[Numpy-discussion] Single precision equivalents of missing C99 functions

Francesc Alted faltet at pytables.org
Mon Jun 1 12:22:03 EDT 2009


In the process of adding single precision support to Numexpr, I'm 
experimenting a divergence between Numexpr and NumPy computations.  It all 
boils down to the fact that my implementation defined single precision 
functions completely.  As for one, consider my version of expm1f:

inline static float expm1f(float x)
    float u = expf(x);
    if (u == 1.0) {
        return x;
    } else if (u-1.0 == -1.0) {
        return -1;
    } else {
        return (u-1.0) * x/logf(u);

while NumPy seems to declare expm1f as:

static float expm1f(float x)
    return (float) expm1((double)x);

This leads to different results on Windows when computing expm1(x) for large 
values of x (like 99.), where my approach returns a 'nan', while NumPy returns 
an 'inf'.  Curiously, on Linux both approaches returns 'inf'.

I suppose that the NumPy crew already experimented this divergence and finally 
used the cast approach for computing the single precision functions.  However, 
this is effectively preventing the use of optimized functions for single 
precision (i.e. double precision 'exp' and 'log' are used instead of single 
precision specific 'expf' and 'logf'), which could perform potentially better.  
So, I'm wondering if it would not be better to use a native implementation 
instead.  Thoughts?


Francesc Alted

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