From MSDN: Short-Circuiting Trade-Offs Short-circuiting can improve performance by not evaluating an expression that cannot alter the result of the logical operation. However, if that expression performs additional actions, short-circuiting skips those actions. For example, if the expression includes a call to a Function procedure, that procedure is not called if the expression is short-circuited, … Read more
Try this: i = 5 ii = 10 if i == 5 and ii == 10: print “i is 5 and ii is 10” Edit: Oh, and you dont need that semicolon on the last line (edit to remove it from my code).
Think of ! (negation operator) as “not”, || (boolean-or operator) as “or” and && (boolean-and operator) as “and”. See Operators and Operator Precedence. Thus: if(!(a || b)) { // means neither a nor b } However, using De Morgan’s Law, it could be written as: if(!a && !b) { // is not a and is … Read more
In the first two cases, the boolean is cast to a number – 1 for true and 0 for false. In the final case, it is a number that is cast to a boolean and any number except for 0 and NaN will cast to true. So your test cases are really more like this: … Read more
The expression x or y evaluates to x if x is true, or y if x is false. Note that “true” and “false” in the above sentence are talking about “truthiness”, not the fixed values True and False. Something that is “true” makes an if statement succeed; something that’s “false” makes it fail. “false” values … Read more
Here’s another way to think about it. Consider four sets: Blue things, red things, big things, and small things. If you intersect the set of all blue things and all small things, you end up with the union of the properties — everything in the set has both the blue property and the small property. … Read more
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
How about: return (returnValue == “1”); or as suggested below: return (returnValue != “0”); The correct one will depend on what you are looking for as a success result.
If the elements are hashable, you could use sets: {a, b} == {y, x}