numpy阵列列表到坐标点列表的转换

使用:

new_list = [tuple(arr[0]) for arr in old_list]

因为列表中的每个元素都是一个具有单行的二维数组。

注意,您的 tuple 对象现在可能包含标量numpy对象 numpy.float64 . 这可能并不理想,在这种情况下,您可以:

new_list = [tuple(map(float, arr[0])) for arr in old_list]

根据保罗·帕泽尔的建议,

In [38]: list_of_arrs = [array([[  72.57218373,  247.1759997 ]]), array([[ 262.51505688,  158.68137977]]),..... ]

In [39]: flat_list = np.ravel(list_of_arrs).tolist()
In [40]: list_of_tuples = list(zip(flat_list[0::2], flat_list[1::2]))

In [41]: list_of_tuples
Out[41]: 
[(72.57218373, 247.1759997),(262.51505688, 158.68137977),....]

再次根据Paul Panzer的建议,为了提高效率,我们可以避免在构建 list_of-tuples 例如:

In [22]: list_of_tuples = list(zip(*2*(iter(flat_list),)))

In [23]: list_of_tuples
Out[23]: [(72.57218373, 247.1759997), (262.51505688, 158.68137977), ...]

计时

# explicit copy
In [36]: %%timeit
    ...: flat_list = 10000*(np.ravel(list_of_arrs).tolist())
    ...: list_of_tuples = list(zip(flat_list[0::2], flat_list[1::2]))
    ...: 
17 ms ± 345 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

# avoiding explicit copy by using an iterator
In [37]: %%timeit
    ...: flat_list = 10000*(np.ravel(list_of_arrs).tolist())
    ...: list_of_tuples = list(zip(*2*(iter(flat_list),)))
    ...: 
14.5 ms ± 267 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)