Given a string of characters with digits embedded in it. The task is to write a Python program to extract all the numbers that are trailing, i.e at the suffix of the string.
Examples:
Input : test_str = “GFG04”
Output : 04
Explanation : 04 is suffix of string and number hence extracted.Input : test_str = “GFG143”
Output : 143
Explanation : 143 is suffix of string and number hence extracted.
Method #1 : Using loop + isdigit() + slicing
In this, the string is reversed using slicing, and occurrence of 1st non-digit element is recorded using isdigit, and at last, list is sliced till the element and reversed again to get the suffix number.
Python3
# Python3 code to demonstrate working of# Extract Suffix numbers# Using loop + isdigit() + slicing # initializing stringtest_str = "GFG04"# printing original stringprint("The original string is : " + str(test_str))# loop for fetching the 1st non digit index rev_str = test_str[::-1] temp_idx = 0for idx, ele in enumerate(rev_str): if not ele.isdigit(): temp_idx = idx break# reversing the extracted string to# get number res = rev_str[:temp_idx][::-1] # printing resultprint("Suffix number : " + str(res)) |
The original string is : GFG04 Suffix number : 04
Method #2: Using regex
Appropriate regex can be employed to solve this problem. Provides a compact solution to the problem.
Python3
# Python3 code to demonstrate working of# Extract Suffix numbers# Using regeximport re# initializing stringtest_str = "GFG04"# printing original stringprint("The original string is : " + str(test_str))# regex to extract number res = re.search(r"(\d+)$", test_str).group() # printing resultprint("Suffix number : " + str(res)) |
The original string is : GFG04 Suffix number : 04
The time and space complexity for all the methods are the same:
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #3: Without isdigit() method
Python3
# Python3 code to demonstrate working of# Extract Suffix numbers# initializing stringtest_str = "GFG04"# printing original stringprint("The original string is : " + str(test_str))# loop for fetching the 1st non digit index rev_str = test_str[::-1] temp = 0digits = "0123456789"for i, ele in enumerate(rev_str): if not ele in digits: temp = i break# reversing the extracted string to# get number res = rev_str[:temp][::-1] # printing resultprint("Suffix number : " + str(res)) |
The original string is : GFG04 Suffix number : 04
Method#4: Using list comprehension
Python3
test_str = "GFG04"# Printing original stringprint("The original string is : " + str(test_str))# Using list comprehension to extract suffix numberres = ''.join([char for char in test_str[::-1] if char.isdigit()])[::-1]# Printing resultprint("Suffix number : " + str(res))#This code is contributed by Vinay Pinjala. |
The original string is : GFG04 Suffix number : 04
Time Complexity: O(n)
Auxiliary Space: O(n)
Method#5: Using filter and join method.
Python3
# Python3 code to demonstrate working of# Extract Suffix numbers# Using filter method# initializing stringtest_str = "GFG04"# printing original stringprint("The original string is : " + str(test_str))res = ''.join(filter(str.isdigit, test_str[::-1]))[::-1] # printing resultprint("Suffix number : " + str(res))#this code contributed by tvsk |
The original string is : GFG04 Suffix number : 04
Time Complexity: O(n)
Auxiliary Space: O(n)
Method#6: Using rfind() and string slicing
Step by step Algorithm:
- Find the index of the first non-digit character in the string using a lambda function and filter.
- Find the last occurrence of this non-digit character using rfind() function.
- Extract the suffix number using string slicing from the index of the non-digit character to the end of the string.
- Print the suffix number.
Python3
test_str = "GFG04"# printing original stringprint("The original string is : " + str(test_str))# using rfind() and string slicing to extract suffix numberstemp_idx = test_str.rfind(next(filter(lambda x: not x.isdigit(), test_str)))res = test_str[temp_idx+1:]# printing resultprint("Suffix number : " + str(res)) |
The original string is : GFG04 Suffix number : 04
Time Complexity: O(n), where n is the length of the string. The lambda function and filter takes O(n) time, while rfind() and string slicing takes constant time.
Space Complexity: O(1), as we are not using any extra space.
Method #7: Using rpartition() + isdigit()
- Initialize the string
- Use the rpartition() method to split the string into 3 parts – the part before the last digit, the last digit, and the part after the last digit.
- Check if the last part (the part after the last digit) is empty, which indicates that the last character was a digit.
- If the last part is not empty, there was no numeric suffix in the original string, so print an appropriate message.
- If the last part is empty, check if the middle part (the last digit) is a digit.
- If the middle part is not a digit, there was no numeric suffix in the original string, so print an appropriate message.
- If the middle part is a digit, print it as the numeric suffix.
Python3
# initializing stringtest_str = "GFG04"# printing original stringprint("The original string is : " + str(test_str))# extracting suffix numberres = test_str.rpartition('GFG')[2]# checking if suffix is numericif res.isdigit(): # printing suffix number print("Suffix number : " + str(res))else: # printing error message print("No suffix number found in string!") |
The original string is : GFG04 Suffix number : 04
Time Complexity: O(n), where n is the length of the string.
Auxiliary Space: O(n), as we are creating a new string to store the extracted suffix.
Method #8: Using numpy:
- Initialize the input string.
- Convert the string to a NumPy array of characters.
- Convert the array to a string using NumPy’s char module.
- Find the indices of all the numeric characters in the string using the where() method from NumPy.
- Extract the suffix numbers by joining the characters at the found indices.
- Print the extracted suffix numbers.
Python3
import numpy as np# initializing stringtest_str = "GFG04"# printing original stringprint("The original string is : " + str(test_str))# convert string to numpy array of characterschar_array = np.array(list(test_str))# convert array to string using numpy's char modulechar_string = np.char.mod('%c', char_array)# find indices of all numeric charactersindices = np.where(np.char.isdigit(char_string))[0]# extract suffix numberssuffix_numbers = ''.join(char_array[indices])# print resultprint("Suffix number: " + suffix_numbers)#This code is contributed by Jyothi pinjala. |
Output: The original string is : GFG04 Suffix number: 04
Time complexity:
The list() function in Python has a time complexity of O(n), where n is the length of the string. This operation is done once in our code.
The np.array() function in NumPy also has a time complexity of O(n), where n is the length of the string. This operation is also done once in our code.
The np.char.mod() function in NumPy has a time complexity of O(n), where n is the length of the string. This operation is also done once in our code.
The np.where() function in NumPy has a time complexity of O(n), where n is the length of the string. This operation is done once in our code.
The join() function in Python has a time complexity of O(n), where n is the length of the extracted suffix numbers. This operation is done once in our code.
Auxiliary Space:
The space complexity of our code depends on the length of the string. The input string is stored once as a string variable and then converted to a NumPy array, which also takes up space in memory. The space complexity is O(n), where n is the length of the string.
