大家好,我正在尝试执行一种算法,以指定顺序对特定项目进行排序。
假设我们有一个项目列表:
[1、2、1、3、3、3、2、2、2、2、2、1、3、2、2、1]
这里有三种不同类型的项目:1、2和3。我想对该列表进行排序,以使一系列遵循相同类型的项目。在排序过程中,我们无法移动该职位的项目,我们只能删除一些项目,结果应该是最长的。
该算法的结果应为:
[1、1、3、3、3、2、2、2、2、2、2、2]
我不知道为什么我的算法不起作用:
# start list
givenList = [1, 2, 1, 3, 3, 3, 2, 2, 2, 2, 2, 1, 3, 2, 2,1]
# whish list (just for an example)
whish = [1, 1, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2]
# list grouped by items ( [ [1], [2], [1], [3, 3, 3], [2, 2, 2, 2, 2] ....])
sortedList = []
# we group the elements
lastElement=0
for currentElement in givenList :
if currentElement != lastElement :
sortedList.append([currentElement])
lastElement=currentElement
else : sortedList[-1].append(currentElement)
# we print the grouped items
for index, element in enumerate(sortedList) :
print("Bloc : ", index , " contient : " , element)
# we sort the same elements by group
result=[]
for index, element in enumerate(sortedList) :
# we pass if it's the first group because he has no backward element
if(index == 0) : continue
# we pass if it's the last group because he has no afterward element
if(index == len(sortedList) - 1) : continue
# backward group
backwardList = sortedList[index - 1]
# current group
currentList = sortedList[index]
# afterward group
forwardList = sortedList[index + 1]
# if the backward groupelement type is the same as the forward
if backwardList[0] == forwardList[0] :
# and if the backwardlist contains more element that the current group
if(len(backwardList) >= len(currentList)) :
# we add the concatenation of the backwards and the forwards group
result.append(backwardList + forwardList)
elif backwardList[0] != forwardList[0] :
# else we just add the current group
result.append(currentList)
# we degroup the grouped and sorted list
resultSorted=[]
for e in result:
for i in e:
resultSorted.append(i)
#
print("#"*20)
print("Given : ", givenList)
print("Whish : ", whish)
print("Result : ", resultSorted)
print("#"*20)
答案 0 :(得分:0)
您可以尝试使用itertools.groupby
来实现:
from itertools import groupby
l = [1, 2, 1, 3, 3, 3, 2, 2, 2, 2, 2, 1, 3, 2, 2, 1]
def get_max(elems, current_elems=None):
if current_elems is None:
current_elems = []
for idx, (color, count) in enumerate(elems):
if color in [c for c, _ in current_elems[:-1]]:
continue
yield current_elems + [(color, count)]
yield from get_max( elems[idx+1:], current_elems + [(color, count)] )
elems = [(v, sum(1 for _ in g)) for v, g in groupby(l)]
l, _ = max(((v, sum(c for _, c in v)) for v in get_max(elems)), key=lambda k: k[1], default=[[], None])
out = []
for v, g in groupby(l, key=lambda k: k[0]):
for _, c in g:
out.extend([v] * c)
print(out)
打印:
[1, 1, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2]
其他测试用例:
l = [1, 2, 3] # [1, 2, 3]
l = [1, 2, 3, 1, 1] # [2, 3, 1, 1]
答案 1 :(得分:0)
最好的解决方案是:
def solve(l):
def aux(i, current, exclude):
if i==len(l):
return []
val = l[i]
if val in exclude:
return aux(i+1, current, exclude)
elif val == current:
return [val] + aux(i+1, current, exclude)
else:
exclude.add(current)
s1 = [val] + aux(i+1, val, exclude)
exclude.remove(current)
s2 = aux(i+1, current, exclude)
if len(s1)>len(s2):
return s1
return s2
return aux(0, -1, set())