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copied!<p>To generate the array, you can use <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.random.choice.html" rel="nofollow"><code>numpy.random.choice</code></a>.</p> <pre><code>np.random.choice([0, 1, 2, 3], size=(10,10), p=[.4, .1, .2, .3]) </code></pre> <p>Then you can convert to coordinates. Note that <code>numpy.random.choice</code> generates a random sample using probabilities <code>p</code>, and thus you are not guaranteed to get the <em>exact</em> proportions in <code>p</code>.</p> <p><strong>Extra</strong></p> <p>If you want to have all the <code>1</code>s on a particular side of the array, you can generate two random arrays and then <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.hstack.html#numpy.hstack" rel="nofollow"><code>hstack</code></a> them. The trick is to slightly modify the probabilities of each number on each side.</p> <pre><code>In [1]: import numpy as np In [2]: rem = .1/3 # amount to de- / increase the probability for non-1s In [3]: A = np.random.choice([0, 1, 2, 3], size=(5, 10), p=[.4-rem, .2, .2-rem, .3-rem]) In [4]: B = np.random.choice([0, 2, 3], size=(5, 10), p=[.4+rem, .2+rem, .3+rem]) In [5]: M = np.hstack( (A, B) ) In [6]: M Out[1]: array([[1, 1, 3, 0, 3, 0, 0, 1, 1, 0, 2, 2, 0, 2, 0, 2, 3, 3, 2, 0], [0, 3, 3, 3, 3, 0, 1, 3, 1, 3, 0, 2, 3, 0, 0, 0, 3, 3, 2, 3], [1, 0, 0, 0, 1, 0, 3, 1, 2, 2, 0, 3, 0, 3, 3, 0, 0, 3, 0, 0], [3, 2, 3, 0, 3, 0, 1, 2, 3, 2, 0, 0, 0, 0, 3, 2, 0, 0, 0, 3], [3, 3, 0, 3, 3, 3, 1, 3, 0, 3, 0, 2, 0, 2, 0, 0, 0, 3, 3, 3]]) </code></pre> <p>Here, because I'm putting <em>all</em> the <code>1</code>s on the left, I double the probability of <code>1</code> and decrease the probability of each number equally. The same logic applies when creating the other side.</p>

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