Initialize the two count variables to 0. 3. Use a for loop to traverse through the characters in the string and increment the first count variable each time a lowercase character is encountered and increment the second count variable each time a uppercase character is encountered.
You can use List comprehensions and sum function to get the total number of upper and lower case letters. def up_low(s): u = sum(1 for i in s if i. isupper()) l = sum(1 for i in s if i. islower()) print( "No.
You can do this with sum
, a generator expression, and str.isupper
:
message = input("Type word: ")
print("Capital Letters: ", sum(1 for c in message if c.isupper()))
See a demonstration below:
>>> message = input("Type word: ")
Type word: aBcDeFg
>>> print("Capital Letters: ", sum(1 for c in message if c.isupper()))
Capital Letters: 3
>>>
You can use re
:
import re
string = "Not mAnY Capital Letters"
len(re.findall(r'[A-Z]',string))
5
Using len
and filter
:
import string
value = "HeLLo Capital Letters"
len(filter(lambda x: x in string.uppercase, value))
>>> 5
I've done some comparisons of the methods above + RE compiled using
Python 3.7.4
For this, I've used the book Alice’s Adventures in Wonderland, by Lewis Carroll from Project Gutenberg.
from urllib.request import urlopen
# Download
text = urlopen('https://www.gutenberg.org/files/11/11-0.txt').read().decode('utf-8')
# Split it into the separate chapters and remove table of contents, etc
sep = 'CHAPTER'
chaps = [sep + ch for ch in text.split('CHAPTER') if len(ch) > 1000]
len(chaps)
Defined all approaches as functions in order to use them in the loop and keep succinct.
import re
import string
def py_isupper(text):
return sum(1 for c in text if c.isupper())
def py_str_uppercase(text):
return sum(1 for c in text if c in string.ascii_uppercase)
def py_filter_lambda(text):
return len(list(filter(lambda x: x in string.ascii_uppercase, text)))
def regex(text):
return len(re.findall(r'[A-Z]',text))
# remove compile from the loop
REGEX = re.compile(r'[A-Z]')
def regex_compiled(text):
return len(REGEX.findall(text))
The results are below.
%%timeit
cnt = [py_isupper(ch) for ch in chaps]
7.84 ms ± 69.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%%timeit
cnt = [py_str_uppercase(ch) for ch in chaps]
11.9 ms ± 94.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%%timeit
cnt = [py_filter_lambda(ch) for ch in chaps]
19.1 ms ± 499 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%%timeit
cnt = [regex(ch) for ch in chaps]
1.49 ms ± 13 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%%timeit
cnt = [regex_compiled(ch) for ch in chaps]
1.45 ms ± 8.69 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
from string import ascii_uppercase
count = len([letter for letter in instring if letter in ascii_uppercase])
This is not the fastest way, but I like how readable it is. Another way, without importing from string and with similar syntax, would be:
count = len([letter for letter in instring if letter.isupper()])
def n_lower_chars(string):
return sum(i.isupper() for i in string)
sums up the numbers of True values in the generator expression
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