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Awesome Team
Philippe Cholet
https://github.com/Philippe-Cholet
Last seen 1 hour 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.
def four_more(i, j, d_1, d_2):
for t in range(1, 4):
if matrix[i][j] != matrix[i + d_1 * t][j + d_2 * t]:
return False
return True
# or
def four_more(i, j, d_1, d_2):
return all(
matrix[i][j] == matrix[i + d_1 * t][j + d_2 *
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`([0-9])\1{3,}` is a simpler pattern.
I would want to simplify what's next too, such as:
for row in matrix:
row = "".join(map(str, row))
if regex.search(row):
print("a")
return True
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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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You can annotate with nearly whatever you want, but here `int` annotation is not useful, and `list` annotation is (useless and) just wrong in the sense that "map" returns an object that is not a list at all.
"max" returns an integer (because "lnumber" is an iterable of integers), hence no need to c
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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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Xs and Os Referee-iskenderunbtr
I just propose a clarification of your code. I did not test it but it should work.
```python
from typing import List
def checkio(game_result: List[str]) -> str:
for xo in 'XO':
# check rows
for row in game_result:
if row == 3 * xo:
return xo
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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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Second_thanks_Phil_15_and_a_hayashi-colinmcnicholl
Following... I understand why you use self.defense=0 on units without this attribute, but it's different to have the attribute = 0, and don't have the attribute. For 8th mission (it's not for today), this wouldn't be a great idea at all. hasattr function is useful on that point.
Anyway, fight funct
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If you are reading this solution and asking yourself why this solution can even be publish (the function signature being clearly incorrect), it's because:
1. The mission (at the very beginning) did not have the second parameter which was latter added (to make it really different from another palind
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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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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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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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**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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