Finding the sum of a nested array
Question
I tried finding the sum of all numbers of a nested array, but I don't get it to work correctly. This is what I tried:
function arraySum(i) {
sum = 0;
for (a = 0; a < i.length; a++) {
if (typeof i[a] == 'number') {
sum += i[a];
} else if (i[a] instanceof Array) {
sum += arraySum(i[a]);
}
}
return sum;
}
When you try it out with the array [[1,2,3],4,5] , it gets 6 as the answer, instead of 15 . Does somebody know where there is a mistake in it?
10 answers
solution1
8 ACCPTED 2013-10-04 12:55:15
The problem with your code is that the sum and a variables are global, instead of local. Because of this you get an infinite loop (a from the first entry in the function is reset by the second entry, so the same elements are processed again).
Fix it by adding var to where sum and a are declared to make them local to the function:
function arraySum(i) {
var sum=0; // missing var added
for(var a=0;a<i.length;a++){ // missing var added
if(typeof i[a]=="number"){
sum+=i[a];
}else if(i[a] instanceof Array){
sum+=arraySum(i[a]);
}
}
return sum;
}
solution2
1 2018-10-17 05:35:37
solution3
1 2020-05-23 07:20:09
I know it's late, but they say "is never late" :)
const sumNestedArray = arr => arr.flat(Infinity).reduce((a,b)=> a+b, 0)
const sumNestedArray = arr => arr.flat(Infinity).reduce((a,b)=> a+b, 0) console.log(sumNestedArray([1,[2], [2, 3, [4]]]))
sumNestedArray([1,[2], [2, 3, [4]]])
solution4
0 2013-10-04 12:46:29
You're missing two var in there. You've implicitly declared sum and a at window scope:
function arraySum(i) {
**var** sum=0;
for(**var** a=0;a<i.length;a++){
if(typeof i[a]=="number"){
sum+=i[a];
}else if(i[a] instanceof Array){
sum+=arraySum(i[a]);
}
}
return sum;
}
solution5
0 2013-10-04 12:52:10
Recurse, for example
function arraySum(x) {
var sum = 0, i;
if (typeof x === 'number')
return x;
else if (x instanceof Array)
for (i = 0; i < x.length; ++i)
sum += arraySum(x[i]);
return sum;
}
arraySum([[1,2,3],4,5]); // 15
I didn't optimise this so that it's clear, you may want some more logic before recursion.
The reason yours isn't working is because you need to var both sum and a .
solution6
0 2013-10-04 13:32:16
First of all why you use 'i' as input to function? we using 'i' to denote running index.. Regarding your question,you want 'a' to be local in your loop so instead of "for(a=0;..." instead write "for(var a=0;"
<html>
<body>
<script>
function NestedArraySummation(arr)
{
var sum=0;
for(var i=0;i<arr.length;i++)
{
if(typeof arr[i]=="number")
sum=sum+arr[i];
else if(arr[i] instanceof Array)
sum=sum+NestedArraySummation(arr[i]);
}
return sum;
}
var MyArray=[1,[2,3],4,10,[1,2]];
var Sum=NestedArraySummation(MyArray);
document.write(Sum);
</script>
</body>
</html>
solution7
0 2018-11-08 03:25:26
This can be done with lodash _.flattenDeep and _.sum :
const arr = [[1, 2, 3], 4, 5]; arraySum(arr); function arraySum(arr) { var arrFlattens = _.flattenDeep(arr); // => [1, 2, 3, 4, 5] console.log(_.sum(arrFlattens)); }
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>
solution8
0 2021-06-14 16:16:35
Here you go:
(My assumption is that you want to parse strings, eg, '13', into numbers in order include them in the sum. If not, just change isNumber to typeof val === 'number' .
function arraySum(arr) {
let sum = 0;
while (arr.length) {
const val = arr.pop();
const isArray = typeof val === 'object' && val.length !== undefined;
const isNumber = !isArray && !isNaN(sum + parseFloat(val));
if (isArray && val.length) {
sum += arraySum(val);
}
else if (isNumber) {
sum += parseFloat(val);
}
}
return sum;
}
console.log(arraySum([])); //0
console.log(arraySum([1, 1, 1, [3, 4, [8]], [4]])); //22
console.log(arraySum([1, 1, [], {}, 'l', 1, [3, 4, [8]], [4]])); //22
solution9
0 2022-04-06 02:32:09
This was posted long ago but it might help others. I did the following and it seems to be working just fine
function countArray (array) { //iterate through the array for(let i = 0;i < array.length; i++){ //check if element of array is an array, and if it is, run method flat() to the whole array. if(Array.isArray(array[i])){ array = array.flat() } } //return the sum of the elements of the array return array.reduce((a,b) => a + b); }
solution10
-1 2016-02-06 11:17:04
from __builtin__ import int
def nestedLists(mylist):
sum = 0
if checkIfAllInt(mylist):
result = addList(mylist)
return result
else:
for value in mylist:
sum = sum + nestedLists(value)
return sum
def addList(listdata):
if(type(listdata) is int):
return listdata
else:
return sum(listdata)
def checkIfAllInt(listdata):
check = True
if(type(listdata) is int):
return check
else:
for value in listdata:
if(not type(value) is int):
check = False
return check
return check
print nestedLists([1,[2,3,[4,5,[6,7]]]])
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.