[英]Finding X values in numpy array and substituting for random value
Consider an list of numpy arrays with values either -1's or 1's allocated in random positions.考虑一个 numpy 数组列表,其中的值在随机位置分配为 -1 或 1。
a = np.array([1,-1,1,1,-1,1,-1,-1,1,-1])
b = np.array([-1,-1,1,-1,1,1,-1,1,-1,-1])
I need to perform operations on these arrays like sum and point wise multiplication.我需要对这些数组执行操作,例如求和和逐点乘法。
For example, after summing 2 arrays i will have a new one with values -2,0 and 2.例如,在对 2 个数组求和后,我将得到一个值为 -2,0 和 2 的新数组。
c = a + b
c = [ 0 -2 2 0 0 2 -2 0 0 -2]
Now i would like to “normalize” it back to -1's and 1's.现在我想将它“规范化”回 -1 和 1。
For the 2's and -2's it is easy:对于 2 和 -2 来说,这很容易:
c[c < 0] = -1
c[c > 0] = 1
The problem is the 0. For them i would like to randomly choose either a -1 or a 1.问题是 0。对于他们,我想随机选择 -1 或 1。
The desired output would be like:所需的输出如下:
c = [ 1 -1 1 -1 -1 1 -1 1 -1 -1]
In generalized terms my question is how to find all N values equal to x, in an array, then substitute each for a random number.概括地说,我的问题是如何在数组中找到等于 x 的所有 N 个值,然后将每个值替换为一个随机数。
My question is how to do this in the most “pythonic”, and fastest, way?我的问题是如何以最“pythonic”和最快的方式做到这一点?
Thank's谢谢
Just Posting the final results from the answers i got so far.只是发布我到目前为止得到的答案的最终结果。 If anyone in the future has a better solution please share it!
如果将来有人有更好的解决方案,请分享它!
I timed the 3 solutions i found and one i did.我对我找到的 3 个解决方案和一个我做了计时。
def Norm1(HV):
HV[HV > 0] = 1
HV[HV < 0] = -1
zind = np.where(HV == 0)[0]
HV[zind] = np.array([np.random.choice([1, -1]) for _ in zind])
return HV
def norm2(HV):
if HV == 0:
return np.random.choice(np.array([-1,1]))
else:
return HV / HV * np.sign(HV)
Norm2 = np.vectorize(norm2)
def Norm3(HV):
HV[HV > 0] = 1
HV[HV < 0] = -1
mask = HV==0;
HV[mask] = np.random.choice((-1,1),HV[mask].shape)
return HV
def generate(size):
return np.random.binomial(1, 0.5, size=size) * 2 - 1
def Norm4(arr):
np.floor_divide(arr, 2, out=arr)
positions = (arr == 0)
size = sum(positions)
np.add.at(arr, positions, generate(size)
The timings were:时间是:
%%timeit
d = Norm1(c)
203 µs ± 5.9 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%%timeit
d = Norm2(c)
33.4 ms ± 1.03 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
%%timeit
d = Norm3(c)
217 µs ± 11.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%%timeit
d = Norm4(c)
21 ms ± 1.23 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
So as it stands it looks like answer 1 and 3 are the best ones.因此,目前看来答案1和3是最好的。 The difference between them looks minimal, but after trying some more runs the number 1 always come slightly on top .
它们之间的差异看起来很小,但是在尝试更多次运行之后,数字 1 总是稍微领先。
Thanks for the Helps guys!感谢各位大侠的帮助! I will add some references to HD computing in the question as this is a core problem in this application so it will be easier for someone to find it if needed.
我将在问题中添加一些对高清计算的引用,因为这是此应用程序中的核心问题,因此如果需要,人们会更容易找到它。
I'm not in any way claiming this is the fastest nor most efficient approach.我不以任何方式声称这是最快的,也最有效的方法。
c = np.array([ 0, -2, 2, 0, 0, 2, -2, 0, 0, -2])
def norm(a):
if a == 0:
return np.random.choice(np.array([-1,1]))
else:
return a / a * np.sign(a)
v_norm = np.vectorize(norm)
norm_arr = v_norm(c)
Result:结果:
In [64]: norm_arr
Out[64]: array([ 1, -1, 1, 1, -1, 1, -1, 1, -1, -1])
You might use:你可能会使用:
>>> c = [0, -2, 2, 0, 0, 2, -2, 0, 0, -2]
>>> c = np.array([0, -2, 2, 0, 0, 2, -2, 0, 0, -2])
>>> zind = np.where(c==0)[0]
>>> c[zind] = np.array([np.random.choice([1, -1]) for _ in zind])
>>> c
array([ 1, -2, 2, -1, -1, 2, -2, -1, 1, -2])
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