I think this is what you’re looking for. List<String> columnData = new List<String>(); using(SqlConnection connection = new SqlConnection(“conn_string”)) { connection.Open(); string query = “SELECT Column1 FROM Table1”; using(SqlCommand command = new SqlCommand(query, connection)) { using (SqlDataReader reader = command.ExecuteReader()) { while (reader.Read()) { columnData.Add(reader.GetString(0)); } } } } Not tested, but this should work fine.
Use lists unless you want some very specific features that are in the C array libraries. python really has three primitive data structures tuple = (‘a’,’b’,’c’) list = [‘a’,’b’,’c’] dict = {‘a’:1, ‘b’: true, ‘c’: “name”} list.append(‘d’) #will add ‘d’ to the list list[0] #will get the first item ‘a’ list.insert(i, x) # Insert an … Read more
Elm added arrays in 0.12.1, and the implementation was massively overhauled in 0.19 to improve correctness and performance. import Array myArray = Array.fromList [1..5] myItem = Array.get 2 myArray Arrays are zero-indexed. Negative indices are not supported currently (bummer, I know). Note that myItem : Maybe Int. Elm does everything it can to avoid runtime … Read more
Here’s another way to do it if you don’t want to change the order of the items: mylist = [1, 4, 0, 3, 2] mylist.remove(max(mylist)) mylist.remove(min(mylist)) Assumes that the high/low don’t have any duplicates in the list, or if there are, that it’s OK to remove only one of them. This will need to do … Read more
As others have said, you are looking for the .AsReadOnly() extension method. However, you should store a reference to the collection instead of creating it during each property access: private readonly List<SomeOtherClass> _items; public WhatClass() { _items = new List<SomeOtherClass>(); this.Items = _items.AsReadOnly(); } public ReadOnlyCollection<SomeOtherClass> Items { get; private set; } This is to … Read more
This can be done with itertools.chain: import itertools l1 = [1, 2, 3, 4] l2 = [5, 6, 7, 8] for i in itertools.chain(l1, l2): print(i, end=” “) Which will print: 1 2 3 4 5 6 7 8 As per the documentation, chain does the following: Make an iterator that returns elements from the … Read more
OK, it’s time for test results of proposed approaches. Here what approaches I have tested (name of each approach is also class name in my sources): NaiveRemoveManyPerformer – ArrayList with iterator and remove – first and naive implementation given in my question. BetterNaiveRemoveManyPerformer – ArrayList with backward iteration and removal from end to front. LinkedRemoveManyPerformer … Read more
Use zip with * and map: >>> map(list, zip(*a)) [[‘AAA’, ‘BBB’, ‘CCC’, ‘DDD’, ‘EEE’, ‘FFF’, ‘GGG’, ‘HHH’, ‘III’], [‘1’, ‘262’, ’86’, ’48’, ‘8’, ’39’, ‘170’, ’16’, ‘4’], [‘1′, ’56’, ’84’, ‘362’, ’33’, ’82’, ‘296’, ’40’, ‘3’], [’10’, ‘238’, ‘149’, ‘205’, ’96’, ’89’, ‘223’, ’65’, ‘5’], [’92’, ‘142’, ’30’, ‘237’, ‘336’, ‘140’, ‘210’, ’50’, ‘2’]] Note … Read more
Filtering an array is tricky if you consider possibility of elements containing spaces (not to mention even “weirder” characters). In particular answers given so far (referring to various forms of ${x[@]//pref*/}) will fail with such arrays. I have investigated this issue somewhat and found a solution however it is not a nice one-liner. But at … Read more
Use itertools.permutations: >>> import itertools >>> stuff = [1, 2, 3] >>> for L in range(0, len(stuff)+1): for subset in itertools.permutations(stuff, L): print(subset) … () (1,) (2,) (3,) (1, 2) (1, 3) (2, 1) (2, 3) (3, 1) …. Help on itertools.permutations: permutations(iterable[, r]) –> permutations object Return successive r-length permutations of elements in the … Read more