Python | Case Counter in String
Last Updated :
23 Mar, 2023
Sometimes, while working with Python String, we can have a problem in which we need to separate the lower and upper case count. This kind of operation can have its application in many domains. Let’s discuss certain ways in which this task can be done.
Method #1: Using map() + sum() + isupper() + islower() The combination of the above functions can be used to perform this task. In this, we separately extract the count using sum() and map() and respective inbuilt functions.
Python3
test_str = "GFG is For GeeKs"
print("The original string is : " + test_str)
res_upper = sum(map(str.isupper, test_str))
res_lower = sum(map(str.islower, test_str))
print("The count of Upper case characters : " + str(res_upper))
print("The count of Lower case characters : " + str(res_lower))
|
Output :
The original string is : GFG is For GeeKs
The count of Upper case characters : 6
The count of Lower case characters : 7
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #2: Using Counter() + isupper() + islower() The combinations of above methods can also be used to perform this particular task. In this, we perform the task of holding count using Counter().
Python3
from collections import Counter
test_str = "GFG is For GeeKs"
print("The original string is : " + test_str)
res = Counter("upper" if ele.isupper() else "lower" if ele.islower()
else " " for ele in test_str)
print("The count of Upper case characters : " + str(res['upper']))
print("The count of Lower case characters : " + str(res['lower']))
|
Output :
The original string is : GFG is For GeeKs
The count of Upper case characters : 6
The count of Lower case characters : 7
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #3: Using ord() function (ASCII Values)
Python3
test_str = "GFG is For GeeKs"
print("The original string is : " + test_str)
res_upper = 0
res_lower = 0
for i in test_str:
if(ord(i) >= ord('A') and ord(i) <= ord('Z')):
res_upper += 1
elif(ord(i) >= ord('a') and ord(i) <= ord('z')):
res_lower += 1
print("The count of Upper case characters : " + str(res_upper))
print("The count of Lower case characters : " + str(res_lower))
|
Output
The original string is : GFG is For GeeKs
The count of Upper case characters : 6
The count of Lower case characters : 7
Time Complexity: O(n)
Auxiliary Space: O(1)
Method #4: Using regex
Python3
import re
test_str = "GFG is For GeeKs"
print("The original string is : " + test_str)
res_upper = len(re.findall(r'[A-Z]',test_str))
res_lower = len(re.findall(r'[a-z]',test_str))
print("The count of Upper case characters : " + str(res_upper))
print("The count of Lower case characters : " + str(res_lower))
|
Output
The original string is : GFG is For GeeKs
The count of Upper case characters : 6
The count of Lower case characters : 7
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #5 : Without any builtin methods
Python3
test_str = "GFG is For GeeKs"
print("The original string is : " + test_str)
res_upper = 0
res_lower = 0
la="abcdefghijklmnopqrstuvwxyz"
ua="ABCDEFGHIJKLMNOPQRSTUVWXYZ"
for i in test_str:
if i in la:
res_lower+=1
elif i in ua:
res_upper+=1
print("The count of Upper case characters : " + str(res_upper))
print("The count of Lower case characters : " + str(res_lower))
|
Output
The original string is : GFG is For GeeKs
The count of Upper case characters : 6
The count of Lower case characters : 7
Time complexity: O(n), where n is the length of the input string, because it has to traverse the string once to count the number of upper and lower case characters.
Auxiliary space: O(1), because it only needs to store the counts of upper and lower case characters, which are constant and don’t change with the size of the input.
Method 6: Using for loop
Python3
test_str = "GFG is For GeeKs"
upper_count = 0
lower_count = 0
for char in test_str:
if char.isupper():
upper_count += 1
elif char.islower():
lower_count += 1
print("The count of Upper case characters : " + str(upper_count))
print("The count of Lower case characters : " + str(lower_count))
|
Output
The count of Upper case characters : 6
The count of Lower case characters : 7
Time Complexity: O(n), where n is the length of test_str
Auxiliary Space: O(1)
Method #7: Using list comprehension
Python3
test_str = "GFG is For GeeKs"
print("The original string is : " + test_str)
res_upper = len()
res_lower = len()
print("The count of Upper case characters : " + str(res_upper))
print("The count of Lower case characters : " + str(res_lower))
|
Output
The original string is : GFG is For GeeKs
The count of Upper case characters : 6
The count of Lower case characters : 7
Time complexity: O(n), where n is the length of the string.
Auxiliary space: O(1), as we are only storing two integer variables.
Method #8:Using collections.defaultdict :
Algorithms :
- Import the defaultdict class from the collections module.
- Initialize a string test_str.
- Create an empty dictionary res_dict using the defaultdict class with the default value as int.
- Loop over each character ele in the test_str.
- If the character is a space, continue to the next iteration of the loop.
- Increment the count of the current character in res_dict.
- Print the count of each character in the string.
Python3
from collections import defaultdict
test_str = "GFG is For GeeKs"
print("The original string is : " + test_str)
res = defaultdict(int)
for char in test_str:
if char.isupper():
res["Upper Case"] += 1
elif char.islower():
res["Lower Case"] += 1
else:
res["Space"] += 1
print("The count of Upper case characters : " + str(res['Upper Case']))
print("The count of Lower case characters : " + str(res['Lower Case']))
|
Output
The original string is : GFG is For GeeKs
The count of Upper case characters : 6
The count of Lower case characters : 7
The time complexity :O(n) because it iterates through the string once and the operations inside the loop take constant time.
The space complexity: O(1) because the only additional data structure used is the defaultdict, which is a constant amount of space regardless of the length of the string
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