Re: Solving Every Sudoku Puzzle

WolframH — Wed, 21 Oct 2009 16:46:38 -0000

In this case (i.e. values[s] is a one-character string), the following statements all have the same effect:

d2, = values[s]
d2 = values[s][0]
d2 = values[s]

However, the last one should clearly be most efficient:
>>> def f(s): d2, = s
...
>>> def g(s): d2 = s[0]
...
>>> def h(s): d2 = s
...
>>> dis.dis(f)
1 0 LOAD_FAST 0 (s)
3 UNPACK_SEQUENCE 1
6 STORE_FAST 1 (d2)
9 LOAD_CONST 0 (None)
12 RETURN_VALUE
>>> dis.dis(g)
1 0 LOAD_FAST 0 (s)
3 LOAD_CONST 1 (0)
6 BINARY_SUBSCR
7 STORE_FAST 1 (d2)
10 LOAD_CONST 0 (None)
13 RETURN_VALUE
>>> dis.dis(h)
1 0 LOAD_FAST 0 (s)
3 STORE_FAST 1 (d2)
6 LOAD_CONST 0 (None)
9 RETURN_VALUE
>>>

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Re: Solving Every Sudoku Puzzle