Python 生成具有特定属性的唯一元组列表的算法
我需要用唯一元组(A,B,C)标记N个对象,其中A 例如:Python 生成具有特定属性的唯一元组列表的算法,python,algorithm,Python,Algorithm,我需要用唯一元组(A,B,C)标记N个对象,其中A
M = 1
N = 4
# As Bs Cs
objects = [(1, 2, 3),
(2, 3, 4),
(3, 4, 5),
(4, 5, 6)]
M = 2
N = 4
objects = [(1, 2, 3),
(1, 2, 4),
(2, 3, 4),
(2, 3, 5)]
# or e.g
objects = [(1, 2, 3),
(2, 3, 4),
(2, 4, 5),
(3, 4, 5)]
M = 3
N = 8
objects = [(1, 2, 3),
(2, 3, 4),
(2, 3, 5),
(2, 4, 5),
(3, 4, 5),
(3, 4, 6),
(3, 5, 6),
(4, 5, 6)]
我提出的程序是一个复杂的if-else怪物:
import sys
# useage: labelme.py <N> <M>
class ObjectListTree(object):
"""Create many possible paths.
Store the parent in each node.
The last nodes are appended to the class wide endnodes.
"""
endnodes = []
def __init__(self, parent, label, counter, n, M, N):
self.parent = parent
self.M = M
self.N = N
self.label = label
self.counter = counter
self.n = n
if n < N:
self.inc_a()
self.inc_b()
self.inc_c()
else:
ObjectListTree.endnodes.append(self)
def inc_a(self):
if self.label[0]+1 < self.label[1]:
if self.counter[1] < self.M:
if self.counter[2] < self.M:
self.plus_1()
else:
self.plus_1_3()
else:
if self.counter[2] < self.M:
self.plus_1_2()
else:
self.plus_all()
elif self.label[1]+1 < self.label[2]:
if self.counter[2] < self.M:
self.plus_1_2()
else:
self.plus_all()
else:
self.plus_all()
def inc_b(self):
if self.counter[0] == self.M:
return
if self.label[1]+1 < self.label[2] and self.counter[2] < self.M:
self.plus_2()
else:
self.plus_2_3()
def inc_c(self):
if self.counter[0] == self.M or self.counter[1] == self.M:
return
else:
self.plus_3()
def plus_all(self):
ObjectListTree(self, (self.label[0]+1, self.label[1]+1, self.label[2]+1),
counter = [1, 1, 1,],
n = self.n+1, N=self.N, M=self.M)
def plus_1_2(self):
ObjectListTree(self, (self.label[0]+1, self.label[1]+1, self.label[2]),
counter = [1, 1, self.counter[2]+1,],
n = self.n+1, N=self.N, M=self.M)
def plus_1_3(self):
ObjectListTree(self, (self.label[0]+1, self.label[1], self.label[2]+1),
counter = [1, self.counter[1]+1, 1,],
n = self.n+1, N=self.N, M=self.M)
def plus_1(self):
ObjectListTree(self, (self.label[0]+1, self.label[1], self.label[2]),
counter = [1, self.counter[1]+1, self.counter[2]+1,],
n = self.n+1, N=self.N, M=self.M)
def plus_2(self):
ObjectListTree(self, (self.label[0], self.label[1]+1, self.label[2]),
counter = [self.counter[0]+1, 1, self.counter[2]+1,],
n = self.n+1, N=self.N, M=self.M)
def plus_2_3(self):
ObjectListTree(self, (self.label[0], self.label[1]+1, self.label[2]+1),
counter = [self.counter[0]+1, 1, 1,],
n = self.n+1, N=self.N, M=self.M)
def plus_3(self):
ObjectListTree(self, (self.label[0], self.label[1], self.label[2]+1),
counter = [self.counter[0]+1, self.counter[1]+1, 1,],
n = self.n+1, N=self.N, M=self.M)
tree = ObjectListTree(parent=None, label=(1, 2, 3), counter = [1,1,1,], n=1, N=int(sys.argv[1]), M=int(sys.argv[2]))
best_path = tree.endnodes[0]
for n in tree.endnodes:
if n.label[2] < best_path.label[2]:
best_path = n
objects = []
while best_path:
objects.append(best_path.label)
best_path = best_path.parent
objects.reverse()
print objects
导入系统
#用途:labelme.py
类ObjectListTree(对象):
“”“创建许多可能的路径。
在每个节点中存储父节点。
最后一个节点附加到类范围的endnodes。
"""
endnodes=[]
定义初始化(自身、父项、标签、计数器、n、M、n):
self.parent=parent
self.M=M
self.N=N
self.label=标签
self.counter=计数器
self.n=n
如果n但我觉得这实际上应该是一些简单的东西,比如将itertools模块中的两个或三个函数组合在一个集合中。有人能看到一个简单的解决方案吗?我认为这段代码满足您的要求,并且总是以尽可能低的C生成解决方案。但是,不完全使用itertools
def generateTuples(N, M):
done = 0
counters = {}
for C in range(3, N + 3):
for B in range(2, C):
for A in range(1, B):
if (counters.get('A%i' % A, 0) < M and
counters.get('B%i' % B, 0) < M and
counters.get('C%i' % C, 0) < M):
yield (A, B, C)
counters['A%i' % A] = counters.get('A%i' % A, 0) + 1
counters['B%i' % B] = counters.get('B%i' % B, 0) + 1
counters['C%i' % C] = counters.get('C%i' % C, 0) + 1
done += 1
if done >= N:
return
for (A, B, C) in generateTuples(8, 3):
print (A, B, C)
def发生器耦合(N,M):
完成=0
计数器={}
对于范围(3,N+3)内的C:
对于范围(2,C)内的B:
对于范围(1,B)内的A:
if(counters.get('A%i'%A,0)=N:
返回
对于(A,B,C)在生成偶(8,3)中:
印刷品(A、B、C)
你听说过itertools.permutations吗?我确实听说过。我只是看不出这有什么帮助。我真的是指最后一句话:它看起来应该是来自itertools的东西,但我看不到。它应该是itertools.product
,但筛选出不符合“最大相同A数”的条目似乎很难。谢谢,这更简单、更快:-)