在这样一个数组中生成一个包含数千个随机位置的数组，例如python中的(3,75)和(56,34)

Question

到目前为止，我已经创建了两个数据点列表，分别标记为x和y。我不知道如何将它们组合成一个数组，但使它们是坐标。

import random
import pylab
x = [(random.randint(0,100)) for i in range(3000)]
y = [(random.randint(0,100)) for j in range(3000)]

Answer 1

可以与zip():

结合使用

zip(x, y)

或者您可以生成数字一起以

开头:

[(random.randint(0,100), random.randint(0,100)) for i in xrange(3000)]

Answer 2

我注意到您正在导入pylab。如果您愿意使用NumPy数组作为坐标，您可以简单地这样写:

import numpy.random
coord = numpy.random.randint(0, 100, (3000, 2))

不需要显式循环或列表推导式。这使得代码比纯Python版本快120倍:

In [6]: %timeit coord = numpy.random.randint(0, 100, (3000, 2))
10000 loops, best of 3: 82.7 us per loop
In [7]: %timeit coord = [(random.randint(0,100), random.randint(0,100)) for i in xrange(3000)]
100 loops, best of 3: 10.3 ms per loop

Answer 3

可以使用zip:

>>> x = [(random.randint(0,100)) for i in range(3000)]
>>> y = [(random.randint(0,100)) for j in range(3000)]
>>> coords = zip(x,y)
>>> len(coords)
3000
>>> type(coords[0])
<type 'tuple'>
>>> coords[0]
(54, 0)
>>> 

Answer 4

您可以将列表作为元组添加到包含坐标的新列表中:

import random
import pylab
XY=[]
x = [(random.randint(0,100)) for i in range(3000)]
y = [(random.randint(0,100)) for j in range(3000)]
for xg, yg in zip(x,y):
    XY.append((xg,yg))