Question

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']

我想检查isl中c中每个项目的每个项目，以便我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"]

可以看出，我也得到了部分字符串匹配。

我试过以下内容：

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

但这只是输出为

out = ["John", "Tamm"]

我也尝试了以下内容：

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

但这仅仅输出＆＃34; John＆＃34;。我如何得到我想要的东西？

Answer 1

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']

如果您只想存储最大匹配，那么您需要以相反的顺序生成子字符串，并在isl找到匹配后立即中断：

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']

Answer 2

好吧我已经拿出来了！一种非常黑客的方式;我自己不喜欢这个方法，但它给了我输出：

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']

Python：检查两个列表之间字符串的部分匹配

2 个答案: