Iterate pairwise through a ruby array [duplicate]
You are looking for each_cons: (1..6).each_cons(2) { |a, b| p a: a, b: b } # {:a=>1, :b=>2} # {:a=>2, :b=>3} # {:a=>3, :b=>4} # {:a=>4, :b=>5} # {:a=>5, :b=>6}
You are looking for each_cons: (1..6).each_cons(2) { |a, b| p a: a, b: b } # {:a=>1, :b=>2} # {:a=>2, :b=>3} # {:a=>3, :b=>4} # {:a=>4, :b=>5} # {:a=>5, :b=>6}
If you have GNU awk, you can use a true multidimensional array. Although this answer uses the split() function, it most certainly doesn’t abuse it. Run like: awk -f script.awk Contents of script.awk: BEGIN { x=SUBSEP a=”Red” x “Green” x “Blue” b=”Yellow” x “Cyan” x “Purple” Colors[1][0] = “” Colors[2][0] = “” split(a, Colors[1], x) … Read more
You can use the += operator to append to an array. args=() for i in “$@”; do args+=(“$i”) done echo “${args[@]}” This shows how appending can be done, but the easiest way to get your desired results is: echo “$@” or args=(“$@”) echo “${args[@]}” If you want to keep your existing method, you need to … Read more
This is working fine for me ([email protected]) var schema = new mongoose.Schema({ factors: [{type: String, enum: [‘1’, ‘2’, ‘3’], required: …}] … }) Note I’m using an Array of Objects
Use join, e.g., tripIds.join(“, “) Tangential If you want to create a list from another list you generally want something like map or collect as opposed to manually creating a list and appending to it, e.g. (untested): def sql = Sql.newInstance(“jdbc:mysql://localhost:3306/steer”, “root”, “”, “com.mysql.jdbc.Driver”) def tripIds = sql.map { it.id } Or if you only … Read more
You have an array of items that implement a protocol. If you don’t tell Swift that this is an AnyObject protocol (or class protocol in earlier versions of Swift), it will assume that it can be implemented by a struct. In a forEach loop, you will essentially have a let variable (called $0 by default) … Read more
What is happening here is, if you have an array of n elements, then the segment tree will have a leaf node for each of these n entries. Thus, we have (n) leaf nodes, and also (n-1) internal nodes. Total number of nodes= n + (n-1) = 2n-1 Now, we know its a full binary … Read more
There is a beautiful algorithm for solving this that works in two passes (total time O(N)) using only constant external space (O(1)). I have an implementation of this algorithm, along with comments including a correctness proof, available here The intuition behind the algorithm is actually quite beautiful. Suppose that you were to have a roomful … Read more
This answer is for the question with no context. I’m adding it because of search results. [System.Byte[]]::CreateInstance([System.Byte],<Length>)
An alternative option is to use Powershell’s ability to assign multiple variables (see this other answer). $arr = 1..5 $first, $rest= $arr $rest 2 3 4 5 It’s been a feature of Powershell for over a decade. I found this functionality from an MSDN blog post: