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Awesome Team
Philippe Cholet
https://github.com/Philippe-Cholet
Last seen 29 minutes ago
Member for 6 years, 2 months, 27 days
Difficulty Advanced
I was a math teacher in France for two years. I'm currently reconverting to IT, I'm here to improve my Python skills, and practice English.
First-shan.u2008
`[x for x in y]` is equivalent to `list(y)` but here you don't even need a list. A string is a good enough iterable: `int(max(str(number)))`
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If you think Army like a list, consider
class Army(list):
#Army is now a list with all methods from list and your mind.
#So no need of init, getitem, len, remove.
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In a more pythonic way...
x,y = max(dict1.keys(),key= lambda x:dict1[x])
# keys method is not really useful, I don't remember any situation where it is.
x, y = max(dict1, key = dict1.get) # your lambda function already exists for dicts.
if len(dict1) == 0:
# or
if not dict1
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I don't see the point of the `~1` instead of `-2` except to make the code a bit weird. Miss I something here?
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if `len(els) < 3` then `range(2, len(els))` is empty so lines 7-8 are not required.
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As I did for "replace first", I add this fourth solution:
```python
# numpy.roll generalize our function.
import functools as fn
import numpy as np
replace_last = fn.partial(np.roll, shift=1)
```
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Intervals are already sorted (see description), so `sorted(set(...))` is a bit useless.
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First, line 10 can be written `return f'({self.x}, {self.y})'`.
Second, I don't really see the point of OOP here, you just need `range(...)`. It's probably because you thought about `chain(*map(...))`. List comprehension is enough here.
Like this (if you want `chain`):
chain.from_iterable(ran
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I would use "for/else" instead of "flag". What do you think?
for i in range(1, 101):
for j in range(4):
if i % prime[j] != attempts[j + 1][0]:
break
else:
return [2, i]
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**EDIT:** in the arguments are of the **same** lengths.
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def navigation(seaside):
for i, row in enumerate(seaside): # n/s coordinate
for j, item in enumerate(row): # w/e coordinate
if item == 'Y':
cY = i, j # now it's a tuple.
elif item == 'C':
cC = i, j
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Possible improvement (decrease the number of possibilities). Between line 27 and 28, put that:
# Only keep keys when the value is useful (not empty sets)
# But put useless cannons (can't shoot any enemy) at North...
reach = [({k: v for k, v in dirs.items() if v} if any(dirs.values())
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Use global variables is usually a bad idea, like here. Honestly it should NOT have pass tests. You only pass them because of "check" part work, and little tests did not warn you about it.
assert isometric_strings('add', 'egg') == True
dict == {'a': 'e', 'd': 'g'}
assert isometric_strin
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width = len(w1[0]) # at begining, can be good for readability. And I use it later.
`range(len(...))` should be forbidden ^^ `enumerate` is good enough.
for row in wl:
for j, ch in enumerate(row):
if ch == '#':
layers[j] += 1
Your way to find weakest spot
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Don't you think you spam published solutions?! 6 (very very similar) published solutions.
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**First**
f2 = lambda x, y: any(True if (x + i, y + j) in X else False ...
# please write it...
f2 = lambda x, y: any((x + i, y + j) in X ...
**Second**, right now for me on checkio, I can't see after 107 characters width without pain ^^
For readability, PEP8 insist on width, width no
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First-MarcAureleCoste
First, you use numpy, I like it but '3rd party' is there for such solutions.
Your `... == 0` is already a boolean, so you can just write
return np.count_nonzero(np_matrix + np_matrix.T) == 0 # True when it's True, False when...
# or
return not np.count_nonzero(np_matrix + np_matrix.T)
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