numpy array filtering twice and replace

Question

I want to execute two filtering and one replacing steps on a large array but I cannot figure a way to do it efficiently without making some copy or for loop.

Example, imagine two numpy arrays of different sizes. I want to apply two filters. The first one to select only elements of b that are inferior to 6 (b') and a second filters to replace the values of b' by the values of a if a is inferior to b':

a = np.array( [ 2,2,2,2,2,2] )
b = np.array( [ 0,1,2,3,4,5,6,7,8,9] )

I apply a first masking by selecting the elements of b inferior to 6 :

m = b < 6

Now, I want to replace the value of b[m] by the minimal value between a and b[m], with expected results in b :

[ 0,1,2,2,2,2,6,7,8,9]

Using :

n = a < b[m]
b[m][n] = a[n]

doesn't work. Probably because of some intermediate array. With

c = np.array( [ 0, 1, 2, 3, 4, 5  ] )

I can directly do :

c[ a < c ] = a [ a < c ]

and it works. Any cool slicing way to do it without making secondary array ? Thanks.

I just found a way : b [ b[b < 6] > a ] = a [ b[ b < 6 ] > a ] but it's not really elegant. Any better idea ? — user3099854
– user3099854, Commented Dec 13, 2013 at 15:26

M4rtini · Accepted Answer · 2013-12-13 15:38:14Z

1

m = b < 6 
b[m] = np.where(b[m]< a,b[m],a)

answered Dec 13, 2013 at 15:38

M4rtini

13.6k4 gold badges37 silver badges42 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

numpy array filtering twice and replace

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related