"not in" 成员运算符结合 "or" 布尔运算符 (Python)

Question

我面临着看似微不足道但非常烦人的难题。在使用 "not in" 运算符结合布尔值 "and" 和 "or" 在以下字典中检查成员资格时：

    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"

就我而言，#4 和#5 在逻辑上是相同的，但显然它们不可能。我查看了官方 Python 参考资料，但它只会增加混乱。有人可以阐明这一点吗？

Answer 1

请注意，not in 的结合力比 and 和 or 都强。所以 #4 按照你的预期被解析，而 #5 等于 if ("name") or ("income" not in data): – 因为非空字符串在 python 中是真实的，这意味着它总是在 "not found" 分支中结束.

"not in" 成员运算符结合 "or" 布尔运算符 (Python)

"not in" membership operator combined with "or" Boolean operator (Python)

python

membership