I want really the vector zz. Unfortunatelly as 5 is present with multiplicity 2 the only index for 5 is 0. I want also 2. How to obtain the desired return ?
So
z=[5,0,5,0,10,13,14,]
zz=[(z.index(v),v) for v in z if v!=0]
show(zz)returns [(0, 5), (0, 5), (4, 10), (5, 13), (6, 14)] when I want [(0, 5), (2, 5), (4, 10), (5, 13), (6, 14)].
It's easier to iterate over the indices (rather than the values) in the first place:
sage: zz = [(k, z[k]) for k in range(len(z)) if z[k] != 0]
sage: zz
[(0, 5), (2, 5), (4, 10), (5, 13), (6, 14)]Or a bit cleaner, using enumerate:
sage: zz = [(k, v) for k, v in enumerate(z) if v != 0]
sage: zz
[(0, 5), (2, 5), (4, 10), (5, 13), (6, 14)]Asked: 3 years ago
Seen: 148 times
Last updated: Dec 14 '21
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