Python 管理多维列表中的值

如何管理多维列表中的字符串值

我知道Python字符串一旦创建就不可变。但我尝试将它从元组转换为列表，并将其从多维列表中删除，然后再添加回去，但仍然没有成功

这就是我所尝试的：

for b_order in list(signal_order_correction):
print repr(b_order)
for a_order in list(signal_order_position):
    print "A order second" + repr(a_order)
    if (a_order[0][0] == b_order[0][0] and a_order[0][1] == b_order[0][1] and a_order[0][4] == b_order[0][4]):
        print "Correcting"
        signal_order_position = signal_order_position.remove(a_order)
        a_order = list(a_order[0])
        a_order[0][5] = str(a_order[0][5])
        # print "repr a"
        # print repr(a_order)
        print repr(a_order[5])
        # a_order[5] = None
        # a_order[6] = None
        # print repr(a_order)
        a_order[0][5] = b_order[0][5]
        a_order[0][6] = b_order[0][6]
        a_order = tuple(a_order)
        print "corrected by order correction"
        signal_order_position = signal_order_position.append(a_order)

但它给了我错误

TypeError: 'str' object does not support item assignment

---

有什么想法吗？

使用删除和附加方法解决了我自己的问题：

        for b_order in list(signal_order_correction):
            print repr(b_order)
            for a_order in list(signal_order_position):
                print "A order second" + repr(a_order)
                if (a_order[0][0] == b_order[0][0] and a_order[0][1] == b_order[0][1] and a_order[0][4] == b_order[0][4]):
                    print "Correcting"
                    signal_order_position.remove(a_order)
                    print "corrected by order correction"
                    signal_order_position.append(b_order)

---

你能举个例子吗？输入和预期输出的类型？Python字符串是不可变的，不能更改。该错误表示您正在尝试更改字符串中的字符，这是不可行的。如果需要，您可以使用字节数组。@TonyTannous我希望我可以在

a\u顺序[0][5]

中从

b\u顺序[0][5]