I dont know python but the pseudocode would look something like this max = –2147483648 for each integer n in collection if n > max max = n It loops through every number in a collection. If the number (n) is bigger than the previous maximum, the maximum is set to be equal to n. … Read more
If you have an array declared like int a[N]; where N is some positive integer value then the valid range of indices to access elements of the array is [0, N). It means that for example this for loop for(i=1;i<=10;i++) { printf(“enter 10 nos. for arr[%d] :”,i); scanf(“%d”,&arr[i]); } must look like for ( i … Read more
You can use the MAX() and MIN() function to get the ids having largest and the id having smallest discount. Select id,discount from customer where discount=(select MAX(discount) from customer) OR discount=(select MIN(discount) from customer); 3 solved select id by the min and max (sql) [closed]
Is something like what I expected standardized in C++17 No. , or proposed for standardization later? There is no such proposal in any of the mailings. Though of course there’s plenty of easy workarounds, so there’s little reason for such a proposal: oy ? std::max(x,*oy) : x; x < oy ? *oy : x *std::max(oint(x), … Read more
The code you posted will give you a NameError: global name ‘numpy’ is not defined because of the way you are importing stuff. Instead of doing from numpy import * just do import numpy Also, the function in numpy is called linspace, not linespace. If you change those two things it will work, but there … Read more
Assuming the table as you give it is in A1:B6 (with headers in row 1): =LOOKUP(1,0/FREQUENCY(0,1/(1+AVERAGEIF(A2:A6,A2:A6,B2:B6))),A2:A6) If more than one company share the highest average score than that occurring first within the list will be returned. Regards 2 solved average the same ones and then find the max on the list
it allows to define a criterion which replaces the < comparison between elements. For instance: >>>l = [“hhfhfhh”,”xx”,”123455676883″] >>>max(l, key=len) ‘123455676883’ returns the longest string in the list which is “123455676883” Without it, it would return “xx” because it’s the highest ranking string according to string comparison. >>>l = [“hhfhfhh”,”xx”,”123455676883″] >>>max(l) ‘xx’ 2 solved What … Read more
def get_smallest_length(x): return [k for k in x.keys() if len(x.get(k))==min([len(n) for n in x.values()])] def get_largest_sum(x): return [k for k in x.keys() if sum(x.get(k))==max([sum(n) for n in x.values()])] x = {‘a’: [4, 2], ‘c’: [4, 3], ‘b’: [3, 4], ‘e’: [4], ‘d’: [4, 3], ‘g’: [4], ‘f’: [4]} print get_smallest_length(x) print get_largest_sum(x) Returns: [‘e’, ‘g’, … Read more
It wasn’t fun to read your query, but I think the problem is here: LEFT JOIN ( SELECT max(notification_date) notification_date, client_id FROM og_ratings WHERE notification_date NOT IN ( SELECT max(notification_date) FROM og_ratings ) if you want the maximum date for every client you need to GROUP BY client_id: SELECT client_id, max(notification_date) notification_date FROM og_ratings GROUP … Read more
Because you are comparing character strings. Consider… max(“apple”,”banana”,”banana2″) #[1] “banana2” max( “1” , “2” , “10” ) #[1] “2” sort( c( “1” , “2” , “10” ) ) #[1] “1” “10” “2” sort( as.integer( c(“1” , “2” , “10” ) ) ) #[1] 1 2 10 max( as.integer( c(“1” , “2” , “10” ) ) … Read more
In general, returning a reference avoids that the return value gets copied, and (if it is not const-qualified) gives you the opportunity to change the “original” value. A return type const T& is often used in conjunction with objects of class type, e.g. std::vector, where copying could result in (unwanted) overhead. In conjunction with elementary … Read more
suggest that if your above SQL statement can query the result. maybe you can just skip the “order by” and simply just select * from (/* your SQL statement without “order by”*/) a order by a.message_id desc sorry that as it is difficult to see your screen capture and help you to resolve the SQL … Read more
In general ORDER BY in a sub-query makes no sense. (It only does when combined with FETCH FIRST/LIMIT/TOP etc.) The solution is to use a correlated sub-query to find the heaviest fish for the “main query”‘s current row’s username, location, species combination. If it’s a tie, both rows will be returned. SELECT * FROM entries … Read more
If you just want the max score per student you can aggregate the rows before joining: select s.id_student, u.name, u.role, s.status, u.no_phone, s.exam_status, sc.score from student s join user u on u.username = s.id_student join ( select Max(score) as score, id_student from score group id_student )sc on sc.id_student = s.id_student where mod(s.id_student, 2) = 0 … Read more
The error is that you are re-using the teleportation_start_position variable. teleportation_start_position = min(teleportation_start_position, teleportation_end_position); Before this line the state of your program is: teleportation_start_position = 8 teleportation_end_position = 2 But after, it is: teleportation_start_position = 2 teleportation_end_position = 2 So in the next line you are doing: teleportation_end_position = max(teleportation_start_position, teleportation_end_position); // teleportation_end_position = max(2, … Read more