def test(): aList.append(3) aList=[1,2] print(aList) test() print(aList) 为什么输出是

[1, 2]和[1, 2, 3]？

这是因为aList在test函数中被修改了。在test函数中，首先将数字3添加到aList中，然后将aList重新赋值为[1,2]。由于列表是可变对象，所以当我们在test函数中修改aList时，它也会影响到在函数外部定义的同一个列表。因此，当我们在函数外部再次打印aList时，它已经被修改为[1, 2, 3]。

相关问题

class Solution: def mergeAlternately(self, word1: str, word2: str) -> str: lista = [] for x,y in zip_longest(word1,word2): if x: lista.append(x) if y: lista.append(y) return ''.join(lista) 的 if x: lista.append(x) if y: lista.append(y)

这段代码中的 `if x: lista.append(x)` 和 `if y: lista.append(y)` 是用来检查变量 `x` 和 `y` 是否为真值（非假值），如果为真，则将它们添加到列表 `lista` 中。

在 Python 中，任何非零、非空、非空字符串、非空列表等都被视为真值，只有零、空、空字符串、空列表等被视为假值。

因此，`if x: lista.append(x)` 和 `if y: lista.append(y)` 语句的作用是判断 `x` 和 `y` 是否存在且不为假值，如果满足条件，则将它们添加到列表 `lista` 中。

修改以下代码，使他避开障碍物：def Astar(Map): s = Point(Map.start, None, None) openlist = [s] closedlist = [] s.get_f(Map.goal) while openlist: n = openlist.pop(0) if n.position[0] == Map.goal[0] and n.position[1] == Map.goal[1]: return n else: closedlist.append(n) subp_list = Point.EXPAND(n, Map) for subp in subp_list: subp.get_f(Map.goal) for openp in openlist: if openp.position[0] == subp.position[0] and openp.position[1] == subp.position[1]: if subp.f < openp.f: openlist.remove(openp) openlist.append(subp) for closedp in closedlist: if closedp.position[0] == subp.position[0] and closedp.position[1] == subp.position[1]: if subp.f < closedp.f: closedlist.remove(closedp) openlist.append(subp) openlist.append(subp) openlist.sort(key= lambda x:x.f)

Sure, here's the modified code that avoids obstacles using A* algorithm:

def Astar(Map):
    s = Point(Map.start, None, None)
    openlist = [s]
    closedlist = []
    s.get_f(Map.goal)
    while openlist:
        n = openlist.pop(0)
        if n.position[0] == Map.goal[0] and n.position[1] == Map.goal[1]:
            return n
        else:
            closedlist.append(n)
            subp_list = Point.EXPAND(n, Map)
            for subp in subp_list:
                if Map.is_obstacle(subp.position[0], subp.position[1]):
                    continue
                subp.get_f(Map.goal)
                for openp in openlist:
                    if openp.position[0] == subp.position[0] and openp.position[1] == subp.position[1]:
                        if subp.f < openp.f:
                            openlist.remove(openp)
                            openlist.append(subp)
                            break
                else:
                    for closedp in closedlist:
                        if closedp.position[0] == subp.position[0] and closedp.position[1] == subp.position[1]:
                            if subp.f < closedp.f:
                                closedlist.remove(closedp)
                                openlist.append(subp)
                                break
                    else:
                        openlist.append(subp)
            openlist.sort(key=lambda x: x.f)

The modification is done by adding a check for obstacles in the sub-points list and skipping those points, as well as updating the open and closed lists based on the new condition.