Python：检查两个列表之间字符串的部分匹配

我有两个列表，如下所示：

c = ['John', 'query 989877 forcast', 'Tamm']
isl = ['My name is Anne Query 989877', 'John', 'Tamm Ju']

我想用

c

中的每个项目检查

isl

中的每个项目，以便获得所有部分字符串匹配。 我需要的输出如下所示：

out = ["john", "query 989877", "tamm"]

 out = []
 for word in c:
    for w in isl:
        if word.lower() in w.lower():
                 out.append(word)

print [word for word in c if word.lower() in (e.lower() for e in isl)]

可以看出，我也得到了部分字符串匹配

我尝试了以下方法：

out = ["john", "query 989877", "tamm"]

 out = []
 for word in c:
    for w in isl:
        if word.lower() in w.lower():
                 out.append(word)

print [word for word in c if word.lower() in (e.lower() for e in isl)]

但这只给了我

out = ["John", "Tamm"]

我还尝试了以下方法：

out = ["john", "query 989877", "tamm"]

 out = []
 for word in c:
    for w in isl:
        if word.lower() in w.lower():
                 out.append(word)

print [word for word in c if word.lower() in (e.lower() for e in isl)]

但这只输出“约翰”。

---

我怎样才能得到我想要的？

也许是这样的：

def get_sub_strings(s):
    words = s.split()
    for i in xrange(1, len(words)+1):      #reverse the order here
        for n in xrange(0, len(words)+1-i):
            yield ' '.join(words[n:n+i])
...             
>>> out = []
>>> for word in c:
    for sub in get_sub_strings(word.lower()):
        for s in isl:
            if sub in s.lower():
                out.append(sub)
...                 
>>> out
['john', 'query', '989877', 'query 989877', 'tamm']

如果只想存储最大的匹配项，则需要按相反顺序生成子字符串，并在

isl

中找到匹配项后立即中断：

def get_sub_strings(s):
    words = s.split()
    for i in xrange(len(words)+1, 0, -1):
        for n in xrange(0, len(words)+1-i):
            yield ' '.join(words[n:n+i])

out = []
for word in c:
    for sub in get_sub_strings(word.lower()):
        if any(sub in s.lower() for s in isl):
            out.append(sub)
            break

print out
#['john', 'query 989877', 'tamm']

---

好吧，我想到了这个！这是一种非常老套的方法；我自己不喜欢这个方法，但它提供了我的输出：

Step1:
in: c1 = []
    for r in c:
       c1.append(r.split()) 
out: c1 = [['John'], ['query', '989877', 'forcast'], ['Tamm']]


Step2:
in: p = []
    for w in isl:
        for word in c1:
            for w1 in word:
                 if w1.lower() in w.lower():
                         p.append(w1)
out: p = ['query', '989877', 'John', 'Tamm']


Step3:
in: out = []
    for word in c:
        t = []
        for i in p:
             if i in word:
                t.append(i)
        out.append(t)
out: out = [['John'], ['query', '989877'], ['Tamm']]

Step4:
in: out_final = []
    for i in out:
        out_final.append(" ".join(e for e in i))
out: out_final = ['John', 'query 989877', 'Tamm']

---

它必须是“查询9877”，还是可以是“查询”，“989877”？是的…我想要所有匹配项（部分和全部），这实际上非常棒！是否要从“输出”列表中删除“查询”和“989877”？因为理想情况下，它们不应该出现在输出中。我坚持这一点的原因是，我以后需要对“out”列表中的所有元素进行计数。如果我像您所示那样保留输出，这将导致错误的答案。@user1452759检查我的第二个解决方案。非常感谢！这太完美了！