Convert Python sequence to NumPy array, filling missing values
You can use itertools.zip_longest:
import itertools
np.array(list(itertools.zip_longest(*v, fillvalue=0))).T
Out:
array([[1, 0],
[1, 2]])
Note: For Python 2, it is itertools.izip_longest.
Here's an almost* vectorized boolean-indexing based approach that I have used in several other posts -
def boolean_indexing(v):
lens = np.array([len(item) for item in v])
mask = lens[:,None] > np.arange(lens.max())
out = np.zeros(mask.shape,dtype=int)
out[mask] = np.concatenate(v)
return out
Sample run
In [27]: v
Out[27]: [[1], [1, 2], [3, 6, 7, 8, 9], [4]]
In [28]: out
Out[28]:
array([[1, 0, 0, 0, 0],
[1, 2, 0, 0, 0],
[3, 6, 7, 8, 9],
[4, 0, 0, 0, 0]])
*Please note that this coined as almost vectorized because the only looping performed here is at the start, where we are getting the lengths of the list elements. But that part not being so computationally demanding should have minimal effect on the total runtime.
Runtime test
In this section I am timing DataFrame-based solution by @Alberto Garcia-Raboso, itertools-based solution by @ayhan as they seem to scale well and the boolean-indexing based one from this post for a relatively larger dataset with three levels of size variation across the list elements.
Case #1 : Larger size variation
In [44]: v = [[1], [1,2,4,8,4],[6,7,3,6,7,8,9,3,6,4,8,3,2,4,5,6,6,8,7,9,3,6,4]]
In [45]: v = v*1000
In [46]: %timeit pd.DataFrame(v).fillna(0).values.astype(np.int32)
100 loops, best of 3: 9.82 ms per loop
In [47]: %timeit np.array(list(itertools.izip_longest(*v, fillvalue=0))).T
100 loops, best of 3: 5.11 ms per loop
In [48]: %timeit boolean_indexing(v)
100 loops, best of 3: 6.88 ms per loop
Case #2 : Lesser size variation
In [49]: v = [[1], [1,2,4,8,4],[6,7,3,6,7,8]]
In [50]: v = v*1000
In [51]: %timeit pd.DataFrame(v).fillna(0).values.astype(np.int32)
100 loops, best of 3: 3.12 ms per loop
In [52]: %timeit np.array(list(itertools.izip_longest(*v, fillvalue=0))).T
1000 loops, best of 3: 1.55 ms per loop
In [53]: %timeit boolean_indexing(v)
100 loops, best of 3: 5 ms per loop
Case #3 : Larger number of elements (100 max) per list element
In [139]: # Setup inputs
...: N = 10000 # Number of elems in list
...: maxn = 100 # Max. size of a list element
...: lens = np.random.randint(0,maxn,(N))
...: v = [list(np.random.randint(0,9,(L))) for L in lens]
...:
In [140]: %timeit pd.DataFrame(v).fillna(0).values.astype(np.int32)
1 loops, best of 3: 292 ms per loop
In [141]: %timeit np.array(list(itertools.izip_longest(*v, fillvalue=0))).T
1 loops, best of 3: 264 ms per loop
In [142]: %timeit boolean_indexing(v)
10 loops, best of 3: 95.7 ms per loop
To me, it seems itertools.izip_longest is doing pretty well!