median

Finding the Median value from a table, Group By Date SQLServer

喜欢而已 提交于 2021-02-17 05:22:10
问题 I have a complicated problem I am trying to solve. Please bear with me and feel free to ask any questions. I am quite new to SQL and having difficulty with this... I need to count the median of a group of values. Now the values are not given in a table. The values are derived from a table based on hourly occurrences grouped by date. Here's the sample table from where data is pooled. CREATE TABLE Table22( Request_Number BIGINT NOT NULL ,Request_Received_Date DATETIME NOT NULL ); INSERT INTO

Calculate Median for each group in AWS Athena table

强颜欢笑 提交于 2021-02-11 15:03:33
问题 Below is the schema for the athena table I wish to calculate median for 'parameter_value' group by standard_lab_parameter_name & units. For this I followed link : https://docs.aws.amazon.com/redshift/latest/dg/r_MEDIAN.html But on running the query select median(parameter_value) from table_name group by standard_lab_parameter_name, units It throws error SYNTAX_ERROR: line 1:8: Function median not registered Any help? Or if some alternative query would be great 回答1: Athena is based on Presto 0

Calculate Median for each group in AWS Athena table

心不动则不痛 提交于 2021-02-11 15:02:30
问题 Below is the schema for the athena table I wish to calculate median for 'parameter_value' group by standard_lab_parameter_name & units. For this I followed link : https://docs.aws.amazon.com/redshift/latest/dg/r_MEDIAN.html But on running the query select median(parameter_value) from table_name group by standard_lab_parameter_name, units It throws error SYNTAX_ERROR: line 1:8: Function median not registered Any help? Or if some alternative query would be great 回答1: Athena is based on Presto 0

Python/Pandas Dataframe replace 0 with median value

↘锁芯ラ 提交于 2021-02-06 20:00:02
问题 I have a python pandas dataframe with several columns and one column has 0 values. I want to replace the 0 values with the median or mean of this column. data is my dataframe artist_hotness is the column mean_artist_hotness = data['artist_hotness'].dropna().mean() if len(data.artist_hotness[ data.artist_hotness.isnull() ]) > 0: data.artist_hotness.loc[ (data.artist_hotness.isnull()), 'artist_hotness'] = mean_artist_hotness I tried this, but it is not working. 回答1: I think you can use mask and

Python/Pandas Dataframe replace 0 with median value

我的梦境 提交于 2021-02-06 19:59:07
问题 I have a python pandas dataframe with several columns and one column has 0 values. I want to replace the 0 values with the median or mean of this column. data is my dataframe artist_hotness is the column mean_artist_hotness = data['artist_hotness'].dropna().mean() if len(data.artist_hotness[ data.artist_hotness.isnull() ]) > 0: data.artist_hotness.loc[ (data.artist_hotness.isnull()), 'artist_hotness'] = mean_artist_hotness I tried this, but it is not working. 回答1: I think you can use mask and

median of selected rows dependent on other columns values [closed]

霸气de小男生 提交于 2021-01-29 10:31:12
问题 Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 12 months ago . Improve this question I have the following data frame (here just a tiny part from a big one) ID= c(1,1,1,2,2,2,2,3,3) week = c(1,1,2,1,1,2,2,1,2) X = c(3.3,4.23,5.6,12,3.1,4.3,5.9,6.1,5.3) Y = c(1.3,2.4,6.8,5.5,4.3,3,6.6,2.6,5.7) TS_DF = data.frame(ID,week,X,Y) I would like to

Pandas groupby and correct with median in new column

扶醉桌前 提交于 2021-01-29 03:09:48
问题 My dataframe look like this Plate Sample LogRatio P1 S1 0.42 P1 S2 0.23 P2 S3 0.41 P3 S4 0.36 P3 S5 0.18 I have calculated the median of each plate (but it's probably not the best idea to start like this) grouped = df.groupby("Plate") medianesPlate = grouped["LogRatio"].median() And I want to add a column on my dataframe CorrectedLogRatio = LogRatio-median(plate) I suppose with : df["CorrectedLogRatio"] = LogRatio-median(plate) To have something like this : Plate Sample LogRatio

ImageMagick: custom rank filter? Like erode or dilate or median but a different rank, e.g. a 'percentile' filter?

假如想象 提交于 2021-01-28 05:40:38
问题 When applying a rank filter with ImageMagick such as erode or delate or median, it takes the minimum ( erode ) or maximum ( dilate ) or middle ( median ) value of all pixels within a certain radius or custom shape around each source pixel. Is it also possible to take a custom rank of the surrounding pixel values? For example when using a 5x5 square, with the erode or dilate or median filters, respectively the lowest, highest, of middle value of the 25 pixels surrounding each pixel is taken.

How can I ignore zeros when I take the median on columns of an array?

六月ゝ 毕业季﹏ 提交于 2021-01-21 04:10:05
问题 I have a simple numpy array. array([[10, 0, 10, 0], [ 1, 1, 0, 0] [ 9, 9, 9, 0] [ 0, 10, 1, 0]]) I would like to take the median of each column, individually, of this array. However, there are a few 0 values in various places which I would like to ignore in the calculation of the medians. To further complicate, I would like to keep the columns with only 0 entries as having the median of 0 . In this manner, those columns would serve as a bit of a place holder, keeping the dimensions of the

How can I ignore zeros when I take the median on columns of an array?

不打扰是莪最后的温柔 提交于 2021-01-21 04:09:05
问题 I have a simple numpy array. array([[10, 0, 10, 0], [ 1, 1, 0, 0] [ 9, 9, 9, 0] [ 0, 10, 1, 0]]) I would like to take the median of each column, individually, of this array. However, there are a few 0 values in various places which I would like to ignore in the calculation of the medians. To further complicate, I would like to keep the columns with only 0 entries as having the median of 0 . In this manner, those columns would serve as a bit of a place holder, keeping the dimensions of the