Question

我已延长this SO问题＆amp;正在比较两个乳胶方程。这是两个二次方程的例子。

eqn1 = "*=\frac{-*\pm\sqrt{*^2-4ac}}{2a}"
eqn2 = "x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}"

我需要比较这些是正确的，因为，而不是x，b，我已经使用了*。我所做的就是将方程式转换为单词列表。

eqn1_word = [*,frac,*,pm,sqrt,*,2,4ac,2a]
eqn2_word = [x,frac,b,pm, sqrt, b, 2, 4ac, 2a]

所以矢量是

eqn1_vec= Counter({'*': 3, 'frac': 1, 'sqrt': 1, '2': 1, '2a': 1, '4ac': 1, 'pm': 1})
eqn2_vec = Counter({'b': 2, 'frac': 1, 'sqrt': 1, '2': 1, '2a': 1, '4ac': 1, 'x': 1, 'pm': 1})

现在我的扩展名是我正在检查eqn1_word中*的百分比，然后检查that回答给出的正常余弦相似度。最后，我添加了两个值，几乎等于1。

这适用于大多数情况（如果一个变量被*替换）。对于eqn1_vec，*值为3，在eqn2_vec中，b = 2，x = 1。

更多说明＆amp;更好的理解请检查这一点。从那个参考，我的代码是这样的。

def get_cosine(self, c_eqn1_eqn, c_eqn2_eqn):
    print 'c_eqn1_eqn = ', c_eqn1_eqn
    print 'c_eqn2_eqn = ', c_eqn2_eqn
    _special_symbol = float(c_eqn1_eqn.count("*"))
    cos_result = 0
    symbol_percentage = 0
    try:
        eqn1_vector = Counter(self.get_word(c_eqn1_eqn))# get word will return word list
        eqn2_vector = Counter(self.get_word(c_eqn2_eqn))
        _words = sum([x for x in eqn1_vector.values()])
        if eqn2_vector.has_key("*"):
            _special_symbol -= eqn2_vector["*"]
        print '_special_symbol = ', _special_symbol
        print '_words @ last = ', _words
        try:
            symbol_percentage = _special_symbol / _words
        except ZeroDivisionError:
            symbol_percentage = 0.0
    except Exception as exp:
        print "Exception at converting equation to vector", exp
        traceback.print_exc()
    else:
        intersection = set(eqn1_vector.keys()) & set(eqn2_vector.keys())
        numerator = sum([eqn1_vector[x] * eqn2_vector[x] for x in intersection])
        _sum1 = sum([eqn1_vector[x]**2 for x in eqn1_vector.keys()])
        _sum2 = sum([eqn2_vector[x]**2 for x in eqn2_vector.keys()])
        denominator = math.sqrt(_sum1) * math.sqrt(_sum2)
        print 'numerator = ', numerator
        print 'denominator = ', denominator
        if not denominator:
            cos_result = 0
        else:
            cos_result = float(numerator) / denominator
        print cos_result
    final_result = float(symbol_percentage) + cos_result
    return final_result if final_result <= 1.0 else 1

问题是当交叉点值很小时，分子变小。我从班上复制了。请忽略自己。

如何解决这个问题。提前致谢。如果有任何错误或我的概念有误，请与我分享。

Answer 1

我找到了解决这个问题的方法。

由于我们可以/不应该增加分子值，我决定改为分母。我的逻辑是，如果eqn2中的*和非交叉值的数量相同，则减小分母值。如果没有，那就放手吧。现在我不必计算＆＃34; *＆＃34;的百分比也没有在余弦结果中加入。

def get_cosine(c_eqn1, c_eqn2):
    _special_symbol = float(c_eqn1.count("*"))
    cos_result = 0
    try:
        eqn1_vector = Counter(get_word(c_eqn1))
        eqn2_vector = Counter(get_word(c_eqn2))
        _special_symbol = 0
        spe_list = list()
        # Storing number of * & the value contains *
        for _val in eqn1_vector.keys():
            if _val.__contains__("*"):
                _special_symbol += eqn1_vector[_val]
                spe_list.append(_val)
        if eqn2_vector.has_key("*"):
            _special_symbol -= eqn2_vector["*"]
    except Exception as exp:
        print "Exception at converting equation to vector", exp
        traceback.print_exc()
    else:
        intersection = set(eqn1_vector.keys()) & set(eqn2_vector.keys())
        numerator = sum([eqn1_vector[x] * eqn2_vector[x]
                         for x in intersection])
        non_intersection_sum = 0
        non_intersection_value = list()
        # storing no of non_matched value
        for _val in eqn2_vector.keys():
            if _val not in intersection:
                non_intersection_sum += eqn2_vector[_val]
                non_intersection_value.append(_val)
        # Join both non intercet lists
        if non_intersection_value:
            non_intersection_value.extend(spe_list)
        # If both non intersect value are not same
        # Empty the list
        if _special_symbol != non_intersection_sum:
            non_intersection_value = list()
        # Cosine similarity formula
        _sum1 = sum([eqn1_vector[x]**2 for x in eqn1_vector.keys() if x not in non_intersection_value])
        _sum2 = sum([eqn2_vector[x]**2 for x in eqn2_vector.keys() if x not in non_intersection_value])
        denominator = math.sqrt(_sum1) * math.sqrt(_sum2)
        if not denominator:
            cos_result = 0
        else:
            cos_result = float(numerator) / denominator
    return cos_result if cos_result <= 1.0 else 1

方程的乳胶代码的余弦相似度

1 个答案: