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copied!<pre><code>import numpy as np a = np.array([[[9, 8, 8], [4, 9, 1]], [[6, 6, 3], [9, 3, 5]]]) ind=(a[:,:,1]<=8) & (a[:,:,1]>=6) a[ind,1]=a[ind,0]*2 print(a) </code></pre> <p>yields</p> <pre><code>[[[ 9 18 8] [ 4 9 1]] [[ 6 12 3] [ 9 3 5]]] </code></pre> <hr> <p>If you wish to check for membership in a set which is not a simple range, then I like both <a href="https://stackoverflow.com/questions/8024305/numpy-checking-if-an-element-in-a-multidimensional-array-is-in-a-tuple/8024538#8024538">mac's idea</a> of using a Python loop and <a href="https://stackoverflow.com/questions/8024305/numpy-checking-if-an-element-in-a-multidimensional-array-is-in-a-tuple/8024510#8024510">bellamyj's idea</a> of using np.in1d. Which is faster depends on the size of <code>check_tuple</code>:</p> <p><strong>test.py:</strong></p> <pre><code>import numpy as np np.random.seed(1) N = 10 a = np.random.randint(1, 1000, (2, 2, 3)) check_tuple = np.random.randint(1, 1000, N) def using_in1d(a): idx = np.in1d(a[:,:,1], check_tuple) idx=idx.reshape(a[:,:,1].shape) a[idx,1] = a[idx,0] * 2 return a def using_in(a): idx = np.zeros(a[:,:,0].shape,dtype=bool) for n in check_tuple: idx |= a[:,:,1]==n a[idx,1] = a[idx,0]*2 return a assert np.allclose(using_in1d(a),using_in(a)) </code></pre> <p>When N = 10, <code>using_in</code> is slightly faster:</p> <pre><code>% python -m timeit -s'import test' 'test.using_in1d(test.a)' 10000 loops, best of 3: 156 usec per loop % python -m timeit -s'import test' 'test.using_in(test.a)' 10000 loops, best of 3: 143 usec per loop </code></pre> <p>When N = 100, <code>using_in1d</code> is much faster:</p> <pre><code>% python -m timeit -s'import test' 'test.using_in1d(test.a)' 10000 loops, best of 3: 171 usec per loop % python -m timeit -s'import test' 'test.using_in(test.a)' 1000 loops, best of 3: 1.15 msec per loop </code></pre>

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