median

问题 How can one find a median of 2 sorted arrays A and B which are of length m and n respectively. I have searched, but most the algorithms assume that both arrays are of same size. I want to know how can we find median if m != n consider example, A={1, 3, 5, 7, 11, 15} where m = 6, B={2, 4, 8, 12, 14} where n = 5 and the median is 7 Any help is appreciated. I am preparing for interviews and i am struggling with this algo right now. 回答1: Here is the JAVA code to find the median of two sorted

问题 This is what I have right now. It just finds the median with an odd amount of numbers. def median(height): height.sort() x = len(height) x -= 1 posn = x // 2 return height[posn] 回答1: "The median is the numeric value separating the higher half of a sample data set from the lower half. The median of a data set can be found by arranging all the values from lowest to highest value and picking the one in the middle. If there is an odd number of data values then the median will be the value in the

问题 I am working on image smoothing using median filter. For that, I am not using the inbuilt functions within the Python library, rather I am writing my own functions. The following code is for calculating the median. def CalcMedian(Image, x, y, gridSize): #x and y are nested loops, that run over the entire image. medianList = [] row, col = Image.shape; k = int(gridSize/2); for i in range(gridSize-1): for j in range(gridSize-1): if (i+x-k)<0 or (j+y-k)<0 or (i+x-k)>row or (j+y-k)>col: break;

问题 My time-series data includes date-time and temperature columns as follows: rn25_29_o: ambtemp dt 1 -1.96 2007-09-28 23:55:00 2 -2.02 2007-09-28 23:57:00 3 -1.92 2007-09-28 23:59:00 4 -1.64 2007-09-29 00:01:00 5 -1.76 2007-09-29 00:03:00 6 -1.83 2007-09-29 00:05:00 I am using median smoothing function to enhance small fluctuations that are caused because of imprecise measurements. unique_timeStamp <- make.time.unique(rn25_29_o$dt) temp.zoo<-zoo(rn25_29_o$ambtemp,unique_timeStamp) m.av<

问题 Sorting five integers using bitwise or comparison operators can be achieved by first getting the highest number then the second highest then the third and so on. Here is my code in getting the highest number: #include <stdio.h> int main() { int a, b, c, d, e; int aa, bb, cc, dd, ee; a = 4; b = 2; c = 5; d = 1; e = 3; aa = (a > b) ? ((a > c) ? ((a > d) ? ((a > e) ? a : e) : ((d > e) ? d : e)) : ((c > d) ? ((c > e) ? c : e) : ((d > e) ? d : e))) : ((b > c) ? ((b > d) ? ((b > e) ? b : e) : ((d >

问题 I have two classes that are structured like this: public class Company { private List<Person> person; ... public List<Person> getPerson() { return person; } ... } public class Person { private Double age; ... public Double getAge() { return age; } ... } Basically the Company class has a List of Person objects, and each Person object can get an Age value. If I get the List of the Person objects, is there a good way to use Java 8 to find the median Age value among all the Person objects (Stream

问题 My professor wrote this median function and I don't understand it very well. Can someone please explain the part about i = len(list)/2 and median = avg() and the else statement? def avg_list(numbers): sum = 0 for num in numbers: sum += num avg = float(sum)/len(numbers) print avg def median(list): list.sort() if len(list)%2 == 0: #have to take avg of middle two i = len(list)/2 median = avg() else: #find the middle (remembering that lists start at 0) i = len(list)/2 median = list return median

问题 I have a data.table called enc.per.day for encounters per day. It has 2403 rows in which a date of service is specified and the number of patients seen on that day. I wanted to see the median number of patients seen on any type of weekday. enc.per.day[,list(patient.encounters=median(n)),by=list(weekdays(DOS))] That line gives an error Error in [.data.table (enc.per.day, , list(patient.encounters = median(n)), : columns of j don't evaluate to consistent types for each group: result for group 4

问题 I am trying to calculate the row median for a data frame df with rowMedians from matrixStats package. Abundance Sample1 Sample2 Sample3 Sample4 Sample5 Sample6 Sample7 Species1 2 4 0 0 0 6 0 Species2 3 5 6 4 0 0 0 Species3 3 7 2 5 8 0 0 Species4 0 0 3 8 0 0 8 Species5 7 5 6 0 0 4 4 Species6 4 2 3 0 0 2 1 I want to calculate the median of every row and append them in a new column. I got an error Argument 'x' must be a vector or matrix so I tried to convert my df to a matrix. The str function

问题 Refering to the solution present at MIT handout I have tried to figure out the solution myself but have got stuck and I believe I need help to understand the following points. In the function header used in the solution MEDIAN -SEARCH (A[1 . . l], B[1 . . m], max(1,n/2 − m), min(l, n/2)) I do not understand the last two arguments why not simply 1, l why the max and min respectively. In the pseduo code if left > right why do we switch A and B arrays if we reach the above condition. Thanking