我认为通过在中点位置逐渐插入来获得最佳分散效果是最好的。从小到大依次应用此方法可以得到最优的分散效果(或接近最优):
首先,您需要一个函数,在目标元素的列表中为m个源元素提供中点插入位置(其中N >= m)。该函数应该从第一、最后和中间三个插入位置开始具有尽可能宽的分散,并从中间使用二分法获取其余插入点。
def iPoints(N,m):
d = N//2
result = [0,N,d]
if m==N: result[1] = N-1
while len(result)<m:
d = max(1,d//2)
for r in result[2:]:
for s in [-1,1]:
p = r+s*d
if p in result : continue
result.append(p)
result = sorted(result[:m])
result = [ p + sum(p>r for r in result[:i]) for i,p in enumerate(result)]
return result
使用这个方法,你可以按照从大到小的顺序遍历集群列表并执行插入操作:最初的回答。
clusterA = ["A", "A", "A", "A", "A", "A", "A", "A"]
clusterB = ["B", "B", "B", "B"]
clusterC = ["C", "C"]
clusters = [clusterA,clusterB,clusterC]
totalSize = sum(map(len,clusters))
order = -1 if all((totalSize-len(c))//(len(c)-1) for c in clusters) else 1
clusters = sorted(clusters,key=lambda c: order*(totalSize-len(c))//(len(c)-1))
merged = clusters[0]
for cluster in clusters[1:]:
target = cluster.copy()
source = merged
if len(source) > len(target):
source,target = target,source
indexes = iPoints(len(target),len(source))
for c,p in zip(source,indexes):
target.insert(p,c)
merged = target
print(merged)
分析这个结果表明,对于这组集群来说,它稍微好一些。不幸的是,它并不能总是提供最优解。"最初的回答"
from statistics import mean
m = "".join(merged)
spreadA = [ d+1 for d in map(len,m.split("A")[1:-1])]
spreadB = [ d+1 for d in map(len,m.split("B")[1:-1])]
spreadC = [ d+1 for d in map(len,m.split("C")[1:-1])]
print("A",spreadA,mean(spreadA))
print("B",spreadB,mean(spreadB))
print("C",spreadC,mean(spreadC))
print("minimum spread",min(spreadA+spreadB+spreadC))
print("average spread", round(mean(spreadA+spreadB+spreadC), 1))
在尝试其他集群大小时,我发现集群处理的顺序很重要。我所使用的顺序基于每个集群的最大范围。如果至少有一个集群比其他集群更大,则按升序排列;否则按降序排列。
Original Answer翻译成"最初的回答"。
clusterA = ["A", "A", "A", "A", "A"]
clusterB = ["B", "B", "B", "B"]
clusterC = ["C", "C"]
B
的取值范围不是(5,2,4)吗? - Mark Setchell3*(n-3)
,其中n是交错序列的长度。这反过来又给出了平均传播范围为3.0
。不确定这是否是某种上限。 - DollarAkshay