为了测试目的,我想要创建一个 M by N
的 numpy 数组,其中有 c
个随机放置的 NaN。
import numpy as np
M = 10;
N = 5;
c = 15;
A = np.random.randn(M,N)
A[mask] = np.nan
我在创建一个具有 c 个真元素的掩码时遇到了问题,或者也许可以直接用索引实现?
replace=False
参数来进行非重复的随机选择,然后将其应用于A
的扁平化版本(使用.ravel()
完成),方法如下所示:np.random.choice
。A.ravel()[np.random.choice(A.size, c, replace=False)] = np.nan
示例运行 -
In [100]: A
Out[100]:
array([[-0.35365726, 0.26754527, -0.44985524, -1.29520237, 2.01505444],
[ 0.01319146, 0.65150356, -2.32054478, 0.40924753, 0.24761671],
[ 0.3014714 , -0.80688589, -2.61431163, 0.07787956, 1.23381951],
[-1.70725777, 0.07856845, -1.04354202, -0.68904925, 1.07161002],
[-1.08061614, 1.17728247, -1.5913516 , -1.87601976, 1.14655867],
[ 1.12542853, -0.26290025, -1.0371326 , 0.53019033, -1.20766258],
[ 1.00692277, 0.171661 , -0.89646634, 1.87619114, -1.04900026],
[ 0.22238353, -0.6523747 , -0.38951426, 0.78449948, -1.14698869],
[ 0.58023183, 1.99987331, -0.85938155, 1.4211672 , -0.43369898],
[-2.15682219, -0.6872121 , -1.28073816, -0.97523148, -2.27967001]])
In [101]: A.ravel()[np.random.choice(A.size, c, replace=False)] = np.nan
In [102]: A
Out[102]:
array([[ nan, 0.26754527, -0.44985524, nan, 2.01505444],
[ 0.01319146, 0.65150356, -2.32054478, nan, 0.24761671],
[ nan, -0.80688589, nan, nan, 1.23381951],
[ nan, nan, -1.04354202, -0.68904925, 1.07161002],
[-1.08061614, 1.17728247, -1.5913516 , nan, 1.14655867],
[ 1.12542853, nan, -1.0371326 , 0.53019033, -1.20766258],
[ nan, 0.171661 , -0.89646634, nan, nan],
[ 0.22238353, -0.6523747 , -0.38951426, 0.78449948, -1.14698869],
[ 0.58023183, 1.99987331, -0.85938155, nan, -0.43369898],
[-2.15682219, -0.6872121 , -1.28073816, -0.97523148, nan]])
您可以使用np.random.shuffle
在新数组上创建掩码:
import numpy as np
M = 10;
N = 5;
c = 15;
A = np.random.randn(M,N)
mask=np.zeros(M*N,dtype=bool)
mask[:c] = True
np.random.shuffle(mask)
mask=mask.reshape(M,N)
A[mask] = np.nan
这将会得到:
[[ 0.98244168 0.72121195 0.99291217 0.17035834 0.46987918]
[ 0.76919975 0.53102064 nan 0.78776918 nan]
[ 0.50931304 0.91826809 0.52717345 nan nan]
[ 0.35445471 0.28048106 0.91922292 0.76091783 0.43256409]
[ 0.69981284 0.0620876 0.92502572 nan nan]
[ nan nan nan 0.24466688 0.70259211]
[ 0.4916004 nan nan 0.94945378 0.73983538]
[ 0.89057404 0.4542628 nan 0.95547377 nan]
[ 0.4071912 0.36066797 0.73169132 0.48217226 0.62607888]
[ 0.30341337 nan 0.75608859 0.31497997 nan]]
random_choice
有一个可选的replace
参数,直接就可以用了! :) - Divakar
np.random.randint(0,high=A.size,size=c)
替换我的应用中的np.random.choice
(如果替换不会真正影响)。但是,ravel()
之后为什么数组没有保持平坦? - Olegnp.random.randint
可能会给你重复的索引,所以我不认为在你的情况下会起作用。关于.ravel()
的事情,它只是一个view
,因此它并没有真正地在内存中扁平化。因此,“扁平化视图”被索引并设置为 NaN,同时仍保持为 2D 数组。 - Divakarravel()
的文档中说“只有在需要时才会创建副本。”。那么我会得到一个扁平化的A
吗? - Olegnp.put
来达到同样的效果。因此,使用它的解决方案将是np.put(A,np.random.choice(A.size, c, replace=False),np.nan)
。 - Divakar