Condition to iterate your loop for(int j=0; j<=99 & a[j] > 1; j++) is j<=99 & a[j]. You initialized j to 0, so j<=99 part is true, but problem lies in second part of this condition. For j=0 condition a[j]>1 becomes a[0]>1, but a[0] is -2, which leaves us with -2>1 which is false. So … Read more
You may start with the largest number first, something like: foreach(number in positiveNumberArray) { if (number > 50) Console.WriteLine(“Out of Range”); else if(number > 40) Console.WriteLine(“Range 40-50”); else if(number > 30) Console.WriteLine(“Range 30-40”); else if(number > 20) Console.WriteLine(“Range 20-30”); else if(number > 10) Console.WriteLine(“Range 10-20”); else Console.WriteLine(“Range 0-10”); } Edit: To count the up the … Read more
Here is a method. We first count the files, then we test the files. @echo off & set cnt=0 for /f %%i in (‘dir /b’) do set /a cnt+=1 if %cnt% equ 4 if exist f1 if exist f2 if exist dir1 if exist dir2 echo All Good! and some light error generations: @echo off … Read more
I quess if( x > y || 10 < 12) { System.out.println(“one of these is true”); } should do the thing. solved is there an equivalent if statement in java as this python one?
The computer must repeat the question till it recieves a valid response. It (output) goes like this indefinitely. Reason : The problem is that you are receiving input only once in your code and then entering into loop to check for the age. since age value is not re-assigned after every iteration, if the first … Read more
You may turn off short-circut evaluation in your compiler. If you work with your own types, you may overload || and &&. Overloaded logical operators are not short-circuted. Both things are really bad. Programmers expect logical operators to behave in certain way and are very likely to get super confused with this unexpected behavior. You … Read more
Are you asking something like the following? Let’s say your initial dataframe is df, which is the following: df A B C 1 2016-02-16 2016-03-21 2016-01-01 2 2016-07-07 2016-06-17 2016-01-31 3 2016-05-19 2016-09-10 2016-03-01 4 2016-01-14 2016-08-21 2016-04-01 5 2016-09-02 2016-06-15 2016-05-01 6 2016-05-09 2016-07-17 2016-05-31 7 2016-06-13 2016-06-23 2016-07-01 8 2016-09-17 2016-03-11 2016-07-31 9 … Read more
Here you are: var array = ‘ABCDEFGHIJKLMNOPQRSTUVWXYZ’.split(”); for (var i = 0; i < array.length; i++) { var str=””; for (var j = 0; j <= i; j++) { if (array[j] == ‘E’) str += ‘3’; else str += array[j]; } document.querySelector(‘span’).innerHTML = document.querySelector(‘span’).innerHTML + str + ‘<br />’; } <span></span> Hope this helps. 9 … Read more
Solution: I put all of the conditional statements inside of if (!allCorrect[i]) so it looks like this if (!allCorrect[i]) { if (int.TryParse(parameters[i], out INT)) { if (_command.ParameterTypes[i] == EVariableType.INT) { allCorrect[i] = true; } else { allCorrect[i] = false; } } } if (!allCorrect[i]) { if (float.TryParse(parameters[i], out FLOAT)) { if (_command.ParameterTypes[i] == EVariableType.FLOAT) { … Read more
I do not know whether this solution is more efficient, but maybe it is more readable and easier to code: #include <iostream> #include <cstdlib> int main() { const int magicNumber = 32640; const int maxOffset = 1920; int n; std::cin >> n; int y = 20; const std::div_t divresult = std::div(n, magicNumber); if (divresult.rem > … Read more
In your second code block you are first checking if the first character is the same as the last. If it isn’t, return false. Else, return true. So this isn’t going to be a valid palindrome check. It only checks if the first and last letters are the same or not and ignores the letters … Read more
Ok your question seems to be particularly unpopular… just for info, have a look at http://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B The operator you are looking for is && but I should not tell you this since in fact as downvotes testify… your question is a bit lazy! I also know that sometimes when you are a noob it can … Read more
You can avoid the loop or the apply statement by vectorizing the ifelse statement. If you define i as the vector from 3:length(E$phone), you can then run the ifelse statement directly. #test data phone<-c(123,123,123,333,333,333,456,456,456,789,789,789,500,500,500) time<-c(“2018-11-06″,”2018-11-06″,”2018-11-06″,”2018-11-09″,”2018-11-09″,”2018-11-09”, “2018-11-07″,”2018-11-07″,”2018-11-07″,”2018-11-05″,”2018-11-05”, “2018-11-05”, “2018-11-06″,”2018-11-06″,”2018-11-06”) time<-as.Date(time) tel<-c(0,0,1,1,0,1,1,1,0,1,1,1,0,0,1) porad<-c(1,2,3,1,2,3,1,2,3,1,2,3,1,2,3) E<-data.frame(phone, time, tel, porad) E$de[1]=ifelse(E$phone[1]==E$phone[2] & E$time[1]==E$time[2] & E$porad[1]==2 & E$tel[1]==1,1,0) E$de[2]=ifelse(E$phone[2]==E$phone[3] & E$time[2]==E$time[3] … Read more
You are always printing something when you are testing, use a temporary variable and print the result after scanning the full list: success = False for i in sheet_data: if (i[0] == “BN”) and (i[1] == “YOU”): found_list.append(i) success = True if success: print(“Success”) else: print(“error”) solved Python else condition prints several times
When comparing values for logic operations, you must use >, <, ==, ===, !=, or !==. You are using a single equals sign, which is not for comparison but for assignment. This is what you are doing $item = ‘b’; // a single equals sign assigns a value if ($item = ‘a’) { // so … Read more