“不在”成员身份运算符与“或”布尔运算符组合（Python）

Question

我面对的可能是一个看似微不足道，但却非常恼人的难题。在下列字典上使用“not in”运算符结合布尔“and”和“or”检查成员身份时：

    data = {"name": "david", "age": 27, "income": "absurd"}

我发现：

#1 - found
if "name" not in data:
    print "not found"
else:
    print "found"

#2 - found
if "name" not in data and "income" not in data:
    print "not found"
else:
    print "found"

#3 - found
if "name" and "income" not in data:
    print "not found"
else:
    print "found"

#4 - found
if "name" not in data or "income" not in data:
    print "not found"
else:
    print "found"

#5 - NOT found (though it should be?)
if "name" or "income" not in data:
    print "not found"
else:
    print "found"

在我看来，4和5在逻辑上是相同的，但显然不可能。我查看了官方的Python参考资料，但这只会增加混乱。有人能解释一下吗？你知道吗

Answer 1

注意，not in比and和or结合更强烈。所以#4会像您所期望的那样被解析，而#5等于if ("name") or ("income" not in data):——因为在python中非空字符串是真实的，这意味着它总是以“找不到”分支结束。你知道吗