40
Last seen 4 days ago
Member for 9 years, 10 months, 19 days
Difficulty Advanced
First-jirka.hrbek
1
Hi,
1. Line 3: `[]` are redundant.
2. Look at `Counter` in `collections` module.
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Dictionary cheat :)
def days_diff(date1, date2):
try:
return {((1982,4,19),(1982,4,22)): 3,
((2014,8,27), (2014,1,1)): 238,
((1,1,1), (9999,12,31)): 3652058,
((9999,12,31), (1,1,1)): 3652058,
((2
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Hi, I've played with your code. [Here](http://www.checkio.org/mission/brackets/publications/suic/python-27/leolos-shortened/) is the result. :)
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Hi, `lambda` is redundant here. _abs()_ is already a function, so you can write:
return sorted(numbers_array, key=abs)
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Hi,
1. look at _sum()_ and _generator expressions_.
2. __bool__ is a _subclass_ of __int__.
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Hi, you don't need that "else cascade" in the end.
1. Instead of `'a' <= x <= 'z'` you can use `x.islower()`.
2. Instead of `'A' <= x <= 'Z'` you can use `x.isupper()`.
3. Instead of `x in b` you can use `x.isdigit()`.
4. Line 14: `if number and u_letter and l_letter:`
etc.
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Hi, I have a few comments:
1. What is the point of using enumerator if you don't use _i_?
2. In fact you do everything twice.
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Hi, you could use _sum()_ with _generator expression_ instead of `for` loop and `result` variable.
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First-mohamed.amer.tarak
1
Hi, Python is not C. I'm not sure if this is speedy, have you measured it?
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First-tehtactics19
Hi, this is one of the standard solutions. It's just. On the other hand you could write it as a one-liner, but it's not necessary. :)
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First-Rahul_Badenkal
Hi,
1. Look at _filter()_ and _str.join()_.
2. You could write `b += a` instead of `b = b + a`.
3. Look at _str.isupper()_.
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First-Rahul_Badenkal
Hi, you could write this
sum(array[::2])
# instead of:
sum(x for x in array[::2])
If `array` is not empty:
array[-1] == array[len(array)-1]
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First-Rahul_Badenkal
Hi, look at _min()_ and _max()_ built-ins. Don't redefine them.
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First-Rahul_Badenkal
Hi,
1. On line 2 _str()_ is redundant as _bin()_ returns __str__.
2. Look at _str.count()_.
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First-Rahul_Badenkal
Hi, look at other solutions. This one is quite awkward. You could write:
b[0:len(b)-1] == b[:len(b)-1]
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