This is a classic mistake. i in your case is already an element from array (i.e. another list), not an index of array (not an int), so You can check the Python tutorial. Also, try doing this: And see what you get! Also I would advise to stick to naming conventions: variables are normally lower-case (volume, not Volume). In this case i is misleading. row or elem would be much more suitable.
The common approach to get a unique collection of items is to use a set. Sets are unordered collections of distinct objects. To create a set from any iterable, you can simply pass it to the built-in set() function. If you later need a real list again, you can similarly pass the set to the list() function. The following example should cover whatever you … Read more
Either: or: Note that this works only for arrays of reference types. For arrays of primitive types, use the traditional way: Update: It is recommended now to use list.toArray(new Foo[0]);, not list.toArray(new Foo[list.size()]);. From JetBrains Intellij Idea inspection: There are two styles to convert a collection to an array: either using a pre-sized array (like … Read more
Given a list of lists t, which means: is faster than the shortcuts posted so far. (t is the list to flatten.) Here is the corresponding function: As evidence, you can use the timeit module in the standard library: Explanation: the shortcuts based on + (including the implied use in sum) are, of necessity, O(T**2) when there are T sublists — as the intermediate … Read more
Using an additional state variable, such as an index variable (which you would normally use in languages such as C or PHP), is considered non-pythonic. The better option is to use the built-in function enumerate(), available in both Python 2 and 3:
Reference: Data Structures > More on Lists Caveats follow Note that while this is perhaps the cleanest way to answer the question as asked, index is a rather weak component of the list API, and I can’t remember the last time I used it in anger. It’s been pointed out to me in the comments that because this answer is heavily … Read more
First, let’s see what this does:Arrays.asList(ia) It takes an array ia and creates a wrapper that implements List<Integer>, which makes the original array available as a list. Nothing is copied and all, only a single wrapper object is created. Operations on the list wrapper are propagated to the original array. This means that if you shuffle the list … Read more
Reference: Data Structures > More on Lists Caveats follow Note that while this is perhaps the cleanest way to answer the question as asked, index is a rather weak component of the list API, and I can’t remember the last time I used it in anger. It’s been pointed out to me in the comments that because this answer is heavily … Read more
Use an Iterator and call remove():
As indicated by the other answers, the error is to due to k = list[0:j], where your key is converted to a list. One thing you could try is reworking your code to take advantage of the split function: Note that if you are using Python 3.x, you’ll have to make a minor adjustment to … Read more