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Awesome Team
Philippe Cholet
https://github.com/Philippe-Cholet
Last seen 38 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_colin_stone_wall-colinmcnicholl
We talked about zip, usefull here to have columns. (there is a little thing to do before)
If you have to count something, use a tool doing it for you. Like str/list/tuple.count( ... )
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No need to put parenthesis around the thing to return, just `return result`, it's not a function.
if ...:
return ...
# else: # no need of else since return stop the function.
return ...
No need to make a list to sum elements
sum([next(n,depth+1,weight*m) for n,m in branch
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There are other ways but I like the use of numpy, especially `reshape` and array of booleans. It's original.
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Given a list of moves _L_, the sequence
(stones(n, L) for n>=0)
is eventually periodic (and very often periodic), and the period depends only on _L_.
Unfortunately, compute this period is not really simple in general. But with it, we can compute _stones(n, L)_ easily for **very large** numb
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Simple-petr.synek.phd
Hi!
Found that weird
marked = marked|set().union(*newly_marked)
# Instead I suggest
marked = marked.union(*newly_marked)
# or if update the set "marked" IN-PLACE is okay (I think it is) :
marked.update(*newly_marked)
and
indexed_circles = [(index, circle) for index, circ
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else after a for loop? I know the thing but useless here.
No need of continue with "!=" test.
if matrix[i][j] != -matrix[j][i]:
return False
Well, then, you should considerate the use of "all(...)"
all(matrix[i][j] == -matrix[j][i] for i in range(len(matrix))
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List comprehension is so... so good...
t = [[- matrix[j][i] for j in p] for i in p]
Then "matrix == t" is already the boolean you want return.
return matrix == t
Finally,
def checkio(matrix):
p = range(len(matrix))
return matrix == [[- matrix[j][i] for j in p] for i i
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One liner-petr.synek.phd
1
Well if `i != 0` then `i - 1 >= 0`
"1" if val == line[max(0, i - 1)] and i != 0 else "0"
str(int(i != 0 and val == line[i - 1]))
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First-colinmcnicholl
1
It seems complicated. But probably clever "sweep line algorithm". Do you think this work for rectangles with float coordinates? Because int coordinates is a big simplification for this problem.
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It's not 3rd party.
Plus
# You do not use the index i but a[i] and b[i].
def hamming(a, b):
return sum(s != t for s, t in zip(str(a), str(b)))
def is_doublet(numbers):
# something with zip too to avoid the use of "i".
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def match(comp, test):
count = 0
# you do not use the index j but comp[j] and test[j]
for c, t in zip(comp, test):
if c == t:
count += 1
# "count == 2" is already True or False, return it
return count == 2
# or
def match
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We can do it in three lines without import math.hypot, but with complex numbers: a little less clear IMO.
grid = lambda N: ([i,j] for i in range(N) for j in range(N))
test = lambda i, j, probes: all(round(abs(i-x+(j-y)*1j)) == d for x,y,d in probes)
checkio = lambda previous: next(c for
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date.datetime-Anton_Prokopovich
abs(d1-d2).days #work so this too
return abs(date(*date1) - date(*date2)).days
# then eventually
days_diff = lambda date1, date2: abs(date(*date1) - date(*date2)).days
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**Few explanations...** (I won't write a mathematical article for it)
_Mathematically:_ P is the transition matrix of the markov chain involved here.
_In english:_ P[i,j] is the probability to go to j from i with one dice roll.
M is the same thing as P with a ban on going through target.
**1/(I-
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`if res[i] != True` then `res[i]` is already `False`. so lines 7 - 11 can be much shorter. Plus two pythonic changes:
def stones(pile, moves):
res = [True] + pile * [False]
for i in range(1, len(res)):
for j in moves:
if i - j > 0 and not res[i - j]:
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First-vladislav.bezugliy
Even if you reinvent the wheel (it's not recommended like Veky said)...
Lines 6-8 can be one simple line:
ints[i], ints[i+1] = ints[i+1], ints[i]
And while loop should be a for loop here.
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Same thing, shorter! Without test of enough letters, and a shorter last part. Not enough readable IMO, that's why I kept my code like it is.
from itertools import product
def hypercube(array, word='hypercube'):
coord_letters = {letter: [] for letter in word}
for i, row in en
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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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No need of sum lists:
Counter(ring for cube in cubes for ring in sides(cube))
I did a sum at first too.
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