[英]Checking line number of permutation occurrence on permutations python
I need a more optimal way of searching for line count for permutations with repeating.我需要一种更优化的方法来搜索行数以进行重复排列。 It works fine with smaller values, but in this case, it needs to go through 26 ^ 12 lines to check correct permutation.
它适用于较小的值,但在这种情况下,它需要通过 26 ^ 12 行来检查正确的排列。 Any help?
有什么帮助吗?
from itertools import product
count = 0
for i in product(list('ABCDEFGHIJKLMNOPQRSTUVWXYZ'), repeat=12):
count += 1
if ''.join(i) == "INTELLIGENCE":
print(count)
Some simple math:一些简单的数学:
>>> sum(26**i * (ord(c) - ord('A')) for i, c in enumerate('INTELLIGENCE'[::-1])) + 1
31302015863412429
Tried it with 'KUBET'
as well, result was 4922080
, same as with your code.也用
'KUBET'
尝试过,结果是4922080
,与您的代码相同。
Alternatively:或者:
count = 0
for c in 'KUBET':
count = 26 * count + ord(c) - ord('A')
count += 1
Another:其他:
>>> table = str.maketrans('ABCDEFGHIJKLMNOPQRSTUVWXYZ', '0123456789ABCDEFGHIJKLMNOP')
>>> int('INTELLIGENCE'.translate(table), 26) + 1
31302015863412429
Slight variation:轻微变化:
>>> int(''.join(chr(ord(c) - (10, 17)[c < 'J']) for c in 'INTELLIGENCE'), 26) + 1
31302015863412429
Yet another:完后还有:
>>> from functools import reduce
>>> reduce(lambda count, c: 26 * count + ord(c) - ord('A'), 'INTELLIGENCE', 0) + 1
31302015863412429
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