To remove the first occurrence of an element, use list.remove: >>> xs = [‘a’, ‘b’, ‘c’, ‘d’] >>> xs.remove(‘b’) >>> print(xs) [‘a’, ‘c’, ‘d’] To remove all occurrences of an element, use a list comprehension: >>> xs = [‘a’, ‘b’, ‘c’, ‘d’, ‘b’, ‘b’, ‘b’, ‘b’] >>> xs = [x for x in xs if … Read more
To get elements which are in temp1 but not in temp2 : In [5]: list(set(temp1) – set(temp2)) Out[5]: [‘Four’, ‘Three’] Beware that it is asymmetric : In [5]: set([1, 2]) – set([2, 3]) Out[5]: set([1]) where you might expect/want it to equal set([1, 3]). If you do want set([1, 3]) as your answer, you can use set([1, … Read more
7 in a Clearest and fastest way to do it. You can also consider using a set, but constructing that set from your list may take more time than faster membership testing will save. The only way to be certain is to benchmark well. (this also depends on what operations you require)
Python 3 for f, b in zip(foo, bar): print(f, b) 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 … Read more
The effects of the three different methods to remove an element from a list: remove removes the first matching value, not a specific index: >>> a = [0, 2, 3, 2] >>> a.remove(2) >>> a [0, 3, 2] del removes the item at a specific index: >>> a = [9, 8, 7, 6] >>> del … Read more
This will convert the dict_keys object to a list: list(newdict.keys()) On the other hand, you should ask yourself whether or not it matters. It is Pythonic to assume duck typing — if it looks like a duck and it quacks like a duck, it is a duck. The dict_keys object can be iterated over just … Read more
In your example, it is because you can’t have a List of a primitive type. In other words, List<int> is not possible. You can, however, have a List<Integer> using the Integer class that wraps the int primitive. Convert your array to a List with the Arrays.asList utility method. Integer[] numbers = new Integer[] { 1, … Read more
To get a new reversed list, apply the reversed function and collect the items into a list: >>> xs = [0, 10, 20, 40] >>> list(reversed(xs)) [40, 20, 10, 0] To iterate backwards through a list: >>> xs = [0, 10, 20, 40] >>> for x in reversed(xs): … print(x) 40 20 10 0
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 … Read more
The easiest way I can think of is to use Linq: List<Order> SortedList = objListOrder.OrderBy(o=>o.OrderDate).ToList();