I believe what I need are two JavaScript functions. I am receiving a comma separated string that holds two types of data: 1) device name followed by 2) numeric value. These two values are separated by a comma, and each set is also separated by a comma. Example string below:
Device_A,5,Device_C,2,Device_A,10,Device_B,8,Device_B,2,Device_C,7
What I want to do is create two separate functions. The first function finds the unique device names and returns just the names in a comma separated string. The second function would calculate the sum of the numeric values for each device. The expected results from the example string above would return:
Function 1 (Device List):
Device_A, Device_B, Device_C
Function 2 (Sums per Device List):
15,10,9
The lists do not need to return in any particular order as long at they both match up. All I have successfully done at this point is return a list of unique values (including numeric values)... I'm stuck on separating the list, but still referring to device name to sum up all of the values.
Thanks in advance. Let me know if you have any questions!
Matt
You could use an object for collecting the names and count.
This edit contains a shared function and two function for the result in equal order.
function getGrouped(data) {
var array = data.split(','),
temp = Object.create(null),
i = 0;
while (i < array.length) {
temp[array[i]] = (temp[array[i]] || 0) + +array[i + 1] || 0;
i += 2;
}
return temp;
}
function getDevices(data) {
var temp = getGrouped(data);
return Object.keys(temp).sort().join();
}
function getCounts(data) {
var temp = getGrouped(data);
return Object.keys(temp).sort().map(function (k) { return temp[k]; }).join();
}
var data = "Device_A,5,Device_C,2,Device_A,10,Device_B,8,Device_B,2,Device_C,7";
console.log(getDevices(data));
console.log(getCounts(data));
When starting out on a problem like this, I think it's wise to not worry about doing it in a single loop or in a fancy one-liner at first.
A) Start out by defining what data structures you need and how to go from one format to another:
Convert my string of data to a list of keys and values
Somehow group these keys and values based on the key
Sum the values for each group
Return a list of all unique keys
Return a list of all summed values
B) Then, try to see if any of the code you've written has the potential be re-used by other parts of your application and refactor accordingly.
C) Finally, assess if there are performance bottle necks and only if there are, optimize for performance.
A. A function for each step:
// 1. From string to array of keys and values
// You already figured this one out. Split by ","!
const namesAndValuesFromString =
str => str.split(",");
// 2. Grouping by key
// Let's first make pairs:
const deviceValuePairs = devicesAndValues => {
let pair = [];
const pairs = [];
devicesAndValues.forEach(x => {
pair.push(x);
if (pair.length === 2) {
pairs.push(pair);
pair = [];
}
});
return pairs;
};
// Key value pairs are a nice starting point for constructing a grouped object:
const kvpsToDeviceValuesObj = kvps => {
const valuesByDevice = {};
kvps.forEach(([key, value]) => {
value = Number(value);
if (!valuesByDevice[key]) {
valuesByDevice[key] = [];
}
valuesByDevice[key].push(value);
});
return valuesByDevice;
};
// 3. Now, we can get to summing the values arrays
const sumValueArrays = valuesByDevice => {
const summedValuesByDevice = {};
// Loop over the objects entries
Object.entries(valuesByDevice).forEach(
([key, values]) => {
summedValuesByDevice[key] = values
.reduce((a, b) => a + b);
}
);
return summedValuesByDevice;
};
// 4. + 5. Now that we have an object with device ids as keys, and summed values inside, we can retrieve the two lists
const getDevices = Object.keys;
const getSums = Object.values;
// Running the code:
const namesAndValues =
namesAndValuesFromString("A,5,C,2,A,10,B,8,B,2,C,7");
console.log(namesAndValues);
const kvps = deviceValuePairs(namesAndValues);
console.log(kvps);
const valuesByDevice = kvpsToDeviceValuesObj(kvps);
console.log(valuesByDevice);
const sumValues = sumValueArrays(valuesByDevice);
console.log(sumValues);
const devices = getDevices(sumValues);
console.log(devices);
const sums = getSums(sumValues);
console.log(sums);
B. Refactoring!
Once you understand each of those steps, you'll start to see things that can be generalized or combined. That's where the fun starts :)
// UTILITIES
const split = del => arr => arr.split(del);
const toPairs = arr => {
let pair = [];
return arr.reduce(
(pairs, x) => {
pair.push(x);
if (pair.length === 2) {
pairs.push(pair);
pair = [];
}
return pairs;
}, []);
};
const sum = (x, y = 0) => +x + y;
const kvpsToGroups = grouper => kvps =>
kvps.reduce(
(groups, [key, value]) => Object.assign(groups, {
[key]: grouper(value, groups[key])
}), {});
// YOUR APP
const sumGrouper = kvpsToGroups(sum);
const dataSplitter = split(",");
const parseData = str => sumGrouper(toPairs(dataSplitter(str)));
// MAIN
const result = parseData("A,5,C,2,A,10,B,8,B,2,C,7");
console.log("devices:", Object.keys(result));
console.log("sums:", Object.values(result));
another way by regexs
let str = "Device_A,5,Device_C,2,Device_A,10,Device_B,8,Device_B,2,Device_C,7", obj = {}
str.match(/(\w+,[0-9]+)/g).forEach((s) => {
s = s.split(',')
obj[s[0]] = (obj[s[0]] || 0) + (Number(s[1]) || 0)
})
console.log(obj)
Something like this should do it:
var input = "Device_A,5,Device_C,2,Device_A,10,Device_B,8,Device_B,2,Device_C,7";
var output = input.split(',').reduce((accumulator, currentValue, currentIndex, array) => {
accumulator[currentValue] = (accumulator[currentValue] || 0)
+ parseInt(array[currentIndex + 1]);
array.splice(0,1);
return accumulator;
}, {});
console.log(Object.keys(output));
console.log(Object.keys(output).map(k => output[k]));
Related
I have a array of string.
let arr=["robin","rohit","roy"];
Need to find all the common character present in all the strings in array.
Output Eg: r,o
I have tried to create a function for above case with multiple loops but i want to know what should be the efficient way to achive it.
Here's a functional solution which will work with an array of any iterable value (not just strings), and uses object identity comparison for value equality:
function findCommon (iterA, iterB) {
const common = new Set();
const uniqueB = new Set(iterB);
for (const value of iterA) if (uniqueB.has(value)) common.add(value);
return common;
}
function findAllCommon (arrayOfIter) {
if (arrayOfIter.length === 0) return [];
let common = new Set(arrayOfIter[0]);
for (let i = 1; i < arrayOfIter.length; i += 1) {
common = findCommon(common, arrayOfIter[i]);
}
return [...common];
}
const arr = ['robin', 'rohit', 'roy'];
const result = findAllCommon(arr);
console.log(result);
const arr = ["roooooobin","rohit","roy"];
const commonChars = (arr) => {
const charsCount = arr.reduce((sum, word) => {
const wordChars = word.split('').reduce((ws, c) => {
ws[c] = 1;
return ws;
}, {});
Object.keys(wordChars).forEach((c) => {
sum[c] = (sum[c] || 0) + 1;
});
return sum;
}, {});
return Object.keys(charsCount).filter(key => charsCount[key] === arr.length);
}
console.log(commonChars(arr));
Okay, the idea is to count the amount of times each letter occurs but only counting 1 letter per string
let arr=["robin","rohit","roy"];
function commonLetter(array){
var count={} //object used for counting letters total
for(let i=0;i<array.length;i++){
//looping through the array
const cache={} //same letters only counted once here
for(let j=0;j<array[i].length;j++){
//looping through the string
let letter=array[i][j]
if(cache[letter]!==true){
//if letter not yet counted in this string
cache[letter]=true //well now it is counted in this string
count[letter]=(count[letter]||0)+1
//I don't say count[letter]++ because count[letter] may not be defined yet, hence (count[letter]||0)
}
}
}
return Object.keys(count)
.filter(letter=>count[letter]===array.length)
.join(',')
}
//usage
console.log(commonLetter(arr))
No matter which way you choose, you will still need to count all characters, you cannot get around O(n*2) as far as I know.
arr=["robin","rohit","roy"];
let commonChars = sumCommonCharacters(arr);
function sumCommonCharacters(arr) {
data = {};
for(let i = 0; i < arr.length; i++) {
for(let char in arr[i]) {
let key = arr[i][char];
data[key] = (data[key] != null) ? data[key]+1 : 1;
}
}
return data;
}
console.log(commonChars);
Here is a 1 liner if anyone interested
new Set(arr.map(d => [...d]).flat(Infinity).reduce((ac,d) => {(new RegExp(`(?:.*${d}.*){${arr.length}}`)).test(arr) && ac.push(d); return ac},[])) //{r,o}
You can use an object to check for the occurrences of each character. loop on the words in the array, then loop on the chars of each word.
let arr = ["robin","rohit","roy"];
const restWords = arr.slice(1);
const result = arr[0].split('').filter(char =>
restWords.every(word => word.includes(char)))
const uniqueChars = Array.from(new Set(result));
console.log(uniqueChars);
trying to find the best way to check for triplicates values inside an array of strings.
I found many stackoverflow solutions for duplicates values which is not the case in here.
This is the farest i could get with solving this and I am not sure if it is the correct way:
const array = [
"peace",
"peace",
"Vrede",
"Patz",
"Salam",
"paz",
"Salam",
"Salam"
];
const findTriplicates = (param) => {
let counts = {};
for (let i = 0; i < param.length; i++) {
if (counts[param[i]]) {
counts[param[i]] += 1;
} else {
counts[param[i]] = 1;
}
}
for (let i in counts) {
if (counts[i] === 3) {
console.log(i + " exists " + counts[i] + " times.");
}
}
};
findTriplicates(array); // Salam exists 3 times.
please don't hesitate to fix my code or to post your solution.
thanks for your support in advance :)
Cheerz!
Your overall idea is good, using Hash Maps (js objects) is the best option for the task.
You can move your "=== 3" check to the first loop and have another object to save triplicates, it will be twice faster.
check this out
const findTriplicates = (param) => {
let values = [...new Set(param)];
let triples = [];
values.forEach(item=>{
let counter = 0;
param.forEach(s=>{
if(s===item) counter++;
})
if(3==counter) triples.push(item);
})
return triples;
};
There is no correct way to do things like this. You can always optimize or sacrifice performance for readability, but that is up to the developer.
I changed nothing about the functionality in findTriplicates, but the code is very different.
findTriplicates2 works a little different but is by no means superior
const array = [
"peace",
"peace",
"Vrede",
"Patz",
"Salam",
"paz",
"Salam",
"Salam"
];
const findTriplicates = (param) => {
let counts = param.reduce((acc, p) => {
acc[p] ? acc[p]++ : acc[p] = 1
return acc;
}, {})
Object.keys(counts).forEach((key) => counts[key] === 3 &&
console.log(`${key} exists 3 times.`)
);
};
findTriplicates(array); // Salam exists 3 times.
const findTriplicates2 = (param) => {
let triples = [...new Set(param)].reduce((acc, item) => {
let counter = param.reduce((acc2, s) => {
if (s === item) acc2++;
return acc2;
}, 0);
if (3 == counter) acc.push(item);
return acc;
}, [])
triples.forEach((triple) => console.log(`${triple} exists 3 times.`));
};
findTriplicates2(array); // Salam exists 3 times.
You create an object to keep count of how many times the string repeats and then iterate through each element in the array while updating the count in the object. Then you filter through that object for any values equal to 3.
const array = [
'peace',
'peace',
'Vrede',
'Patz',
'Salam',
'paz',
'Salam',
'Salam',
];
// Object to keep count of each word
const countObj = {};
// Iterate through the array to get the count for each word
array.forEach((element) => {
// Does word exist in the count object? if so -> add 1 to its count, else -> add word and count 1 to the object
countObj[element] ? (countObj[element] += 1) : (countObj[element] = 1);
});
// Filter out keys that appear exactly 3 times and print them them
const filteredArray = Object.keys(countObj).filter(
(key) => countObj[key] === 3
);
console.log(filteredArray); // Salam
I have a string in the form
Key=asdf, num=90, Key=ert, num=20, Key=yged, num=20, Key=kned, num=35
I have to filter only Key num pairs which has value 20 and store them into a Key Value pair such that Key=ert, num=20 will be first record and Key=yged, num=20 will be second record so on. How can I use Map in JavaScript so that always first value will go as key and second will go as value and form pairs in this case. I have used the following :
var dataString = JSON.parse(data).data;
var arrayVal = new Array();
arrayVal = dataString.split(', ');
for(a in arrayVal){
console.log(arrayVal[a]);
}
Array.map probably isn't the best tool for the job here. Array.reduce would be a better approach. Since you know what your delimiters are (namespaces and equal signs), you can logically split things up so that you know that every other iteration will give you a key/value. You can then create a mechanism to track what the last key is so you can map the value to it.
const str = 'Key=asdf, num=90, Key=ert, num=20, Key=yged, num=20, Key=kned, num=35';
const arr = str.split(', ').reduce( (acc, curr) => {
const entry = curr.split('=');
const key = entry[0];
const val = entry[1];
if (key === 'Key') {
acc['last_key'] = val
acc[val] = null;
} else if (key === 'num') {
acc[acc['last_key']] = val;
}
return acc;
}, {});
delete arr.last_key;
console.log(arr);
Here you go. It's kinda ugly.
let result = dataString.split(', ')
.map((value, index, array) =>
value === 'num=20' && array[index - 1])
.filter(x => x);
console.log(result);
Here's my say
const dataString = JSON.parse(data).data;
const arrayVal = dataString.split(', ');
const obj = {}; // initialize the object
for (let i = 0; i < arrayVal.length; i += 2) { // Iterating over two elements for the key and value
const key = arrayVal[i].split('=')[1];
const value = arrayVal[i + 1].split('=')[1];
obj[key] = value;
}
console.log(obj);
I am having some trouble rationalising and aggregating an array of objects in javascript.
Given the array
[{"description":"Bright","size":"2XL","price":10.99},{"description":"Bright","size":"XL","price":10.99},{"description":"Bright","size":"L","price":9.99},{"group":"Foos","description":"Dull","size":"XL","price":9.99},{"description":"Dull","size":"L","price":8.99},{"description":"Dull","size":"2XL","price":9.99},{"description":"Shiny","size":"XL","price":9.99},{"description":"Shiny","size":"S","price":8.99},{"description":"Shiny","size":"3XL","price":10.3},{"description":"Shiny","size":"2XL","price":9.99}]
I am trying to convert it to an array in the format (actual values may be wrong here).
[{"descriptions":"Shiny, Bright, Dull","sizeRange":"S - L","price":8.99},{"descriptions":"Shiny, Bright, Dull","sizes":"XL - 2XL","price":9.99},{"descriptions":"Dark","sizes":"S - 2XL","price":10.99}]
That is - I wish to group each set of items by price, showing the descriptions and size ranges for them.
So far this is what I have, and it seems to be working, but it just seems very cumbersome. Really I'd be more than happy to use something like lodash or underscore if it would help to rationalise the code a bit rather than using native JS.
function groupBy (array, key) {
return array.reduce(function(value, property) {
(value[property[key]] = value[property[key]] || []).push(property);
return value;
}, {});
};
function unique(array) {
return Array.from(new Set(array));
};
function getRanges(data)
{
var result = [];
// simple map of sizes from smallest to largest to use for sorting
var sizeSort = {'S':1, 'M':2, 'L':3, 'XL':4, '2XL':5, '3XL':6, '4XL':7, '5XL':8};
// group the remaining variants by price
var group = groupBy(data, 'price');
// for each variant price group
for(var price in group) {
var item = {};
item.price = price;
// get the range of sizes sorted smallest to largest
var sizes = unique(group[price].map(function(i) {
return i.size;
})).sort(function(a, b) {
return sizeSort[a] - sizeSort[b];
});
// Add single size, or first and last size.
item.sizes = (sizes.length === 1) ?
sizes.shift() :
sizes.shift() + ' - ' + sizes.pop();
// Add the descriptions as alphabetically sorted CSV
item.description = unique(group[price].map(function(i) {
return i.description;
})).sort().join(", ");
result.push(item);
}
return result;
}
Here is a version using lodash..
I think it looks more rational..
function calc(data) {
var sizeSort = {'S':1, 'M':2, 'L':3, 'XL':4, '2XL':5,
'3XL':6, '4XL':7, '5XL':8};
return _.chain(data).
groupBy('price').
map(function(f){
var sizes = _.chain(f).map('size').uniq().
sortBy(function (a) { return sizeSort[a] }).value();
return {
price: _.head(f).price,
description: _.chain(f).map('description').uniq().join(',').value(),
size: sizes.length === 1 ? _.first(sizes) : _.join([_.first(sizes),_.last(sizes)], ' - ')
}
}).
sortBy(['price']).
value();
}
//put data at end, so not having to scroll down to see code
var data = [{"description":"Bright","size":"2XL","price":10.99},{"description":"Bright","size":"XL","price":10.99},{"description":"Bright","size":"L","price":9.99},{"group":"Foos","description":"Dull","size":"XL","price":9.99},{"description":"Dull","size":"L","price":8.99},{"description":"Dull","size":"2XL","price":9.99},{"description":"Shiny","size":"XL","price":9.99},{"description":"Shiny","size":"S","price":8.99},{"description":"Shiny","size":"3XL","price":10.3},{"description":"Shiny","size":"2XL","price":9.99}];
console.log(calc(data));
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.16.3/lodash.js"></script>
A vanilla JavaScript solution (with ES6 template strings)
/*
Some boilerplate functions. Listed underscore/lodash functions that
could replace them above
*/
// _.mapObject(object, reducer)
function reduceValues(object, reducer) {
let newObject = {}
for (var property in object) {
if (object.hasOwnProperty(property)) {
newObject[property] = reducer(object[property])
}
}
return newObject
}
// _.groupBy
function groupBy(arr, key) {
let reducer = (grouped, item) => {
let group_value = item[key]
if (!grouped[group_value]) {
grouped[group_value] = []
}
grouped[group_value].push(item)
return grouped
}
return arr.reduce(reducer, {})
}
// _.values
function objectValues(object) {
let values = []
for (var property in object) {
if (object.hasOwnProperty(property)) {
values.push(object[property])
}
}
return values
}
/*
Shirt specific functions and data
*/
// Mapping of shirts to their order.
let sizesToNumbers = {'S':1, 'M':2, 'L':3, 'XL':4, '2XL':5, '3XL':6, '4XL':7, '5XL':8};
// Create an intermediate summary with data instead of strings.
// This makes processing easier to write and reason about
function reduceShirtsToSummary(shirts) {
let reducer = (summary, shirt) => {
summary['descriptions'].add(shirt['description'])
let shirtSize = shirt['size']
if (!summary['smallestSize'] || sizesToNumbers[shirtSize] < sizesToNumbers[summary['smallestSize']]) {
summary['smallestSize'] = shirtSize
}
if (!summary['largestSize'] || sizesToNumbers[shirtSize] > sizesToNumbers[summary['largestSize']]) {
summary['largestSize'] = shirtSize
}
summary['prices'].push(shirt['price'])
return summary
}
return shirts.reduce(reducer, {'descriptions': new Set(), 'prices': []})
}
// Convert the shirt summary data into the "labelized" version with strings in the example
function labelizeShirtSummary(shirtSummary) {
let labelizedShirtSummary = {}
labelizedShirtSummary['descriptions'] = Array.from(shirtSummary['descriptions']).join(', ')
labelizedShirtSummary['price'] = shirtSummary['prices'][0]
labelizedShirtSummary['sizes'] = `${shirtSummary['smallestSize']} - ${shirtSummary['largestSize']}`
return labelizedShirtSummary
}
let grouped = groupBy(shirts, 'price')
let groupedAndSummarized = reduceValues(grouped, reduceShirtsToSummary)
let labelizedSummaries = objectValues(groupedAndSummarized).map(labelizeShirtSummary)
// Gives desired output
console.log(labelizedSummaries)
I have an array with X number of items. Each has variables separated by a pipe character. In a loop I can split on the pipe to get the second item; but how do I splice to remove the duplicate.
"Sometext|22621086|address|333629dc87894a7ea7df5291fa6d1836|PC_E|1803"
"Sometext2|22622138|working|d3e70175ffe942568cd21f1cf96f4d63|PC_E|1803"
"Sometext3|22622138|working|851946e6325445da99c113951590f714|PC_E|1803"
Results should be this.
"Sometext|22621086|address|333629dc87894a7ea7df5291fa6d1836|PC_E|1803"
"Sometext2|22622138|working|d3e70175ffe942568cd21f1cf96f4d63|PC_E|1803"
Note that the duplicate 22622138 is a random number so the solution needs to work for any number in this location (it's always in the arr[1] position).
This is what I tried:
$.each(arr_transcript, function (i, e) {
if (e.length != 0) {
var arr = e.split("|")
var i = arr_transcript.indexOf(arr[1]);
if (i != -1) {
arr_transcript.splice(i, 1);
}
}
});
Here's a generic function:
function uniqBy(a, key) {
let seen = new Set();
return a.filter(item => {
let k = key(item);
return !seen.has(k) && seen.add(k);
});
};
var data = [
"Sometext|22621086|address|333629dc87894a7ea7df5291fa6d1836|PC_E|1803",
"Sometext2|22622138|working|d3e70175ffe942568cd21f1cf96f4d63|PC_E|1803",
"Sometext3|22622138|working|851946e6325445da99c113951590f714|PC_E|1803"
];
var result = uniqBy(data, item => item.split('|')[1]);
console.log(result)
See here for more info.
Create a map of the numbers you want to check against, and then filter based on that
var arr_transcript = [
"Sometext|22621086|address|333629dc87894a7ea7df5291fa6d1836|PC_E|1803",
"Sometext2|22622138|working|d3e70175ffe942568cd21f1cf96f4d63|PC_E|1803",
"Sometext3|22622138|working|851946e6325445da99c113951590f714|PC_E|1803"
];
var map = arr_transcript.map(function(text) {
return text.split('|')[1];
});
var filtered = arr_transcript.filter(function(item, index) {
return index === map.lastIndexOf( map[index] );
});
console.log(filtered)