[ad_1] You could use a simple list comprehension here : S = [x[0] for x in content_E.most_common(3)] or, if you need to generate from E, then: S = [x[0] for x in E] 0 [ad_2] solved get most common keys from list
[ad_1] What if x is something besides 0-9? You don’t have a return statement for that case. Add it at the bottom, below your last else if: return “Other”; 1 [ad_2] solved Missing return statement } in java error [duplicate]
[ad_1] There are no ready-made components in UIKit specifically for onboarding views. You’ll have to roll your own. [ad_2] solved iPhone Onboarding screen in Swift [closed]
[ad_1] Because created_at is a DateTime object which includes both a Date and Time value, then your Message.find_by(created_at: Date.today.beginning_of_week) # Message Load (6.2ms) SELECT “messages”.* FROM “messages” WHERE “messages”.”created_at” = $1 LIMIT $2 [[“created_at”, “2018-01-29”], [“LIMIT”, 1]] … will try to find a record at exactly 2018-01-29 00:00:00 which is a Message record that is … Read more
[ad_1] Using calendar built-in library, you can handle it like this: import calendar s_day = raw_input() days = list(calendar.day_name) # days.index(‘Saturday’) will print 5 print days.index(s_day) Like this, the raw input must be in the correct format. Such as ‘Monday’ not ‘MOnday’ [ad_2] solved Python – How to determine number of day of week given … Read more
[ad_1] You gan get all the screens from this code: let screens = NSScreen.screens And set wallpaper By This code let newWallpaperURL = URL(/* … */) for i in screens { try! NSWorkspace.shared().setDesktopImageURL(newWallpaperURL, for: i, options: [:]) } [ad_2] solved how to change the wallpaper using OC or Swift on mac [closed]
[ad_1] You can use .isin() directly in pandas filtering – recent_indicators_filtered = recent_indicators[recent_indicators[‘CountryCode’].isin(developed_countries)] Also, you can come up with a boolean column that says True if developed – recent_indicators[‘Developed’] = recent_indicators[‘CountryCode’].isin(developed_countries) [ad_2] solved How to check whether data of a row is in list, inside of np.where()?
[ad_1] Based on Nacho’s idea, this is my solution (not ideal, but I’ll take it for now): [TestCase(new int[] { 0, 1, 2, 3 }, true)] [TestCase(new int[] { 0, 3, 4 }, false)] [TestCase(new int[] { 4, 4, 4 }, true)] [TestCase(new int[] { 6, 6, 8, 8 }, true)] [TestCase(new int[] { 5, … Read more
[ad_1] I was going to recreate the project from scratch, but decided to start by recreating the files in the project first. I copied the code in LoginVC, deleted the file, created a new LoginVC, pasted the code back in, and the whole project runs as expected now. Breakpoints and code execute as expected throughout … Read more
[ad_1] fmt.Println is a variadic function whose arguments are generic interfaces. Any type can fulfill this interfere, including strings and floats. The second example works for this reason. The first example, however, involves the binary operator +. As https://golang.org/ref/spec#Operators says, binary operators work in identical types. This means you can’t “add” a float to a … Read more
[ad_1] You are passing by reference $a to $b with the & symbol. That means if you change the value of $b you are changing $a too. In other words $b has the memory address of $a [ad_2] solved please explain following out put [closed]
[ad_1] It depends on the type of Pivot of course. The easiest, standard pivot should look like this one: high,low,close are High(_Symbol,PERIOD_D1,1) etc. double values[] is a buffer of 9 pivot values double p = (high+low+close)/3; values[0+4] = p; values[1+4] = 2*p-low; values[2+4] = p+high-low; values[3+4] = 2*p+high-2*low; values[4+4] = 2*p-high; values[5+4] = p+low-high; values[6+4] … Read more
[ad_1] The comparison function allows you to create custom logic to determine whether the two entries are the same or not. The keys in these two arrays look totally different, because they are in different languages. $data1 = [ ‘red’ => true, ‘yellow’ => true, ‘green’ => true, ‘blue’ => true, ‘black’ => true, ]; … Read more
[ad_1] The problem could be solved using recursion. For example, the code below prints exactly the required matrix for a given n. import java.util.Scanner; public class Main { public static void main(final String[] args) { final Scanner scanner = new Scanner(System.in); final int n = scanner.nextInt(); final int[][] matrix = create(1, (int) Math.pow(2, n)); print(matrix); … Read more
[ad_1] From Google Play Terms of Service Limits on access on Devices. Google may from time to time place limits on the number of Devices and/or software applications you may use to access Content (for more information, please visit the Help link for the relevant Content within Google Play). Google may record and store the … Read more