Question

我在21个范围内找到了可以满足a*a + b*b == c*c的组合号，但是我不想重复组合结果。那么如何对其重新编程，而不显示重复的组合号

nums = range(1, 21)
trips = [(a, b, c) for a in nums for b in nums for c in nums if a*a + b*b == c*c]
print(trips)

Answer 1

一个简单的解决方案是使用集合和排序来丢弃重复的元素：

nums = range(1, 21)
trips = sorted({(tuple(sorted((a, b))), c) for a in nums for b in nums for c in nums if a*a + b*b == c*c})
print(trips)
# [((3, 4), 5), ((5, 12), 13), ((6, 8), 10), ((8, 15), 17), ((9, 12), 15), ((12, 16), 20)]

如果要展平元组：

trips = [(a, b, c) for (a, b), c in trips]

但是，明智的做法是不要迭代a和b的重复组合。您可以这样做：

nums = range(1, 21)
trips = [(a, b, c) for a in nums for b in nums[a:] for c in nums if a*a + b*b == c*c]
print(trips)
# [(3, 4, 5), (5, 12, 13), (6, 8, 10), (8, 15, 17), (9, 12, 15), (12, 16, 20)]

或使用itertools：

from itertools import combinations

nums = range(1, 21)
trips = [(a, b, c) for a, b in combinations(nums, 2) for c in nums if a*a + b*b == c*c]
print(trips)
# [(3, 4, 5), (5, 12, 13), (6, 8, 10), (8, 15, 17), (9, 12, 15), (12, 16, 20)]

实际上也很简单。

Answer 2

更长的代码，但是效率更高：我们生成a和b的组合而没有重复项，并使用字典来检查a**2 + b**2是否是我们要查找的值之一，这比签入列表要快。

from itertools import combinations

maxvalue = 21

roots = {x**2:x for x in range(1, maxvalue)}

solutions = []
for a, b in combinations(range(1, maxvalue), 2):
    c2 = a**2 + b**2
    if c2 in roots:
        solutions.append((a, b, roots[c2]))

print(solutions)

# [(3, 4, 5), (5, 12, 13), (6, 8, 10), (8, 15, 17), (9, 12, 15), (12, 16, 20)]

Answer 3

我建议您使用Counter，然后根据需要打印不重复的项目。

这是一个开始：

import collections

nums = range(1, 21)
trips = [(a, b, c) for a in nums for b in nums for c in nums if a*a + b*b == c*c]
counted_trips = collections.Counter(trips)
print(counted_trips)

# Then you can only print trips that have occurrency == 1 (unique)
for trip, occurrency in counted_trips.items():
    if occurrency == 1:
        print(trip)

输出：

Counter({(3, 4, 5): 1, (8, 6, 10): 1, (15, 8, 17): 1, (12, 5, 13): 1, (5, 12, 13): 1, (12, 16, 20): 1, (16, 12, 20): 1, (9, 12, 15): 1, (12, 9, 15): 1, (4, 3, 5): 1, (6, 8, 10): 1, (8, 15, 17): 1})

(8, 6, 10)
(15, 8, 17)
(9, 12, 15)
(16, 12, 20)
(12, 9, 15)
(8, 15, 17)
(12, 16, 20)
(4, 3, 5)
(3, 4, 5)
(12, 5, 13)
(6, 8, 10)
(5, 12, 13)

Answer 4

只需依次整理a，b，c，您可以在a<b<c

中添加if

如何重新编程不显示重复组合？

4 个答案: