integer dot product (Trac #1192) #1790

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1. thouis commented on Oct 19, 2012

   

   thouis

   on Oct 19, 2012

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   @rgommers wrote on 2011-03-01

   Your example works fine with int32, see below, so your assertion it works only with floats is false. If one uses int8 the expected return values are also of dtype int8. If you use int8 arrays, it's often because you want to use integer arithmetic with its own rules. So I think this ticket can be closed.

   In [3]: from numpy.random import randint
   
   In [4]: a = randint(0, 2, 5000).astype(np.int8)
   
   In [5]: a
   Out[5]: array([1, 1, 1, ..., 0, 1, 0], dtype=int8)
   
   In [6]: b = randint(0, 2, 5000).astype(np.int8)
   
   In [8]: np.dot(a, b); sum(a*b)
   Out[8]: -13
   Out[8]: 1267
   
   In [10]: a = randint(0, 2, 5000).astype(np.int32)
   
   In [11]: b = randint(0, 2, 5000).astype(np.int32)
   
   In [12]: np.dot(a, b); sum(a*b)
   Out[12]: 1255
   Out[12]: 1255
   
2. thouis commented on Oct 19, 2012

   

   thouis

   on Oct 19, 2012

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   @bsouthey wrote on 2011-03-30

   (Any examples must use numpy's functions.)
   This is invalid because dot keeps the input dtype causing an overflow. This can be seen by using the dtype argument:

   np.sum(a*b).dtype # dtype('int64') on 64-bit linux 
   np.sum(a*b, dtype=numpy.int8).dtype # dtype('int8') but wrong answer
   

   By default numpy.sum (and probably Python's sum) uses the platform default and will give the correct answer until that precision overflows.
3. thouis commented on Oct 19, 2012

   

   thouis

   on Oct 19, 2012

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   @rgommers wrote on 2011-03-30

   Not sure if you are saying the same thing as I did here, expecting a floating point warning, or yet something else?
4. thouis

   closed this as completedon Oct 19, 2012