使用python

Question

我有一个像这样的文件[samplefile.txt]-

dummy line 1
dummy line 2
create_list animal -list     
  tiger
create_list bird -list        
  crow
dummy line 3
create_list car -list        
  ford/mustang
dummy line 4
dummy line 5
create_list truck
  -list ford/F150
create_list human -list {
  jack
  john}

我想得到这样的输出- ['tiger', 'crow', 'ford/mustang', 'ford/F150']

我正在使用的代码-

import re

def extractListItems(File):
    f = open(File,"r")
    content = f.read()

    list_items = []

    for m in re.finditer(r'(?:\bcreate_list+\s+\S+\s+-list+\s*)((?:\b\S+\b(?:\s*))+)(?:\n)', content):
            list_items.extend(re.split(r'\s+', m.group(1).strip()))

    print (list_items)
    f.close()

extractListItems("samplefile.txt")

要获得所需的输出，我需要进行哪些修改？

编辑-根据文件内容和所需的输出进行了更改。

Answer 1

我认为您要匹配的单词总是缩进的。

import re
regex = r"^\s+(\S.*)"

test_str = ("dummy line 1\n"
    "dummy line 2\n"
    "create_list animal -list     \n"
    "  tiger\n"
    "create_list bird -list        \n"
    "  crow\n"
    "dummy line 3\n"
    "create_list car -list        \n"
    "  ford/mustang\n"
    "dummy line 4\n"
    "dummy line 5")

match = re.findall(regex, test_str, re.MULTILINE)
print(match)

输出

['tiger', 'crow', 'ford/mustang']

Answer 2

如果您的内容总是以空格开头，请尝试

import re

def extractListItems(File):
    f = open(File,"r")
    content = f.read().splitlines()
    pattern=lambda data:len(re.findall(r"^\s+",data))
    lst=[c.strip() for c in content if pattern(c)>0]
    print(lst)
    f.close()

extractListItems("samplefile.txt")

输出

['tiger', 'crow', 'ford/mustang']

OR

 pattern=lambda data:re.findall(r"^[\s].*",data)
 lst=[val[0].strip() for val in map(pattern,content) if val]

结果

 ['tiger', 'crow', 'ford/mustang']