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?
最佳答案
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
关于python - 什么更快 : Python 3's ' len' or numpys shape?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35547853/