Question

我无法修改汉明距离算法以便以两种方式影响我的数据

如果为小写字母切换大写字母，则将.5添加到汉明距离，除非它位于第一个位置。
例子包括：＆＃34;杀手＆＃34;和＆＃34;杀手＆＃34;距离为0＆＃34;杀手＆＃34;和＆＃34; KiLler＆＃34;汉明距离为.5。＆＃34;滑稽＆＃34;和FAnny＆＃34;距离为1.5（不同字母为1，不同大写为0.5）。
使b和d（以及他们的大写同行）被视为同一件事

这是我发现的代码，它构成了基本的Hamming程序

def hamming_distance(s1, s2):
    assert len(s1) == len(s2)
    return sum(ch1 != ch2 for ch1, ch2 in zip(s1, s2))

if __name__=="__main__":
    a = 'mark'
    b = 'Make'
    print hamming_distance(a, b)

欢迎任何建议！

Answer 1

这是一个简单的解决方案。当然，它可以针对更好的性能进行优化。

注意：我使用Python 3，因为Python 2 will retire soon。

def hamming_distance(s1, s2):
    assert len(s1) == len(s2)
    # b and d are interchangeable
    s1 = s1.replace('b', 'd').replace('B', 'D')
    s2 = s2.replace('b', 'd').replace('B', 'D')
    # add 1 for each different character
    hammingdist = sum(ch1 != ch2 for ch1, ch2 in zip(s1.lower(), s2.lower()))
    # add .5 for each lower/upper case difference (without first letter)
    for i in range(1, len(s1)):
        hammingdist += 0.5 * (s1[i] >= 'a' and s1[i] <= 'z' and\
                              s2[i] >= 'A' and s2[i] <= 'Z' or\
                              s1[i] >= 'A' and s1[i] <= 'Z' and\
                              s2[i] >= 'a' and s2[i] <= 'z')
    return hammingdist

def print_hamming_distance(s1, s2):
    print("hamming distance between", s1, "and", s2, "is",
          hamming_distance(s1, s2))

if __name__ == "__main__":
    assert hamming_distance('mark', 'Make') == 2
    assert hamming_distance('Killer', 'killer') == 0
    assert hamming_distance('killer', 'KiLler') == 0.5
    assert hamming_distance('bole', 'dole') == 0
    print("all fine")
    print_hamming_distance("organized", "orGanised")
    # prints: hamming distance between organized and orGanised is 1.5

对汉明距离/编辑距离的修改

1 个答案: