efficient loop over numpy array
<p>Versions of this question have already been asked but I have not found a satisfactory answer.</p>

<p><strong>Problem</strong>: given a large numpy vector, find indices of the vector elements which are duplicated (a variation of that could be comparison with tolerance). </p>

<p>So the problem is ~O(N^2) and memory bound (at least from the current algorithm point of view). I wonder why whatever I tried Python is 100x or more slower than an equivalent C code.</p>

<pre><code>import numpy as np
N = 10000
vect = np.arange(float(N))
vect[N/2] = 1
vect[N/4] = 1
dupl = []
print("init done")
counter = 0
for i in range(N):
    for j in range(i+1, N):
        if vect[i] == vect[j]:
            dupl.append(j)
            counter += 1

print("counter =", counter)
print(dupl)
# For simplicity, this code ignores repeated indices 
# which can be trimmed later. Ref output is
# counter = 3
# [2500, 5000, 5000]
</code></pre>

<p>I tried using numpy iterators but they are even worse (~ x4-5)
<a href="http://docs.scipy.org/doc/numpy/reference/arrays.nditer.html" rel="noreferrer">http://docs.scipy.org/doc/numpy/reference/arrays.nditer.html</a></p>

<p>Using N=10,000 I'm getting 0.1 sec in C, 12 sec in Python (code above), 40 sec in Python using np.nditer, 50 sec in Python using np.ndindex. I pushed it to N=160,000 and the timing scales as N^2 as expected.</p>