Sure. An iterator is just an implementation of the java.util.Iterator interface. If you’re using an existing iterable object (say, a LinkedList) from java.util, you’ll need to either subclass it and override its iterator function so that you return your own, or provide a means of wrapping a standard iterator in your special Iterator instance (which has the advantage of being more broadly used), … Read more
They are three separate errors: won’t work because end() returns a an iterator, and object is a pointer. Those are generally distinct types (A vector can use raw pointers as iterators, but not all implementations do it. Other containers require special iterator types). doesn’t work because you take the address of the returned iterator. That … Read more
An Iterable is a simple representation of a series of elements that can be iterated over. It does not have any iteration state such as a “current element”. Instead, it has one method that produces an Iterator. An Iterator is the object with iteration state. It lets you check if it has more elements using hasNext() and move to the next element … Read more
I take it your code is giving you the error message, TypeError: zip argument #1 must support iteration. You are getting this error due to the expression zip(float(price), Bids). This simplified code demonstrates the error: The Python 2.x zip() built-in library function requires all its arguments to be iterables. float(price) is not an iterable. If … Read more
I’ve never heard of a best practice for this specific question. However, one best practice in general is to use the simplest solution that solves the problem. In this case the array-style access (or c-style if you want to call it that) is the simplest way to iterate while having the index value available. So … Read more
const_iterators don’t allow you to change the values that they point to, regular iterators do. As with all things in C++, always prefer const, unless there’s a good reason to use regular iterators (i.e. you want to use the fact that they’re not const to change the pointed-to value).
Python 3 zip stops when the shorter of foo or bar stops. In Python 3, zip returns an iterator of tuples, like itertools.izip in Python2. To get a list of tuples, use list(zip(foo, bar)). And to zip until both iterators are exhausted, you would use itertools.zip_longest. Python 2 In Python 2, zip returns a list of tuples. This is fine when foo and bar are not massive. If they are both massive then forming zip(foo,bar) is … Read more
If you have access to C++11 you can use range-based for loops Otherwise you should use begin() and end() You can also use std::begin and std::end (these require C++11 as well) begin will return an iterator to the first element in your vector. end will return an iterator to one element past the end of your vector. So the order in which you get the elements iterating … Read more
numpy.linspace() gives you a one-dimensional NumPy array. For example: Therefore: cannot work. You would need some kind of two-dimensional array with two elements in the second dimension: Now you can do this:
The index can only be used for containers that support random access – direct access to a given position. The iterator offers a unified way to access any collection/data structure. The flexibility when refactoring your code is immense.