Is it possible to use numpy.nanargmin, so that it returns numpy.nan, on columns where there are only nans in them. Right now, it raises a ValueError, when that happens. And i cant use numpy.argmin, since that will fail when there are only a few nans in the column.
http://docs.scipy.org/doc/numpy/reference/generated/numpy.nanargmin.html says that the ValueError is raised for all-nan slices. In that case, i want it to return numpy.nan (just to further mask the "non-data" with nans)
this next bit does this, but is super-slow and not really pythonic:
for i in range(R.shape[0]): bestindex = numpy.nanargmin(R[i,:]) if(numpy.isnan(bestindex)): bestepsilons[i]=numpy.nan else: bestepsilons[i]=epsilon[bestindex]
This next bit works too, but only if no all-nan columns are involved:
ar = numpy.nanargmin(R, axis=1) bestepsilons = epsilon[ar]
So ideally i would want this last bit to work with all-nan columns as well
>>> def _nanargmin(arr, axis): ... try: ... return np.nanargmin(arr, axis) ... except ValueError: ... return np.nan
Demo:
>>> a = np.array([[np.nan]*10, np.ones(10)]) >>> _nanargmin(a, axis=1) nan >>> _nanargmin(a, axis=0) array([1, 1, 1, 1, 1, 1, 1, 1, 1, 1])
Anyway, it's unlikely to be what you want. Not sure what exactly you are after. If all you want is to filter away the nans, then use boolean indexing:
>>> a[~np.isnan(a)] array([ 1., 1., 1., 1., 1., 1., 1., 1., 1., 1.]) >>> np.argmin(_) 0
EDIT2: Looks like you're after the masked arrays:
>>> a = np.vstack(([np.nan]*10, np.arange(10), np.arange(11, 1, -1))) >>> a[2, 4] = np.nan >>> m = np.ma.masked_array(a, np.isnan(a)) >>> np.argmin(m, axis=0) array([1, 1, 1, 1, 1, 1, 2, 2, 2, 2]) >>> np.argmin(m, axis=1) array([0, 0, 9])
Found a solution:
# makes everything nan to start with bestepsilons1 = numpy.zeros(R.shape[0])+numpy.nan # finds the indices where the entire column would be nan, so the nanargmin would raise an error d0 = numpy.nanmin(R, axis=1) # on the indices where we do not have a nan-column, get the right index with nanargmin, and than put the right value in those points bestepsilons1[~numpy.isnan(d0)] = epsilon[numpy.nanargmin(R[~numpy.isnan(d0),:], axis=1)]
This basically is a workaround, by only taking the nanargmin on the places where it will not give an error, since at those places we want the resulting index to be a nan anyways
