vector<T>::const_iterator first = myVec.begin() + 100000; vector<T>::const_iterator last = myVec.begin() + 101000; vector<T> newVec(first, last); It’s an O(N) operation to construct the new vector, but there isn’t really a better way.
Off the top of my head, no. I think the best you could do is something like this: def loop(f,n): for i in xrange(n): f() loop(lambda: <insert expression here>, 5) But I think you can just live with the extra i variable. Here is the option to use the _ variable, which in reality, is … Read more
What does it mean for the ranges to overlap? It means there exists some number C which is in both ranges, i.e. x1 <= C <= x2 and y1 <= C <= y2 To avoid confusion, considering the ranges are: [x1:x2] and [y1:y2] Now, if we are allowed to assume that the ranges are well-formed … Read more
You can either use: [x / 10.0 for x in range(5, 50, 15)] or use lambda / map: map(lambda x: x/10.0, range(5, 50, 15))
Let’s say you want to scale a range [min,max] to [a,b]. You’re looking for a (continuous) function that satisfies f(min) = a f(max) = b In your case, a would be 1 and b would be 30, but let’s start with something simpler and try to map [min,max] into the range [0,1]. Putting min into … Read more
For each selection, you could serialize the selected range to character offsets and deserialize it again on reload using something like this: Demo: http://jsfiddle.net/WeWy7/3/ Code: var saveSelection, restoreSelection; if (window.getSelection && document.createRange) { saveSelection = function(containerEl) { var range = window.getSelection().getRangeAt(0); var preSelectionRange = range.cloneRange(); preSelectionRange.selectNodeContents(containerEl); preSelectionRange.setEnd(range.startContainer, range.startOffset); var start = preSelectionRange.toString().length; return { start: … Read more
Just define the variable as a variant, and make them equal: Dim DirArray As Variant DirArray = Range(“a1:a5”).Value No need for the Array command.
I had such a query with a calendar application I once wrote. I think I used something like this: … WHERE new_start < existing_end AND new_end > existing_start; UPDATE This should definitely work ((ns, ne, es, ee) = (new_start, new_end, existing_start, existing_end)): ns – ne – es – ee: doesn’t overlap and doesn’t match (because … Read more
Is there a method I can use to convert the whole range in one line? Yes you can convert without looping. Try this Sub Sample() [A1:A20] = [INDEX(UPPER(A1:A20),)] End Sub Alternatively, using a variable range, try this: Sub Sample() Dim rng As Range Set rng = Range(“A1:A20”) rng = Evaluate(“index(upper(” & rng.Address & “),)”) End … Read more
All of the following examples use var str = “Hello, playground” Swift 4 Strings got a pretty big overhaul in Swift 4. When you get some substring from a String now, you get a Substring type back rather than a String. Why is this? Strings are value types in Swift. That means if you use … Read more