I have this string:
item_type = 'a, a, b'
How can I count the values so I would get something like this:
number_of_a = 2
number_of_b = 1
I tried something like below but got some JavaScript for input string "[object Object]" error
if(item_type != null) {
item_type = item_type.split(",");
item_type.forEach(function(x) {
number_of_a[x] = (number_of_a[x] || 0) + 1;
});
}
reduce is your friend, as with any grouping operation
const item_type = 'a, a, b';
const result = item_type.split(", ").reduce ( (a,i) => {
a[i] = (a[i] +1 || 1)
return a;
},{})
console.log(result);If we are interested in only alphabets, improving the @Tamas Szoke answer.
const str = 'a, a, b'
const chars = {}
for (let char of str) {
if(char.toUpperCase() != char.toLowerCase()){
chars[char] = chars[char] + 1 || 1;
}
}
console.log(chars)If there could be only a or b you can do it that way:
const item_type = 'a, a, b'
let number_of_a = 0
let number_of_b = 0
if (item_type != null) {
item_type
.split(',')
.map(el => el.trim()) //remove whitespaces from each element
.forEach(el => el === 'a' ? number_of_a++ : number_of_b++)
}
console.log(`a: ${number_of_a}`)
console.log(`b: ${number_of_b}`)Otherwise you should use an array or an object:
const item_type = 'a, a, b'
const occurrences = {}
if (item_type != null) {
item_type
.split(',')
.map(el => el.trim()) //remove whitespaces from each element
.forEach(el => {
if (!occurrences[el]) {
occurrences[el] = 1
} else {
occurrences[el] += 1
}
})
}
for (const [key, value] of Object.entries(occurrences)) {
console.log(`${key}: ${value}`)
}