zip is its own inverse! Provided you use the special * operator. >>> zip(*[(‘a’, 1), (‘b’, 2), (‘c’, 3), (‘d’, 4)]) [(‘a’, ‘b’, ‘c’, ‘d’), (1, 2, 3, 4)] The way this works is by calling zip with the arguments: zip((‘a’, 1), (‘b’, 2), (‘c’, 3), (‘d’, 4)) … except the arguments are passed to … Read more
You can use the List.addAll() method. It accepts a Collection as an argument, and your set is a Collection. List<String> mainList = new ArrayList<String>(); mainList.addAll(set); EDIT: as respond to the edit of the question. It is easy to see that if you want to have a Map with Lists as values, in order to have … Read more
Quite a few problems with your code: On Arrays.asList returning a fixed-size list From the API: Arrays.asList: Returns a fixed-size list backed by the specified array. You can’t add to it; you can’t remove from it. You can’t structurally modify the List. Fix Create a LinkedList, which supports faster remove. List<String> list = new LinkedList<String>(Arrays.asList(split)); … Read more
Use: if my_item in some_list: … Also, inverse operation: if my_item not in some_list: … It works fine for lists, tuples, sets and dicts (check keys). Note that this is an O(n) operation in lists and tuples, but an O(1) operation in sets and dicts.
If you’re using .Net 3+, you can use Linq. List<T> withDupes = LoadSomeData(); List<T> noDupes = withDupes.Distinct().ToList();
variable = [] Now variable refers to an empty list*. Of course this is an assignment, not a declaration. There’s no way to say in Python “this variable should never refer to anything other than a list”, since Python is dynamically typed. *The default built-in Python type is called a list, not an array. It … Read more
>>> import string >>> string.ascii_lowercase ‘abcdefghijklmnopqrstuvwxyz’ >>> list(string.ascii_lowercase) [‘a’, ‘b’, ‘c’, ‘d’, ‘e’, ‘f’, ‘g’, ‘h’, ‘i’, ‘j’, ‘k’, ‘l’, ‘m’, ‘n’, ‘o’, ‘p’, ‘q’, ‘r’, ‘s’, ‘t’, ‘u’, ‘v’, ‘w’, ‘x’, ‘y’, ‘z’] Alternatively, using range: >>> list(map(chr, range(97, 123))) [‘a’, ‘b’, ‘c’, ‘d’, ‘e’, ‘f’, ‘g’, ‘h’, ‘i’, ‘j’, ‘k’, ‘l’, ‘m’, ‘n’, … Read more
Use the list constructor: >>> list(“foobar”) [‘f’, ‘o’, ‘o’, ‘b’, ‘a’, ‘r’] list builds a new list using items obtained by iterating over the input iterable. A string is an iterable — iterating over it yields a single character at each iteration step.
With rbind do.call(rbind.data.frame, your_list) Edit: Previous version return data.frame of list‘s instead of vectors (as @IanSudbery pointed out in comments).
Do this: list(map(chr,[66,53,0,94])) In Python 3+, many processes that iterate over iterables return iterators themselves. In most cases, this ends up saving memory, and should make things go faster. If all you’re going to do is iterate over this list eventually, there’s no need to even convert it to a list, because you can still … Read more