Initialise numpy array of unknown length

Question

I want to be able to \'build\' a numpy array on the fly, I do not know the size of this array in advance.

For example I want to do something like this:

Answer 1

For posterity, I think this is quicker:

a = np.array([np.array(list()) for _ in y])

You might even be able to pass in a generator (i.e. [] -> ()), in which case the inner list is never fully stored in memory.

---

Responding to comment below:

>>> import numpy as np
>>> y = range(10)
>>> a = np.array([np.array(list) for _ in y])
>>> a
array([array(<type 'list'>, dtype=object),
       array(<type 'list'>, dtype=object),
       array(<type 'list'>, dtype=object),
       array(<type 'list'>, dtype=object),
       array(<type 'list'>, dtype=object),
       array(<type 'list'>, dtype=object),
       array(<type 'list'>, dtype=object),
       array(<type 'list'>, dtype=object),
       array(<type 'list'>, dtype=object),
       array(<type 'list'>, dtype=object)], dtype=object)

Answer 2

a = np.empty(0)
for x in y:
    a = np.append(a, x)

Answer 3

Build a Python list and convert that to a Numpy array. That takes amortized O(1) time per append + O(n) for the conversion to array, for a total of O(n).

    a = []
    for x in y:
        a.append(x)
    a = np.array(a)

Answer 4

You can do this:

a = np.array([])
for x in y:
    a = np.append(a, x)

Answer 5

Since y is an iterable I really do not see why the calls to append:

a = np.array(list(y))

will do and it's much faster:

import timeit

print timeit.timeit('list(s)', 's=set(x for x in xrange(1000))')
# 23.952975494633154

print timeit.timeit("""li=[]
for x in s: li.append(x)""", 's=set(x for x in xrange(1000))')
# 189.3826994248866