If you’re already on EL 3.0 (Tomcat 8+, WildFly 8+, GlassFish 4+, Payara 4+, TomEE 7+, etc), which supports new operations on collection objects, you can use ${[…]} syntax to construct a list, and ${{…}} syntax to construct a set. <c:set var=”alphabet” value=”${[‘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’]}” scope=”application” /> If you’re not on EL 3.0 yet, use the ${fn:split()} … Read more
Dynamic programming? Given an array A[0..n], let M(i) be the optimal solution using the elements with indices 0..i. Then M(-1) = 0 (used in the recurrence), M(0) = A[0], and M(i) = max(M(i – 1), M(i – 2) + A[i]) for i = 1, …, n. M(n) is the solution we want. This is O(n). … Read more
Xcode 11 • Swift 5.1 extension Collection { func choose(_ n: Int) -> ArraySlice<Element> { shuffled().prefix(n) } } Playground testing var alphabet = [“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”] let shuffledAlphabet = alphabet.shuffled() // “O”, “X”, “L”, “D”, “N”, “K”, “R”, “E”, “S”, “Z”, “I”, “T”, “H”, “C”, “U”, “B”, “W”, “M”, “Q”, “Y”, “V”, “A”, “G”, “P”, “F”, “J”] … Read more
Here is a possible implementation in Swift using binary search (from http://rosettacode.org/wiki/Binary_search#Swift with slight modifications): extension Array { func insertionIndexOf(_ elem: Element, isOrderedBefore: (Element, Element) -> Bool) -> Int { var lo = 0 var hi = self.count – 1 while lo <= hi { let mid = (lo + hi)/2 if isOrderedBefore(self[mid], elem) { … Read more
You need to use cell arrays. If the number of iterations are known beforehand, I suggest you preallocate: N = 10; names = cell(1,N); for i=1:N names{i} = ‘string’; end otherwise you can do something like: names = {}; for i=1:10 names{end+1} = ‘string’; end
Here’s my favorite implementation of binary search. It’s useful not only for finding the element but also for finding the insertion index. Details about assumed sorting order (ascending or descending) and behavior with respect to equal elements are controlled by providing a corresponding predicate (e.g. { $0 < x } vs { $0 > x … Read more
Additional observations: x(1.2:3) should theoretically be interpreted as: subsref(x, substruct(‘()’,1.2:3)). However, as mentioned in the question, “No rounding takes place when the indexing array is a standard array“, which causes the explicit subscripted reference to fail. This suggests that a mechanism similar to logical short-circuiting or perhaps multithreaded partitioning (where intermediate variable are “not really … Read more
You can use the filter function to leave only those keywords contained in content: val match = keywords.filter { it in content } Here match is a List<String>. If you want to get an array in the result, you can add .toTypedArray() call. in operator in the expression it in content is the same as … Read more
The simplest thing would be: update table set array = array_append(array, ‘element’) where … or perhaps use the || operator: update table set array = array || ‘element’ where … Both of those are equivalent to the more common set n = n + 11 for numbers. Depending on the types involved, you might need … Read more
Sort within the same array variable Sort functions bellow are exactly the same, the only difference how short and expressive they are: Full declaration: myArr.sort { (lhs: EntryStruct, rhs: EntryStruct) -> Bool in // you can have additional code here return lhs.deadline < rhs.deadline } Shortened closure declaration: myArr.sort { (lhs:EntryStruct, rhs:EntryStruct) in return lhs.deadline … Read more