sorting

问题 I know that this question has been asked several times. But none of the answers match my case. I've a pandas dataframe with columns,department and employee_count. I need to sort the employee_count column in descending order. But if there is a tie between 2 employee_counts then they should be sorted alphabetically based on department. Department Employee_Count 0 abc 10 1 adc 10 2 bca 11 3 cde 9 4 xyz 15 required output: Department Employee_Count 0 xyz 15 1 bca 11 2 abc 10 3 adc 10 4 cde 9 This

问题 I am having a problem sorting Swedish strings. I am having problems with the following characters: v, w, å, ä, ö. new[] { "ö", "ä", "å", "wa", "va", "wb", "vb", "a" } .OrderBy(x => x, new CultureInfo("sv-SE").CompareInfo.GetStringComparer(CompareOptions.None)) Expected: a, va, vb, wa, wb, å, ä, ö Actual: a, va, wa, vb, wb, å, ä, ö Is the there any option to make it sort the strings as expected? 回答1: As a work-around I have switched to the culture se-SE (Sami) just for sorting alphabetic

问题 I was given this as an exercise. I could of course sort a list by using sorted() or other ways from Python Standard Library, but I can't in this case. I think I'm only supposed to use reduce() . from functools import reduce arr = [17, 2, 3, 6, 1, 3, 1, 9, 5, 3] sorted_arr = reduce(lambda a,b : (b,a) if a > b else (a,b), arr) The error I get: TypeError: '>' not supported between instances of 'tuple' and 'int' Which is expected, because my reduce function inserts a tuple into the int array,

问题 I am taking an algorithms course and there I saw that the time complexity of counting sort is O(n+k) where k is the range of numbers and n is the input size. My question is, when the difference between k and n is too much, such as when k=O(n 2 )or O(n 3 ), can we say that the complexity is O(n 2 ) or O(n 3 )? Then in this case counting sort is not a wise approach and we can use merge sort. Am I right? 回答1: Yes, you are exactly right on all counts. Furthermore, we can make stronger statements:

问题 I am taking an algorithms course and there I saw that the time complexity of counting sort is O(n+k) where k is the range of numbers and n is the input size. My question is, when the difference between k and n is too much, such as when k=O(n 2 )or O(n 3 ), can we say that the complexity is O(n 2 ) or O(n 3 )? Then in this case counting sort is not a wise approach and we can use merge sort. Am I right? 回答1: Yes, you are exactly right on all counts. Furthermore, we can make stronger statements:

问题 I am taking an algorithms course and there I saw that the time complexity of counting sort is O(n+k) where k is the range of numbers and n is the input size. My question is, when the difference between k and n is too much, such as when k=O(n 2 )or O(n 3 ), can we say that the complexity is O(n 2 ) or O(n 3 )? Then in this case counting sort is not a wise approach and we can use merge sort. Am I right? 回答1: Yes, you are exactly right on all counts. Furthermore, we can make stronger statements:

问题 I have a vector of class pointers std::vector<Square*> listSquares . I want to sort it with one of the attributes of the class as the key. This is what I'm doing bool compById(Square* a, Square* b) { return a->getId() < b->getId(); } std::sort(listSquares.begin(), listSquares.end(), compById) but the compiler says: error: no matching function for call to 'sort(std::vector::iterator, std::vector::iterator, <unresolved overloaded function type>)' what am I doing wrong here? 回答1: In order to use

问题 I have a vector of class pointers std::vector<Square*> listSquares . I want to sort it with one of the attributes of the class as the key. This is what I'm doing bool compById(Square* a, Square* b) { return a->getId() < b->getId(); } std::sort(listSquares.begin(), listSquares.end(), compById) but the compiler says: error: no matching function for call to 'sort(std::vector::iterator, std::vector::iterator, <unresolved overloaded function type>)' what am I doing wrong here? 回答1: In order to use

问题 I have a vector of class pointers std::vector<Square*> listSquares . I want to sort it with one of the attributes of the class as the key. This is what I'm doing bool compById(Square* a, Square* b) { return a->getId() < b->getId(); } std::sort(listSquares.begin(), listSquares.end(), compById) but the compiler says: error: no matching function for call to 'sort(std::vector::iterator, std::vector::iterator, <unresolved overloaded function type>)' what am I doing wrong here? 回答1: In order to use

问题 I am making an uno card game with structure of card with fields: struct card { int rank char *color char *action. } I am able to make it sort by color with insertion. I wonder how could I first sort the array of uno card by color, then sort the rank of each color. 回答1: Thanks to Bo Persoson, this is the solution to my question void sort(card *a, int length) { int j; for (int i = 1; i < length; i++) { j = i; while (j > 0 && a[j].color < a[j - 1].color || (a[j].color == a[j - 1].color && a[j]