Question

我有两个元组列表如下：

actions = [(2, 1), (1, 0), 'lift 4', (2, 1), 'down 2', (1, 2), 'lift 4', 'down 2']
costs = [(2, 1, 3), (1, 0, 3), (2, 1, 3), (1, 2, 3)]

这是我想要的最终结果：

final_list = [(2, 1, 3), (1, 0, 3),'lift 4', (2, 1, 3),'down 2', (1, 2, 3),'lift 4','down 2' ]

我在这做什么..元组的前两个元素对应于（x，y）坐标..然后字符串是＆＃34;动作＆＃34;我接受那个坐标。现在，在＆＃34;成本＆＃34; list ..基于运动（对角线运动成本为3，而线性运动成本为2），我将成本附加到元组，忽略了动作。

如何合并这两个列表，以便我得到结果＆＃34; final_list＆＃34;

Answer 1

一种解决方案是将costs简单地视为查找表。第一，将其转换为方便的dict：

cost_dict = {i[:2] : i for i in costs}

{(1, 2): (1, 2, 3), (1, 0): (1, 0, 3), (2, 1): (2, 1, 3)}

然后有条件地转换actions中的值：

final_list = [cost_dict[i] if isinstance(i, tuple) else i for i in actions]

[(2, 1, 3), (1, 0, 3), 'lift 4', (2, 1, 3), 'down 2', (1, 2, 3), 'lift 4', 'down 2']

Answer 2

相当直接的方法：

costs.reverse()
final_list = []
for action in actions:
    if isinstance(action, tuple):
        # assume that costs in the order as actions
        final_list.append(costs.pop())
    else:
        final_list.append(action)
final_list

Answer 3

首先创建一个字典以将操作映射到成本：

actions = [(2, 1), (1, 0), 'lift 4', (2, 1), 'down 2', (1, 2), 'lift 4', 'down 2']
costs = [(2, 1, 3), (1, 0, 3), (2, 1, 3), (1, 2, 3)]
cost_map = {cost[:2] : cost for cost in costs}

现在，您可以使用dict.get()和默认值来处理操作列表：

final_list = [cost_map.get(action, action) for action in actions]
>>> final_list
[(2, 1, 3), (1, 0, 3), 'lift 4', (2, 1, 3), 'down 2', (1, 2, 3), 'lift 4', 'down 2']

Answer 4

假设元组actions和costs列表的位置已映射并且按顺序排列，为了获得所需的结果，您可以创建一个中间列表来存储{{1}的索引tuple列表中的对象：

actions

然后根据以前的索引将值从>>> actions = [(2, 1), (1, 0), 'lift 4', (2, 1), 'down 2', (1, 2), 'lift 4', 'down 2'] >>> costs = [(2, 1, 3), (1, 0, 3), (2, 1, 3), (1, 2, 3)] >>> tuple_position = [i for i, a in enumerate(actions) if isinstance(a, tuple)] >>> tuple_position [0, 1, 3, 5]列表更新到costs列表：

actions

从第一个列表中获取相应的值并添加到python中的另一个列表

4 个答案: