[solved] list of list with one or more empty one

[paul18fr]

Hi

Without using a loop, is there a way to find an empty list as showing in the following example?

The idea is to find the location and to replace it by a value (such as -1 or zero for example) prior to chnage it to an array

import numpy as np

M = [[1], [], [5]]
Mprime = np.asarray(M, dtype=np.uint32)

logical error:

Error:
ValueError: setting an array element with a sequence. The requested array has an inhomogeneous shape after 1 dimensions. The detected shape was (3,) + inhomogeneous part.

Thanks for your suggestion

Paul

[Gribouillis]

Without using a loop

Why not using a loop?

You can try itertools, for example

>>> from itertools import filterfalse
>>> M = [[1], [], [5]]
>>> a = next(filterfalse(None, M))
>>> a
[]
>>> a.append(0)
>>> M
[[1], [0], [5]]

I'm not sure you gain anything.

[paul18fr]

Why not using a loop?

I'm using a list comprehension so far, and in a list of more than a million of components, performance falls of about 25% ; it looks like:

Mprime = [[0] if n == [] else n for n in M]

Let me trying itertools

Paul

[Gribouillis]

Mprime = [[0] if n == [] else n for n in M]

Try this perhaps

Mprime = [n or [0] for n in M]

or if you dont mind having shared lists

x = [0]
Mprime = [n or x for n in M]

Or don't build a list, use a generator

from itertools import chain
x = [0]
Mprime =  Mprime = np.fromiter(chain.from_iterable(n or x for n in M), dtype=np.uint32)
Mprime = Mprime.reshape((Mprime.size, 1))

[paul18fr]

Thanks Gribouillis for the support and for all advices, it has been helpfull.

[jeffreyteague]

Nice, thanks for share with us. Beacause i have same problem