22
Last seen 9 hours ago
Member for 2 years, 3 months, 1 day
Difficulty Easy
I'm here to learn / improve my Python skills and help other people if possible.
Thank you for showing me `setdefault`: As fas as I remember I have not used it before. I like this solution. But I'm unsure if two loops were needed. Maybe one loop combined with del will be faster.
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Thank you for teaching me something about '~'. I immediately searched for more. Really interesting. Thank you.
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Well, it seems to work. But if you would have used set() and intersection this would have been much easier. Also no loops would have been needed.
Hint from description:
> Here you can learn how to work with strings and **sets**.
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Good solution. ... and I have counted spaces. ;-) Maybe I should search for a fitting library next time, first.
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Interesting way to solve this with many "<" and ">". :-) I've learned something. Thank you.
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I've never used map before. After doing some reseach, I understand how this is working. So basically this "converts the string into a array of integers and get the biggest of it". Good Job. :-)
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Cool way to solve this with bool as index value. I really have learned something. Thank you!
But the mission seem to have changed a bit. When checking out your solution I got:
Traceback (most recent call last):
File "12-sum-by-type.py", line 42, in
assert sum_by_types([]
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Nice. It's similar to my solution, but I have not seen "//" before and have used int() for that. I have learned something. :-) Thank you.
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Thank you! That teached me a bit more about "lambda". So you just have to replace the complete:
def nearest_square(number):
...
with just:
nearest_square = lambda x: round(x**0.5)**2
which replaces the original function. Nice. :-)
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Nice. And so I learned something new about the string module and how to use it this way. :-) I like it. Thank you for this lines.
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Well, that's a nice and clean solution. Thank you for showing it to us. :-)
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I tried something similar at first. However, cases like
'12 34 567'
(multiple space between numbers) are not taken into account. So it's easier just to compare the text against it's uppered text.
return text == text.upper()
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