40
Last seen 4 days ago
Member for 9 years, 10 months, 19 days
Difficulty Advanced
A few comments:
1. In fact lines 5-7 are superfluous as a, b and c don't have to be sorted.
2. Then:
a, b, c = length
is the same as
a = length[0]
b = length[1]
c = length[2]
3. Consider creating a separate function from:
round(degrees(acos((b * b + c * c - a * a)
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Good job! There's no redundancy. And it's even safer than it has to be, as all inputs in the tests are already sorted :)
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You don't have to use third variable two swap values of two variables. You can simply write:
if n < m:
m, n = n, m
instead of:
if n < m:
tmp = n
n = m
m = tmp
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You can use the [str.rjust](https://docs.python.org/3/library/stdtypes.html?highlight=str.rjust#str.rjust) method for this:
fill = abs(len(n) - len(m))*'0'
if len(n) < len(m):
n = fill + n
else:
m = fill + m
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_nbin_ and _mbin_ are both _str_ which is iterable there for using list() in lines 10 and 11 is superfluous. And also you can write:
size = max(len(nbin), len(mbin))
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Even the last (Solve)-Yoratheon
Few tips:
array[len(array) - 1] == array[-1] # there's no need to write len(array)
n = n + 1 --> n += 1
0 % 2 == 0 # you can omit n == 0 in your if clause
The code in else branch doesn't have to be in else branch.
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Index Power (Solution)-Yoratheon
The return None in your code is never reached, you may remove it.
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First-I_s_a_a_c_123
Hi,
1. Line 10 is redundant.
2. You use `len(data)/2` three times, so it's good candidate for a variable.
3. Line 3 is more readable this way, but you could also write it like this:
if len(data) % 2:
...
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Hi, this:
for i in range(len(data)):
if data.count(data[i]) > 1:
data_1.append(data[i])
i = i + 1
else:
i = i + 1
# could be replaced with:
for i in data:
if data.count(i) > 1:
data_l.append(i)
But you can use _f
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I'm nitpicking (so don't take it too serious): Romanian numerals? :) rumuński != rzymski.
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Hi, I like lambdas but in this case I agree with @PaulBrown. One tip to make code a bit more readable:
Separate long comment from code, e. g.:
# returns sum of 2 rows starting from column fromcol,
# where second one is multiplied by mult
addrows = lambda row1, row2, mult=1, fromco
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Hi, I would shorted valid like this:
return all(0 <= spot[i] < len(dfs.matrix[i]) for i in (0, 1))
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Patrick-pdtpatrick
Hi,
1. You could use `int.__mul__` instead of `lambda`.
2. `new_number` is redundant. You can put that comprehension in `reduce` directly.
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Patrick-pdtpatrick
Hi,
1. `== True` is redundant.
2. `...if...else...` on line 5 is redundant:
new_sentence = [x.isalpha() for x in sentence.split()]
# or you could use map
new_sentence = map(str.isalpha, sentence.split())
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