优化列表的python列表上的最小操作 [英] Optimise a minimum operation on a python list of list

问题描述

我有一个列表列表，其中每个列表都对应于图形的边界点.我必须提取四个点的列表，每个点定义图形的封闭矩形.例如，我有这个列表

I have a list of list where each list corresponds to the boundary point of a figure. I have to extract a list of four points, each defining the enclosing rectangle of the figures. For example, I have this list

[[[2,5],[3,4],[5,8],[5,6],[5,9]],
 [[11,14],[12,15],[16,17]],
 ...
]

此处每个列表都定义了图形边界.我要得到的是作为

Here each list defines a figure boundary. what I have to get is list of four points as

[[min_x, min_y], [min_x, max_y], [max_x, max_y], [max_x, min_y]]

即

[[[2,4],[2,9],[5,9],[5,4]],
 [[11,14], [11, 17], [16,17], [16,14]]
 ...
]

我已经使用python循环完成了此任务，该循环非常正常. 这是代码.

I have done this using python loop which is perfectly working. Here is the code.

cleaned_contours = list()
for cur_cont in contours:
    min_x, min_y = cur_cont.min(axis=0).flatten()
    max_x, max_y = cur_cont.max(axis=0).flatten()

    cleaned_contours.append(np.array([[min_x, min_y], [min_x, max_y], [max_x, max_y], [max_x, min_y]]))

有没有一种方法可以执行此操作而无需使用循环或列表理解. 我正在使用python3.

Is there a way I can do this without using loops or list comprehension. I am using python3.

推荐答案

如果愿意，您可以使用理解力(仍然是循环+丑陋的):

You can use comprehension if you like (still a loop + ugly):

a = [[[2,5],[3,4],[5,8],[5,6],[5,9]], [[11,14],[12,15],[16,17]]]

[[[j[0], j[1]], [j[0], j[3]], [j[2], j[3]], [j[2], j[1]]] for j in [np.array(i).min(0).tolist() + np.array(i).max(0).tolist() for i in a]]

#[[[2, 4], [2, 9], [5, 9], [5, 4]],
# [[11, 14], [11, 17], [16, 17], [16, 14]]]

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