如何返回可修改的别名python函数?

时间:2018-05-04 07:20:47

标签: python function reference return alias

所以我有一个4维数组,一个非常简单的函数来帮助从所述数组中获取元素。

Board=[[
[['K','D','S']  ,  ['T','R','T'] ,  ['X','B','R']],
[['P','P','P']  ,  ['P','P','P'] ,  ['P','P','P']],
[['0','0','0']  ,  ['0','0','0'] ,  ['0','0','0']] 
],[
[['0','0','0']  ,  ['0','0','0'] ,  ['0','0','0']],
[['0','0','0']  ,  ['0','0','0'] ,  ['0','0','0']],
[['0','0','0']  ,  ['0','0','0'] ,  ['0','0','0']] 
],[
[['0','0','0']  ,  ['0','0','0'] ,  ['0','0','0']],
[['p','p','p']  ,  ['p','p','p'] ,  ['p','p','p']],
[['j','r','x']  ,  ['t','r','t'] ,  ['s','d','d']]
]]
def GrabAPeice(title):
    BigLet={'A':0,'B':1,'C':2,'D':3,'E':4}
    SmallLet={'a':0,'b':1,'c':2,'d':3,'e':4}
    BigNum={'5':0,'4':1,'3':2,'2':3,'1':4}
    SmallNum={'5':0,'4':1,'3':2,'2':3,'1':4}
    return Board[BigNum[title[2]]][SmallNum[title[4]]][BigLet[title[0]]][SmallLet[title[1]]]

其中title将分别为'Aa1:2'

唯一的问题是当我尝试使用该功能进行分配时,

GrabAPeice('Bb2:2')='Q'它不允许我分配值。

是否可以让函数在python中向另一个对象返回引用或别名?

我知道get和set函数,我真的不会使用它们。

0 个答案:

没有答案