exp() overflow error python 3

Question

I tried various solutions for below, but I still get the errors as described:

log1p(1 + math.exp(comp * -1))

Error: OverflowError: math range error

So I changed it to: log1p(1 + np.exp(comp * -1)) Now I get error : RuntimeWarning: overflow encountered in exp

So again based on some suggestion on previous questions asked I changed it to: log1p(1 + np.exp((comp * -1), dtype=np.float256))

Now my error is : module 'numpy' has no attribute 'float256'

Any other suggestions? Please help thanks!

EDIT: X -> Input feature array of 'N' rows and 'm' features. w -> weight vector of size 'm'

    for rowIndex in range(len(X)):
        val1 = np.sum(np.dot(X[rowIndex], w))
        val2 = y[rowIndex]
        comp = np.dot(val2, val1)
        loss = loss + log1p(1 + np.exp((comp * -1)))

What is the value of comp? Also, what you are attempting to compute is ln(2 + exp(-comp)). Is this really what you want? — jmd_dk
– jmd_dk, Commented Mar 27, 2018 at 18:25
'comp' is just some value which could go upto 6000 in some cases... 'log1p(1 + math.exp(comp * -1))' is what I want to do... — user3868051
– user3868051, Commented Mar 27, 2018 at 18:30
Do you realise that log1p(x) = log(x + 1)? Also, log1p(1 + math.exp(comp * -1)) works fine for various values of comp, including 6000. Please provide a complete example demonstrating the problem. — jmd_dk
– jmd_dk, Commented Mar 27, 2018 at 18:33
I note that already for comp = 40, the exp part can be treated as zero: log1p(1+exp(-40))-log(2) is zero. — Arndt Jonasson
– Arndt Jonasson, Commented Mar 27, 2018 at 18:34
Can comp be negative? What you describe doesn't seem possible otherwise. And if comp is large and negative, then the 1 + ... will be insignificant, so your expression log1p(1 + math.exp(comp * -1)) will be simply -comp, to within numerical error. (Also: what @jmd_dk said: are you sure you want log(1 + 1 + math.exp(...))? — Mark Dickinson
– Mark Dickinson, Commented Mar 27, 2018 at 19:23

user3868051 · Accepted Answer · 2018-03-28 02:48:56Z

0

I replaced the code as below:

loss = loss - log1p(expit(val))

Basically I rearranged my code to be able to use the expit function...

answered Mar 28, 2018 at 2:48

user3868051

1,2693 gold badges26 silver badges49 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Mark Dickinson Over a year ago

That doesn't look like the same thing at all. Note that it's not true that log1p(thing) = -log1p(1/thing). Are you sure you mean log1p and not log?

Collectives™ on Stack Overflow

exp() overflow error python 3

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related