Why does ddof=2 and ddof=3 give the same result?
np.var([1, 2, 3], ddof=0)
0.66666666666666663
np.var([1, 2, 3], ddof=1)
1.0
np.var([1, 2, 3], ddof=2)
2.0
np.var([1, 2, 3], ddof=3)
2.0
np.var([1, 2, 3], ddof=4)
-2.0
I expected NaN for ddof=3.
On Fri, Dec 10, 2010 at 4:42 PM, Keith Goodman kwgoodman@gmail.com wrote:
Why does ddof=2 and ddof=3 give the same result?
np.var([1, 2, 3], ddof=0)
0.66666666666666663
np.var([1, 2, 3], ddof=1)
1.0
np.var([1, 2, 3], ddof=2)
2.0
np.var([1, 2, 3], ddof=3)
2.0
np.var([1, 2, 3], ddof=4)
-2.0
I expected NaN for ddof=3.
It's a floating point calculation, so I would expect np.inf
Josef
NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
On Fri, Dec 10, 2010 at 2:26 PM, josef.pktd@gmail.com wrote:
On Fri, Dec 10, 2010 at 4:42 PM, Keith Goodman kwgoodman@gmail.com wrote:
Why does ddof=2 and ddof=3 give the same result?
np.var([1, 2, 3], ddof=0)
0.66666666666666663
np.var([1, 2, 3], ddof=1)
1.0
np.var([1, 2, 3], ddof=2)
2.0
np.var([1, 2, 3], ddof=3)
2.0
np.var([1, 2, 3], ddof=4)
-2.0
I expected NaN for ddof=3.
It's a floating point calculation, so I would expect np.inf
Right, NAFN (F=Finite). Unless, of course, the numerator is zero too.
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Keith Goodman