boolean-logic

问题 How to create a logical OR with logical AND, and logical NOT? 回答1: Check De Morgans's laws. You are looking for the Substitution form . P OR Q = NOT( (NOT P) AND (NOT Q) ) 回答2: It's De Morgan's Law: A OR B = NOT ( NOT A AND NOT B ) Truth table for A OR B: A B X 0 0 0 0 1 1 1 0 1 1 1 1 Truth table for the De Morgan equivalent: A B !A !B (!A AND !B) !(!A AND !B) 0 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 0 1 1 1 0 0 0 1 回答3: Like not (not x and not y) ? 回答4: Pretty simple: A || B = !(!A && !B) 回答5: Using

问题 Is one more readable than the other? At first, I disliked the false === approach but as I see it more and more often, I'm warming up to it. I'm pretty sure they return identical results. 回答1: I greatly prefer false === $var Namely because sometimes you are only using equality and not looking for identity. In which case you write false == $var But sometimes you aren't at the top of your game, and might write false = $var which will give an immediate error, and let's you fix it right away.

问题 I just want to flip a boolean based on what it already is. If it's true - make it false. If it's false - make it true. Here is my code excerpt: switch(wParam) { case VK_F11: if (flipVal == true) { flipVal = false; } else { flipVal = true; } break; case VK_F12: if (otherVal == true) { otherValVal = false; } else { otherVal = true; } break; default: break; } 回答1: You can flip a value like so: myVal = !myVal; so your code would shorten down to: switch(wParam) { case VK_F11: flipVal = !flipVal;

问题 I am filtering rows in a dataframe by values in two columns. For some reason the OR operator behaves like I would expect AND operator to behave and vice versa. My test code: import pandas as pd df = pd.DataFrame({'a': range(5), 'b': range(5) }) # let's insert some -1 values df['a'][1] = -1 df['b'][1] = -1 df['a'][3] = -1 df['b'][4] = -1 df1 = df[(df.a != -1) & (df.b != -1)] df2 = df[(df.a != -1) | (df.b != -1)] print pd.concat([df, df1, df2], axis=1, keys = [ 'original df', 'using AND (&)',

问题 I was playing around in jsfiddle.net and I'm curious as to why this returns true? if(0 < 5 < 3) { alert("True"); } So does this: if(0 < 5 < 2) { alert("True"); } But this doesn't: if(0 < 5 < 1) { alert("True"); } Is this quirk ever useful? 回答1: Order of operations causes (0 < 5 < 3) to be interpreted in javascript as ((0 < 5) < 3) which produces (true < 3) and true is counted as 1, causing it to return true. This is also why (0 < 5 < 1) returns false, (0 < 5) returns true, which is

问题 I know the rules for && and || but what are & and | ? Please explain these to me with an example. 回答1: Those are the bitwise AND and bitwise OR operators. int a = 6; // 110 int b = 4; // 100 // Bitwise AND int c = a & b; // 110 // & 100 // ----- // 100 // Bitwise OR int d = a | b; // 110 // | 100 // ----- // 110 System.out.println(c); // 4 System.out.println(d); // 6 Thanks to Carlos for pointing out the appropriate section in the Java Language Spec (15.22.1, 15.22.2) regarding the different

问题 i have three boolean value returning from the method , i want to check condition like this :First It will check all the three boolean value Scenario 1: if locationMatch, matchCapacity, filterMatchStatus then statement return true value. Scenario 2: if locationMatch, matchCapacity, filterMatchStatus if any boolean is false then it return false value I tried like this but , it is returning true if any boolean value is true public boolean matchFilter(FilterTruck filter){ boolean locationMatch =

问题 I do not fully understand the difference between a bitwise operation and a logical operation. Please see the bitwise operation below: x=7 and 15 I understand that x will equal 7 in this case after inspecting each bit individually. How does boolean logic work at a lower level. I believe that a Boolean is a 32 Bit data type (I could be wrong). Are boolean literals (TRUE and FALSE) treated as single bits? 回答1: This is generally how these things work. Bitwise operations perform Boolean operations

问题 I have been trying to get this to work for a while now, but still not finding a way. I am trying to compute the Look ahead estimate density of a piecewise gaussian function. I'm trying to estimate the stationary distribution of a piecewise normally distributed function. is there a way to avoid the error type: Error-type: the truth value of an array with more than one element is ambiguous. Use a.any() or a.all(). for instance y=np.linspace(-200.0,200.0,100) and x = np,linspace(-200.0,200.0,100

问题 I have a certain string k that may or many not a key of a dict reversedmapping , and another dict spendDict where it may also be a key. Why does the following check succeed (i.e. run to completion), for a value of k that is not a key of reversedmapping : if (k not in reversedmapping) or (reversedmapping[k] not in spendDict) whereas I get a KeyError (for k) when I change it to logical AND: if (k not in reversedmapping) and (reversedmapping[k] not in spendDict) : And how do I rewrite the AND