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6894325
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2011-08-01T04:51:28.517
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2013-06-22T15:14:14.050
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2017-05-23T12:19:50.857
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6894251
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What are you describing is called rejection sampling, and is a bad idea. For example if you wanted only numbers <code>>0.999</code>, it would take 1000 times longer to generate them. <hr> Best way The correct way to do this is to use another sampling technique, such as using the inverse of the CDF (cumulative density function), and doing <code>[inverseCDF(random()) for _ in range(10)]</code>. In your case, this would be, like GaretJax suggests, <code>[0.5+random()/2 for _ in range(10)]</code> <hr> Intent I think your intent was being able to further filter based on the entire expression of a list comprehension. In that case, you would put your iterable (here with parentheses rathern than [...] for laziness, though it doesn't matter) inside another comprehension: <pre><code>[r for r in (random.random() for x in range(10)) if r>0.5] </code></pre> For a better solution along these lines that can handle rejection, see the "Solution with list comprehensions only" section or <a href="https://stackoverflow.com/a/6894840/711085">eryksun's answer</a>. <hr> Solution with generators (see below for a list-comprehension only solution) You can do this for any linear range. If your rejection function is arbitrarily complex however, a general way to do rejection sampling would be as follows. (Again, a very bad idea unless you know what you're doing and efficiency doesn't matter.) <pre><code>from random import random from itertools import * def randoms(): while True: yield random() def rejectionSample(pred, n): return islice(filter(pred, randoms()), n) </code></pre> Example: <pre><code>>>> print( list(rejectionSample(lambda x:x>0.5)) ) [0.6656564857979361, 0.9850389778418555, 0.9607471536139308, 0.9191328900300356, 0.810783093197139] </code></pre> You could also do something like: <pre><code>def rejectionSample(pred, n): count = 0 while count<n: r = random() if pred(r): yield r count += 1 </code></pre> <hr> Solution with list comprehensions only However since you want to use list comprehensions only, this means the expression-part of your comprehension cannot fail, so you have to somehow embed a while loop in the comprehension. This is impossible to do with a single lambda function alone, but we can pull it off as long as we have some recursive/loopy primitive, for example... <pre><code>[next(filter(pred,randoms())) for _ in range(10)] </code></pre> (If you really wanted a one-liner list comprehension, <code>randoms()</code> can be rewritten as <code>(random() for _ in count())</code>.) Again, this is unnecessary if you easily find the analytic inverse cumulative distribution function for your particular distribution. <hr> edit: I take that back... it... is possible... with just lambdas... DEAR GOD HAVE MERCY WHAT HORRORS HAVE I UNLEASHED UPON THE WORLD NOOOOOO <pre><code>[ (lambda f:f(f,random()))(lambda self,r:r if r>0.5 else self(self,random())) for _ in range(10) ] </code></pre>
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1. POProduce 10 (or less) uniformly distributed random floats using a list comprehension by filtering based on random value
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1. POProduce 10 (or less) uniformly distributed random floats using a list comprehension by filtering based on random value
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1. COIf you're already implementing the randoms function, you can easily implement the filtering there or use my approach and split the next/filter statement up into different generators for better readability.
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