如何在Python中获得两个变量的逻辑异或?
例如，我有两个变量，我希望它们是字符串。我想测试它们中只有一个包含True值(不是None或空字符串):
str1 = raw_input("Enter string one:")
str2 = raw_input("Enter string two:")
if logical_xor(str1, str2):
print "ok"
else:
print "bad"
^操作符似乎是按位的，并不是在所有对象上都定义:
>>> 1 ^ 1
0
>>> 2 ^ 1
3
>>> "abc" ^ ""
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for ^: 'str' and 'str'
当前回答
按位异或已经内置在Python的operator模块中(与^ operator相同):
from operator import xor
xor(bool(a), bool(b)) # Note: converting to bools is essential
其他回答
奖励线程:
阳极的想法……您只需尝试(可能是)python表达式«is not»，以获得逻辑«xor»的行为
真值表是:
>>> True is not True
False
>>> True is not False
True
>>> False is not True
True
>>> False is not False
False
>>>
对于你的例子字符串:
>>> "abc" is not ""
True
>>> 'abc' is not 'abc'
False
>>> 'abc' is not ''
True
>>> '' is not 'abc'
True
>>> '' is not ''
False
>>>
然而;正如上面所指出的，这取决于你想要抽出任意一对字符串的实际行为，因为字符串不是bolean…甚至更多:如果你«潜入Python»你会发现«“and”和“or”的特殊性质» http://www.diveintopython.net/power_of_introspection/and_or.html
抱歉我的书面英语，它不是我天生的语言。
的问候。
假设A和B是bool。
A is not B
这里建议的一些实现在某些情况下会导致操作数的重复求值，这可能会导致意想不到的副作用，因此必须避免。
也就是说，一个返回True或False的xor实现相当简单;如果可能的话，返回其中一个操作数的方法要棘手得多，因为对于应该选择哪个操作数没有共识，特别是当有两个以上的操作数时。例如，xor(None， -1， []， True)是否应该返回None，[]或False?我敢打赌，每个答案对某些人来说都是最直观的。
对于True-或false -结果，有多达五种可能的选择:返回第一个操作数(如果它在value中匹配最终结果，否则布尔值)，返回第一个匹配(如果至少存在一个，否则布尔值)，返回最后一个操作数(如果…Else…)，返回最后一次匹配(if…Else…)，或者总是返回boolean。总共是5 ** 2 = 25种xor口味。
def xor(*operands, falsechoice = -2, truechoice = -2):
"""A single-evaluation, multi-operand, full-choice xor implementation
falsechoice, truechoice: 0 = always bool, +/-1 = first/last operand, +/-2 = first/last match"""
if not operands:
raise TypeError('at least one operand expected')
choices = [falsechoice, truechoice]
matches = {}
result = False
first = True
value = choice = None
# avoid using index or slice since operands may be an infinite iterator
for operand in operands:
# evaluate each operand once only so as to avoid unintended side effects
value = bool(operand)
# the actual xor operation
result ^= value
# choice for the current operand, which may or may not match end result
choice = choices[value]
# if choice is last match;
# or last operand and the current operand, in case it is last, matches result;
# or first operand and the current operand is indeed first;
# or first match and there hasn't been a match so far
if choice < -1 or (choice == -1 and value == result) or (choice == 1 and first) or (choice > 1 and value not in matches):
# store the current operand
matches[value] = operand
# next operand will no longer be first
first = False
# if choice for result is last operand, but they mismatch
if (choices[result] == -1) and (result != value):
return result
else:
# return the stored matching operand, if existing, else result as bool
return matches.get(result, result)
testcases = [
(-1, None, True, {None: None}, [], 'a'),
(None, -1, {None: None}, 'a', []),
(None, -1, True, {None: None}, 'a', []),
(-1, None, {None: None}, [], 'a')]
choices = {-2: 'last match', -1: 'last operand', 0: 'always bool', 1: 'first operand', 2: 'first match'}
for c in testcases:
print(c)
for f in sorted(choices.keys()):
for t in sorted(choices.keys()):
x = xor(*c, falsechoice = f, truechoice = t)
print('f: %d (%s)\tt: %d (%s)\tx: %s' % (f, choices[f], t, choices[t], x))
print()
有时我发现自己使用1和0而不是布尔值True和False。在这种情况下，xor可以定义为
z = (x + y) % 2
它有如下的真值表:
x
|0|1|
-+-+-+
0|0|1|
y -+-+-+
1|1|0|
-+-+-+
当你知道XOR是做什么的时候就很简单了:
def logical_xor(a, b):
return (a and not b) or (not a and b)
test_data = [
[False, False],
[False, True],
[True, False],
[True, True],
]
for a, b in test_data:
print '%r xor %s = %r' % (a, b, logical_xor(a, b))
