-1

sample data

 array_data = [
  [6326586, 0.4126],
  [6326586, 0.48],
  [6326586, 0.45],
  [6326586, 0.35],
  [6326586, 0.2685],
  [6326614, 0.4008],
  [6326614, 0.1],
  [6326614, 0.074],
  [6327407, 0.066],
  [6327408, 0.1],
  [6344999, 0.1572],
  [6344999, 0.003],
  [6394500, 0.2112]
]

I have a nested array if the first index value is the same as the other value then sum of these values as show below result

I need an expected array

new_array_data = [
  [6326586, 1.9611],
  [6326614,0.5748],
  [6327407, 0.066],
  [6327408, 0.1],
  [6344999, 0.1602],
  [6394500, 0.2112],
]

Thanks

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2 Answers 2

Reset to default
2

Using Map and Array#Reduce in Vanilla JS

const 
  arr = [[6326586,.4126],[6326586,.48],[6326586,.45],[6326586,.35],[6326586,.2685],[6326614,.4008],[6326614,.1],[6326614,.074],[6327407,.066],[6327408,.1],[6344999,.1572],[6344999,.003],[6394500,.2112]], 

  res = Array.from(arr.reduce((m, [k, v]) => m.set(k, (m.get(k) || 0) + v), new Map()))

console.log(res)

Using Object and Array#Reduce in Vanilla JS

** Go through the comments before using this

const 
  arr = [[6326586,.4126],[6326586,.48],[6326586,.45],[6326586,.35],[6326586,.2685],[6326614,.4008],[6326614,.1],[6326614,.074],[6327407,.066],[6327408,.1],[6344999,.1572],[6344999,.003],[6394500,.2112]], 
  
  res = Object.values(arr.reduce((o, [k, v]) => (o[k] ??= [k, 0], o[k][1] += v, o), {}));

console.log(res)

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5 Comments

With object as accumulator, the output has strings and not numbers in 0 index
@adiga Thanks a lot for mentioning this, I completely missed this. I have currently mapped the result, is there any other better option?
@SomShekharMukherjee Its helpful for me thanks
You could change it to how Sebastian Richner's has handled it like o[k] ??= [k, 0], o[k][1] += v and use Object.values() instead of entries. But, object accumulator are a bit unreliable. If you had an entry say, ["constructor",.4126], it will check if constructor exists on the object and it will return Object.prototype.constructor and try to append to that function
@adiga Again an interesting point +1. So, when I am checking if k is in o, for "constructor" it becomes true and then it starts adding to the function. Have I understood it correctly? But I hope OP will not have strings in keys.
2

You can do this in plain JavaScript, you don't need jQuery. You can use Array.reduce as follows:

const array_data = [
  [6326586, 0.4126],
  [6326586, 0.48],
  [6326586, 0.45],
  [6326586, 0.35],
  [6326586, 0.2685],
  [6326614, 0.4008],
  [6326614, 0.1],
  [6326614, 0.074],
  [6327407, 0.066],
  [6327408, 0.1],
  [6344999, 0.1572],
  [6344999, 0.003],
  [6394500, 0.2112]
]

const result = array_data.reduce((acc, [key, value]) => {
  if (acc[key]) {
    acc[key][1] += value;
  } else {
    acc[key] = [key, 0];
  }
  return acc;
}, {});

console.log(Object.values(result));

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2 Comments

The output has strings and not numbers in 0 index
You're right, check the updated answer.

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