On Wed, Oct 24, 2012 at 4:47 AM, Robert Kern <robert.k...@gmail.com> wrote:
> How about this?
>
>
> def nancumsum(x):
> nans = np.isnan(x)
> x = np.array(x)
> x[nans] = 0
> reset_idx = np.zeros(len(x), dtype=int)
> reset_idx[nans] = np.arange(len(x))[nans]
> reset_idx = np.maximum.accumulate(reset_idx)
> cumsum = np.cumsum(x)
> cumsum = cumsum - cumsum[reset_idx]
> return cumsum
>
Thank you for putting in the time to look at this.
It doesn't work for the first group of numbers if x[0] is non-zero. Could
perhaps concatenate a np.nan at the beginning to force a reset and adjust
the returned array to not include the dummy value...
def nancumsum(x):
x = np.concatenate(([np.nan], x))
nans = np.isnan(x)
x = np.array(x)
x[nans] = 0
reset_idx = np.zeros(len(x), dtype=int)
reset_idx[nans] = np.arange(len(x))[nans]
reset_idx = np.maximum.accumulate(reset_idx)
cumsum = np.cumsum(x)
cumsum = cumsum - cumsum[reset_idx]
return cumsum[1:]
>>> a
array([ 4., 1., 2., 0., 18., 5., 6., 0., 8., 9.],
dtype=float32)
If no np.nan, then 'nancumsum' and 'np.cumsum' should be the same...
>>> np.cumsum(a)
array([ 4., 5., 7., 7., 25., 30., 36., 36., 44., 53.],
dtype=float32)
>>> nancumsum(a)
array([ 4., 5., 7., 7., 25., 30., 36., 36., 44., 53.])
>>> a[3] = np.nan
>>> np.cumsum(a)
array([ 4., 5., 7., nan, nan, nan, nan, nan, nan, nan],
dtype=float32)
>>> nancumsum(a)
array([ 4., 5., 7., 0., 18., 23., 29., 29., 37., 46.])
Excellent!
Kindest regards,
Tim
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