median

问题 I'm reposting the question asked here hoping maybe to get a little more visibility. This is a question concerning Lumley's survey package for R. Specifically, its handling of interpolation for median estimation, after several hours of looking into the matter. I'm using a svyrep design which has the following form: design <- svydesign(id = ~id_directorio, strata = ~estrato, weights = ~f_pers, check.strata = TRUE, data = datos) options(survey.lonely.psu="remove") set.seed(234262762) SB2K_2 = as

问题 I'm reposting the question asked here hoping maybe to get a little more visibility. This is a question concerning Lumley's survey package for R. Specifically, its handling of interpolation for median estimation, after several hours of looking into the matter. I'm using a svyrep design which has the following form: design <- svydesign(id = ~id_directorio, strata = ~estrato, weights = ~f_pers, check.strata = TRUE, data = datos) options(survey.lonely.psu="remove") set.seed(234262762) SB2K_2 = as

问题 I am trying to do a percentile over a column using a Window function as below. I have referred here to use the ApproxQuantile definition over a group. val df1 = Seq( (1, 10.0), (1, 20.0), (1, 40.6), (1, 15.6), (1, 17.6), (1, 25.6), (1, 39.6), (2, 20.5), (2 ,70.3), (2, 69.4), (2, 74.4), (2, 45.4), (3, 60.6), (3, 80.6), (4, 30.6), (4, 90.6) ).toDF("ID","Count") val idBucketMapping = Seq((1, 4), (2, 3), (3, 2), (4, 2)) .toDF("ID", "Bucket") //jpp import org.apache.spark.sql.Column import org

问题 I am trying to do a percentile over a column using a Window function as below. I have referred here to use the ApproxQuantile definition over a group. val df1 = Seq( (1, 10.0), (1, 20.0), (1, 40.6), (1, 15.6), (1, 17.6), (1, 25.6), (1, 39.6), (2, 20.5), (2 ,70.3), (2, 69.4), (2, 74.4), (2, 45.4), (3, 60.6), (3, 80.6), (4, 30.6), (4, 90.6) ).toDF("ID","Count") val idBucketMapping = Seq((1, 4), (2, 3), (3, 2), (4, 2)) .toDF("ID", "Bucket") //jpp import org.apache.spark.sql.Column import org

问题 I need to return a median of only a certain category on a spread sheet. Example Below Airline 5 Auto 20 Auto 3 Bike 12 Airline 12 Airline 39 ect. How can I write a formula to only return a median value of the Airline Categories. Similar to Average if, only for median. I cannot re-arrange the values. Thank you! 回答1: Assuming your categories are in cells A1:A6 and the corresponding values are in B1:B6, you might try typing the formula =MEDIAN(IF($A$1:$A$6="Airline",$B$1:$B$6,"")) in another

问题 How can I use AWK to compute the median of a column of numerical data? I can think of a simple algorithm but I can't seem to program it: What I have so far is: sort | awk 'END{print NR}' And this gives me the number of elements in the column. I'd like to use this to print a certain row (NR/2) . If NR/2 is not an integer, then I round up to the nearest integer and that is the median, otherwise I take the average of (NR/2)+1 and (NR/2)-1 . 回答1: This awk program assumes one column of numerically

问题 How can I use AWK to compute the median of a column of numerical data? I can think of a simple algorithm but I can't seem to program it: What I have so far is: sort | awk 'END{print NR}' And this gives me the number of elements in the column. I'd like to use this to print a certain row (NR/2) . If NR/2 is not an integer, then I round up to the nearest integer and that is the median, otherwise I take the average of (NR/2)+1 and (NR/2)-1 . 回答1: This awk program assumes one column of numerically

问题 How can I use AWK to compute the median of a column of numerical data? I can think of a simple algorithm but I can't seem to program it: What I have so far is: sort | awk 'END{print NR}' And this gives me the number of elements in the column. I'd like to use this to print a certain row (NR/2) . If NR/2 is not an integer, then I round up to the nearest integer and that is the median, otherwise I take the average of (NR/2)+1 and (NR/2)-1 . 回答1: This awk program assumes one column of numerically

问题 I'm a beginner in R and I would like to know how to do the following task: I want to replace the missing values of my dataset by the median for all the columns of my dataset. However, for each column, I want the median of a certain category (depending on another column).My dataset is as follows structure(list(Country = structure(1:5, .Label = c("Afghanistan", "Albania", "Algeria", "Andorra", "Angola"), class = "factor"), CountryID = 1:5, Continent = c(1L, 2L, 3L, 2L, 3L), Adolescent.fertility

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