Python - 查找字符串中第一次出现的字符串列表的索引位置

Question

我想搜索一些文本，找到一组字符串的第一次出现的索引（比如“ - >”或“--x”或“--XX”）一旦找到，我需要知道从哪里开始找到的字符串的位置，以及找到的特定字符串（更具体地说是已识别字符串的长度）

这就是我到目前为止......但还不够。 请帮忙。

arrowlist = {"->x","->","->>","-\","\\-","//--","->o","o\\--","<->","<->o"}
def cxn(line,arrowlist):
   if any(x in line for x in arrowlist):
      print("found an arrow {} at position {}".format(line.find(arrowlist),2))
   else:
      return 0 

也许正则表达式会更容易，但我真的很挣扎，因为箭头列表可能是动态的，箭头字符串的长度也可以变化。

谢谢！

Answer 1

我喜欢这个解决方案，灵感来自这篇文章：

如何在列表推导中使用重新匹配对象

import re

arrowlist = ["xxx->x", "->", "->>", "-\"","\\-"," // --","x->o", "-> ->"]

lines = ["xxx->x->->", "-> ->", "xxx->x", "xxxx->o"]

def filterPick(list,filter):
    return [(m.group(), item_number, m.start()) for item_number,l in enumerate(list) for m in (filter(l),) if m]


if __name__ == '__main__':

    searchRegex = re.compile(r''+ '|'.join(arrowlist) ).search
    x = filterPick(lines, searchRegex)
    print(x)

输出显示：

[('xxx->x', 0, 0), ('->', 1, 0), ('xxx->x', 2, 0), ('x->o', 3, 3)]

第一个数字是列表索引，第二个是字符串的起始索引。

Answer 2

继你的例子的逻辑之后，这成为找到“第一个”匹配箭头并打印它的位置的最便捷的方法。 但是，集合的顺序不是FIFO，所以如果你想保留顺序，我建议用一个列表代替一个箭头列表，这样就可以保留顺序。

    arrowlist = {"->x","->", "->>", "-\\", "\\-","//--","->o","o\\--","<->","<->o"}
    def cxn(line, arrowlist):
       try:
           result = tuple((x, line.find(x)) for x in arrowlist if x in line)[0]
           print("found an arrow {} at position {} with length {}".format(result[0], result[1], len(result[0])))

       # Remember in general it's not a great idea to use an exception as
       # broad as Exception, this is just for example purposes.
       except Exception:
          return 0

如果您正在寻找提供的字符串（行）中的第一个匹配项，您可以这样做：

arrowlist = {"->x","->", "->>", "-\\", "\\-","//--","->o","o\\--","<->","<->o"}

def cxn(line, arrowlist):
   try:
       # key first sorts on the position in string then shortest length 
       # to account for multiple arrow matches (i.e. -> and ->x)
       result = sorted([(x, line.find(x)) for x in arrowlist if x in line], key=lambda r: (r[1],len(r[0])))[0]
       # if you would like to match the "most complete" (i.e. longest-length) word first use:
       # result = sorted([(x, line.find(x)) for x in arrowlist if x in line], key=lambda r: (r[1], -len(r[0])))[0]
       print("found an arrow {} at position {} with length {}".format(result[0], result[1], len(result[0])))

   except Exception:
      return 0

或者，如果您可以访问标准库，则可以使用operator.itemgetter获得几乎相同的效果，并通过较少的函数调用获得效率：

from operator import itemgetter

arrowlist = {"->x","->", "->>", "-\\", "\\-","//--","->o","o\\--","<->","<->o"}

def cxn(line, arrowlist):
   try:
       # key first sorts on the position in string then alphanumerically 
       # on the arrow match (i.e. -> and ->x matched in same position
       # will return -> because when sorted alphanumerically it is first)
       result = sorted([(x, line.find(x)) for x in arrowlist if x in line], key=(itemgetter(1,0)))[0]
       print("found an arrow {} at position {} with length {}".format(result[0], result[1], len(result[0])))

   except Exception:
      return 0

***注意：我使用的箭头列表与您的示例略有不同，因为您提供的箭头列表似乎正在弄乱默认代码格式（可能是因为引用关闭问题）。 请记住，您可以在前面添加一个带有'r'的字符串： r"Text that can use special symbols like the escape \\and\\ be read in as a 'raw' string literal\\" 。 有关原始字符串文字的更多信息， 请参阅此问题 。

Answer 3

你可以做点什么

count = 0
for item in arrowlist:
    count += 1
    if item in line:
        print("found an arrow {} at position {}".format(item,count))

Answer 4

想要发布我想出的答案（来自反馈的组合），你可以看到，这个结果 - 它是非常冗长和非常低效的将返回在正确的位置索引处找到的正确的箭头字符串。 -

arrowlist = ["xxx->x", "->", "->>", "xxx->x","x->o", "xxx->"]
doc =""" @startuml
    n1 xxx->xx n2 : should not find
    n1 ->> n2 : must get the third arrow
    n2  xxx-> n3 : last item
    n3   -> n4 : second item
    n4    ->> n1 : third item"""

def checkForArrow(arrows,line):
    for a in arrows:
        words = line.split(' ')
        for word in words:
            if word == a:
                return(arrows.index(a),word,line.index(word))

for line in iter(doc.splitlines()):
    line = line.strip()
    if line != "":
        print (checkForArrow(arrowlist,line))

返回以下结果:(箭头列表中的项目索引，找到的字符串，行中文本的索引位置）

None
None
(2, '->>', 3)
(5, 'xxx->', 4)
(1, '->', 5)
(2, '->>', 6)