numbers).sort((a, b) => a - b);
const middle = Math.floor(sorted.length / 2);
if (sorted.length % 2 === 0) {
return (sorted[middle - 1] + sorted[middle]) / 2;
}
return sorted[middle];
}
function findMedian(arr) {
arr.sort((a, b) => a - b)
let i = Math.floor(arr.length / 2)
return arr[i]
}
let result = findMedian([0, 1, 2, 4, 6, 5, 3])
console.log(result)
I am trying to write a function that will produce the factorial of a provided integer and then reduce the factorial array (by multiplying each array element).
For example:
factor(5) >>> [1, 2, 3, 4, 5] >>> 1 * 2 * 3 * 4 * 5 >>> 120
var array = [ ];
function factor(num) {
for (i = 1; i <= num; i++) {
array.push(i);
}
array.reduce(function(a, b) {
return a * b;
})
};
factor(5);
However, it keeps returning undefined.
I think it has to do with the formatting of the function, but I'm not sure.
Try to pass the initial value for reduce,
array.reduce(function(a, b) {
return a * b;
},1);
Also return the reduced value of the array from the function,
function factor(num) {
for (i = 1; i <= num; i++) {
array.push(i);
}
return array.reduce(function(a, b) {
return a * b;
},1)
};
console.log(factor(5));
Your problem that your function does not actually return any value, and the "return" value of such a function is undefined. Let's look at why this is happening:
function factor(num) {
for (i = 1; i <= num; i++) {
array.push(i);
}
array.reduce(function(a, b) {
return a * b; // <-- Here is probably your misunderstanding
});
};
That return that I marked returns from the function that you pass to reduce(), but there is no return from factor().
.reduce() is not necessary to return expected results. Try utilizing single for loop
var array = [];
function factor(num) {
for (var i = 1, res = 1; i < num; array[i - 1] = i, res *= ++i);
return res
};
console.log(
factor(3)
, factor(4)
, factor(5)
, factor(6)
, array
)
using Array indexes and reduce if anybody doesnt want to fill their arrays with ordered numbers.
num = 10;
Array(num).fill(null).reduce((acc, val, i) => {
if (i == 0) {
return 1;
}
return acc * (i + 1);
}, 1)
function factor(num) {
return Array(num)
.fill(null)
.map((item, index) => index + 1)
.reduce((pv, cv) => pv * cv);
}
console.log(factor(3), factor(4), factor(5), factor(6));
How to add value of equals index of an array to another?
I know I can do something like this:
var i = 0,
a = [1, 2, 3],
b = [4, 5, 6],
total = [];
while (i < a.length) {
total[i] = a[i] + b[i];
i = i + 1;
}
console.log(total); //[5, 7, 9]
But is there a standard or faster way of doing this?
(In my specific case, a.length === b.length in all cases).
You might find this a bit more compact:
a.map(function(v, i) { return v + b[i]; })
It will be even more compact in ES6:
a.map((v, i) => v + b[i])
We can write our own version of map which loops over two arrays in parallel:
function mapTwo(arr1, arr2, fn, ctxt) {
return arr1.map(function(v, i) {
return fn.call(ctxt, v, arr2[i], i);
};
}
Now we can solve the original problem with
mapTwo(a, b, function(v1, v2) { return v1 + v2; }))
Or, if we really wanted to, let's write a map which iterates over many arrays. It will invoke the callback with an array of values, one from each input array.
function mapMany(arrays, fn, ctxt) {
fn = fn.bind(ctxt);
return arrays[0].map(function(_, i) {
return fn(arrays.map(function(array) { return array[i]; }), i);
};
}
Now we can do
mapMany([a,b], sum);
where sum is
function sum(values) {
return values.reduce(function(result, val) { return result+val; });
}
We could also write a more general function, called zip, which combines the elements at each index in the input arrays into an array, so returning an array of arrays. We will now handle the case where the arrays are of different length;
function zip(arrays) {
var result = [],
max = Math.max.apply(0, arrays.map(function(array) { return array.length; }));
for (var i=0; i<max; i++) {
result.push(arrays.map(function(array) { return array[i]; });
}
return result;
}
Now we can create the pair-wise sums with
zip([a,b]).map(sum)
No, there is not really a 'standard' way of doing this. Maybe libraries as LoDash can help you out here. Specifically take a look at the .zip() function