Numpy.trim_zeros for structured array without creating new array -

is possible trim 0 'records' of structured numpy array without copying it; i.e. free allocated memory 'unused' 0 entries @ beginning or end; actually, interested in trimming zeros @ end.

there builtin function numpy.trim_zeros() 1d arrays. return value:

returns:

trimmed : 1-d array or sequence

the result of trimming input. input data type preserved.

however, can't whether not create copy , frees memory. not proficient enough tell source code behaviour.

more specifically, have following code:

import numpy edges = numpy.zeros(3, dtype=[('i', 'i4'), ('j', 'i4'), ('length', 'f4')]) # fill first 2 records sensible data: edges[0]['i'] = 0 edges[0]['j'] = 1 edges[0]['length'] = 2.0 edges[1]['i'] = 1 edges[1]['j'] = 2 edges[1]['length'] = 2.0 # list memory adress , size edges.__array_interface__ edges = numpy.trim_zeros(edges)  # not work structured array edges.__array_interface__ 

update

my question 'twofold':

1) builtin function frees memory or copy array?

answer: copies creates slice (=view); [ipython console] import numpy; numpy?? (see resize numpy array smaller size without copy , view onto numpy array?)

2) solution have similar functionality structured arrays?

answer:

begin=(edges!=numpy.zeros(1,edges.dtype)).argmax() end=len(edges)-(edges!=numpy.zeros(1,edges.dtype))[::-1].argmax() # 1) create slice without copy no memory free goodedges=edges[begin:end] # 2) or copy , free memory (temporary both arrays exist) goodedges=edges[begin:end].copy() del edges 

imho, there 2 problem.

• first, trim_zeros function doesn't recognize zeroes on composite dtype.

you can locate them begin=(edges!=zeros(1,edges.dtype)).argmax() , end=len(edges)-(edges!=zeros(1,edges.dtype))[::-1].argmax(). goodedges=edges[begin:end] interresting data.

• second, trim_zeros function doesn't free memory:

returns ------- trimmed : 1-d array or sequence. result of trimming input. input data type preserved.

so think must manually : goodedges=edges[begin:end].copy();del edges.