softmax function implementation in js

softmax.js

// Fork & examples for the one-line version by @vladimir-ivanov:
//let softmax = (arr) => (index) => Math.exp(arr[index]) / arr.map(y => Math.exp(y)).reduce((a, b) => a + b);
//
// Also see comments for improvements

function softmax(arr) {
    return arr.map(function(value,index) { 
      return Math.exp(value) / arr.map( function(y /*value*/){ return Math.exp(y) } ).reduce( function(a,b){ return a+b })
    })
}

example1=[ 0.9780449271202087,
  0.01590355671942234,
  0.0019390975357964635,
  0.0015482910675927997,
  0.0012942816829308867,
  0.0006004497990943491,
  0.0004827099328394979,
  0.0001868270628619939 ]
softmax1=softmax(example1)

example2= [ 
  { prob: 0.32289665937423706, cat: '25_32' },
  { prob: 0.15404804050922394, cat: '38_43' },
  { prob: 0.03673655539751053, cat: '4_6' },
  { prob: 0.01545996405184269, cat: '48_53' },
  { prob: 0.011709162034094334, cat: '15_20' },
  { prob: 0.008010754361748695, cat: '8_13' },
  { last: true, prob: 0.0054732030257582664, cat: '60+' } ].map(function(v){return v.prob})
softmax2=softmax(example2)

example3=[ { prob: 0.125, cat: '25_32' },
  { prob: 0.125, cat: '38_43' },
  { prob: 0.125, cat: '15_20' },
  { prob: 0.125, cat: '8_13' },
  { prob: 0.125, cat: '4_6' },
  { prob: 0.125, cat: '48_53' },
  { prob: 0.125, cat: '60+' },
  { prob: 0.125, cat: '0_2' } ].map(function(v){return v.prob})
softmax3=softmax(example3)

Nice job. I was looking for a couple of practical examples and this helps!

elated to know it helped. thanks

Nice job.

I see two problems with the code.
(1) It is slow (repeating calculating the denominator for the same result)
(2) Math.exp(v) could return Infinity (when v is large), softmax would result in NaN

Here's a solution:

function softmax(arr) {
    const C = Math.max(...arr);
    const d = arr.map((y) => Math.exp(y - C)).reduce((a, b) => a + b);
    return arr.map((value, index) => { 
        return Math.exp(value - C) / d;
    })
}

Late to the party but here is my solution inspired by @enobufs's but without repeating the exponentiation.

function softmax(logits) {
    const maxLogit = Math.max(...logits);
    const scores = logits.map(l => Math.exp(l - maxLogit));
    const denom = scores.reduce((a, b) => a + b);
    return scores.map(s => s / denom);
}

More efficient version.

1. Avoid using Math.max(...arr) which will cause RangeError: Maximum call stack size exceeded
2. No need in additional loop to calculate denominator.

// data could be both array or object
function softmax(data, from = 0, to = data.length) {
  let max = -Infinity; // Math.max(...data) vould crash on large array
  for (let id = from; id < to; id++) {
    if (max < data[id]) {
      max = data[id];
    }
  }
  // No need to use reduce, just sum the exps in the loop
  let sumOfExp = 0;
  const result = Array.isArray(data) ? [] : {};
  for (let id = from; id < to; id++) {
    result[id] = Math.exp(data[id] - max);
    sumOfExp += result[id];
  }
  // Finally divide by the sum of exps
  for (let id = from; id < to; id++) {
    result[id] = result[id] / sumOfExp;
  }

  return result;
}

If you still want to use prettier version you could just replace Math.max(...arr) with reduce.

function softmax(logits) {
  //find max in logits using reduce
  const maxLogit = logits.reduce((a, b) => Math.max(a, b), -Infinity);
  const scores = logits.map((l) => Math.exp(l - maxLogit));
  const denom = scores.reduce((a, b) => a + b);
  return scores.map((s) => s / denom);
}

By any chance, does anyone have an example of the softmax derivative ? Or is able to tell me if this one is correct:

function derivative(arr) {
  const values = softmax(arr);

  return arr.map((x, i) => {
    return values[i] * (values[i] - (i === 0 ? 1 : 0));
  });
}

Math.exp(x) can grows fast, if you use very large number for x, this may cause exponential overflows infinity. For example,

Math.exp(1000)
will cause overflow. Here is the implementation that used by PyTorch, Tensorflow, Cuda

softmax(logits) {
  const highest = Math.max(...logits);
  const shifted = logits.map(score => Math.exp(score - highest));
  const total = shifted.reduce((acc, val) => acc + val, 0);
  return shifted.map(prob => prob / total);
}

Because the softmax formula is invariant under constant shifts, that means mathematically, subtracting the same value from all inputs doesn’t change the output