The algorithm to sort this problem is simple, just like you said in your question description, sort characters in each word first, then sort these sorted-word again. Like this: $ echo heya64 this is21 a good89 day91 | perl -anE ‘say(join ” “, sort(map { join “”, sort split // } @F))’ 12is 19ady 46aehy … Read more
For Java 8 use a lambda expression and the Arrays sort function: Users[] someUsers = {user1, user2, user3}; Arrays.sort(someUsers, (a,b) -> a.firstName.compareTo(b.firstName)); For previous versions use a Collection Comparator: Users[] someUsers = {user1, user2, user3}; List<Users> userList = new ArrayList<Users>(Arrays.asList(someUsers)); Comparator<Users> comparator = new Comparator<Users>() { @Override public int compare(Users left, Users right) { return … Read more
As per assuming few conditions like provided array size and array data type to be integer you can use the following program for reference. This program takes the no. of elements and array as input. #include<stdio.h> #include<limits.h> int main(){ int n; int a[100]; printf(“Size of the array:”); scanf(“%d”, &n); for(int i = 0; i < … Read more
Take a look at array_slice $onlyTenResults = array_slice($myArray, 40, 10); // take 10 items starting from 40 1 solved Grab every 10 results from array PHP
SELECT DISTINCT y.* FROM my_table x JOIN my_table y ON y.category = x.category ORDER BY y.title=”dog” DESC , y.category; or something like that 1 solved Sort by field values in MySQL Query [closed]
Maybe you could use one of the containers, e.g. vector. you could push_back all the moves into this vector until home is reached. The size of this vector will be the number of moves. You can setup a counter array[400] for counting the number of moves into same coordinates. For each move, – `push_back` the … Read more
Yes, instead of having an array of primitive types (like ints) or Strings, define your own class and then create an array of that. Then you can have whatever information you want in there. Alternatively, if the array must be an array of primitives or Strings because some other method requires it to be in … Read more
Just had to limit the inner loop to <9. The fixed code: for(int i=0;i<10;i++) //1st array, ascending { for(int j=0;j<9;j++) { if(array1[j]>array1[j+1]) { int temp=array1[j]; array1[j]=array1[j+1]; array1[j+1]=temp; } } } //Over for(int i=0;i<10;i++) //2nd array, descending { for(int j=0;j<9;j++) { if(array2[j]<array2[j+1]) { int temp=array2[j]; array2[j]=array2[j+1]; array2[j+1]=temp; } } } //Over Thank you guys! solved Sorting … Read more
dictionaries are unordered (well pre 3.6 at least), instead sort the items d = {3: ‘__init__’, 5: ‘other’, 7: ‘hey ‘, 11: ‘hey’} print(sorted(d.items(),key=lambda item:(item[0].strip(),item[1]))) # output => [(3, ‘__init__’), (7, ‘hey ‘), (11, ‘hey’), (5, ‘other’)] if you really want it as a dict (for reasons i cant fathom) and you are using 3.6+ … Read more
If you do not care about destroying arr2, neither about having a deep copy of arr1[0], a simple unshift() can do that: const arr1 = [[1, 2, 3, 4], [5, 6, 7, 8]]; const arr2 = [[‘some1’, ‘some2’], [‘some3’, ‘some4’]]; arr2.unshift(arr1[0]); console.log(JSON.stringify(arr2)); Of course those are really some conditions which may not fit your case. … Read more
You should change Collections.sort(array, new j()); to Arrays.sort(array, new j()); solved Java error :- “no suitable method found for sort(Student[],j)”
You can do in following way: a.sort { return $0.last as! Int > $1.last as! Int } Don’t forget to add additional checks while using this code, case where the last item is not an integer or there is an array not in the expected format. Otherwise, it will lead to a crash. 0 solved … Read more
after converting to datetime pd.to_datetime(df[‘date’]) you can create a separate column with the hour in it, e.g. df[‘Hour’] = df.date.dt.hour and then sort by it df.sort_values(‘Hour’) EDIT: As you want to sort by time you can instead of using hour, put the timestamp part into a ‘time’ column. In order to get times between 9 … Read more
Use the comparator, and then, the Float Comparator: Arrays.sort(strengths, new Comparator<String>() { @Override public int compare(String o1, String o2) { return Float.compare(Float.valueOf(o1.split(” “)[0]), Float.valueOf(o2.split(” “)[0])); } }); 0 solved Sort string array by first part of each element in java
Sorting list items considering unit’s place. You can try: list1 = [1236,985,150,641876,167] sorted(list1, key=lambda n: n%10) [150, 985, 1236, 641876, 167] solved How to sort sum digits from the following list [closed]