在Python阵列内的随机数 [英] Random numbers within Array in Python
问题描述
我是新来的Python。在阅读,请提及关于如何改善我的Python code任何其他建议。
问:如何生成Python中的数8xn维数组包含随机数? 的约束是,此数组中的每一列必须包含8无需更换吸引了来自设置整数[1,8] 的。更具体地说,当N = 10,我想是这样的。
[6. 2. 3. 4. 5. 7. 5. 7. 8. 4]
[1. 4. 5. 5. 4. 4. 8 5 7。5.]
[7. 3. 8. 8. 3. 8 7。3 6. 7.]
[3. 6. 7. 1. 5. 6. 2. 1. 5. 1.]
[8 1。4 3 8 2. 3. 4. 3. 3.]
[5 8 1 7 1 3 6 8。1。6]
[4. 5. 2. 6. 2. 1. 1. 6 4 2]
[2 7 6。2. 6. 7. 4. 2。2 8]
要做到这一点我用下面的办法:
进口numpy.random
进口numpy的
高清rand_M(N):
M = numpy.zeros(形状=(8,N))
对于在范围I(0,N):
M [:,我] = numpy.random.choice(8,大小= 8,更换= FALSE)+ 1
回报中号
在实践中N值将〜1E7。上述算法是O(n)在时间和大约需要0.38秒当N = 1E3。因此,当N = 1E7是1小时〜时间(即3800秒)。必须有一个更有效的方式。
定时功能
从timeit进口计时器
T =定时器(拉姆达:rand_M(1000))
打印(t.timeit(5))
0.3863314103162543
创建特定形状的随机排列,然后沿排序,你想保留的限制,从而给我们一个量化的,非常有效的解决方案轴。这将在此基础上 智能答案
以<一个href=\"http://stackoverflow.com/questions/29156823/matlab-randomly-permuting-columns-differently\"><$c$c>MATLAB随机置换列不同 。这里的执行 -
样运行 -
在[122]:N = 10在[123]:np.argsort(np.random.rand(8,N),轴= 0)+1
出[123]:
阵列([[7,3,5,1,1,5,2,4,1,4]
[8,4,3,2,2,8,5,5,6,2],
[1,2,4,6,5,4,4,3,4,7],
[5,6,2,5,8,2,7,8,5,8],
[2,8,6,3,4,7,1,1,2,6]
[6,7,7,8,6,6,3,2,7,3]
[4,1,1,4,3,3,8,6,8,1],
[3,5,8,7,7,1,6,7,3,5],DTYPE = int64类型)
运行测试 -
在[124]:高清sortbased_rand8(N):
...:返回np.argsort(np.random.rand(8,N),轴= 0)+1
...:
...:DEF rand_M(N):
...:M = np.zeros(形状=(8,N))
......因为我在范围(0,N):
...:M [:,我] = np.random.choice(8,大小= 8,更换= FALSE)+ 1
...:返回中号
...:在[125]:N = 5000在[126]:%timeit sortbased_rand8(N)
100圈,最好的3:每循环1.95毫秒在[127]:%timeit rand_M(N)
1循环,最好的3:每循环233毫秒
因此,等待一个 120X
加速比!
I'm new to Python. While reading, please mention any other suggestions regarding ways to improve my Python code.
Question: How do I generate a 8xN dimensional array in Python containing random numbers? The constraint is that each column of this array must contain 8 draws without replacement from the integer set [1,8]. More specifically, when N = 10, I want something like this.
[[ 6. 2. 3. 4. 7. 5. 5. 7. 8. 4.]
[ 1. 4. 5. 5. 4. 4. 8. 5. 7. 5.]
[ 7. 3. 8. 8. 3. 8. 7. 3. 6. 7.]
[ 3. 6. 7. 1. 5. 6. 2. 1. 5. 1.]
[ 8. 1. 4. 3. 8. 2. 3. 4. 3. 3.]
[ 5. 8. 1. 7. 1. 3. 6. 8. 1. 6.]
[ 4. 5. 2. 6. 2. 1. 1. 6. 4. 2.]
[ 2. 7. 6. 2. 6. 7. 4. 2. 2. 8.]]
To do this I use the following approach:
import numpy.random
import numpy
def rand_M(N):
M = numpy.zeros(shape = (8, N))
for i in range (0, N):
M[:, i] = numpy.random.choice(8, size = 8, replace = False) + 1
return M
In practice N will be ~1e7. The algorithm above is O(n) in time and it takes roughly .38 secs when N=1e3. The time therefore when N = 1e7 is ~1hr (i.e. 3800 secs). There has to be a much more efficient way.
Timing the function
from timeit import Timer
t = Timer(lambda: rand_M(1000))
print(t.timeit(5))
0.3863314103162543
Create a random array of specified shape and then sort along the axis where you want to keep the limits, thus giving us a vectorized and very efficient solution. This would be based on this smart answer
to MATLAB randomly permuting columns differently
. Here's the implementation -
Sample run -
In [122]: N = 10
In [123]: np.argsort(np.random.rand(8,N),axis=0)+1
Out[123]:
array([[7, 3, 5, 1, 1, 5, 2, 4, 1, 4],
[8, 4, 3, 2, 2, 8, 5, 5, 6, 2],
[1, 2, 4, 6, 5, 4, 4, 3, 4, 7],
[5, 6, 2, 5, 8, 2, 7, 8, 5, 8],
[2, 8, 6, 3, 4, 7, 1, 1, 2, 6],
[6, 7, 7, 8, 6, 6, 3, 2, 7, 3],
[4, 1, 1, 4, 3, 3, 8, 6, 8, 1],
[3, 5, 8, 7, 7, 1, 6, 7, 3, 5]], dtype=int64)
Runtime tests -
In [124]: def sortbased_rand8(N):
...: return np.argsort(np.random.rand(8,N),axis=0)+1
...:
...: def rand_M(N):
...: M = np.zeros(shape = (8, N))
...: for i in range (0, N):
...: M[:, i] = np.random.choice(8, size = 8, replace = False) + 1
...: return M
...:
In [125]: N = 5000
In [126]: %timeit sortbased_rand8(N)
100 loops, best of 3: 1.95 ms per loop
In [127]: %timeit rand_M(N)
1 loops, best of 3: 233 ms per loop
Thus, awaits a 120x
speedup!
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