range

What is the best possible way to reduce an array of ranges in javascript. For example I have 1-3,4-5,10-12,2-4 the result I need for this is 1-5, 10-12 What is the best way to tackle this problem ? I would first create another array with no duplicates, storing the numbers that are covered by the ranges: 1-3 covers 1, 2, 3 --> [1, 2, 3] 4-5 covers 4, 5 --> [1, 2, 3, 4, 5] 10-12 covers 10, 11, 12 --> [1, 2, 3, 4, 5, 10, 11, 12] 2-4 covers 2, 3, 4 --> [1, 2, 3, 4, 5, 10, 11, 12] Then, sort the array: [1, 2, 3, 4, 5, 10, 11, 12] // nothing changed in this example Finally, rebuild the ranges,

What is the meaning of 'infinity' in PostgreSQL range types? Is there any difference between specifying infinity or -infinity as a bound, or NULL ? I.e. is infinity an explicit form of specifying that the range bound is infinite, whereas NULL would implicit specify an infinite bound range? See the following examples: SELECT tstzrange('-infinity','infinity') && tstzrange(NULL, NULL); ?column? ---------- t SELECT tstzrange('2013-01-01 00:00:00+01', '2013-02-01 00:00:00+01') && tstzrange(NULL, '2013-03-01 00:00:00+01'); ?column? ---------- t SELECT tstzrange('2013-01-01 00:00:00+01', '2013-02-01

Thanks to this question and answer posted by Tim Down, I made a function to get the word preceding caret in a "contenteditable" div. Here's a fiddle , and here's the function: function getWordPrecedingCaret (containerEl) { var preceding = "", sel, range, precedingRange; if (window.getSelection) { sel = window.getSelection(); if (sel.rangeCount > 0) { range = sel.getRangeAt(0).cloneRange(); range.collapse(true); range.setStart(containerEl, 0); preceding = range.toString(); } } else if ((sel = document.selection) && sel.type != "Control") { range = sel.createRange(); precedingRange = range

I am trying to check if a date of format mm.dd.yyyy is greater than today and less than the date after 6 months from today. Here is my code: var isLinkExpiryDateWithinRange = function(value) { var monthfield = value.split('.')[0]; var dayfield = value.split('.')[1]; var yearfield = value.split('.')[2]; var inputDate = new Date(yearfield, monthfield - 1, dayfield); var today = new Date(); today = new Date(today.getFullYear(), today.getMonth(), today.getDate()); alert(inputDate > today);//alert-> true var endDate = today; endDate.setMonth(endDate.getMonth() + 6); alert(inputDate > today);//alert

In Awk, the range pattern is not an expression, so canot use the "!" to not it. so how to implement it (Printing all contents EXCEPT matching range pattern using awk)? e.g. $cat 1.t abd hfdh # fafa deafa 123 # end the result I wanted: cat 1.t abd hfdh end EDIT: I gave an impertinent example. the endpattern should be different with the startpattern because I just have not test this. That's My fault. At the same time, I want to operate the range pattern and the not range pattern differently. So sed is not my choice. you just gave a tricky (I don't know I should call it good or bad ^_^ ) example.

I'm new to the Scala language. I need Range for Long type. I need a List of [1, 2, 3 ... 10000000] with step 1. If I use until/to I get an error because of using Long instead of Int. I try to write simple function which expects a start, an end and and an empty List and generates a List of [start .. end]. Here is my function: def range_l(start : Long, end : Long, list : List[Long]) : List[Long] = { if (start == end){ val add_to_list = start :: list return add_to_list } else { val add_to_list = start :: list range_l(start + 1, end, add_to_list) } } If I call it like: range_l(1L, 1000000L, List()

This question already has an answer here: Python 3 turn range to a list 8 answers Having a beginner issue with Python range. I am trying to generate a list, but when I enter: def RangeTest(n): # list = range(n) return list print(RangeTest(4)) what is printing is range(0,4) rather than [0,1,2,3] What am I missing? Thanks in advance! TerryA You're using Python 3, where range() returns an "immutable sequence type" instead of a list object (Python 2). You'll want to do: def RangeTest(n): return list(range(n)) If you're used to Python 2, then range() is equivalent to xrange() in Python 2. By the