相关疑难解决方法(0)

如何访问索引本身以获取如下列表？

ints = [8, 23, 45, 12, 78]
for i in ints:
    print('item #{} = {}'.format(???, i))

当我使用循环遍历它时for,如何访问循环索引,在这种情况下从1到5？

我已经检查了一些想法和原因,下面针对这个问题进行了调查...... "太多值解包"异常 (Stefano Borini的解释)

但是在这里,我正在迭代列表作为理解列表并将结果移动到列表中......!所以输入的数量读取输出变量的数量,即tempList......

然后,这个过程出了什么问题？!

def DoProcess(self, myList):
    tempList = []
    tempList = [[x,y,False] for [x,y] in myList]
    return tempList

编辑1:myList是列表,就像[[x1, y1], [x2, y2], [x3, y3], [x4 y4]].

class Agent(object):
    def __init__(self, point = None):
        self.locationX = point.x
        self.locationY = point.y

    def __iter__(self):
        return self

    def __next__(self):
        return [self.locationX, self.locationY]

    def __getItem__(self):
        return [self.locationX, self.locationY]

    def GenerateAgents(self, numberOfAgents):
        agentList = []
        while len(agentList) < numberOfAgents:

            point = Point.Point()
            point.x = random.randint(0, 99)
            point.y …