Given a String, extract all the numbers that are percentages.
Input : test_str = ‘neveropen 20% is 100% way to get 200% success’ Output : [‘20%’, ‘100%’, ‘200%’] Explanation : 20%, 100% and 200% are percentages present. Input : test_str = ‘neveropen is way to get success’ Output : [] Explanation : No percentages present.
Method #1 : Using findall() + regex
In this, we employ appropriate regex having “%” symbol in suffix and use findall() to get all occurrences of such numbers from String.
Python3
# Python3 code to demonstrate working of # Extract Percentages from String # Using regex + findall()import re# initializing stringstest_str = 'neveropen is 100 % way to get 200 % success'# printing original stringprint("The original string is : " + str(test_str))# getting required result from string res = re.findall('\d*%', test_str)# printing result print("The percentages : " + str(res)) |
The original string is : neveropen is 100% way to get 200% success The percentages : ['100%', '200%']
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #2 : Using re.sub() + split()
In this, we perform split of all words, and then from words that have %, we remove all non-numeric strings. This can be buggy in cases, we have different ordering of % and numbers in string.
Python3
# Python3 code to demonstrate working of # Extract Percentages from String # Using re.sub() + split()import re# initializing stringstest_str = 'neveropen is 100 % way to get 200 % success'# printing original stringprint("The original string is : " + str(test_str))# extracting wordstemp = test_str.split()# using res = []for sub in temp: if '%' in sub: # replace empty string to all non-number chars res.append(re.sub(r'[^\d, %]', '', sub))# printing result print("The percentages : " + str(res)) |
The original string is : neveropen is 100% way to get 200% success The percentages : ['100%', '200%']
Time complexity: O(n), where n is the length of the test_list. The re.sub() + split() takes O(n) time
Auxiliary Space: O(n), extra space of size n is required
