如何修复mathematica中的列表分配
我想在 Mathematica 中执行以下操作
Table[p[i], {i, -3, 0}] = Flatten[{Table[0, {i, -3, -1}], 1}]
，但出现错误：
Set::write: Tag Table in Table[p[i], {i, -3, 0}] is Protected.
但是，这样做完全没问题：
{p[-3], p[-2], p[-1], p[0]} = Flatten[{Table[0, {i, -3, -1}], 1}]
非常感谢！
I want to do the following in Mathematica
Table[p[i], {i, -3, 0}] = Flatten[{Table[0, {i, -3, -1}], 1}]
But I got an error:
Set::write: Tag Table in Table[p[i], {i, -3, 0}] is Protected.
However, it is perfectly fine to do:
{p[-3], p[-2], p[-1], p[0]} = Flatten[{Table[0, {i, -3, -1}], 1}]
Many thanks!
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强制 LHS 评估为可分配给的片段：
Evaluate[Table[p[i], {i, -3, 0}]] = Flatten[{Table[0, {i, -3, -1}] , 1}]
Force the LHS to evaluate into pieces that can be assigned to:
Evaluate[Table[p[i], {i, -3, 0}]] = Flatten[{Table[0, {i, -3, -1}], 1}]
它不起作用的原因是
Set具有属性HoldFirst。这意味着Set[a,stuff]将符号a而不是a的值传递给Set函数。至于为什么它有这个属性，问问自己：当你执行Set[a,stuff]时，你是否要将stuff赋值给符号a>，或者a的值？在您的示例中，
a保存一个变量名称表，因此您需要a的值，而HoldFirst很烦人。但是，大多数时候a的值类似于5，并且您希望a=stuff分配stuff到符号a，而不是值5绕过 Holding 属性的常见模式如下：
The reason it doesn't work is because
Sethas attributeHoldFirst. It means thatSet[a,stuff]passes symbolainstead of value ofatoSetfunction. As to why it has this attribute, ask yourself: when you doSet[a,stuff], do you want to assignstuffto symbola, or to the value ofa?In your example,
aholds a table of variable names, so you want the value ofaandHoldFirstis annoying. However, most of the timeawill have a value like5and you wanta=stuffto assignstuffto symbola, not to the value5A common pattern to get around Holding attributes is the following: