Numeric equivalent of understanding if / else

Question

Are there a few ways to do

n = [xt if x > 0 else x for x in nps] 

similar to this

 n = np.array(a) n[np.abs(n) < t] = 0 

Is something like this possible?

 n[n > 0] = nt 

Answer 1

Verification failed, but try

 np.where(n > 0, n - t, n) 

See documentation