median

问题 This question already has answers here : Closed 8 years ago . Possible Duplicate: Function to Calculate Median in Sql Server I have a table like this: SET ANSI_NULLS ON GO SET QUOTED_IDENTIFIER ON GO SET ANSI_PADDING ON GO CREATE TABLE [dbo].[cars]( [id] [int] IDENTITY(1,1) NOT NULL, [sp] [int] NOT NULL, [dst] [int] NOT NULL, [type] [varchar](10) NULL, CONSTRAINT [PK_id] PRIMARY KEY CLUSTERED ( [id] ASC )WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW

问题 In R, how can I generate N numbers that have a mean of X and a median of Y (at least close to). Or perhaps more generally, is there an algorithm for this? 回答1: There is an infinite number of solutions. Approximate algorithm: Generate n/2 numbers below the median Generate n/2 numbers above the median Add you desired median and check Add one number with enough weight to satisfy your mean -- which you can solve Example assuming you want a median of zero and a mean of twenty: R> set.seed(42) R>

问题 I am trying to calculate the median of individual columns in R and then subtract the median value with every value in the column. The problem that I face here is I have N/A's in my column that I dont want to remove but just return them without subtracting the median. For example ID <- c("A","B","C","D","E") Point_A <- c(1, NA, 3, NA, 5) Point_B <- c(NA, NA, 1, 3, 2) df <- data.frame(ID,Point_A ,Point_B) Is it possible to calculate the median of a column having N/A's? My resulting output would

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 4 years ago . I have a data frame (data) with 53 columns (ID column plus 52 numeric values). I can add a column with mean with the following: data$mean <- rowMeans(data[,2:51],na.rm = TRUE) # add mean of rows But, I cannot do the same for median with the following: data$medians <- rowMedians(month.sum[,2:51],na.rm = TRUE) #

问题 A quite confusing thing is what I got: The Median Absolute Deviation output of the following vector is vec = c( -5.665488 ,3.963051, 14.14956, 0, -5.665488) > mad(vec) [1] 8.399653 However, if I compute that I got the following value: Median absolute deviation = 5.665488 which is equal to the value of the computation I have found online as well: http://www.miniwebtool.com/median-absolute-deviation-calculator/ How can the difference between the calculated value of mine and the website and the

问题 I need to calculate median value of a numeric sequence in Google BigQuery efficiently. Is the same possible? 回答1: Yeah it's possible with PERCENTILE_CONT window function. Returns values that are based upon linear interpolation between the values of the group, after ordering them per the ORDER BY clause. must be between 0 and 1. This window function requires ORDER BY in the OVER clause. So an example query would be like (the max() is there just to work across the group by but it's not being

问题 I have a list of numbers [1, 2, 3, 4, 5, 6, 7] and I want to have a function to return the interquartile range of this list of numbers. The interquartile range is the difference between the upper and lower quartiles. I have attempted to calculate the interquartile range using NumPy functions and using Wolfram Alpha. I find all of the answers, from my manual one, to the NumPy one, tothe Wolfram Alpha, to be different. I do not know why this is. My attempt in Python is as follows: >>> a = numpy

问题 Following the answer found here, https://stackoverflow.com/a/10931091/1311773, I am attempting to implement two heaps so I can calculate a running median. I'm unfamiliar with heaps, and am not sure where to begin implementing this function described here. http://programmingpraxis.com/2012/05/29/streaming-median/ My goal is to create a small test program that calculates running medians efficiently, so as the list grows the median doesn't need to be recalculated from scratch. Using two heaps, I

问题 Hello folks, I am new to R and I am trying to compute median profit for a particular country in a data frame.I tried below one but it doesn't work for me. data("Forbes2000", package = "HSAUR") median(Forbes2000[,"sales","country"="United States"]) 回答1: median(Forbes2000$sales[Forbes2000$country == "United States"]) Though it's hard to be certain without knowing what your data frame looks like. If you want to get a data.frame with the median of every country instead of just one, you could do:

问题 I have the following sql query which gives me the total h_time grouped by month, week and day. Instead I want the median h_time for month, week and day. How do I do that in Oracle SQL? SELECT DAY, MEDIAN(H_TIME) AS HANDLE_TIME FROM( select MONTH, WEEK, DAY, CASE WHEN C.JOINED IS NOT NULL THEN (NVL(C.TOTAL_TALK,0) + NVL(C.TOTAL_HOLD,0) + (NVL((C.DATETIME - C.START_DATETIME)*86400,0)) )/86400 ELSE 0 END AS H_TIME from TABLE1 C LEFT JOIN TABLE2 S ON S.ID = C.ID where c.direct = 'Inbound' ) where