Question

我面对的似乎是微不足道的，但却是非常讨厌的难题。在使用“not in”运算符检查成员资格时，结合以下字典中的Boolean“and”和“or”：

    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"

对于我的钱＃4和＃5在逻辑上是相同的，但显然它们不可能。我查看了官方的Python参考资料，但它只是增加了混乱。有人会对此有所了解吗？

Answer 1

请注意，import numpy as np import matplotlib.pyplot as plt # Example labelled arrays a and b input_a = np.array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 0], [0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 0], [0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 0], [0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 0], [0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 0], [0, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 0], [0, 0, 3, 3, 3, 3, 2, 2, 2, 2, 0, 0], [0, 0, 3, 3, 3, 3, 2, 2, 2, 2, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]) input_b = np.array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 1, 1, 1, 3, 3, 3, 3, 3, 0, 0], [0, 0, 1, 1, 1, 3, 3, 3, 3, 3, 0, 0], [0, 0, 1, 1, 1, 2, 2, 2, 2, 2, 0, 0], [0, 0, 1, 1, 1, 2, 2, 2, 2, 2, 0, 0], [0, 0, 1, 1, 1, 2, 2, 2, 2, 2, 0, 0], [0, 0, 1, 1, 1, 2, 2, 2, 2, 2, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]) # Plot inputs plt.imshow(input_a, cmap="spectral", interpolation='nearest') plt.imshow(input_b, cmap="spectral", interpolation='nearest') # Desired output, union of a and b output = np.array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 1, 1, 1, 2, 3, 3, 3, 3, 0, 0], [0, 0, 1, 1, 1, 2, 3, 3, 3, 3, 0, 0], [0, 0, 1, 1, 1, 4, 7, 7, 7, 7, 0, 0], [0, 0, 5, 5, 5, 6, 7, 7, 7, 7, 0, 0], [0, 0, 5, 5, 5, 6, 7, 7, 7, 7, 0, 0], [0, 0, 5, 5, 5, 6, 7, 7, 7, 7, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]) # Plot desired output plt.imshow(output, cmap="spectral", interpolation='nearest')比not in和and的绑定更强。因此＃4会按照您的预期进行解析，而＃5等于or - 因为非空字符串在python中是真的，这意味着它总是在“未找到”分支中结束。

“不在”成员运算符与“或”布尔运算符（Python）结合使用

1 个答案: