Convert matlab statement to python -

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• numpy: find first index of value fast 14 answers

i should convert following matlab statement in python:

lam = find(abs(b)>tol,1,'first')-1;  

thanks in advance

numpy doesn't have capability first matching one, not yet anyway.

so, need matching ones np.where replace find , numpy.abs replacement abs. thus, -

import numpy np  lam = np.where(np.abs(a)>tol)[0][0]-1 

thus, getting matching indices np.where(np.abs(a)>tol)[0], getting first index indexing first element np.where(np.abs(a)>tol)[0][0] , subtracting 1 it, did on matlab.

sample run -

on matlab :

>> b = [-4,6,-7,-2,8]; >> tol = 6; >> find(abs(b)>tol,1,'first')-1 ans =      2 

on numpy :

in [23]: = np.array([-4,6,-7,-2,8])  in [24]: tol = 6  in [25]: np.where(np.abs(a)>tol)[0][0]-1 out[25]: 1 # 1 lesser matlab vesion because of 0-based indexing 

for performance, suggest using np.argmax give index of first matching one, rather computing matching indices np.where(np.abs(a)>tol)[0], -

in [40]: np.argmax(np.abs(a)>tol)-1 out[40]: 1