-1

编辑

我试过了

self.openlist.sort(key = lambda d: (d['f']))

import operator
self.openlist.sort(key = operator.itemgetter('f'))

并且都因 AttributeError 而崩溃:节点实例没有属性“ getitem

结束编辑

这可能是由于我对 Python 的一般不熟悉,但我无法通过搜索找到我的问题的答案,所以我在这里问。

我正在编写一个 A* 程序来查找城市之间的最短路径。

我将每一步我可以前往的所有城市都存储在一个列表 self.openlist 中,当我处理它们时,我会为它们分配所有子值,例如 self.openlist.g、self.openlist.h、self。 openlist.f 等。

我需要按 f 子值从最低到最高对 self.openlist 进行排序。

另外,我有很多不必要的打印,只是为了帮助我在调试程序时跟踪程序的位置。

任何提示将不胜感激。下面是我的代码:

import adata

class AS:
def __init__(self, startcity, goalcity, tracefunc):
    self.startcity = startcity
    self.goalcity = goalcity
    self.tracefunc = tracefunc
    self.openlist = [Node([startcity])]
    self.closedlist = []
    self.openlist[0].name = startcity
    self.openlist[0].path = startcity

def astar_run(self, dbg=""):
    while self.openlist:
        if self.openlist[0].name == self.goalcity:
            print self.openlist[0].path
            return self.openlist[0]
        else:
            print "Current city", self.openlist[0].name
            c2 = Roads(self.openlist[0].name)
            templist = []
            print "c2 contains", c2
            print "Templist is", templist
            x = 1
            print "Length of c2", len(c2)
            for i in range(0, len(c2)):
                print "Test point", x
                print "i value is now:", i
                print c2[i]
                templist.append(Node(c2[i]))
                templist[i].name = c2[i]
                templist[i].g = Distance(self.openlist[0].name, c2[i]) + self.openlist[0].g
                templist[i].h = Distance(c2[i], self.goalcity)
                templist[i].f = templist[i].g + templist[i].h
                templist[i].path = self.openlist[0].path + ", to " + c2[i]
                self.openlist.append(templist[i])
                print "x value is now:", x
                x = x + 1
                print self.openlist[i].name
        self.closedlist.append(self.openlist[0].name)
        del self.openlist[0]
        p = 0
        q = len(self.openlist)
        while p in range(0, q):
            print "Openlist", p, "is", self.openlist[p].name
            if self.openlist[p].name in self.closedlist:
                print "Deleting", self.openlist[p]
                del self.openlist[p]
                p = p - 1
                q = len(self.openlist)
            p = p + 1
                #print "Openlist", p, "is now", self.openlist[p].name
        print "Closedlist is", self.closedlist




def Roads(city):
    c1 = adata.roadlist(city)
    return c1

def Distance(city1, city2):
    c3 = adata.dist(city1, city2)
    return c3



class Node:
def __init__(self, path=[], f=0, g=0, h=0):
    self.path = path[:]
    self.name = []
    self.f = f
    self.g = g
    self.h = h

另外,我意识到我在 def 上的标签已经丢失,将代码转移到这个网站,我很抱歉。但是,所有这些都在我的 py.

提前致谢!

4

1 回答 1

1

thing.fthing['f']做不同的事情。利用

self.openlist.sort(key=operator.attrgetter('f'))
于 2013-09-23T04:56:26.097 回答