Python - 字符串列表中单词的开始和结束索引

给定一个字符串，我们的任务是编写一个Python程序，从字符串提取另一个列表的单词的所有元素的开始和结束索引。

Input : test_str = “gfg is best for all CS geeks and engineering job seekers”, check_list = [“geeks”, “engineering”, “best”, “gfg”]

Output : {‘geeks’: [23, 27], ‘engineering’: [33, 43], ‘best’: [7, 10], ‘gfg’: [0, 2]}

Explanation : “geeks” starts from index number 23 till 27, hence the result.

Input : test_str = “gfg is best for all CS geeks and engineering job seekers”, check_list = [“geeks”, “gfg”]

Output : {‘geeks’: [23, 27], ‘gfg’: [0, 2]}

Explanation : “geeks” starts from index number 23 till 27, hence the result.

方法 #1：使用循环+ index() + len()

在这里，循环用于从列表中获取每个元素。 index() 获取初始索引， len() 获取字符串list 中所有元素的最后一个索引。

Python3

# Python3 code to demonstrate working of
# Start and End Indices of words from list in String
# Using loop + index() + len()
  
# initializing string
test_str = "gfg is best for all CS geeks and engineering job seekers"
  
# printing original string
print("The original string is : " + str(test_str))
  
# initializing check_list 
check_list = ["geeks", "engineering", "best", "gfg"]
  
res = dict()
for ele in check_list :
    if ele in test_str:
          
        # getting front index 
        strt = test_str.index(ele)
          
        # getting ending index
        res[ele] = [strt, strt + len(ele) - 1]
  
# printing result
print("Required extracted indices  : " + str(res))

输出：

The original string is : gfg is best for all CS geeks and engineering job seekers

Required extracted indices  : {‘geeks’: [23, 27], ‘engineering’: [33, 43], ‘best’: [7, 10], ‘gfg’: [0, 2]}

方法 #2：使用字典理解+ len() + index()

在这里，我们执行类似于上述函数的任务，但结果字典的构建是使用字典理解使用速记完成的。

蟒蛇3

# Python3 code to demonstrate working of
# Start and End Indices of words from list in String
# Using dictionary comprehension + len() + index()
  
# initializing string
test_str = "gfg is best for all CS geeks and engineering job seekers"
  
# printing original string
print("The original string is : " + str(test_str))
  
# initializing check_list
check_list = ["geeks", "engineering", "best", "gfg"]
  
# Dictionary comprehension to be used as shorthand for
# forming result Dictionary
res = {key: [test_str.index(key), test_str.index(key) + len(key) - 1]
       for key in check_list if key in test_str}
  
# printing result
print("Required extracted indices  : " + str(res))

输出：

The original string is : gfg is best for all CS geeks and engineering job seekers

Required extracted indices  : {‘geeks’: [23, 27], ‘engineering’: [33, 43], ‘best’: [7, 10], ‘gfg’: [0, 2]}