我有一个数组a = [[1,2,3,4,5,6,7,8,9,10],[4,1,6,2,3,5,8,9,7,10]], where lets say a1 = [1,2,3,4,5,6,7,8,9,10] and a2 = [4,1,6,2,3,5,8,9,7,10]
,从中构造了循环置换。请注意,a1是一个排序的数组。就我而言,周期是;
c = [[4, 2, 1], [6, 5, 3], [8, 9, 7], [10]]
lets say c1 = [4, 2, 1]
c2 = [6, 5, 3]
c3 = [8, 9, 7]
c4 = [10]
我有一个方法可以在给定的排列中给出所有循环,但是从中构造新的数组似乎很复杂。在python3中实现它的任何想法将不胜感激。
获得周期;
import numpy as np
import random
def cx(individual):
c = {i+1: individual[i] for i in range(len(individual))}
cycles = []
while c:
elem0 = next(iter(c)) # arbitrary starting element
this_elem = c[elem0]
next_item = c[this_elem]
cycle = []
while True:
cycle.append(this_elem)
del c[this_elem]
this_elem = next_item
if next_item in c:
next_item = c[next_item]
else:
break
cycles.append(cycle)
return cycles
aa = cx([4,1,6,2,3,5, 8,9,7,10])
print("array: ", aa)
答案 0 :(得分:1)
您可以使用itertools.permutations
来获取a
项的不同排列,然后使用itertools.cycle
循环浏览将a
子列表项映射到其索引的字典,并用映射压缩c
的子列表,以产生遵循循环dict指定的索引的序列:
a = [[1,2,3,4,5,6,7,8,9,10],[4,1,6,2,3,5,8,9,7,10]]
c = [[4, 2, 1], [6, 5, 3], [8, 9, 7], [10]]
from itertools import cycle, permutations
print([[d[i] for i in range(len(d))] for l in permutations(a) for d in ({p[n]: n for s, p in zip(c, cycle({n: i for i, n in enumerate(s)} for s in l)) for n in s},)])
这将输出:
[[1, 2, 6, 4, 3, 5, 7, 8, 9, 10], [4, 1, 3, 2, 5, 6, 8, 9, 7, 10]]