[SciPy-dev] modulo operation and new scipy core

Travis Oliphant oliphant at ee.byu.edu
Wed Oct 12 03:46:11 CDT 2005

Arnd Baecker wrote:

>one thing which I find irritating is the behaviour of
>the modulo operation for arrays:
>In [1]: from scipy import *
>In [2]: -0.4 % 1.0
>Out[2]: 0.59999999999999998
>In [3]: x=arange(-0.6,1.0,0.1)
>In [4]: x%1.0
>array([ -6.00000000e-01,  -5.00000000e-01,  -4.00000000e-01,
>        -3.00000000e-01,  -2.00000000e-01,  -1.00000000e-01,
>         1.11022302e-16,   1.00000000e-01,   2.00000000e-01,
>         3.00000000e-01,   4.00000000e-01,   5.00000000e-01,
>         6.00000000e-01,   7.00000000e-01,   8.00000000e-01,
>         9.00000000e-01])
>Even worse (IMHO): take a scalar (I know it is still an array,
>but it does not look like one ;-) from the array
Actually it is a real scalar (it's just using the array math right now).

>In [5]: x[2]
>Out[5]: -0.39999999999999997
>In [6]: x[2] % 1.0
>Out[6]: -0.39999999999999997
>It seems that for arrays % behaves like `fmod` and not like `mod`.
Yes, that has been the behavior of Numeric.  There is the mod function for
arrays.   Should we switch that?   It will cause a couple of 
incompatibilities if people relied on the old (arguably) non-standard 

>I find this confusing as it is in contrast to the
>python 2.4 documentation:
>"5.6. Binary arithmetic operations"
>   """The % (modulo) operator yields the remainder from the division
>      of the first argument by the second. [...]
>      The arguments may be floating point numbers, e.g.,
>      3.14%0.7 equals 0.34 (since 3.14 equals 4*0.7 + 0.34.)
>      The modulo operator always yields a result with the same sign as
>      its second operand (or zero); the absolute value of the result
>      is strictly smaller than the absolute value of the second
>      operand."""
>Would it be possible for the new scipy core that % behaves
>the same (standard python) way for scalars and for arrays?
I would do this.  What do others think?


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