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@rodpoblete
Created September 27, 2019 02:47
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Tome el bloque de texto proporcionado y strip el espacio en blanco en ambos extremos. Dividir el texto por nueva línea (\n). Recorre las líneas, para cada línea: quite los espacios iniciales, compruebe si el primer carácter está en minúsculas, si e
text = """
The Zen of Python, by Tim Peters
Beautiful is better than ugly.
Explicit is better than implicit.
Simple is better than complex.
Complex is better than complicated.
Flat is better than nested.
Sparse is better than dense.
Readability counts.
Special cases aren't special enough to break the rules.
Although practicality beats purity.
Errors should never pass silently.
Unless explicitly silenced.
In the face of ambiguity, refuse the temptation to guess.
There should be one-- and preferably only one --obvious way to do it.
Although that way may not be obvious at first unless you're Dutch.
Now is better than never.
Although never is often better than *right* now.
If the implementation is hard to explain, it's a bad idea.
If the implementation is easy to explain, it may be a good idea.
Namespaces are one honking great idea -- let's do more of those!
"""
vowels = 'aeiou'
def strip_vowels(text: str) -> (str, int):
"""Replace all vowels in the input text string by a star
character (*).
Return a tuple of (replaced_text, number_of_vowels_found)
So if this function is called like:
strip_vowels('hello world')
... it would return:
('h*ll* w*rld', 3)
The str/int types in the function defintion above are part
of Python's new type hinting:
https://docs.python.org/3/library/typing.html"""
new_str = []
chars = list(text)
num_vowels = 0
for c in chars:
if c.lower() in vowels:
c = '*'
num_vowels += 1
new_str.append(c)
return print(''.join(new_str), num_vowels)
#
# once you get into regular expressions you can rewrite this with one line:
# import re
# return re.subn(f'[{vowels}]', '*', text, flags=re.I)
strip_vowels(text)
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