Mathematica 模块与使用或块 - 指南，使用经验法则? [英] Mathematica Module versus With or Block - Guideline, rule of thumb for usage?

问题描述

Leonid 在他的书的第四章中写道:...模块、块和与.这些结构在 Mathematica Book 和 Mathematica Help 中有详细解释，所以我在这里只说几句...."

Leonid wrote in chapter iv of his book : "... Module, Block and With. These constructs are explained in detail in Mathematica Book and Mathematica Help, so I will say just a few words about them here. ..."

从我读到的(能够找到的)来看，我仍然一无所知.对于打包的函数，我(简单地)使用 Module，因为它可以工作并且我知道它的构造.虽然它可能不是最好的选择.我并不完全清楚(从文档中)何时、何地或为什么使用 With(或 Block).

From what I have read ( been able to find ) I am still in the dark. For packaged functions I ( simply ) use Module, because it works and I know the construct. It may not be the best choice though. It is not entirely clear to me ( from the documentation ) when, where or why to use With ( or Block ).

推荐答案

Block 和 Module 之间更实际的区别可以在这里看到:

A more practical difference between Block and Module can be seen here:

Module[{x}, x]
Block[{x}, x]
(*
-> x$1979
   x
*)

所以如果你想返回例如x，你可以使用Block.例如，

So if you wish to return eg x, you can use Block. For instance,

Plot[D[Sin[x], x], {x, 0, 10}]

不起作用；要使其工作，可以使用

does not work; to make it work, one could use

Plot[Block[{x}, D[Sin[x], x]], {x, 0, 10}]

(当然这并不理想，这只是一个例子).

(of course this is not ideal, it is simply an example).

另一个用途是类似于 Block[{$RecursionLimit = 1000},...]，它临时改变了 $RecursionLimit (Module不会起作用，因为它重命名了 $RecursionLimit).

Another use is something like Block[{$RecursionLimit = 1000},...], which temporarily changes $RecursionLimit (Module would not have worked as it renames $RecursionLimit).

还可以使用 Block 来阻止对某事的评估，例如

One can also use Block to block evaluation of something, eg

Block[{Sin}, Sin[.5]] // Trace
(*
-> {Block[{Sin},Sin[0.5]],Sin[0.5],0.479426}
*)

即，它返回 Sin[0.5] 仅在 Block 完成执行后才计算.这是因为 Block 中的 Sin 只是一个符号，而不是正弦函数.你甚至可以做类似的事情

ie, it returns Sin[0.5] which is only evaluated after the Block has finished executing. This is because Sin inside the Block is just a symbol, rather than the sine function. You could even do something like

Block[{Sin = Cos[#/4] &}, Sin[Pi]]
(*
-> 1/Sqrt[2]
*)

(使用 Trace 来查看它是如何工作的).所以你也可以使用 Block 在本地重新定义内置函数:

(use Trace to see how it works). So you can use Block to locally redefine built-in functions, too:

Block[{Plus = Times}, 3 + 2]
(*
-> 6
*)

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