python - 从列表中提取出发和到达

Question

我正在尝试从结构和长度可变的列表中提取一些参数。基本上，这些参数是路线的出发地址和到达地址。此列表是根据自然语言中的一个句子构建的，因此它不遵循任何特定模板:

1st example : ['go', 'Buzenval', 'from', 'Chatelet']
2nd example : ['How', 'go', 'street', 'Saint', 'Augustin', 'from', 'Buzenval']
3rd example : ['go', 'from', '33', 'street', 'Republique', 'to', '12','street','Napoleon']

我已经设法为每种情况创建了另一个非常相似的列表，除了出发和到达被实际的词“出发”和“到达”代替。通过上面的例子，我得到:

1st example : ['go', 'arrival', 'from', 'departure']
2nd example : ['How', 'go', 'arrival', 'from', 'departure']
3rd example : ['go', 'from', 'departure', 'to', 'arrival']

现在我有了这两种列表，我想确定出发和到达:

1rst example : departure = ['Chatelet'], arrival = ['Buzenval']
2nd example : departure =  ['Buzenval'], arrival = ['street','Saint','Augustin']
3rd example : departure = ['33','street','Republique'], arrival = ['12','street','Napoleon']

基本上，参数是两个列表中不同的所有内容，但我需要确定哪个是出发，哪个是到达。我认为 Regex 可以在这方面帮助我，但我不知道如何。

感谢您的帮助!

Answer 1

我找到了解决您的三个示例的方法。你应该改变的一件事是变量名，我不知道如何命名它们。 _{(这是旧版本，速度慢且难以理解。后面的版本更好)}

def extract_places(names, modes):
    keywords = set(modes).intersection(names)
    extracted = [[] for _ in modes]
    j = 0
    for i, mode in enumerate(modes):
        if mode.lower() in keywords:
            if mode.lower() != names[j].lower():
                while mode.lower() != names[j].lower():
                    extracted[i - 1].append(names[j])
                    j += 1
            else:
                extracted[i].append(names[j])
                j += 1
        else:
            if names[j].lower() not in keywords:
                while j < len(names) and names[j].lower() not in keywords:
                    extracted[i].append(names[j])
                    j += 1

    extracted = dict(zip(modes, extracted))
    return extracted["arrival"], extracted["departure"]

我找到了另一种方法，可能更容易理解。但这种方式比第一种方式快十倍，因此您可能想要使用它。

def partition(l, word): # Helper to split a list or tuple at an specific element
    i = l.index(word)
    return l[:i], l[i + 1:]

def extract_places(names, modes):
    keywords = set(modes).intersection(names)
    mapped = [(modes, names)]
    for word in keywords:
        new_mapped = []
        for mode,name in mapped:
            if word in mode:
                m1, m2 = partition(mode, word)
                n1, n2 = partition(name, word)
                if m1:
                    new_mapped.append((m1, n1))
                if m2:
                    new_mapped.append((m2, n2))
            else:
                new_mapped.append((mode,name))
        mapped = new_mapped
    mapped = {m[0]: n for m, n in mapped}
    return mapped['arrival'], mapped['departure']

两种方式完全一样:

for example in ((['go', 'Buzenval', 'from', 'Chatelet'],
                 ['go', 'arrival', 'from', 'departure']
                 ),
                (['How', 'go', 'street', 'Saint', 'Augustin', 'from', 'Buzenval'],
                 ['How', 'go', 'arrival', 'from', 'departure']
                 ),
                (['go', 'from', '33', 'street', 'Republique', 'to', '12', 'street', 'Napoleon'],
                 ['go', 'from', 'departure', 'to', 'arrival']
                 )):
    print(extract_places(*example))

两者的打印:

(['Buzenval'], ['Chatelet'])
(['street', 'Saint', 'Augustin'], ['Buzenval'])
(['12', 'street', 'Napoleon'], ['33', 'street', 'Republique'])