Question

我有一个像[3,10,4,3,9,15,6,13]这样的列表，我希望找到两个不超过一圈的系列/序列，并通过获取最大最低值来获得最大值。它们必须是顺序的，因此您不能从1中减去第3项。但是您可以从3中减去索引1。

所以在我的例子中你会得到对[3,10]和[3,15]。你怎么能以程序化的方式做到这一点。这是我到目前为止所得到的。

蟒：

l = [3,10,4,3,9,15,6,13]
pair1=[max(l), min(l)]
l.remove(max(l))
l.remove(min(l))
pair2=[max(l), min(l)]

当然不能做我想做的事，但我不确定如何继续寻找两对。找到一对上面的代码工作正常，但不是两个，因为你通常会得到重叠的序列。

Answer 1

写这个很有趣：

import itertools as it

l = [3,10,4,3,9,15,6,13]

r=sorted(
    map(lambda x: (x[0:2], x[2:4]),
        sorted(it.imap(
            lambda x: (min(x[0:2]), max(x[0:2]), min(x[2:4]), max(x[2:4])),
            it.imap(lambda x: (l[x[0]], l[x[1]], l[x[2]], l[x[3]]),
                    it.combinations(range(len(l)), 4))),
               key=lambda x: -(x[1]-x[0]+x[3]-x[2])))[0],
    key=lambda x: x[0]-x[1])

print(r)

Answer 2

可能不是那么优雅，但试试这个：

l = [3,10,4,3,9,15,6,13]

# list to store diffs
diff=[]
# calculate diffs
for i in range(0, len(l)-1):
    diff.append(l[i+1]-l[i])

# list to store diff sequences
results=[]

# Findinf the sequences
curr=0;
while curr<=(len(diff)-1):

    # list to store positive differences
    result=[]

    # llop while the difference is positive
    while diff[curr]>0 :

        # if it is a first element
        if len(result) == 0:
            # in 0th place store the diff sum
            result.append(diff[curr])
            # in 1st place store the sequence start index 
            result.append(curr)
            # in 2nd place store the sequence end index
            result.append(curr+1)
        else:
            # summ the sequnce diff
            result[0]+=diff[curr]
            # update sequence end index
            result[2]=curr+1
        # move to next element
        curr+=1
        # if reached end of array, break the loop 
        if curr>(len(diff)-1):
            break
    # if there was a sequence store it             
    if (len(result) > 0):
        results.append(result)
    else:
        # move to next element in case of there was no sequence
        curr+=1

# sort the results by sum in reverse order of the 0th element(sum) and normal order of 1st element(start index)
results.sort(key=lambda x: x[0]*10-x[1], reverse=True)
# print first 2 results
for i in range(0,2):
    print "[%d,%d]" % (l[results[i][1]],l[results[i][2]])

至少对于您的输入，它会打印想要的结果：）

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2 个答案: