40
Last seen 6 days ago
Member for 9 years, 10 months, 13 days
Difficulty Advanced
Hi, I like you solution except from it's repetitive nature, which you can avoid:
base = "I"*data
rs = (("I"*5, "V"),
("V"*2, "X"),
("X"*5, "L"),
("L"*2, "C"),
("C"*5, "D"),
("D"*2, "M"),
("DCCCC", "CM"),
("CCCC", "CD"),
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Hi, two comments:
1. You could "precompile" the regexes and used the compiled version in `re.search`.
2. Instead of `(c for c in word)` you could write `iter()`.
```python
import re
def is_stressful(subj):
matchers = [
re.compile('+[.!-]*'.join(iter(word)), re.I)
for word in
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my own ugly code, publish it to supervise myself-Akemi_Homura
4
4
Hi, you could clean it up a bit. E. g.:
1. Remove the print statements
2. int(a, 10) == int(a) so you can omit the ", 10"
3. Try to look at [this](https://docs.python.org/2/library/stdtypes.html#truth-value-testing).
# Because [] == False == 0:
if stack != []: == if stack:
if sta
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2-liner: verify me-przemyslaw.daniel
2
Hi, nice one. You can write it like this:
```python
verify_anagrams = lambda *a: list.__eq__(*map(f, a))
# or this
from operator import eq
...
verify_anagrams = lambda *a: eq(*map(f, a))
```
Regards,
suic
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Hi, I have some comments:
1. __str__ is not a good name as [__str__](https://docs.python.org/2/library/functions.html#str) is a built-in python class
2. Instead of defining str variable and checking s in str you could use the islower() method the __str__ class. Check help(str.islower).
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Hi, one of the most important and powerful feature of Python is `in`:
for word in words:
now = 0;
word = word.lower();
wordLength = len(word);
while now < textLength - wordLength:
if text[now:wordLength+now] == word:
count+=1;
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Super Simple Regex Solution-AaronCritchley
2
1
Hi, you can do it without _re_. [_filter()_](https://docs.python.org/3/library/functions.html?highlight=filter#filter) is perfect for this:
return "".join(filter(str.isupper, text))
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Hi, few things:
1. Commenting obvious things isn't a good practice.
2. `str` is iterable so there's no need to convert it to `list`.
3. `1` is the [identity element](https://en.wikipedia.org/wiki/Identity_element) of multiplication. (see below)
4. Look at `functools.reduce`.
Regards,
suic
```pyt
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First-yarmak_vladislav
1
Hi, there's three time the same pattern on line 7, try to replace this:
[acos((a**2 + c**2 - b**2)/(2*a*c)), acos((b**2 + c**2 - a**2)/(2*b*c)), acos((a**2 + b**2 - c**2)/(2*a*b))]]
with list comprehension.
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First-CraigFranklin
1
Hi, let me have some comments:
You don't need:
1. the else branches at all.
2. "== True" as test is a bool.
3. the test variable as you can use any as a condition.
4. letts as you can use data instead.
5. the parentheses on lines 5-7.
After this basic cleanup your code looks like this:
def c
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One line method-titaneric
1
1
Hi, look at [extended slices](https://docs.python.org/2/whatsnew/2.3.html#extended-slices):
```python
array[::2]
# is the same as
[num for i, num in enumerate(array) if i % 2 == 0]
```
Regards,
suic
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Hi, that `if` is redundant:
return (pattern, command) in [(42,"12a0b3e4"), ...]
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Hi, you can shorten your code like this:
def checkio(data):
data.sort()
length = len(data)
if length%2 == 0:
return (data[int((length/2)-1)] + data[int(length/2)])/2
else:
return data[int((length-1)/2)]
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First-EvanRavenelle
1
1
Hi,
1. lines 9, 11 are redundant. _filter()_ and Counter() can do the job for you.
2. You can replace line 13-15 with a generator expression.
def checkio(text):
d = Counter(filter(str.isalpha, text.lower()))
return min(t for t in d if d[t] >= d[max(d, key=d.get)])[0]
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BFS tree-titaneric
1
1
Hi,
1. `list` isn't a good object good object for representing queues as `lst.pop(0)` is `O(n)` (see [this](https://wiki.python.org/moin/TimeComplexity)). You can use `collections.deque` instead.
2. There are dictionary comprehensions in Python too :)
```python
from collections import deque
def
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One line-titaneric
1
1
Hi, `[2:]` is redundant as `bin()` returns a string in `0x...`. Regards, suic
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This is so funny :)
1. "all the magic" (including `itertools.groupby`) on one-line.
2. wrapped by a typical newbie `if...else...` anti-pattern.
```python
return (
subj.isupper() # bool
or subj.endswith('!!!') # bool
or any(...) # bool
)
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