Here’s another version if you can’t work with the other answer: awk -vval=2.1 ‘{ # set “val” to the new value for column 3 on first two lines if(NR==1 || NR==2) { # if it’s the first or second line $3=val; # set column 3 to val $2=$3*2.9 # set column 2 to column 3 … Read more
It is only run after the nested loop. Looking at the output from the post you referenced, the program starts with 1, prints one 1 without a newline, then exits the nested loop, then prints the newline. Then it enters the nested loop with 2, loops twice (prints a 2 without a newline, then prints … Read more
I believe you want is: Multiply each key with 2. Not the square of it. d = {2: (1,2), 8: (2,4), 30: (10,3)} d = {k*2: v for (k, v) in d.items()} print(d) OUTPUT: {4: (1, 2), 16: (2, 4), 60: (10, 3)} solved How to multiply dictionary keys? [duplicate]
You shouldn’t use nested for-each loops here. Try an indexed for on the listMain and store the result of each iteration in a kind of tuple. This tuple could be stored in a Map. As your friend mentioned, you could use a HashMap for this. Something like this should do the trick: ArrayList<Main>listMain = mainDAO.getlAllMain(Connection … Read more
You could try something like this: var data = [ { user: ‘VAY9090’, value: [ ‘KL65’ ] }, { user: ‘VAY9090’, value: [ ‘KL6I’ ] }, { user: ‘VAY9092’, value: [ ‘KLMF’ ] }, { user: ‘VAY9092’, value: [ ‘KLMQ’ ] }, { user: ‘VAY9090’, value: [ ‘KLMR’ ] }, { user: ‘BTD9891’, value: [ … Read more
Introduction This question is about whether it is possible to create a for loop for an array of objects, removing duplicate values and adding the values which consist of the same user. The answer to this question is yes, it is possible to create a for loop for an array of objects, removing duplicate values … Read more
What have you tried? Here’s one possible solution: class ForDemo { public static void main(String[] args){ for(int i=1; i<11; i++){ if(i != 5){ System.out.println(“Count is: ” + i); } } } } I added a check in there: if(i != 5) which executes the code inside only if the condition is true (ie: i is … Read more
I’m not comfortable with Django, so I show a couple of option in plain Python, given the collections: something1 = [1,2,3,4] something2 = [1,2,3,4,5,6,7,8,9,10] You can access objects by index (not the same as database index): i = 1 for e1 in something1: print(e1) i += 1 for i2 in range(i,len(something2)): print(something2[i2]) Or slice the … Read more
Do you mean this? m.value1= Arrays.asList(11,12); m1.value2 = Arrays.asList(1,2,3,4,5); for(Integer i : m.value1) { for(Integer j : m1.value2){ System.out.println(“(“+i+”,”+j+”)”); } } 5 solved Using for loop arraylist in Java
I think the function you are looking for is IndexOf(). This is the equivalent of your find call above: foreach (var uniqueKey in message) { var num = CHARACTER.IndexOf(uniqueKey); if (num >= 0) { … } } 4 solved C# for loop and encryption [closed]
You don’t want to use any looping structure like for or while. Instead just make a function that gets invoked with the index to start it off, and then inside the callback, it increments the index and recursively invokes the same function. when the index gets to 3, set it back to 0 before the … Read more
Just check where you are in your loop to know if you need to print the comma. public static void Test() { for (int i = 49; i >= 1; i–) { Console.WriteLine(i + (i != 1 ? “,” : “”)); } Console.ReadLine(); } solved Loop to write a cimma separated list of items [closed]
When i is 1, you print it with the printf statement. Then i gets incremented (via the ++ operator in your printf statement). Then i /= 2 is performed, which results to i = 2 / 2 which results to 1. This satisfies your condition i >= 1, making it an infinite loop. solved What … Read more
For that you can use the modulo operator. public class Main { /** http://stackoverflow.com/q/36053971/6077352 */ public static void main(String[] args) { int input = 123456789; while (input > 0) { if (input % 2 == 0) { System.out.println(input % 10 + ” is even”); } else { System.out.println(input % 10 + ” is odd”); } … Read more
I finally came to the bottom of this, I figured I could create as many new column I wanted in order to calculate what I had to so that’s what I did. I think this is the best way to do it but am open to suggestions if there are other ways or even faster. … Read more