Here's the numpy.ndarray I've got:
a=[[[ 0.01, 0.02 ]], [[ 0.03, 0.04 ]]]
and I want it to convert to
a = [(0.01, 0.02), (0.03, 0.04)]
Currently I just use the following for loop but I'm not sure whether it's efficient enough:
b = []
for point in a:
b.append((point[0][0], point[0][1]))
print(b)
I've found somewhat a similar question but there're no tuples so a suggested ravel().tolist() approach didn't work for me here.
# initial declaration
>>> a = np.array([[[ 0.01, 0.02 ]], [[ 0.03, 0.04 ]]])
>>> a
array([[[0.01, 0.02]],
[[0.03, 0.04]]])
# check the shape
>>> a.shape
(2L, 1L, 2L)
# use resize() to change the shape (remove the 1L middle layer)
>>> a.resize((2, 2))
>>> a
array([[0.01, 0.02],
[0.03, 0.04]])
# faster than a list comprehension (for large arrays)
# because numpy's backend is written in C
# if you need a vanilla Python list of tuples:
>>> list(map(tuple, a))
[(0.01, 0.02), (0.03, 0.04)]
# alternative one-liner:
>>> list(map(tuple, a.reshape((2, 2))))
...
2L there? In Python 3 all integers are long integers, so I would propose to remove those Ls as to not confuse new Python programmers into thinking this is "good style" or even "required".
You can use list comprehension, they are faster than for loops
a = np.array([[[ 0.01, 0.02 ]], [[ 0.03, 0.04 ]]])
print([(i[0][0], i[0][1]) for i in a]) # [(0.01, 0.02), (0.03, 0.04)]
alternatively:
print([tuple(l[0]) for l in a]) # [(0.01, 0.02), (0.03, 0.04)]
tuple(i[0]).