作者veriaw (ver)
看板Python
標題[問題] node, actions, visited=fringe.pop()
時間Wed Mar 23 11:48:19 2016
不好意思怕有人誤會
得先說明這是作業,但沒要求解釋以下的問題
最近在寫berkelyey pacman
有3個步驟有點不太明白
想請問
(1)
node, actions, visited = fringe.pop()
一次把右邊的pop assign給左邊的多個variables是什麼意思呢?
(2)
fringe.push((coord, actions+[direction], visited+[node]))
push定義如下,我的理解是依序把coord, actions+[direction], visited+[node]
加入到list中,不知是否正確呢?
(3)
for coord, direction, steps in problem.getSuccessors(node):
我的程度只到 for i in list:
有點不太明白for 後面接多個variables跑的意思是什麼
#Code:
def depthFirstSearch(problem)
fringe = util.Stack()
fringe.push( (problem.getStartState(), [], []) )
while not fringe.isEmpty():
node, actions, visited = fringe.pop()
for coord, direction, steps in problem.getSuccessors(node):
if not coord in visited:
if problem.isGoalState(coord):
return actions + [direction]
fringe.push((coord, actions + [direction], visited + [node] ))
return []
#util
class Stack:
"A container with a last-in-first-out (LIFO) queuing policy."
def __init__(self):
self.list = []
def push(self,item):
"Push 'item' onto the stack"
self.list.append(item)
def pop(self):
"Pop the most recently pushed item from the stack"
return self.list.pop()
def isEmpty(self):
"Returns true if the stack is empty"
return len(self.list) == 0
--
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※ 編輯: veriaw (140.112.49.174), 03/23/2016 12:22:29
→ uranusjr: 這三個其實是同一個概念: tuple, 第一個叫做 unpacking 03/23 12:37
→ uranusjr: on assignment, 就是直接把回傳的 tuple 在 = 時展開 03/23 12:38
→ uranusjr: 在 for 迴圈放多個變數也是這個概念的變形 03/23 12:38
推 caim0725: 每迭代一次就會產生三個元素 再一一指派給左側 03/23 23:20
推 ResolaQQ: 2的理解如果我沒誤會的話,是錯誤的 03/24 02:36
→ ResolaQQ: 白話點,1的意思是 03/24 02:37
→ ResolaQQ: 寫 a, b, c = (1, 2, 3),則 a = 1, b = 2, c = 3 03/24 02:38
→ ResolaQQ: 2我猜你以為是 03/24 02:39
→ ResolaQQ: list.append(a), list.append(b), list.append(c) 03/24 02:40
→ ResolaQQ: 但實際上是 list.append( (a, b, c) ) 03/24 02:40
→ ResolaQQ: 前面是加入三個物件,後面是加入一個物件但它有三個資料 03/24 02:41
→ ResolaQQ: 3同1,把i代換成1的例子就對了 03/24 02:42
→ veriaw: 非常感謝3位的協助 03/26 23:13
→ veriaw: 我的誤解完全解開了~~ 03/26 23:19