immutability

This question already has an answer here: Why does list.append evaluate to false in a boolean context? [duplicate] 7 answers I am trying to understand how extend works in Python and it is not quite doing what I would expect. For instance: >>> a = [1, 2, 3] >>> b = [4, 5, 6].extend(a) >>> b >>> But I would have expected: [4, 5, 6, 1, 2, 3] Why is that returning a None instead of extending the list? The extend() method appends to the existing array and returns None . In your case, you are creating an array — [4, 5, 6] — on the fly, extending it and then discarding it. The variable b ends up with

I'm using quite a few immutable collections and I'm curious how to deserialize them using Gson. As nobody answered and I've found the solution myself, I'm simplifying the question and presenting my own answer. I had two problems: How to write a single Deserializer working for all ImmutableList<XXX> ? How to register it for all ImmutableList<XXX> ? Update: There's https://github.com/acebaggins/gson-serializers which covers many guava collections: ImmutableList ImmutableSet ImmutableSortedSet ImmutableMap ImmutableSortedMap How to write a single Deserializer working for all ImmutableList? The

I came across a problem that required iterating over an array in pairs. What's the best way to do this? Or, as an alternative, what's the best way of transforming an Array into an Array of pairs (which could then be iterated normally)? Here's the best I got. It requires output to be a var , and it's not really pretty. Is there a better way? let input = [1, 2, 3, 4, 5, 6] var output = [(Int, Int)]() for i in stride(from: 0, to: input.count - 1, by: 2) { output.append((input[i], input[i+1])) } print(output) // [(1, 2), (3, 4), (5, 6)] // let desiredOutput = [(1, 2), (3, 4), (5, 6)] // print

Does the const keyword in JavaScript create an immutable reference to immutable data structures? [I'm assuming that immutable data structures exist in JavaScript.] For string it appears to do so: var x = "asdf"; const constantX = x; alert("before mutation: " + constantX); x = "mutated" alert("after mutation: " + constantX); output: before mutation: asdf after mutation: asdf http://jsfiddle.net/hVJ2a/ Yes, it does create an immutable reference, but it is not been standardized and is not supported in all browsers. See this MDN article on the const keyword for details and compatability. However,

I know all the basic rules to make our class immutable but I am a little confused when there is another class reference. I know if there is collection instead of Address then we can make use of Collections.unmodifiableList(new ArrayList<>(modifiable)); and then we can make our class immutable. But in below case I am still unable to get the concept. public final class Employee{ private final int id; private Address address; public Employee(int id, Address address) { this.id = id; this.address=address; } public int getId(){ return id; } public Address getAddress(){ return address; } } public

I have a dataframe with Multiindex and would like to modify one particular level of the Multiindex. For instance, the first level might be strings and I may want to remove the white spaces from that index level: df.index.levels[1] = [x.replace(' ', '') for x in df.index.levels[1]] However, the code above results in an error: TypeError: 'FrozenList' does not support mutable operations. I know I can reset_index and modify the column and then re-create the Multiindex, but I wonder whether there is a more elegant way to modify one particular level of the Multiindex directly. Thanks to @cxrodgers's