22
Dahlia Ramm
Last seen 1 year ago
Member for 8 years, 1 month, 3 days
Difficulty Normal
Nice one!
Why are u using _**key=lambda x: abs(x)**_ instead of _**key=abs**_ ?
Instead of casting "numbers_array" to a list, i guess udid this to use list.sort(),
u could have used the bult-in function sorted().
sorted() also takes key functions **and** it can deal with any iterable as it returns
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Line 2:
if not len(data): return 0
Calling len(data) is unnecessary!
if not data: return 0
does the same, and follows PEP8.
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First-basketbal44champ
Line 3:
n = str(bin(num))[2:]
bin() already returns a string, so there is no need to cast it!
Slicing away "0b" also makes no sence as it never effects the result.
Instead of using a for-loop to count the unities, you could return n.count("1")
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Why are u doing this?
> list(zip(*data))
already returns the transposed matrix as a 2D-List.
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first fizzbuzz-phyrenight
Line 6:
if (number%5) == 0 and (number%3) == 0:
the least common multiple of 2 and 5 is 15, so:
if number & 15 == 0:
will give the correct result with less code.
The "elif" in line 8 and 10 should be "if", because if the expression before evaluates to true
they will never be reached.
Lin
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I like it, even though its a bit complicated.
But I had no idea that's it possible to use tuple as index for dictionary... learned something.
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First-flyingmonkey132
Nice!
But in line 10:
fib_seq = fibo(20)
you are generating the first 20 fibonacci numbers, that's more you ever need
Ge.nerating the next Fibonacci number only if you need it makes more sense.
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You are generating more fibonacci numbers then you might need, I think it makes more sense to only generate the next fibonacci number it there is need to do so.
Is a dictionary the proper type to store your result?
The older the ghost gets, the more kv pairs are in your dict, even thoungh will only
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What's bad about this solution?
When the ghosts age is 55 the while-loop will be executed 55 times...
When the ghosts age is 4689 the while-loop will be executed 4689 times...
Needs a second try...
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1-Vladyslav
1
Line 2:
... lambda numbers_array: numbers_array if (numbers_array > 0) else -numbers_array)...
Have a look at the [built-in function abs()](https://docs.python.org/3.6/library/functions.html#abs). ;)
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"Non-unique Elements" / SevenDayEngineer-sevendayengineer
1
Line 5:
counted = data.count(i) # count the number of occurences of unique item
How often can a **unique** number occur?
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return bin(number)[2:].count('1')
You are slicing away the first 2 chracters of the string bin(number) returns.
Is that necessary? will one of these character ever be "1"?
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numpy.bitwise_xor-greg.p.mueller
1
Is the numpy.bitwise_xor() more effective then the bitwise xor operator "^"?
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Line 19:
return str(answer)
do you need to type cast answer to string here?
Line 11:
if number % 3 == 0 and number % 5 == 0:
when you use the least common multiple of 3 and 5 you can get rid of the "and" comparison.
if number % 15 == 0:
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First-MilfordCubicle
1
Clear!
Line 6:
if number % 3 == 0 and number % 5 == 0:
what about the least common multiple of 3 and 5?
if number % 15 == 0:
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Dumb solution-sirajul_islam_anik
1
Line 3:
if len(splitted) == 0:
this "if" is redundant, len(splitted) can never be 0.
Besides that if you ever want to check whether a sequence is empty or not use:
if sequence:
do something
if the sequence is empty, this "if" evaluates to **False** otherwise to **True**
Line 14 to
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Clear!
Instead of type casting the tuple to a list manually you could have used the [built-in function sorted()](https://docs.python.org/3.6/library/functions.html#sorted).
You can pass iterable to this function and it returns a list.
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