从numpy 2D数组中随机选择特定百分比的细胞

Question

I have foll. 2D numpy array:

array([[[-32768, -32768, -32768, ..., -32768, -32768, -32768],
        [-32768, -32768, -32768, ..., -32768, -32768, -32768],
        [-32768, -32768, -32768, ..., -32768, -32768, -32768],
        ..., 
        [-32768, -32768, -32768, ..., -32768, -32768, -32768],
        [-32768, -32768, -32768, ..., -32768, -32768, -32768],
        [-32768, -32768, -32768, ..., -32768, -32768, -32768]]], dtype=int16)

with following unique values:

array([-32768,    401,    402,    403,    404], dtype=int16)

Is there a way I can create a new array where 10% of the cells with 401 are changed to 500? I can use np.random.random_sample() to start but not how to select specified percentage of cells (eg 10% from this numpy 2D array)

Answer 1

Denote your array by a . Then this will do the job:

locs=np.vstack(np.where(a==401)).T
n=len(locs)
changes_loc=np.random.permutation(locs)[:n//10]
a[changes_loc[:,0],changes_loc[:,1]]=500

Here is a small example (with different numbers and 25% instead of 10%, just to depict the behavior):

a=np.array([[1,2,3],[4,1,5],[1,1,7]])
locs=np.vstack(np.where(a==1)).T
n=len(locs)
changes_loc=np.random.permutation(locs)[:n//4]
a[changes_loc[:,0],changes_loc[:,1]]=70

the result is

array([[70,  2,  3],
       [ 4,  1,  5],
       [ 1,  1,  7]])