After reading a certain question on the Stack Overflow website, I tried to write a solution to the problem just for fun. I'm, however, left with the nagging feeling there is a beautiful one-liner that can be used instead.
The question's premises:
Create a function that will receive a string. The function will count each vowel in the string, and return a count of all the vowels, whether found or not, each in a tuple pair. Each tuple parir will be stored in a list.
Example:
>>> vowels_finder("This has some vowels")
[('a', 1), ('o', 2), ('u', 0), ('e', 2), ('i', 1)] # tuple pair of each vowel.
>>>
My attempt:
def vowels_finder(s):
vowels = {'a':0, 'e':0, 'i':0, 'o':0, 'u':0}
for el in s:
if el in {'a', 'e', 'i', 'o', 'u'}:
vowels[el]+=1
vowels = [(key, pair) for key, pair in vowels.items()]
return vowels
My code above is not commented, but I'm confident that its brevity will allow me to pass on this.
Questions:
- Is there _any way, besides using a python library, that this can be condensed into a one-liner?
- Would there be a way to not have to convert the
vowelskey's back into tuple pairs, and just have them be tuples in the beginning. eg:vowels = [('a', 0), ('e', 0), ('i', 0), ('o', 0), ('u', 0)]?
language: Python 3.4
After reading a certain question on the Stack Overflow website, I tried to write a solution to the problem just for fun. I'm, however, left with the nagging feeling there is a beautiful one-liner that can be used instead.
The question's premises:
Create a function that will receive a string. The function will count each vowel in the string, and return a count of all the vowels, whether found or not, each in a tuple pair. Each tuple parir will be stored in a list.
Example:
>>> vowels_finder("This has some vowels")
[('a', 1), ('o', 2), ('u', 0), ('e', 2), ('i', 1)] # tuple pair of each vowel.
>>>
My attempt:
def vowels_finder(s):
vowels = {'a':0, 'e':0, 'i':0, 'o':0, 'u':0}
for el in s:
if el in {'a', 'e', 'i', 'o', 'u'}:
vowels[el]+=1
vowels = [(key, pair) for key, pair in vowels.items()]
return vowels
My code above is not commented, but I'm confident that its brevity will allow me to pass on this.
Questions:
- Is there _any way, besides using a python library, that this can be condensed into a one-liner?
- Would there be a way to not have to convert the
vowelskey's back into tuple pairs, and just have them be tuples in the beginning. eg:vowels = [('a', 0), ('e', 0), ('i', 0), ('o', 0), ('u', 0)]?
language: Python 3.4
After reading a certain question on the Stack Overflow website, I tried to write a solution to the problem just for fun. I'm, however, left with the nagging feeling there is a beautiful one-liner that can be used instead.
The question's premises:
Create a function that will receive a string. The function will count each vowel in the string, and return a count of all the vowels, whether found or not, each in a tuple pair. Each tuple parir will be stored in a list.
Example:
>>> vowels_finder("This has some vowels")
[('a', 1), ('o', 2), ('u', 0), ('e', 2), ('i', 1)] # tuple pair of each vowel.
>>>
My attempt:
def vowels_finder(s):
vowels = {'a':0, 'e':0, 'i':0, 'o':0, 'u':0}
for el in s:
if el in {'a', 'e', 'i', 'o', 'u'}:
vowels[el]+=1
vowels = [(key, pair) for key, pair in vowels.items()]
return vowels
My code above is not commented, but I'm confident that its brevity will allow me to pass on this.
Questions:
- Is there _any way, besides using a python library, that this can be condensed into a one-liner?
- Would there be a way to not have to convert the
vowelskey's back into tuple pairs, and just have them be tuples in the beginning. eg:vowels = [('a', 0), ('e', 0), ('i', 0), ('o', 0), ('u', 0)]?
language: Python 3.4
After reading a certain question on the Stack Overflow website, I tried to write a solution to the problem just for fun. I'm, however, left with the nagging feeling there is a beautiful one-liner that can be used instead.
The question's premises:
Create a function that will receive a string. The function will count each vowel in the string, and return a count of all the vowels, whether found or not, each in a tuple pair. Each tuple parir will be stored in a list.
Example:
>>> vowels_finder("This has some vowels")
[('a', 1), ('o', 2), ('u', 0), ('e', 2), ('i', 1)] # tuple pair of each vowel.
>>>
My attempt:
def vowels_finder(s):
vowels = {'a':0, 'e':0, 'i':0, 'o':0, 'u':0}
for el in s:
if el in {'a', 'e', 'i', 'o', 'u'}:
vowels[el]+=1
vowels = [(key, pair) for key, pair in vowels.items()]
return vowels
My code above is not commented, but I'm confident that its brevity will allow me to pass on this.
Questions:
- Is there any way_any way, besides
(削除) using a builtin function or method (削除ここまで), Ausing a python library function, that this can be condensed into a one-liner? - Would there be a way to not have to convert the
vowelskey's back into tuple pairs, and just have them be tuples in the beginning. eg:vowels = [('a', 0), ('e', 0), ('i', 0), ('o', 0), ('u', 0)]?
language: Python 3.4
After reading a certain question on the Stack Overflow website, I tried to write a solution to the problem just for fun. I'm, however, left with the nagging feeling there is a beautiful one-liner that can be used instead.
The question's premises:
Create a function that will receive a string. The function will count each vowel in the string, and return a count of all the vowels, whether found or not, each in a tuple pair. Each tuple parir will be stored in a list.
Example:
>>> vowels_finder("This has some vowels")
[('a', 1), ('o', 2), ('u', 0), ('e', 2), ('i', 1)] # tuple pair of each vowel.
>>>
My attempt:
def vowels_finder(s):
vowels = {'a':0, 'e':0, 'i':0, 'o':0, 'u':0}
for el in s:
if el in {'a', 'e', 'i', 'o', 'u'}:
vowels[el]+=1
vowels = [(key, pair) for key, pair in vowels.items()]
return vowels
My code above is not commented, but I'm confident that its brevity will allow me to pass on this.
Questions:
- Is there any way, besides
(削除) using a builtin function or method (削除ここまで), A library function that this can be condensed into a one-liner? - Would there be a way to not have to convert the
vowelskey's back into tuple pairs, and just have them be tuples in the beginning. eg:vowels = [('a', 0), ('e', 0), ('i', 0), ('o', 0), ('u', 0)]?
language: Python 3.4
After reading a certain question on the Stack Overflow website, I tried to write a solution to the problem just for fun. I'm, however, left with the nagging feeling there is a beautiful one-liner that can be used instead.
The question's premises:
Create a function that will receive a string. The function will count each vowel in the string, and return a count of all the vowels, whether found or not, each in a tuple pair. Each tuple parir will be stored in a list.
Example:
>>> vowels_finder("This has some vowels")
[('a', 1), ('o', 2), ('u', 0), ('e', 2), ('i', 1)] # tuple pair of each vowel.
>>>
My attempt:
def vowels_finder(s):
vowels = {'a':0, 'e':0, 'i':0, 'o':0, 'u':0}
for el in s:
if el in {'a', 'e', 'i', 'o', 'u'}:
vowels[el]+=1
vowels = [(key, pair) for key, pair in vowels.items()]
return vowels
My code above is not commented, but I'm confident that its brevity will allow me to pass on this.
Questions:
- Is there _any way, besides using a python library, that this can be condensed into a one-liner?
- Would there be a way to not have to convert the
vowelskey's back into tuple pairs, and just have them be tuples in the beginning. eg:vowels = [('a', 0), ('e', 0), ('i', 0), ('o', 0), ('u', 0)]?
language: Python 3.4
After reading a certain question on the Stack Overflow website, I tried to write a solution to the problem just for fun. I'm, however, left with the nagging feeling there is a beautiful one-liner that can be used instead.
The question's premises:
Create a function that will receive a string. The function will count each vowel in the string, and return a count of all the vowels, whether found or not, each in a tuple pair. Each tuple parir will be stored in a list.
Example:
>>> vowels_finder("This has some vowels")
[('a', 1), ('o', 2), ('u', 0), ('e', 2), ('i', 1)] # tuple pair of each vowel.
>>>
My attempt:
def vowels_finder(s):
vowels = {'a':0, 'e':0, 'i':0, 'o':0, 'u':0}
for el in s:
if el in {'a', 'e', 'i', 'o', 'u'}:
vowels[el]+=1
vowels = [(key, pair) for key, pair in vowels.items()]
return vowels
My code above is not commented, but I'm confident that its brevity will allow me to pass on this.
Questions:
- Is there any way, besides using a builtin function or method
(削除) using a builtin function or method (削除ここまで), A library function that this can be condensed into a one-liner? - Would there be a way to not have to convert the
vowelskey's back into tuple pairs, and just have them be tuples in the beginning. eg:vowels = [('a', 0), ('e', 0), ('i', 0), ('o', 0), ('u', 0)]?
language: Python 3.4
After reading a certain question on the Stack Overflow website, I tried to write a solution to the problem just for fun. I'm, however, left with the nagging feeling there is a beautiful one-liner that can be used instead.
The question's premises:
Create a function that will receive a string. The function will count each vowel in the string, and return a count of all the vowels, whether found or not, each in a tuple pair. Each tuple parir will be stored in a list.
Example:
>>> vowels_finder("This has some vowels")
[('a', 1), ('o', 2), ('u', 0), ('e', 2), ('i', 1)] # tuple pair of each vowel.
>>>
My attempt:
def vowels_finder(s):
vowels = {'a':0, 'e':0, 'i':0, 'o':0, 'u':0}
for el in s:
if el in {'a', 'e', 'i', 'o', 'u'}:
vowels[el]+=1
vowels = [(key, pair) for key, pair in vowels.items()]
return vowels
My code above is not commented, but I'm confident that its brevity will allow me to pass on this.
Questions:
- Is there any way, besides using a builtin function or method, that this can be condensed into a one-liner?
- Would there be a way to not have to convert the
vowelskey's back into tuple pairs, and just have them be tuples in the beginning. eg:vowels = [('a', 0), ('e', 0), ('i', 0), ('o', 0), ('u', 0)]?
language: Python 3.4
After reading a certain question on the Stack Overflow website, I tried to write a solution to the problem just for fun. I'm, however, left with the nagging feeling there is a beautiful one-liner that can be used instead.
The question's premises:
Create a function that will receive a string. The function will count each vowel in the string, and return a count of all the vowels, whether found or not, each in a tuple pair. Each tuple parir will be stored in a list.
Example:
>>> vowels_finder("This has some vowels")
[('a', 1), ('o', 2), ('u', 0), ('e', 2), ('i', 1)] # tuple pair of each vowel.
>>>
My attempt:
def vowels_finder(s):
vowels = {'a':0, 'e':0, 'i':0, 'o':0, 'u':0}
for el in s:
if el in {'a', 'e', 'i', 'o', 'u'}:
vowels[el]+=1
vowels = [(key, pair) for key, pair in vowels.items()]
return vowels
My code above is not commented, but I'm confident that its brevity will allow me to pass on this.
Questions:
- Is there any way, besides
(削除) using a builtin function or method (削除ここまで), A library function that this can be condensed into a one-liner? - Would there be a way to not have to convert the
vowelskey's back into tuple pairs, and just have them be tuples in the beginning. eg:vowels = [('a', 0), ('e', 0), ('i', 0), ('o', 0), ('u', 0)]?
language: Python 3.4