equality

I'm learning python on codecademy and my current task is this: Write a function, shut_down, that takes one parameter (you can use anything you like; in this case, we'd use s for string). The shut_down function should return "Shutting down..." when it gets "Yes" , "yes" , or "YES" as an argument, and "Shutdown aborted!" when it gets "No" , "no" , or "NO" . If it gets anything other than those inputs, the function should return "Sorry, I didn't understand you." Seemed easy to me but somehow I still can't do it. My code I made to test the function: def shut_down(s): if s == "Yes" or s == "yes" or

I saw a couple of questions here on SO, with ansers including the function isEqual: instead of the standard == . So far, I have only learned to use the == , so I'm wondering what's better to use, what are the pros and cons of each? When should you use them? Thank you. They do different things; so you need to use the appropriate one: Consider, if you will: NSString *a = @"Hello!"; NSString *b = a; NSString *c = [a mutableCopy]; if (a == b) NSLog(@"This prints"); if (b == c) NSLog(@"This doesn't"); if ([a isEqual:c]) NSLog(@"This does"); In other words; == merely checks if two pointers point to

I noticed I can use the == operator to compare all the native data types (integers, strings, booleans, floating point numbers etc) and also lists, tuples, sets and dictionaries which contain native data types. In these cases the == operator checks if two objects are equal. But in some other cases (trying to compare instances of classes I created) the == operator just checks if the two variables reference the same object (so in these cases the == operator is equivalent to the is operator) My question is: When does the == operator do more than just comparing identities? EDIT: I'm using Python 3

This question already has an answer here: Javascript: operator overloading 4 answers Overloading Arithmetic Operators in JavaScript? 11 answers Is it possible to define custom operators between instances of a type in JavaScript? For example, given that I have a custom vector class, is it possible to use vect1 == vect2 to check for equality, whilst the underlying code would be something like this? operator ==(a, b) { return a.x == b.x && a.y == b.y && a.z == b.z; } (This is nonsense of course.) I agree that the equal function on the vector prototype is the best solution. Note that you can also

Assuming dictionary keys and values have their equals and hash methods implemented correctly, what is the most succinct and efficient way to test for equality of two dictionaries? In this context two dictionaries are said to be equal if they contain the same set of keys (order not important), and for every such key, they agree on the value. here are some ways i came up with (there are probably many more): public bool Compare1<TKey, TValue>( Dictionary<TKey, TValue> dic1, Dictionary<TKey,TValue> dic2) { return dic1.OrderBy(x => x.Key). SequenceEqual(dic2.OrderBy(x => x.Key)); } public bool

A few weeks ago, I have read this thread Is < faster than <=? about comparison operators in C . It was said that there is no difference in the performance between < and <= as they are interpreted as same/similar machine commands. At the same time, in our company's "best practices", it was said that we should always use "===" to compare things instead of "==". So, I started to wonder if this is always appropriate as I am used to using the "==" and "typeof ... == " and do not want to change my way of writing :-] Note that this is in the context of JavaScript. So, I have a little research and

Consider this code: >>> x = "google" >>> x is "google" True >>> x = "google.com" >>> x is "google.com" False >>> Why is it like that? To make sure the above is correct, I have just tested on Python 2.5.4, 2.6.5, 2.7b2, Python 3.1 on windows and Python 2.7b1 on Linux. It looks like there is consistency across all of them, so it's by design. Am I missing something? I just find it out that from some of my personal domain filtering script failing with that. is verifies object identity, and any implementation of Python, when it meets literal of immutable types, is perfectly free to either make a