median

问题 I have two vectors. One that actually holds the data (let's say floats) and one that holds the indices. I want to pass at nth_element the indices vector, but I want the comparison to be done by the vector that actually holds the data. I was thinking about a functor, but this provides only the () operator I guess. I achieved that by making the data vector a global one, but of course that's not desired. std::vector<float> v; // data vector (global) bool myfunction (int i,int j) { return (v[i]<v

问题 I wrote some code that returns the median of an unsorted odd numbered array, but it does not return the median of an even numbered array. I know that in order to find the median of an even numbered array, you have to take the middle two numbers of the array, average them, and that's the median. I can't translate that into usable code. Aside from the obvious verbosity of this code, the issue seems to be with lines 7-8 and I don't see why. I prefer hints to answers, but if you rather post some

问题 I am required to implement median filtering in MATLAB for images. However, I'm not allowed to use the medfilt2 or ordfilt2 functions in MATLAB. We also have recently started learning MATLAB. Is there any code available for the median filter or Gaussian filter available? 回答1: NB: This assumes that the Image Processing Toolbox is installed. The basic premise behind median filtering is to analyze pixel neighbourhoods in your image, sort their intensities, then choose the middle intensity as the

问题 I have implemented a 2D median filter in CUDA and the whole program is shown below. #include "cuda_runtime.h" #include "cuda_runtime_api.h" #include "device_launch_parameters.h" #include <iostream> #include <fstream> #include <iomanip> #include <windows.h> #include <io.h> #include <stdio.h> #include<conio.h> #include <cstdlib> #include "cstdlib" #include <process.h> #include <stdlib.h> #include <malloc.h> #include <ctime> using namespace std; #define MEDIAN_DIMENSION 3 // For matrix of 3 x 3.

问题 This question was migrated from Cross Validated because it can be answered on Stack Overflow. Migrated 7 years ago . Does somebody know whether there is a sliding window method in R for 2d matrices and not just vectors. I need to apply median function to an image stored in matrix 回答1: The function focal() in the excellent raster package is good for this. It takes several arguments beyond those shown in the example below, and can be used to specify a non-rectangular sliding window if that's

问题 As of right now, my functioin finds the median of 3 numbers and sorts them, but it always makes three comparisons. I'm thinking I can use a nested if statement somewhere so that sometimes my function will only make two comparisons. int median_of_3(int list[], int p, int r) { int median = (p + r) / 2; if(list[p] > list[r]) exchange(list, p, r); if(list[p] > list[median]) exchange(list, p, median); if(list[r] > list[median]) exchange(list, r, median); comparisons+=3; // 3 comparisons for each

问题 Pyspark API provides many aggregate functions except the median. Spark 2 comes with approxQuantile which gives approximate quantiles but exact median is very expensive to calculate. Is there a more Pyspark way of calculating median for a column of values in a Spark Dataframe? 回答1: Here is an example implementation with Dataframe API in Python (Spark 1.6 +). import pyspark.sql.functions as F import numpy as np from pyspark.sql.types import FloatType Let's assume we have monthly salaries for

问题 My dataset is as following: salary number 1500-1600 110 1600-1700 180 1700-1800 320 1800-1900 460 1900-2000 850 2000-2100 250 2100-2200 130 2200-2300 70 2300-2400 20 2400-2500 10 How can I calculate the median of this dataset? Here's what I have tried: x <- c(110, 180, 320, 460, 850, 250, 130, 70, 20, 10) colnames <- "numbers" rownames <- c("[1500-1600]", "(1600-1700]", "(1700-1800]", "(1800-1900]", "(1900-2000]", "(2000,2100]", "(2100-2200]", "(2200-2300]", "(2300-2400]", "(2400-2500]") y <-

问题 I'm working on a quicksort-variant implementation based on the Select algorithm for choosing a good pivot element. Conventional wisdom seems to be to divide the array into 5-element blocks, take the median of each, and then recursively apply the same blocking approach to the resulting medians to get a "median of medians". What's confusing me is the choice of 5-element blocks rather than 3-element blocks. With 5-element blocks, it seems to me that you perform n/4 = n/5 + n/25 + n/125 + n/625 +

问题 Closed . This question needs to be more focused. It is not currently accepting answers. Want to improve this question? Update the question so it focuses on one problem only by editing this post. Closed 6 years ago . I have recently found out that there exists a method called nth_element in the STL. To quote the description: Nth_element is similar to partial_sort, in that it partially orders a range of elements: it arranges the range [first, last) such that the element pointed to by the