Use pos = my_list.index(‘asd’) nearest = my_list[pos + 1] Note pos is 3 for the 4th element as Python is 0- based. Note avoid using list for variables as this name has a special meaning in Python. 1 solved How…
Finding the nearest element in a list can be a tricky task. Fortunately, there are a few methods that can be used to make the process easier. In this article, we will discuss how to find the nearest element in…
you can use an expression bodied method. static double Stirling(long n) => Math.Sqrt((2 * n + 1.0 / 3.0) * Math.PI) * Math.Pow(n, n) / Math.Exp(n); This is only possible when the method executes a single statement, and in this…
Thanks for all responses The Lambda expression query is correct only. db.Companies.where(Company => (Compare(Convert(Company.Name), “Test”) > 0)) db.Companies.where(Company => (Compare(Convert(Company.Name), “Test”) >= 0)) I changed the “right arg” value to “Test”. I have one record ‘Name’ with Test. executed the…
I think the reason is pretty clear. ints.forEach((i) -> { System.out.print(ints.get(i-1) + ” “); }); Translates approximately to: for (Integer i : ints) { System.out.println(ints.get(i – 1) + ” “); } Which will cause IndexOutOfBoundsExceptions because i refers to the…
[&](int n) {} means that in the lambda block you capture every variable from the scope by reference in contrast for example to [=](int n) {} where you have an access by value. You could also specify excactly what variable…
You would use an Anonymous Inner Class, as Java 8 lambda expressions are essentially syntatical sugar which do nearly the same thing. That would look something like this. files.addAll(Arrays.asList(folder.listFiles(new FileFilter(){ @Override public boolean accept(File f) { return f.getName().endsWith(CustomConstantsRepository.FILE_EXT_DAT) && f.getName().startsWith(fileName)));…
Resolved. The method was called recursively, creating the parameter and the expression and returned to himself. In this process the parameters were removed from memory, as they had already been used believed not to have problems. But they need to…
Does the String in Map<String, List<A>> related to the output? I assume that you want to get every pair <F, E> from the original map right? So this might help Map<String, List<A>> input; input.values().stream() .flatMap(Collection::stream) .map(a -> a.getListB()) // extract…
I believe this does what you want… arr = [[2,6,8],[1,4,7],[3,3,4],[2,4,9],[3,3,7]] foo = lambda arr, evenOdd, noDoubles: [ i for i in arr if not (evenOdd and (all(k % 2 == 1 for k in i) or all(k%2 == 0 for…
I have updated the function… check now? navigator.geolocation.getCurrentPosition(function(position){ var long = position.coords.longitude; var lat = position.coords.latitude; var theDateC = new Date(); var D = (367*theDateC.getFullYear())-(parseInt((7/4)*(theDateC.getFullYear+parseInt((theDateC.getMonth()+9)/12))))+parseInt(275*(theDateC.getMonth()/9))+theDateC.getDate()-730531.5; var L = 280.461+0.9856474*D; var M = 357.528+0.9856003*D; var Lambda = L +1.915*Math.sin(M)+0.02*Math.sin(2*M); var Obliquity…
If I understand the question, you want a method that takes any number of arguments, and returns an object that will create a range of numbers when the [] method is used. This method takes any number of arguments (using…
@StevenRumbalski has given the right answer i.e., output = filter(lambda k: 5 <= dict1[k] <= 8, dict1) solved Use filter method and find the keys with values between 5 to 8 from the given dictionary
Since you have a List<object> that actually contains List<List<Player>> and you want to get a list of players in one big list. Unfortunately for some external reason you are unable to change the API to give you sensibly typed data,…
The comparator seems correct. The problem seems to be in your filter clause, where you compare the event id to the device id lastDeviceEvent = deviceEvents .stream() .filter (o -> o.getDeviceId().equals(deviceId)) // Original code used getId() .sorted(comparing((DeviceEvent de) -> de.getId()).reversed())…