Develop Reference

how to get a cumulative total array from 3 two dimensional arrays

I have 3 two dimensional arrays as given below which are series data to plot lines on a graph with the key being the timestamp.
const arr1 = [[1641013200000,1881],[1643691600000,38993],[1646110800000,41337],[1648785600000,78856],[1651377600000,117738],[1654056000000,119869],[1656648000000,157799],[1659326400000,196752],[1662004800000,199061],[1664596800000,237034],[1667275200000,239153],[1669870800000,269967]]
const arr2 = [[1641013200000,1302],[1643691600000,3347],[1646110800000,4754],[1648785600000,6948],[1651377600000,9725],[1654056000000,11314],[1656648000000,13787],[1659326400000,16666],[1662004800000,18370],[1664596800000,20876],[1667275200000,22384],[1669870800000,23560]]
const arr3 = [[1643691600000,67350],[1648785600000,134700],[1651377600000,202148],[1654056000000,202270],[1656648000000,269843],[1659326400000,337346],[1662004800000,337470],[1664596800000,404861],[1667275200000,404889],[1669870800000,472239]]
I want to plot another series line which gives the cumulative total of all three arrays values
(Note: if a timestamp is not present in either of the arrays, add the previous index value)
const totalArray = [
[1641013200000,3183],[1643691600000, 109690],[1646110800000, 113441],[1648785600000, 220504],
[1651377600000, 329611],[1654056000000, 333453],[1656648000000, 441429],[1659326400000, 550764],
[1662004800000, 554901],[1664596800000, 662771],[1667275200000, 666426],[1669870800000, 765766]
]
I have tried this, but some values are incorrect due to the timestamp not being present in either one
Approach:
const arr1 = [
[1641013200000, 1881],
[1643691600000, 38993],
[1646110800000, 41337],
[1648785600000, 78856],
[1651377600000, 117738],
[1654056000000, 119869],
[1656648000000, 157799],
[1659326400000, 196752],
[1662004800000, 199061],
[1664596800000, 237034],
[1667275200000, 239153],
[1669870800000, 269967]
];
const arr2 = [
[1641013200000, 1302],
[1643691600000, 3347],
[1646110800000, 4754],
[1648785600000, 6948],
[1651377600000, 9725],
[1654056000000, 11314],
[1656648000000, 13787],
[1659326400000, 16666],
[1662004800000, 18370],
[1664596800000, 20876],
[1667275200000, 22384],
[1669870800000, 23560]
];
const arr3 = [
[1643691600000, 67350],
[1648785600000, 134700],
[1651377600000, 202148],
[1654056000000, 202270],
[1656648000000, 269843],
[1659326400000, 337346],
[1662004800000, 337470],
[1664596800000, 404861],
[1667275200000, 404889],
[1669870800000, 472239]
];
const calculateTotal = () => {
var ret;
for (let a3 of arr3) {
var index = arr1.map(function(el) {
return el[0];
}).indexOf(a3[0]);
console.log(index);
if (index === -1) {
ret = arr1[index][0];
console.log(ret);
}
}
let unsortedArr = arr1.concat(arr2, arr3);
var sortedArray = unsortedArr.sort((a, b) => a[0] - b[0]);
var added = addArray(sortedArray);
console.log("Curent Output: " + JSON.stringify(added));
}
const addArray = (tuples) => {
var hash = {},
keys = [];
tuples.forEach(function(tuple) {
var key = tuple[0],
value = tuple[1];
if (hash[key] === undefined) {
keys.push(key);
hash[key] = value;
} else {
hash[key] += value;
}
});
return keys.map(function(key) {
return ([key, hash[key]]);
});
}
calculateTotal();
Is it possible to achieve this?

In your code there is this:
if (index === -1) {
ret = arr1[index][0];
But that assignment will fail as arr1[-1] is not defined.
Then when you do:
let unsortedArr = arr1.concat(arr2, arr3);
...you end up with an array that does not have the knowledge to use default values (from a previous index) when any of the three arrays has a "missing" time stamp.
I would suggest this approach:
Collect all the unique timestamps (from all arrays) into a Map, and associate arrays to each of those keys: these will be empty initially.
Populate those arrays with the timestamps from the original arrays
Get the sorted list of entries from that map
Fill the "gaps" by carrying forward values to a next array when the corresponding slot is undefined. At the same time sum up these values for the final output.
Here is an implementation:
function mergeArrays(...arrays) {
const map = new Map(arrays.flatMap(arr => arr.map(([stamp]) => [stamp, []])));
arrays.forEach((arr, i) => {
for (const [timeStamp, value] of arr) {
map.get(timeStamp)[i] = value;
}
});
const state = Array(arrays.length).fill(0);
return Array.from(map).sort(([a], [b]) => a - b).map(([timeStamp, arr], i) =>
[timeStamp, state.reduce((sum, prev, j) => sum + (state[j] = arr[j] ?? prev), 0)]
);
}
// Example run
const arr1 = [[1641013200000,1881],[1643691600000,38993],[1646110800000,41337],[1648785600000,78856],[1651377600000,117738],[1654056000000,119869],[1656648000000,157799],[1659326400000,196752],[1662004800000,199061],[1664596800000,237034],[1667275200000,239153],[1669870800000,269967]];
const arr2 = [[1641013200000,1302],[1643691600000,3347],[1646110800000,4754],[1648785600000,6948],[1651377600000,9725],[1654056000000,11314],[1656648000000,13787],[1659326400000,16666],[1662004800000,18370],[1664596800000,20876],[1667275200000,22384],[1669870800000,23560]];
const arr3 = [[1643691600000,67350],[1648785600000,134700],[1651377600000,202148],[1654056000000,202270],[1656648000000,269843],[1659326400000,337346],[1662004800000,337470],[1664596800000,404861],[1667275200000,404889],[1669870800000,472239]];
const result = mergeArrays(arr1, arr2, arr3);
console.log(result);

Summing numbers from 1st array if 2nd array match

I have 2 arrays that looks like this:
amountArray = ["200","150","500","100"];
currencyArray = ["EUR","EUR","USD","USD"];
I'd like to sum amounts from 1st array if 2nd array match.
Result i'd like to get :
totalAmount = "350 EUR | 600 USD";

You could take a Map for collecting same currencies and get the joined values.
let amounts = ["200", "150", "500", "100"],
currencies = ["EUR", "EUR", "USD", "USD"],
result = Array
.from(
currencies.reduce(
(m, c, i) => m.set(c, (m.get(c) || 0) + +amounts[i]),
new Map
),
([k, v]) => [v, k].join(' ')
)
.join(' | ');
console.log(result);

Store your data in a hashmap with the currencies as keys. Then while looping through your amounts, if the key exists, add to the existing sum.
At the end, convert back to an array and print.
const amountArray = ["200","150","500","100"];
const currencyArray = ["EUR","EUR","USD","USD"];
const result = {};
amountArray.forEach((amt, idx) => {
const amountInt = parseInt(amt, 10);
const currency = currencyArray[idx];
const existingTotal = result[currency] || 0;
result[currency] = existingTotal + amountInt;
});
const resultArray = Object.keys(result).map(key => `${result[key]} ${key}`);
const totalAmount = resultArray.join(' | ');
console.log(totalAmount);

You could do this
amountArray = ["200","150","500","100"];
currencyArray = ["EUR","EUR","USD","USD"];
var res = {}
currencyArray.forEach((elem, index)=>{
res[elem] = res[elem] ? parseInt(res[elem]) + parseInt( amountArray[index]) : parseInt(amountArray[index])
});
console.log(res);
var totalAmount = '';
for(var key in res){
totalAmount += ` ${res[key]} ${key} |`;
}
console.log(totalAmount.substr(0, totalAmount.length-1))

If possible, you can creat a class that contains 2 fields, 1 is amount, 1 is corresponding currency. Then you can group by currency and then do the sum up

amountArray = ["200","150","500","100"];
currencyArray = ["EUR","EUR","USD","USD"];
var totalAmount = [];
var result = amountArray.reduce(function(result, field, index) {
//console.log(field);
if(!(currencyArray[index] in result)){
//console.log("afaff");
result[currencyArray[index]] = 0;
}
result[currencyArray[index]] = result[currencyArray[index]] + parseInt(field);
//console.log(result)
return result;
}, {})
console.log(totalAmount);
//totalAmount = "350 EUR | 600 USD";

You could use reduce function to get the desired result.
let amountArray = ["200","150","500","100"];
let currencyArray = ["EUR","EUR","USD","USD"];
let result = currencyArray.reduce((acc,c,i) => {
if(acc.hasOwnProperty(c)){
return{
...acc,
[c]:parseInt(acc[c])+parseInt(amountArray[i])
}
}else{
return{
...acc,
[c]:amountArray[i]
}
}
},{})
console.log(result)

Use forEach to go over both arrays and build one object with accumulated values.
Then use map and join to make required string.
amountArray = ["200", "150", "500", "100"];
currencyArray = ["EUR", "EUR", "USD", "USD"];
const res = {};
currencyArray.forEach(
(key, i) => (res[key] = (res[key] ?? 0) + Number(amountArray[i]))
);
const str = Object.entries(res)
.map(([key, sum]) => `${sum} ${key}`)
.join(" | ");
console.log(str);

I have used java to solve this..
String[] amountArray = {"200","150","500","100"};
String[] currencyArray = {"EUR","EUR","USD","USD"};
HashMap<String, Integer> map = new HashMap<>();
for(int i =0; i < currencyArray.length;i++)
{
Integer n = Integer.parseInt(amountArray[i]);
Integer old = map.get(currencyArray[i]);
if(old == null)
{
old = new Integer(0);
}
Integer val = n+old;
map.put(currencyArray[i], val);
}

A hashmap solution. The first part forms the hashmap which uses currency as key and amount array as value. The second part constructs the string result. The time complexity is O(n^2), space complexity is O(n). n is the length of amountArray or currencyArray.
const amountArray = ["200","150","500","100"];
const currencyArray = ["EUR","EUR","USD","USD"];
function getTotalAmount() {
// --- First Part ---- //
const map = new Map()
const uniqueCurrencyArray = [];
for (let i = 0; i < currencyArray.length; i++) {
if (!uniqueCurrencyArray.includes(currencyArray[i])) {
uniqueCurrencyArray.push(currencyArray[i]);
}
}
for(const currency of uniqueCurrencyArray) {
const result = []
for(const [index, cur] of currencyArray.entries()) {
if(cur === currency) {
result.push(amountArray[index])
}
}
map.set(currency, result)
}
// --- Second Part -- //
let finalResult = ""
for(const key of map.keys()) {
if(finalResult !== "") {
finalResult += " | "
}
const amountArr = map.get(key)
let totalAmount = 0
for(const amount of amountArr) {
totalAmount += parseInt(amount, 10)
}
finalResult += `${totalAmount} ${key}`
}
return finalResult
}
console.log(getTotalAmount())

Convert an array which contains strings into key value pairs

I have a string "101-2000-10-102-2000-15" which I have to map as key: 101 values: {2000, 10}.
With the below code I am able to get the output as 101 => 2000 but I am unable to the one remaining value.
This is the Code:
let myString = "101-2000-10-102-2000-15"
let strArray = myString.match(/[^-]+-[^-]+-[^-]+/g);
console.log(strArray);
let compartmentMap = new Map(strArray.map(x => x.split("-")));
console.log(compartmentMap);
My input: "101-2000-10-102-2000-15"
Desired Output: {101 => {2000,10}, 102 => {2000,15}}

You need to get an array of values as well.
let myString = "101-2000-10-102-2000-15"
let strArray = myString.match(/[^-]+-[^-]+-[^-]+/g);
console.log(strArray);
let compartmentMap = new Map(strArray.map(x => {
const [k, ...v] = x.split("-");
return [k, v];
}));
console.log(Array.from(compartmentMap));

I think I'd be fairly pedestrian about it:
const result = new Map();
const rex = /(\d+)-(\d+)-(\d+)/g;
let match;
while ((match = rex.exec(myString)) !== null) {
result.set(match[1], [match[2], match[3]]);
}
That's assuming you want the 2000, 10 part as an array.
Live Example:
const myString = "101-2000-10-102-2000-15"
const result = new Map();
const rex = /(\d+)-(\d+)-(\d+)/g;
let match;
while ((match = rex.exec(myString)) !== null) {
result.set(match[1], [match[2], match[3]]);
}
console.log([...result.entries()]);
Or with more meaningful names via named capture groups and destructuring:
const result = new Map();
const rex = /(?<key>\d+)-(?<value1>\d+)-(?<value2>\d+)/g;
let match;
while ((match = rex.exec(myString)) !== null) {
const {key, value1, value2} = match.groups;
result.set(key, [value1, value2]);
}
Live Example:
const myString = "101-2000-10-102-2000-15"
const result = new Map();
const rex = /(?<key>\d+)-(?<value1>\d+)-(?<value2>\d+)/g;
let match;
while ((match = rex.exec(myString)) !== null) {
const {key, value1, value2} = match.groups;
result.set(key, [value1, value2]);
}
console.log([...result.entries()]);
Or with the new matchAll and destructuring:
const rex = /(\d+)-(\d+)-(\d+)/g;
const result = new Map(
[...myString.matchAll(rex)].map(
([, key, value1, value2]) => [key, [value1, value2]]
)
);
Live Example:
const myString = "101-2000-10-102-2000-15"
const rex = /(\d+)-(\d+)-(\d+)/g;
const result = new Map(
[...myString.matchAll(rex)].map(
([, key, value1, value2]) => [key, [value1, value2]]
)
);
console.log([...result.entries()]);

One approach would be to just split the string, and then perform a reduce() operation for every three elements:
const s = '101-2000-10-102-2000-15';
const result = s.split('-').reduce((r, v, i, a) =>
i % 3 ? r : {...r, [v]: a.slice(i + 1, i + 3)}, {});
console.log(result);

Slight change to your code, use reduce instead of map.
let myString = "101-2000-10-102-2000-15";
let strArray = myString.match(/[^-]+-[^-]+-[^-]+/g);
console.log(strArray);
let compartmentMap = strArray.reduce((acc, curr) => {
const [a, b, c] = curr.split("-");
return Object.assign(acc, { [a]: [b, c] });
}, {});
console.log(compartmentMap);

Actually, I figured a better which gives out a Map with exact key-value pairs. All I need is to pass a Map and My array to it and it spits out the map with the key-values.
const myString = '101-2000-10-102-2000-15';
let strArray = myString.match(/[^-]+-[^-]+-[^-]+/g);
console.log(strArray);
const compartmentMap = (someMap, someArray) => (someArray.map(x => { console.log(x)
const [a, b, c] = x.split("-");
someMap.set(a, {b,c});
}));
const x = new Map();
compartmentMap(x, strArray);
console.log(x);

How to create an object/array from a key/value string made up with '/'

Take the following string:
/foo/1/bar/2/cat/bob
I need to parse this into an object or array which ends up being:
foo = 1
bar = 2
cat = bob

var sample = "/foo/1/bar/2/cat/bob".substring(1);
var finalObj = {};
var arr = sample.split('/');
for(var i=0;i<arr.length;i=i+2){
finalObj[arr[i]] = arr[i+1];
}
console.log(finalObj);

const str = '/foo/1/bar/2/cat/bob/test/'
const parts = str.split('/')
.filter(val => val !== '')
const obj = {}
for (let ii = 0; ii < parts.length; ii+=2) {
const key = parts[ii]
const value = parts[ii+1]
obj[key] = !isNaN(value) ? Number(value) : value
}
console.log(obj)

Regular expressions is the tool of choice for all kinds of parsing:
str = '/foo/1/bar/2/cat/bob'
obj = {};
str.replace(/(\w+)\/(\w+)/g, (...m) => obj[m[1]] = m[2]);
console.log(obj);

Just for the fun of it,
let str = "/foo/1/bar/2/cat/bob",
arr = str.split("/"),
obj = {};
arr.shift();
while (arr.length) obj[arr.shift()] = arr.shift();

Get the index of a multidimensional array with the value of a given string in javascript

I have this array,
var arr = [["present",["John","Josh","Jay"]],["absent",["May","Mary","Mary Jane"]]];
var arr2 = [["J",["John","Josh","Jay"]],["M",["May","Mary","Mary Jane"]]];
And I want to get the data[0] of the array that have value of "Mary".
So in my example, value that I will get is "absent".
I want also to get the index of the array arr2 that have value of "Josh".
So in my second array, value that I will get is 0.
If possible to use underscore js, we can use it.
I tried to use _.contains() but failed.
Also these array is used in knockout js.

One other way of doing this job could be as follows;
var a1 = [["present",["John","Josh","Jay"]],["absent",["May","Mary","Mary Jane"]]],
a2 = [["J",["John","Josh","Jay"]],["M",["May","Mary","Mary Jane"]],["S",["Sally","Sam","Sammy Davis"]]],
getStatus = (a,n) => a.find(e => e[1].indexOf(n) !== -1)[0],
getIndex = (a,n) => a.findIndex(e => e[1].indexOf(n) !== -1);
console.log(getStatus(a1,"Mary"));
console.log(getIndex(a2,"Sammy Davis"));

var arr = [["present",["John","Josh","Jay"]],["absent",["May","Mary","Mary Jane"]]];
var arr2 = [["J",["John","Josh","Jay"]],["M",["May","Mary","Mary Jane"]]];
arr.forEach(function(e,i,a){
if(e[1].indexOf("Mary")>-1){
console.log(e[0])
}
});
arr2.forEach(function(e,i,a){
if(e[1].indexOf("Josh")>-1){
console.log(e[0])
}
});

First using filter() and second findIndex()
var arr = [["present",["John","Josh","Jay"]],["absent",["May","Mary","Mary Jane"]]];
var result1 = arr.filter(x => x[1].indexOf("Mary") !== -1)[0][0];
console.log(result1); // absent
var arr2 = [["J",["John","Josh","Jay"]],["M",["May","Mary","Mary Jane"]]];
var result2 = arr.findIndex(x => x[1].indexOf("Josh") !== -1);
console.log(result2); // 0

If your data set isn't really huge you can store two map objects in memory to quickly access values. Note that this won't work for duplicate names.
The benefit of this approach is that you only have to loop through each array once. If you use any method based on indexOf, you'll loop through your data every time you retrieve a value.
var arr = [["present",["John","Josh","Jay"]],["absent",["May","Mary","Mary Jane"]]];
var arr2 = [["J",["John","Josh","Jay"]],["M",["May","Mary","Mary Jane"]]];
var makeKeyMap = function(arr) {
return arr.reduce(function(map, data) {
data[1].forEach(function(key) {
map[key] = data[0];
});
return map;
}, {});
};
var makeIndexMap = function(arr) {
return arr.reduce(function(map, data, index) {
data[1].forEach(function(key) {
map[key] = index;
});
return map;
}, {});
};
var arrMap = makeKeyMap(arr);
var arr2Map = makeIndexMap(arr2);
console.log(arrMap["Mary"]);
console.log(arr2Map["Josh"]);
Edit: a performance test
var myTestData = createTestData();
var randomNameToFind = (function() {
var namesToFind = ["Aileen","Christina","Donna","Judith","Mandy","Sandra","Dawn","Tracey","Mhairi","Victoria","Carolyn","Gayle","Maria","Valerie"];
return function() {
return namesToFind[Math.floor(Math.random() * namesToFind.length)];
}
}());
console.log("Finding the number index for a random name out of 800 names, 10000 times:");
console.time("using index of approach");
var usingIndexOf = (a,n) => a.findIndex(e => e[1].indexOf(n) !== -1);
var results = [];
for (var i = 0; i < 10000; i += 1) {
results.push(usingIndexOf(myTestData, randomNameToFind()));
}
console.timeEnd("using index of approach");
console.time("using map approach");
var makeIndexMap = function(arr) {
return arr.reduce(function(map, data, index) {
data[1].forEach(function(key) {
map[key] = index;
});
return map;
}, {});
};
var myMap = makeIndexMap(myTestData);
results = [];
for (var j = 0; j < 10000; j += 1) {
results.push(myMap[randomNameToFind()]);
}
console.timeEnd("using map approach");
console.log("index map size: " + sizeof(myMap) + " bytes");
// Random data generation code below
function createTestData() {
var names = ["Nicola","Karen","Fiona","Susan","Claire","Sharon","Angela","Gillian","Julie","Michelle","Jacqueline","Amanda","Tracy","Louise","Jennifer","Alison","Sarah","Donna","Caroline","Elaine","Lynn","Margaret","Elizabeth","Lesley","Deborah","Pauline","Lorraine","Laura","Lisa","Tracey","Carol","Linda","Lorna","Catherine","Wendy","Lynne","Yvonne","Pamela","Kirsty","Jane","Emma","Joanne","Heather","Suzanne","Anne","Diane","Helen","Victoria","Dawn","Mary","Samantha","Marie","Kerry","Ann","Hazel","Christine","Gail","Andrea","Clare","Sandra","Shona","Kathleen","Paula","Shirley","Denise","Melanie","Patricia","Audrey","Ruth","Jill","Lee","Leigh","Catriona","Rachel","Morag","Kirsten","Kirsteen","Katrina","Joanna","Lynsey","Cheryl","Debbie","Maureen","Janet","Aileen","Arlene","Zoe","Lindsay","Stephanie","Judith","Mandy","Jillian","Mhairi","Barbara","Carolyn","Gayle","Maria","Valerie","Christina","Marion","Nicola","Karen","Susan","Claire","Fiona","Angela","Sharon","Gillian","Julie","Jennifer","Michelle","Louise","Lisa","Amanda","Donna","Tracy","Alison","Elaine","Jacqueline","Sarah","Caroline","Elizabeth","Laura","Lynn","Deborah","Lesley","Margaret","Joanne","Pauline","Lorraine","Carol","Kirsty","Yvonne","Lorna","Emma","Lynne","Tracey","Heather","Catherine","Pamela","Helen","Linda","Jane","Anne","Kerry","Suzanne","Wendy","Victoria","Diane","Mary","Dawn","Clare","Gail","Paula","Ann","Shona","Hazel","Christine","Andrea","Samantha","Marie","Lynsey","Sandra","Denise","Lee","Kelly","Gayle","Debbie","Jill","Kathleen","Patricia","Joanna","Catriona","Shirley","Ruth","Zoe","Leigh","Rachel","Melanie","Kirsteen","Aileen","Christina","Janet","Katrina","Stephanie","Audrey","Kirsten","Arlene","Maureen","Morag","Marion","Mhairi","Allison","Cheryl","Maria","Kim","Anna","Lindsay","Rebecca","Katherine","Mandy","Nicola","Karen","Claire","Angela","Fiona","Susan","Jennifer","Julie","Gillian","Michelle","Sharon","Sarah","Louise","Donna","Laura","Amanda","Alison","Lisa","Caroline","Kirsty","Jacqueline","Elaine","Lesley","Lynn","Deborah","Elizabeth","Joanne","Emma","Tracy","Lorraine","Lynne","Margaret","Heather","Carol","Lorna","Pauline","Kelly","Helen","Catherine","Linda","Victoria","Suzanne","Kerry","Pamela","Lee","Wendy","Jane","Yvonne","Tracey","Anne","Clare","Mary","Diane","Christine","Lynsey","Samantha","Shona","Andrea","Marie","Gail","Melanie","Dawn","Ann","Paula","Jill","Ruth","Leigh","Hazel","Debbie","Joanna","Denise","Lindsay","Gayle","Patricia","Catriona","Kathleen","Sandra","Leanne","Stephanie","Rachel","Katrina","Shirley","Kirsteen","Janet","Arlene","Zoe","Jillian","Anna","Judith","Mhairi","Natalie","Audrey","Carolyn","Morag","Aileen","Cheryl","Rebecca","Allison","Barbara","Mandy","Claire","Nicola","Karen","Angela","Gillian","Fiona","Jennifer","Laura","Susan","Julie","Michelle","Lisa","Sharon","Louise","Sarah","Tracy","Donna","Kelly","Kirsty","Amanda","Alison","Joanne","Caroline","Emma","Jacqueline","Elaine","Elizabeth","Lynne","Lesley","Deborah","Kerry","Victoria","Carol","Catherine","Lynn","Pauline","Margaret","Lorna","Lynsey","Lorraine","Linda","Suzanne","Tracey","Heather","Yvonne","Jane","Dawn","Mary","Helen","Anne","Wendy","Lee","Pamela","Jill","Lindsay","Clare","Christine","Diane","Leigh","Samantha","Shona","Joanna","Ruth","Debbie","Gail","Marie","Andrea","Paula","Kathleen","Catriona","Katrina","Denise","Melanie","Ann","Sandra","Gayle","Hazel","Jillian","Stephanie","Rachel","Kim","Natalie","Katherine","Patricia","Leanne","Cheryl","Mhairi","Morag","Arlene","Zoe","Kathryn","Aileen","Ashley","Judith","Anna","Frances","Janet","Lucy","Vicky","Christina","Kirsten","Rebecca","Nicola","Claire","Laura","Karen","Michelle","Louise","Jennifer","Fiona","Lisa","Gillian","Angela","Julie","Susan","Sarah","Kelly","Donna","Sharon","Emma","Caroline","Alison","Joanne","Tracy","Kirsty","Lynne","Amanda","Elaine","Jacqueline","Lesley","Kerry","Elizabeth","Lynn","Margaret","Deborah","Catherine","Heather","Lorna","Yvonne","Carol","Lorraine","Suzanne","Lynsey","Victoria","Helen","Linda","Pauline","Dawn","Anne","Jane","Tracey","Clare","Mary","Diane","Jill","Denise","Lee","Leanne","Christine","Shona","Pamela","Samantha","Paula","Joanna","Debbie","Stacey","Hazel","Cheryl","Lindsay","Gail","Rachel","Marie","Ann","Catriona","Andrea","Ruth","Kathryn","Katrina","Mhairi","Wendy","Leigh","Gayle","Melanie","Sandra","Stephanie","Anna","Jillian","Amy","Carolyn","Patricia","Carrie","Natalie","Kathleen","Lyndsey","Ashley","Rebecca","Vicky","Christina","Lindsey","Katherine","Arlene","Sara","Laura","Claire","Nicola","Lisa","Louise","Michelle","Fiona","Karen","Gillian","Jennifer","Emma","Angela","Susan","Kelly","Julie","Donna","Sarah","Kirsty","Sharon","Joanne","Amanda","Tracy","Alison","Elizabeth","Caroline","Elaine","Jacqueline","Lynne","Leanne","Deborah","Lesley","Lorraine","Victoria","Lynn","Pamela","Kerry","Lynsey","Lorna","Carol","Margaret","Heather","Helen","Catherine","Suzanne","Tracey","Yvonne","Cheryl","Linda","Pauline","Debbie","Jane","Dawn","Clare","Lindsay","Mary","Shona","Anne","Rachel","Jill","Christine","Natalie","Samantha","Diane","Lee","Wendy","Joanna","Paula","Marie","Ann","Denise","Catriona","Gayle","Hazel","Kathleen","Stacey","Gail","Ashley","Andrea","Ruth","Anna","Jillian","Leigh","Katrina","Stephanie","Mhairi","Katherine","Sandra","Lyndsey","Christina","Lucy","Patricia","Carrie","Rebecca","Kathryn","Lyndsay","Melanie","Amy","Sara","Arlene","Kirsten","Laura","Claire","Lisa","Nicola","Louise","Karen","Fiona","Jennifer","Michelle","Emma","Sarah","Gillian","Kelly","Susan","Angela","Donna","Kirsty","Julie","Pamela","Joanne","Caroline","Amanda","Tracy","Sharon","Lynne","Elaine","Deborah","Jacqueline","Alison","Lynsey","Victoria","Kerry","Leanne","Lorraine","Lesley","Elizabeth","Lorna","Catherine","Lynn","Suzanne","Heather","Helen","Lindsay","Margaret","Clare","Cheryl","Debbie","Pauline","Dawn","Carol","Mary","Natalie","Linda","Jane","Diane","Stacey","Carrie","Yvonne","Rebecca","Christine","Marie","Charlene","Rachel","Anne","Tracey","Jill","Samantha","Ashley","Paula","Joanna","Stephanie","Andrea","Shona","Denise","Anna","Hazel","Katrina","Ruth","Gayle","Lee","Sara","Catriona","Kathryn","Leigh","Mhairi","Wendy","Amy","Jillian","Katherine","Gail","Linsey","Christina","Lucy","Melanie","Sandra","Ann","Kathleen","Shelley","Kirsten","Kim","Lyndsey","Laura","Claire","Lisa","Nicola","Emma","Louise","Jennifer","Michelle","Sarah","Fiona","Karen","Gillian","Kirsty","Donna","Kelly","Pamela","Susan","Julie","Angela","Amanda","Lynsey","Sharon","Lynne","Deborah","Joanne","Victoria","Caroline","Alison","Leanne","Gemma","Elaine","Jacqueline","Lesley","Elizabeth","Lorraine","Kerry","Heather","Debbie","Catherine","Lynn","Lorna","Tracy","Suzanne","Yvonne","Cheryl","Natalie","Margaret","Lindsay","Diane","Helen","Pauline","Ashley","Rachel","Clare","Carol","Christine","Linda","Dawn","Rebecca","Stephanie","Jill","Tracey","Jane","Stacey","Paula","Shona","Anna","Charlene","Anne","Marie","Catriona","Samantha","Joanna","Ruth","Andrea","Mary","Denise","Kim","Mhairi","Hazel","Lauren","Amy","Kathryn","Carrie","Lyndsey","Lucy","Gail","Katherine","Christina","Linsey","Wendy","Katrina","Kimberley","Ann","Lee"];
var nameMap = names.reduce((map, n) => {
map[n[0]] = map[n[0]] || [];
map[n[0]].push(n);
return map;
}, {});
var testData = Object.keys(nameMap)
.sort()
.reduce((res, k) => {
res.push([k, nameMap[k]]);
return res;
}, []);
return testData;
};
<script src="http://code.stephenmorley.org/javascript/finding-the-memory-usage-of-objects/sizeof.compressed.js"></script>