这是我得到的numpy.ndarray:
a=[[[ 0.01, 0.02 ]], [[ 0.03, 0.04 ]]]
我想将其转换为
a = [(0.01, 0.02), (0.03, 0.04)]
目前,我仅使用以下for循环,但不确定其效率是否足够:

b = []
for point in a:
   b.append((point[0][0], point[0][1]))
print(b)

我找到了a similar question,但是没有元组,因此建议的ravel().tolist()方法对我来说不起作用。

最佳答案

# 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))))
...

关于python - 如何有效地将numpy ndarray转换为元组列表?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/55336870/

10-15 14:00