numpy division with RuntimeWarning invalid value encountered in double scalars

Question

I wrote the following script   import numpy  d   numpy array   1089  1093    e   numpy array   1000  4443    answer   numpy exp -3   d  answer1   numpy exp -3   e  res   answer sum   answer1 sum   print res   But I got this result and with the error occurred   nan C  Users Desktop test py 16  RuntimeWarning  invalid value encountered in double scalars   res   answer sum   answer1 sum     It seems to be that the input element were too small that python turned them to be zeros  but indeed the division has its result   How to solve this kind of problem

User · Answer

You can use np logaddexp  which implements the idea in  gg349 s answer    In  33   d   np array   1089  1093     In  34   e   np array   1000  4443     In  35   log res   np logaddexp -3 d 0 0   -3 d 0 1   - np logaddexp -3 e 0 0   -3 e 0 1    In  36   log res Out 36   -266 99999385580668  In  37   res   exp log res   In  38   res Out 38   1 1050349147204485e-116   Or you can use scipy special logsumexp   In  52   from scipy special import logsumexp  In  53   res   np exp logsumexp -3 d  - logsumexp -3 e    In  54   res Out 54   1 1050349147204485e-116

User · Answer

You can t solve it  Simply answer1 sum    0  and you can t perform a division by zero    This happens because answer1 is the exponential of 2 very large  negative numbers  so that the result is rounded to zero   nan is returned in this case because of the division by zero   Now to solve your problem you could    go for a library for high-precision mathematics  like mpmath  But that s less fun  as an alternative to a bigger weapon  do some math manipulation  as detailed below  go for a tailored scipy numpy function that does exactly what you want  Check out  Warren Weckesser answer    Here I explain how to do some math manipulation that helps on this problem  We have that for the numerator   exp -x  exp -y    exp log exp -x  exp -y                      exp log exp -x   1 exp -y x                       exp log exp -x    log 1 exp -y x                      exp -x   log 1 exp -y x      where above x 3  1089 and y 3  1093  Now  the argument of this exponential is  -x   log 1 exp -y x     -x   6 1441934777474324e-06  For the denominator you could proceed similarly but obtain that log 1 exp -z k   is already rounded to 0  so that the argument of the exponential function at the denominator is simply rounded to -z -3000  You then have that your result is  exp -x   log 1 exp -y x    exp -z    exp -x z log 1 exp -y x                                         exp -266 99999385580668    which is already extremely close to the result that you would get if you were to keep only the 2 leading terms  i e  the first number 1089 in the numerator and the first number 1000 at the denominator    exp 3  1089-1000   exp -267    For the sake of it  let s see how close we are from the solution of Wolfram alpha  link    Log  exp -3 1089  exp -3 1093     exp -3 1000  exp -3 4443    - gt  -266 999993855806522267194565420933791813296828742310997510523   The difference between this number and the exponent above is  1 7053025658242404e-13  so the approximation we made at the denominator was fine   The final result is    exp -266 99999385580668    1 1050349147204485e-116   From wolfram alpha is  link   1 105034914720621496      10 -116   Wolfram alpha    and again  it is safe to use numpy here too

[python] numpy division with RuntimeWarning: invalid value encountered in double_scalars

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