python - 什么更快 : Python 3's ' len' or numpys shape?

Question

我有一个坐标数组:

points = [x,y]

具有(numpy)尺寸/形状:(18, 1, 2)

在 matlab 中，要初始化一个 1 数组以用“3”索引这些点，我可以这样做:

A = ones(size(points,1),1)*3'

我怎样才能以最快的方式使用 python3 和 numpy 做到这一点？

Answer 1

    def time_compare(self):
        loops = 100000000
        start = time.time()
        for i in range(loops):
            self.value_map.shape[0]
            self.value_map[0].shape[0]
            self.value_map[0][0].shape[0]
            self.value_map[0][0][0].shape[0]
        end = time.time()
        timed = (end - start)
        print("shape={}".format(timed))

        start = time.time()
        for i in range(loops):
            len(self.value_map)
            len(self.value_map[0])
            len(self.value_map[0][0])
            len(self.value_map[0][0][0])
        end = time.time()
        timed = (end - start)
        print("len={}".format(timed))

shape=102.26551818847656
len=87.99720764160156

len 比 shape 快