具有非标准评估的自定义函数(行为类似于 Table)

问题描述

我想要一个函数 AnyTrue[expr,{i,{i1,i2,...}}] 来检查 expr 是否为 True for any i1,i2... 它应该就像 AnyTrue 是 Table 后跟 Or@@%，区别在于它只计算 expr 直到找到第一个 True.

I'd like a function AnyTrue[expr,{i,{i1,i2,...}}] which checks if expr is True for any of i1,i2... It should be as if AnyTrue was Table followed by Or@@%, with the difference that it only evaluates expr until first True is found.

短路部分是可选的，我真正想知道的是模拟Table的非标准评估序列的正确方法.

Short-circuiting part is optional, what I'd really like to know is the proper way to emulate Table's non-standard evaluation sequence.

11/14 更新

这是迈克尔的解决方案，您可以使用它来链接所有"和存在"检查

Here's a solution due to Michael, you can use it to chain "for all" and "there exists" checks

SetAttributes[AllTrue, HoldAll];
SetAttributes[AnyTrue, HoldAll];
AllTrue[{var_Symbol, lis_List}, expr_] :=
  LengthWhile[lis,
    TrueQ[ReleaseHold[Hold[expr] /. HoldPattern[var] -> #]] &] ==
   Length[lis];
AnyTrue[{var_Symbol, lis_List}, expr_] :=
  LengthWhile[lis,
    Not[TrueQ[ReleaseHold[Hold[expr] /. HoldPattern[var] -> #]]] &] <
   Length[lis];
AllTrue[{a, {1, 3, 5}}, AnyTrue[{b, {2, 4, 5}}, EvenQ[a + b]]]
AnyTrue[{a, {1, 3, 5}}, AllTrue[{b, {2, 4, 5}}, EvenQ[a + b]]]

推荐答案

这个怎么样?

SetAttributes[AnyTrue, HoldAll];

AnyTrue[expr_, {var_Symbol, lis_List}] :=
  LengthWhile[lis,
    Not[TrueQ[ReleaseHold[Hold[expr] /. HoldPattern[var] -> #]]] &
  ] < Length[lis]

通过 LengthWhile 包括短路，并在必要时保留所有内容，以便事情按预期工作，var 具有函数外的值:

Includes short-circuiting via LengthWhile and keeps everything held where necessary so that things work as expected with var has a value outside the function:

In[161]:= x = 777;

In[162]:= AnyTrue[Print["x=", x]; x == 3, {x, {1, 2, 3, 4, 5}}]
During evaluation of In[162]:= x=1
During evaluation of In[162]:= x=2
During evaluation of In[162]:= x=3
Out[162]= True

内置的 Or 也是短路的，因为它的价值.(但我意识到用例如 Table 来构建未评估的术语是一种痛苦):

The built-in Or is short-circuiting, too, for what it's worth. (but I realize building up the unevaluated terms with e.g. Table is a pain):

In[173]:= Or[Print[1];True, Print[2];False]
During evaluation of In[173]:= 1
Out[173]= True

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