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82 lines
1.7 KiB
Python
82 lines
1.7 KiB
Python
# Copyright (c) 2005-2009 Bryce "Zooko" Wilcox-O'Hearn
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# mailto:zooko@zooko.com
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# http://zooko.com/repos/pyutil
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# http://pypi.python.org/pypi/pyutil
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# http://allmydata.org/trac/pyutil
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# Permission is hereby granted, free of charge, to any person obtaining a copy
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# of this work to deal in this work without restriction (including the rights
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# to use, modify, distribute, sublicense, and/or sell copies).
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"""
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A few commonly needed functions.
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"""
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import math
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def div_ceil(n, d):
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"""
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The smallest integer k such that k*d >= n.
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"""
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return (n/d) + (n%d != 0)
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def next_multiple(n, k):
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"""
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The smallest multiple of k which is >= n.
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"""
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return div_ceil(n, k) * k
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def pad_size(n, k):
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"""
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The smallest number that has to be added to n so that n is a multiple of k.
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"""
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if n%k:
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return k - n%k
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else:
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return 0
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def is_power_of_k(n, k):
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return k**int(math.log(n, k) + 0.5) == n
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def next_power_of_k(n, k):
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if n == 0:
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x = 0
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else:
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x = int(math.log(n, k) + 0.5)
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r = k**x
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if k**x < n:
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return k**(x+1)
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else:
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return k**x
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def ave(l):
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return sum(l) / len(l)
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def log_ceil(n, b):
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"""
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The smallest integer k such that b^k >= n.
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log_ceil(n, 2) is the number of bits needed to store any of n values, e.g.
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the number of bits needed to store any of 128 possible values is 7.
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"""
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p = 1
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k = 0
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while p < n:
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p *= b
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k += 1
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return k
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def log_floor(n, b):
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"""
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The largest integer k such that b^k <= n.
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"""
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p = 1
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k = 0
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while p <= n:
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p *= b
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k += 1
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return k - 1
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def round_sigfigs(f, n):
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fmt = "%." + str(n-1) + "e"
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return float(fmt % f)
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