[SciPy-User] scipy.stats.nanstd, bias and ddof
Fri Jan 15 12:48:11 CST 2010
On Fri, Jan 15, 2010 at 1:07 PM, Keith Goodman <email@example.com> wrote:
> By default np.std and scipy.std normalize by N. But scipy.stats.nanstd
> normalizes by N-1.
>>> x = np.random.rand(4)
>>> scipy.stats.nanstd(x, bias=True)
> Can the default for nanstd be changed to bias=True? Or would that break code?
> Even better I guess would be to replace the bias keyword with ddof as
> used in np.std and scipy.std. So
> if bias:
> m2c = m2 / n
> m2c = m2 / (n - 1.)
> in scipy.stats.nanstd would become
> m2c = m2 / (n - ddof)
> For me it doesn't matter if the default ddof is 0 or 1. But it is nice
> when all std functions use the same default.
I agree with the consistency across function argument. But changing
the degrees of freedom will affect user code, and we would have to go
through a warning period, and maybe add the ddof argument in the
meantime. (but having both bias and ddof as arguments would be a bit
Or maybe numpy should get a nanmean and nanvar, nanstd, similar to
nansum ? Then it would be easier to depreciate like the other stats
functions that moved to numpy.
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