Switching 0 to 1 and 1 to 0
Using not.
1
2
3
4
| flip = lambda x: int(not x)
a, b = 1, 0
flip(a), flip(b)
|
(0, 1)
Using xor.
1
2
3
4
| flip = lambda x: x ^ 1
a, b = 1, 0
flip(a), flip(b)
|
(0, 1)
'5'
If-else logic in append statement
1
2
3
4
5
6
7
| small = [1, 2]
large = [11, 12]
for x in [3, 4, 13, 14]:
(small if x < 10 else large).append(x)
small, large
|
([1, 2, 3, 4], [11, 12, 13, 14])
Indexing with the unary invert operator
1
2
3
4
5
| def is_palindromic(string):
return all(string[i] == string[~i] for i in range(len(string) // 2))
is_palindromic("kayak"), is_palindromic("world")
|
(True, False)
What’s happening here? ~ is the bitwise unary invert operator, which, for an integer x, returns -(x+1) (docs, to understand what’s going on, start here).
(-2, -3, -13, 11)
This allows us to step through an array from the outside in.
1
2
3
| a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
for i in range(5):
print(a[i], a[~i])
|
1 10
2 9
3 8
4 7
5 6
Using an iterator to eliminate leading zeroes in arrays
(From Elements of Programming Interviews in Python problem 5.3)
1
2
| a = [0, 0, 0, 1, 0, 2, 3]
a[next(i for i, x in enumerate(a) if x != 0) :]
|
[1, 0, 2, 3]
What’s happening here? Just like for a list extension, Python creates an object containing all elements that meet the condition and then iterates through them.
1
| [i for i in a if i > 0]
|
[1, 2, 3]
1
2
3
| iterator = (i for i in a if i > 0)
for item in iterator:
print(item)
|
1
2
3
Using next once thus returns the first item that meets the condition. In the original example we thus get the index of the first non-zero item, which is 3.
1
| next(i for i, x in enumerate(a) if x != 0)
|
3
The rest of the syntax produces a common list slice of the form a[3:], which gets us what we want. Really rather clever.
Sources