Question

我正在处理字典问题，遇到障碍，所以我有一个3D列表，其中包含我用作参考的坐标，还有一个字典，其键与列表中的坐标匹配，这是我需要的，如果键和坐标匹配，然后将值附加到另一个3D列表中。我几乎知道该怎么做，但是我没有得到想要的东西，这是我尝试过的：

localhost/AmazonPay/Client.php

这给了我正确的值，但是顺序有点混乱，尽管它是3D列表，但分隔与参考列表不同，这应该是我想要的输出：

reference = [[[2, 3], [2, 4], [3, 2], [4, 2]], 
             [[2, 3], [3, 2], [3, 4], [4, 3]], 
             [[2, 3], [3, 2], [3, 4], [4, 3]]]

mydict = {(2, 3): [5, 1], (2, 4): [14, 16], (3, 2): [19, 1], (3, 4): [14, 30], (4, 2): [16, 9], (4, 3): [6, 2]}

aux = [[tuple(j) for j in i] for i in reference] #This transform the 3D list to tuples to match the keys

print(aux)

aux = [[(2, 3), (2, 4), (3, 2), (4, 2)], [(2, 3), (3, 2), (3, 4), (4, 3)], [(2, 3), (3, 2), (3, 4), (4, 3)]]

aux_list = []
    for key, value in mydict.items():
        final_list =[]
        for i in aux:
            for j in i:                    #If the key matches the list of tuples then append the value
                if j == key:
                    aux_list.append(value)
        final_list.append(aux_list)

print(final_list)

final_list = [[[5, 1], [5, 1], [5, 1], [14, 16], [19, 1], [19, 1], [19, 1], [14, 30], [14, 30], [16, 9], [6, 2], [6, 2]]]

这只是一个例子，我相信绝对应该有一个简单的方法来做到这一点，但是我是一个新手，所以我不确定问题出在哪里，因此任何帮助或参考都将不胜感激，谢谢你好厉害！

Answer 1

reference = [[[2, 3], [2, 4], [3, 2], [4, 2]],
             [[2, 3], [3, 2], [3, 4], [4, 3]],
             [[2, 3], [3, 2], [3, 4], [4, 3]]]

mydict = {(2, 3): [5, 1], (2, 4): [14, 16], (3, 2): [19, 1], (3, 4): [14, 30], (4, 2): [16, 9], (4, 3): [6, 2]}


out = [[mydict.get(tuple(v), v) for v in row] for row in reference]

from pprint import pprint
pprint(out)

打印：

[[[5, 1], [14, 16], [19, 1], [16, 9]],
 [[5, 1], [19, 1], [14, 30], [6, 2]],
 [[5, 1], [19, 1], [14, 30], [6, 2]]]

将字典的键与参考列表进行匹配并获取值

1 个答案: