证明'或'关键字的行为的输出是正确的(Justify the output of bahavior of 'or' keyword)

证明'或'关键字的行为的输出是正确的(Justify the output of bahavior of 'or' keyword)

请考虑以下声明：

问题1：

>>> a="abc" >>> 'd' or 'e' in a 'd'

有人请解释一下。 我期待一个对错......

问题2：

>>> print any(c in a for c in 'da') True

这里发生了什么事 ？ 如果我这样做，

>>> (c in a for c in 'da') <generator object <genexpr> at 0x011E4300>

正如你所看到的，它给出了生成器对象......“任何”（方法，函数??）在这里扮演什么角色？ 结果呢？

问题3：

>>> Pattern="sdfdfg" >>> if '\\'or '^' or '.' in Pattern: print "yes" else:print "no" yes

这个“是”究竟是怎么回事？

告诉我一个人有光的...........

Consider the statement below:

Prob 1:

>>> a="abc" >>> 'd' or 'e' in a 'd'

Someone please explain this . I was expecting a True or False ...

Prob 2:

>>> print any(c in a for c in 'da') True

Whats happening here ? If i do this ,

>>> (c in a for c in 'da') <generator object <genexpr> at 0x011E4300>

As you can see , it gives generator object...What role does 'any' (method,function ??) play here? And the result ?

Prob 3:

>>> Pattern="sdfdfg" >>> if '\\'or '^' or '.' in Pattern: print "yes" else:print "no" yes

How on earth is this "YES" ??

Show me the light someone plz...........

最满意答案

您看到的结果是or和in运算符的优先级与您预期的不同。

'd' or 'e' in a

这应该被理解为'd' or ('e' in a) ，即2个操作数or是'd'和('e' in a) 。 因为d在python中被认为是真值 ， or 没有查看它的下一个操作数 ，所以只返回它的第一个操作数'd' 。 请注意，这意味着or不仅仅是逻辑OR，它可以处理其他类型，并返回其他类型。

print any(c in a for c in 'da')

请阅读此print any((c in a) for c in 'da') ，即，检查'da'的元素，测试该字母是否也在a ，然后查看其中任何a是否为真。 你<generator object <genexpr> at 0x011E4300>看到<generator object <genexpr> at 0x011E4300>是Python没有向你展示生成器的元素，除非它有理由跨过它们。 如果您想查看单个元素，请写：

[(c in a) for c in 'da'] if '\\'or '^' or '.' in Pattern:

再次，请阅读此if ('\\' or '^') or ('.' in Pattern:) 。 由于'\\'为真，结果是正确的。

优先表

The results you are seeing are the result of the precedence of the or and in operators being different than you expect.

'd' or 'e' in a

This should be read as 'd' or ('e' in a), that is, the 2 operands to or are 'd' and ('e' in a). Since d is considered a true value in python, or doesn't look at its next operand, and just returns its first operand, 'd'. Note that this means that or isn't just a logical OR, it can deal with other types, and return other types.

print any(c in a for c in 'da')

Read this print any((c in a) for c in 'da'), ie, go over the elementns of 'da', test if that letter is also in a, and then see if that held true for any of them. The reason you see <generator object <genexpr> at 0x011E4300> is that Python does not show you the elements of a generator unless it has cause to step over them. If you want to see the individual elements, write:

[(c in a) for c in 'da'] if '\\'or '^' or '.' in Pattern:

Again, read this if ('\\' or '^') or ('.' in Pattern:). Since '\\' is true, the result is true.

Table of precedence