Python – 测试子串顺序

给定两个字符串，检查子字符串字符是否以正确的顺序出现在字符串中。

Input : test_str = ‘geeksforgeeks’, K = ‘sees’
Output : True
Explanation : “s” after that “ee” and then “s” is present in order in string 1.

Input : test_str = ‘geeksforgeeks’, K = ‘seef’
Output : False
Explanation : Unordered String.

方法 #1：使用 join() + 生成器表达式 + in运算符

在此，我们检查我们使用 join() 连接子字符串中出现的所有字符，然后使用 in运算符检查子字符串是否存在。

Python3

# Python3 code to demonstrate working of 
# Test substring order
# Using join() + in operator + generator expression
  
# initializing string
test_str = 'geeksforgeeks'
  
# printing original string
print("The original string is : " + str(test_str))
  
# initializing substring
K = 'seek'
  
# concatenating required characters 
temp = lambda sub: ''.join(chr for chr in sub if chr in set(K))
  
# checking in order
res = K in temp(test_str)
  
# printing result 
print("Is substring in order : " + str(res))

输出

The original string is : geeksforgeeks
Is substring in order : True

方法 #2：使用 all() + next() + 生成器表达式

在此，我们使用 next() 和生成器表达式获取仅包含子字符串字符的字符串，为了检查顺序，对子字符串中的每个字符使用 all() 操作。

Python3

# Python3 code to demonstrate working of 
# Test substring order
# Using all() + next() + generator expression
  
# initializing string
test_str = 'geeksforgeeks'
  
# printing original string
print("The original string is : " + str(test_str))
  
# initializing substring
K = 'seek'
  
# concatenating required characters using next()
# all() used to test order
test_str = iter(test_str)
res = all(next((ele for ele in test_str if ele == chr), None) is not None for chr in K)
  
# printing result 
print("Is substring in order : " + str(res)) 

输出

The original string is : geeksforgeeks
Is substring in order : True