[ad_1] ord() function will return the ascii value of a character. The 2nd point says that you create a list of length 26 (as english letters are 26) and each entry will have the frequency of that letter. Eg: ‘a’ = 0; ‘b’ = 1;… ‘z’: 25 ord(‘a’) gives 97 – ASCII value of ‘a’. … Read more
[ad_1] public static void main(String[] args) { Scanner numPlayers = new Scanner(System.in); ArrayList<Player> playerList = new ArrayList<>(); int input = numPlayers.nextInt(); for (int i = 0; i < input; i++){ System.out.println(“what is player ” + (i + 1) + ” name?”); String playerName = numPlayers.next(); playerList.add(new Player(playerName)); } } You should have declared scanner object … Read more
[ad_1] First put all files in a single folder. filenames <- list.files(pattern = “.csv”) all <- lapply(filenames, function(name) { readr:: read_csv(name) }) Based on your description the list should have a length of 300, each item containing a dataframe. If you want to bind the rows of all dataframes you can use dplyr’s bind_rows(). Below … Read more
[ad_1] I would probably consider doing this another way… Have an empty hashmap<String sub, int multiplier> Read in the name Generate the subname Fetch from or create subname in hashmap If new, set multiplier to 1 If exists, increment multiplier Return String.format(“%s%3d”, subname, multiplier x 5).toUpperCase() I would give you code but writing the above … Read more
[ad_1] Create a while loop like this. def run() clear() #check_win() should return true if there is a winner or false if not while(check_win()==False): #if there is nobody win, continue this loop if turn == 0: # X’s move runX() turn = 1 # the next turn will be O’s elif turn == 1: # … Read more
[ad_1] You have not called print method from main method. One more mistake is in your code that, you mentioned about 3 times of index variable and in your code you are taking cube of index variable. public class progprblm5{ public static void main(String []args){ double alpha[] = new double[50]; for(int i =0;i<25;i++){ alpha[i]= i*i; … Read more
[ad_1] Print ‘#’ if red else print ‘.’. If encounted sequence red, not red then print ‘.’ for the rest of the array: prev = None it = iter(data) for point in it: if point == ‘red’: print ‘#’, else: print ‘.’, if prev == ‘red’: # encounted [‘red’, ‘blank’] break prev = point for … Read more
[ad_1] Presumably MAC_ADRESSES is not an array of pointers. NULL is a pointer (normally (void *)0 in C), so you can’t assign it to a non-pointer variable. Edit: Since your definition is char MAC_ADRESSES[MAX_LINES][100], you have a 2D array, not an array of pointers. You can’t store NULL in this array. You can wipe out … Read more
[ad_1] while (true) { System.out.println(“What type of Employee? Enter ‘o’ for Office ” + “Clerical, ‘f’ for Factory, or ‘s’ for Saleperson. Enter ‘x’ to exit.” ); String response= inp.nextLine().toLowerCase(); // validate input, do you switch statement, return; on x } 1 [ad_2] solved Trying to sentinel loop this program [closed]
[ad_1] my solution. I like just using the ES6(es2015+) way. const test_array = [ { “name”: “AnyManagedFundsRow”, “columnMeta”: { “a0”: “STRING”, “a1”: “STRING”, “a2”: “STRING”, “a3”: “DATE”, “a4”: “DATE”, “a5”: “DOUBLE”, “a6”: “INT” }, “rows”: [ [ “华夏基金管理有限公司”, “华夏大中华企业精选灵活配置混合(QDII)”, “其他型基金”, “2016-01-20”, “”, 21.877086009428236, 65135 ], [ “华夏基金管理有限公司”, “华夏大盘精选混合”, “混合型基金”, “2015-09-01”, “2017-05-02”, 10.307680340705128, 2944 ] ] … Read more
[ad_1] I have a feeling that you are requiring a formula to get number of iteration the nested loops will. for( i = 1; i <= n; i++ ) for ( j = i+1; j <= n; j++ ) for( k = j+1; k <= n; k++ ) For those loops, the number of iterations … Read more
[ad_1] In short: probably Matlab and Python achieve similar speeds in loops. The sources that you found are likely comparing old versions of Matlab, as @Eitan mentioned the new versions have proper Just In Time mechanisms which speed simple loops up a lot. There will be loops that python does faster, and those that Matlab … Read more
[ad_1] I’m taking a gamble by answering this.. but I’m assuming that you want to keep on adding two indexes within your array that are next to each other, to see if they’d equal any value in your second array? Also, I have no idea why you’re using a nested loop. You only need to … Read more
[ad_1] Each time you initializing var part=[] that it will make the array empty . Only the last value will be inside it . So initialize before the .each method , not inside in it . 1 [ad_2] solved 0 index is undefined or just missing [closed]
[ad_1] Try this let newArray = []; carArray.forEach(function(car) { let arrayKey = Object.keys(car)[0] if (newArray.length > 0) { newArray.find(function (data) { if(!data.hasOwnProperty(Object.keys(data)[0])) { newArray.push(car); } }) } else newArray.push(car) }); 6 [ad_2] solved Javascript Loop over array of objects and get unique values