问题描述
在ruby中，计算两个无符号整数(例如汉明距离)之间的位差的最有效方法是什么?
In ruby, what is the most efficient way to calculate the bit difference between two unsigned integers (e.g. the hamming distance)?
例如，我有整数a = 2323409845和b =1782647144.
Eg, I have integer a = 2323409845 and b = 1782647144.
它们的二进制表示形式是:
Their binary representations are:
a = 10001010011111000110101110110101
b = 01101010010000010000100101101000
a&之间的位差b是17 ..
The bit difference between the a & b is 17..
我可以对它们执行逻辑XOR，但这会给我一个不同的整数！= 17，然后我必须遍历结果的二进制表示并计算1的数目.
I can do a logical XOR on them, but that will give me a different integer != 17, I would then have to iterate through the binary representation of the result and tally the # of 1s.
计算位差的最有效方法是什么?
What is the most efficient way to calculate the bit difference?
现在，计算许多整数序列的位差的答案是否改变?例如.给定2个无符号整数序列:
Now, does the answer change for calculating the bit difference of sequences of many ints? E.g. given 2 sequences of unsigned integers:
x = {2323409845,641760420,509499086....}
y = {uint,uint,uint...}
计算两个序列之间的位差的最有效方法是什么?
What is the most efficient way to calculate the bit difference between the two sequences?
您是要遍历整个序列，还是有一种更快的方法来一次计算整个序列的差异?
Would you iterate through the sequence, or is there a faster way to calculate the difference across the entire sequence at once?
推荐答案
您可以使用Ruby中优化的String函数来进行位计数，而不是使用纯算术.有了一些快速的基准测试，结果大约快了6倍.
You can make use of the optimized String functions in Ruby to do the bit counting, instead of pure arithmetic. It turns out to be about 6 times faster with some quick benchmarking.
def h2(a, b)
(a^b).to_s(2).count("1")
end
h1是正常的计算方式，而h2将xor转换为字符串，并计算"1"的数量
h1 is the normal way to calculate, while h2 converts the xor into a string, and counts the number of "1"s
基准:
ruby-1.9.2-p180:001:0>> def h1(a, b)
ruby-1.9.2-p180:002:1*> ret = 0
ruby-1.9.2-p180:003:1*> xor = a ^ b
ruby-1.9.2-p180:004:1*> until xor == 0
ruby-1.9.2-p180:005:2*> ret += 1
ruby-1.9.2-p180:006:2*> xor &= xor - 1
ruby-1.9.2-p180:007:2*> end
ruby-1.9.2-p180:008:1*> ret
ruby-1.9.2-p180:009:1*> end
# => nil
ruby-1.9.2-p180:010:0>> def h2(a, b)
ruby-1.9.2-p180:011:1*> (a^b).to_s(2).count("1")
ruby-1.9.2-p180:012:1*> end
# => nil
ruby-1.9.2-p180:013:0>> h1(2323409845, 1782647144)
# => 17
ruby-1.9.2-p180:014:0>> h2(2323409845, 1782647144)
# => 17
ruby-1.9.2-p180:015:0>> quickbench(10**5) { h1(2323409845, 1782647144) }
Rehearsal ------------------------------------
2.060000 0.000000 2.060000 ( 1.944690)
--------------------------- total: 2.060000sec
user system total real
1.990000 0.000000 1.990000 ( 1.958056)
# => nil
ruby-1.9.2-p180:016:0>> quickbench(10**5) { h2(2323409845, 1782647144) }
Rehearsal ------------------------------------
0.340000 0.000000 0.340000 ( 0.333673)
--------------------------- total: 0.340000sec
user system total real
0.320000 0.000000 0.320000 ( 0.326854)
# => nil
ruby-1.9.2-p180:017:0>>
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