I've managed to construct a matrix with this piece of code
c_bed = np.append(np.array([1, 2, 3]), np.nan).reshape(4, 1)
c_bath = np.array([1, 1, 2, 2], dtype=np.float).reshape(4, 1)
ds = np.append(c_bed, c_bath, axis=1)
which gives
array([[ 1., 1.],
[ 2., 1.],
[ 3., 2.],
[nan, 2.]])
the output is exactly what i want though, I am wondering is there a better way to construct this matrix?
How about using zip_longest
from itertools import zip_longest
np.array(list(zip_longest([1,2,3],[1,1,2,2],fillvalue=np.nan)))
Out[228]:
array([[ 1., 1.],
[ 2., 1.],
[ 3., 2.],
[nan, 2.]])
fillvalue=np.nan and you get a numerical array.
np.array(list(zip_longest([1,2,3],[1,1,2,2])),'d') also works.Is there any reason not to use this matrix = numpy.array([[1, 1], [2, 1], [3, 2], [numpy.nan, 2]])?
If you have
beds = [1, 2, 3]
baths = [1, 1, 2, 2]
data = (beds, baths)
You can do as follows:
ds = np.full((max(map(len, data)), 2), np.nan)
ds[:len(beds), 0] = beds
ds[:len(baths), 1] = baths
Disclaimer: the following may well be the shortest method, but it is most certainly not the healthiest. I wouldn't use it in production code.
np.c_[[2,4,6,np.nan],2:6]//2
# array([[ 1., 1.],
# [ 2., 1.],
# [ 3., 2.],
# [nan, 2.]])
