filter(lambda prime: all(prime%num for num in range(2, prime)), range(2,1000))
或
reduce(lambda i,n: i if 0 in [n%x for x in i] else i+[n] , xrange(2,1000), [])
python 求1000以内的质数
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转载自scm002.iteye.com/blog/2258544
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