I think that most people consider something like isPrimaryColor = [“Red”, “Blue”, “Yellow”].contains(someColor) to be sufficiently clear that they don’t need extra syntax for this.
It really depends on what you’re trying to accomplish. If you have no else clause then if(!doSomething()) seems fine. However, if you have if(!doSomething()) { … } else { // do something else } I’d probably reverse that logic to remove the ! operator and make the if clause slightly more clear.
I believe the explanation has to do with how older programming languages (WordBasic and early VBA) stored the integer values of True and False. In those days, True = -1 and False = 0. Newer programming languages still use 0 for False, but 1 for True. The majority of Word’s Boolean type properties continue to … Read more
XNOR is simply equality on booleans; use A == B. This is an easy thing to miss, since equality isn’t commonly applied to booleans. And there are languages where it won’t necessarily work. For example, in C, any non-zero scalar value is treated as true, so two “true” values can be unequal. But the question … Read more
Yes there is a better way! Just use the & element-wise logical and operator: d.bar & d.foo 0 True 1 False 2 False dtype: bool
Try this: mask = (foo < 40) | (foo > 60) Note: the __or__ method in an object overloads the bitwise or operator (|), not the Boolean or operator.
Use the ~ operator: DECLARE @MyBoolean bit SET @MyBoolean = 0 SET @MyBoolean = ~@MyBoolean SELECT @MyBoolean
short answer: YES Ref: http://docs.scipy.org/doc/numpy/reference/generated/numpy.invert.html Notice: Computes the bit-wise NOT of the underlying binary representation of the integers in the input arrays. This ufunc implements the C/Python operator ~. and bitwise_not is an alias for invert: >> np.bitwise_not is np.invert >> True
The corresponding operator is |: df[(df < 3) | (df == 5)] would elementwise check if value is less than 3 or equal to 5. If you need a function to do this, we have np.logical_or. For two conditions, you can use df[np.logical_or(df<3, df==5)] Or, for multiple conditions use the logical_or.reduce, df[np.logical_or.reduce([df<3, df==5])] Since the … Read more