Question

我有一个坐标数组如下：

[array([[  72.57218373,  247.1759997 ]]), array([[ 262.51505688,  158.68137977]]), array([[ 244.53157113,  100.05665591]]), array([[ 245.30108919,   88.97536043]]), array([[ 43.60839312,  82.11394031]]), array([[ 260.62555476,  161.252084  ]]), array([[ 234.46830867,  269.07508699]]), array([[ 284.51718016,  142.55126301]]), array([[ 195.30252058,   84.7559498 ]]), array([[ 141.08621668,  276.6625877 ]]), array([[ 294.18468717,  131.91470926]]), array([[ 35.46840935,  51.14890688]]), array([[  36.15758189,  131.19703618]]), array([[ 175.95018737,  216.52553084]]), array([[  62.99149427,  217.7800738 ]]), array([[ 245.81708339,   10.06949286]]), array([[ 137.31450413,  145.10778886]]), array([[ 197.76975618,   75.96150192]]), array([[  17.87096457,  120.88350573]]), array([[ 262.89229921,  287.40287273]]), array([[ 177.51716203,  101.5694418 ]]), array([[  97.76530613,  190.06919232]]), array([[ 229.31904186,   97.92874288]])]

我想将此转换为点数，以便我的最终列表应为

形式

[(72.57218373,  247.1759997), (262.51505688,  158.68137977), ( 244.53157113,  100.05665591), (245.30108919,   88.97536043), (43.60839312,  82.11394031), (260.62555476,  161.252084), ...]

有人可以帮我解决这个问题吗？

谢谢，

Answer 1

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行的二维数组。

注意，您的tuple对象现在将包含标量numpy对象，可能是numpy.float64。这可能不太理想，在这种情况下，您可以这样做：

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

Answer 2

以Paul Panzer的建议为基础，

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时制作切片的明确副本：

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)

numpy数组列表，用于协调点转换列表

2 个答案:

计时