python - Find number of zeros before non-zero in a numpy array -

i have numpy array a. return number of zeros before non-zero in a in efficient way in loop.

if a = np.array([0,1,2]) np.nonzero(a)[0][0] returns 1. if a = np.array([0,0,0]) doesn't work (i answer 3 in case). , if big , first non-zero near beginning seems inefficient.

here's iterative cython version, may best bet if serious bottleneck

# saved file count_leading_zeros.pyx import numpy np cimport numpy np cimport cython  dtype = np.int ctypedef np.int_t dtype_t  @cython.boundscheck(false) def count_leading_zeros(np.ndarray[dtype_t, ndim=1] a):     cdef int elements = a.size     cdef int = 0     cdef int count = 0     while < elements:         if a[i] == 0:             count += 1         else:             return count         += 1     return count 

this similar @mtrw's answer indexing @ native speeds. cython bit sketchy there may further improvements made.

a quick test of extremely favourable case ipython few different methods

in [1]: import numpy np  in [2]: import pyximport; pyximport.install() out[2]: (none, <pyximport.pyximport.pyximporter @ 0x53e9250>)  in [3]: import count_leading_zeros  in [4]: %paste def count_leading_zeros_python(x):     ctr = 0     k in x:         if k == 0:             ctr += 1         else:             return ctr     return ctr ## -- end pasted text -- in [5]: = np.zeros((10000000,), dtype=np.int)  in [6]: a[5] = 1  in [7]:   in [7]: %timeit np.min(np.nonzero(np.hstack((a, 1)))) 10 loops, best of 3: 91.1 ms per loop  in [8]:   in [8]: %timeit np.where(a)[0][0] if np.shape(np.where(a)[0])[0] != 0  else np.shape(a)[0] 10 loops, best of 3: 107 ms per loop  in [9]:   in [9]: %timeit count_leading_zeros_python(a) 100000 loops, best of 3: 3.87 µs per loop  in [10]:   in [10]: %timeit count_leading_zeros.count_leading_zeros(a) 1000000 loops, best of 3: 489 ns per loop 

however i'd use if had evidence (with profiler) bottleneck. many things may seem inefficient never worth time fix.