在字符串中查找特定单词的位置

Posted 2021-03-27

我有一个基因列表，我需要确定列表中的基因是否存在于“文章标题”中，如果存在，则在句子中找到基因的开始和结束位置。

开发的代码确实识别基因并检测基因在句子中的位置。但是，我需要帮助找到基因的起始位置和终点位置

doc = tree.getroot()
 for ArticleTitle in doc.iter('ArticleTitle'):
    file1 = (ET.tostring(ArticleTitle, encoding='utf8').decode('utf8'))
    filename = file1[52:(len(file1))]
    Article= filename.split("<")[0]
    # print(Article)
    # print(type(Article))
    title= Article.split()
    gene_list = ["ABCD1","ADA","ALDOB","APC","ARSB","ATAD3B","AXIN2","BLM","BMPR1A","BRAF","BRCA1"] 
    for item in title:
        for item1 in gene_list:
            if item == item1:
                str_title= ' '.join(title)
                print(str_title)
                print("Gene Found: " + item)
                index= title.index(item)
                print("Index of the Gene :" +str(index))

                result = 0
                for char in str_title:
                    result +=1
                print(result)

目前的输出是：

Healthy people 2000: a call to action for ADA members.
Gene Found: ADA
Index of the Gene :8
54

预期产出是：

Healthy people 2000: a call to action for ADA members.
Gene Found: ADA
Index of the Gene :8
Gene start position: 42
Gene End postion:  45

开始和结束位置也应该计算单词之间的空格。

答案

可以使用正则表达式

l=["ABCD1","ADA","ALDOB","APC","ARSB"]
l='|'.join(l)
test_string='Healthy people 2000: a call to action for ADA members.'
pos=0
for i in test_string.split():
    m=re.search(l,i)
    if m:
        gene=m.group(0)
        start=test_string.find(gene)
        end=start+len(gene)
        print(start,end,gene,pos)
    pos+=1

产量

(42, 45, 'ADA', 8)

没有字符串中实际位置的较短解决方案可能是

l=["ABCD1","ADA","ALDOB","APC","ARSB"]
l='|'.join(l)
test_string='Healthy people 2000: a call to action for ADA members.'

[(m.start(),m.group(0),m.end()) for m in re.finditer(l,test_string)]

另一答案

我们也可以使用Flashtext

from flashtext import KeywordProcessor

kpo = KeywordProcessor(case_sensitive=True)

gene_list = ["ABCD1","ADA","ALDOB","APC","ARSB","ATAD3B","AXIN2","BLM","BMPR1A","BRAF","BRCA1"] 

for word in gene_list:
    kpo.add_keyword(word)

kpo.extract_keywords("Healthy people 2000: a call to action for ADA members.",span_info=True)
#o/p --> [('ADA', 42, 45)]