用R中的数据表计算数列分位数的问题_R_Datatable_Quantile

用R中的数据表计算数列分位数的问题

用R中的数据表计算数列分位数的问题,r,datatable,quantile,R,Datatable,Quantile,我需要为数据表R中的每一行计算数字序列的分位数 Table: 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 2018 2019 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA

我需要为数据表R中的每一行计算数字序列的分位数

Table:  

       2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 2018 2019
         NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA
       11.7 10.7 10.8 11.8 12.2 13.8  7.0 10.2 11.2  6.8  7.4  9.1  9.5  9.4  9.3 15.6 11.3 13.0 10.9 10.5
         NA   NA  9.5 11.3 16.6 12.2   NA   NA 69.6   NA   NA 12.4 10.8 10.5  8.8  9.9   NA  7.7 12.1   NA
        9.1  8.7 29.9 23.1 18.3 23.5 21.5 23.0 18.2 28.8 39.9 16.4 16.9 23.4 18.8 31.9 26.2 22.4 29.2 25.2
       14.7 17.5 21.1 19.4 20.0 14.5 14.1 12.6  9.9 12.6  6.4  9.6 18.5 14.3 26.2 10.7  6.4  6.9  7.1  9.0

我想计算上面表格中每一行的分位数。请参阅下面的代码，但我需要将每行的值放置在“输出”中

一个选项是按行分组

year_cols <- as.character(2000:2019)
Table[, c("10%", "25%", "50%", "75%", "100%") := 
   as.list(quantile(unlist(.SD), na.rm = TRUE, 
       c(0.1,0.25,0.5,0.75,1.0))), by = seq_len(nrow(Table)), 
        .SDcols = year_cols]
Table
#   2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 2018 2019   10%
#1:   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA    NA
#2: 11.7 10.7 10.8 11.8 12.2 13.8  7.0 10.2 11.2  6.8  7.4  9.1  9.5  9.4  9.3 15.6 11.3 13.0 10.9 10.5  7.36
#3:   NA   NA  9.5 11.3 16.6 12.2   NA   NA 69.6   NA   NA 12.4 10.8 10.5  8.8  9.9   NA  7.7 12.1   NA  8.87
#4:  9.1  8.7 29.9 23.1 18.3 23.5 21.5 23.0 18.2 28.8 39.9 16.4 16.9 23.4 18.8 31.9 26.2 22.4 29.2 25.2 15.67
#5: 14.7 17.5 21.1 19.4 20.0 14.5 14.1 12.6  9.9 12.6  6.4  9.6 18.5 14.3 26.2 10.7  6.4  6.9  7.1  9.0  6.85
#      25%   50%    75% 100%
#1:     NA    NA     NA   NA
#2:  9.375 10.75 11.725 15.6
#3:  9.800 11.05 12.250 69.6
#4: 18.275 23.05 26.850 39.9
#5:  9.450 13.35 17.750 26.2

数据

对不起。我不想要“分位数”列，但请查看“输出”表。我希望prob值作为列标题，并为每行计算分位数值。“输出”表第二行中的分位数值是使用第二行中的值计算的。非常感谢您的快速回复。这就是我一直在寻找的答案。
Output:

2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 2018 2019 10%  25%  50%  75%  100% 
  NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA  NA  NA   NA   NA   NA
11.7 10.7 10.8 11.8 12.2 13.8  7.0 10.2 11.2  6.8  7.4  9.1  9.5  9.4  9.3 15.6 11.3 13.0 10.9 10.5 7.36 9.37 10.75 11.72 15.60 
  NA   NA  9.5 11.3 16.6 12.2   NA   NA 69.6   NA   NA 12.4 10.8 10.5  8.8  9.9   NA  7.7 12.1   NA
 9.1  8.7 29.9 23.1 18.3 23.5 21.5 23.0 18.2 28.8 39.9 16.4 16.9 23.4 18.8 31.9 26.2 22.4 29.2 25.2
14.7 17.5 21.1 19.4 20.0 14.5 14.1 12.6  9.9 12.6  6.4  9.6 18.5 14.3 26.2 10.7  6.4  6.9  7.1  9.0

year_cols <- as.character(2000:2019)
Table[, c("10%", "25%", "50%", "75%", "100%") := 
   as.list(quantile(unlist(.SD), na.rm = TRUE, 
       c(0.1,0.25,0.5,0.75,1.0))), by = seq_len(nrow(Table)), 
        .SDcols = year_cols]
Table
#   2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 2018 2019   10%
#1:   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA    NA
#2: 11.7 10.7 10.8 11.8 12.2 13.8  7.0 10.2 11.2  6.8  7.4  9.1  9.5  9.4  9.3 15.6 11.3 13.0 10.9 10.5  7.36
#3:   NA   NA  9.5 11.3 16.6 12.2   NA   NA 69.6   NA   NA 12.4 10.8 10.5  8.8  9.9   NA  7.7 12.1   NA  8.87
#4:  9.1  8.7 29.9 23.1 18.3 23.5 21.5 23.0 18.2 28.8 39.9 16.4 16.9 23.4 18.8 31.9 26.2 22.4 29.2 25.2 15.67
#5: 14.7 17.5 21.1 19.4 20.0 14.5 14.1 12.6  9.9 12.6  6.4  9.6 18.5 14.3 26.2 10.7  6.4  6.9  7.1  9.0  6.85
#      25%   50%    75% 100%
#1:     NA    NA     NA   NA
#2:  9.375 10.75 11.725 15.6
#3:  9.800 11.05 12.250 69.6
#4: 18.275 23.05 26.850 39.9
#5:  9.450 13.35 17.750 26.2

library(matrixStats)
Table[, c("10%", "25%", "50%", "75%", "100%") := 
    as.data.frame(rowQuantiles(as.matrix(.SD), na.rm = TRUE,
     probs = c(0.1,0.25,0.5,0.75,1.0))), .SDcols = as.character(year_cols)]



Table
#   2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 2018 2019   10%
#1:   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   NA    NA
#2: 11.7 10.7 10.8 11.8 12.2 13.8  7.0 10.2 11.2  6.8  7.4  9.1  9.5  9.4  9.3 15.6 11.3 13.0 10.9 10.5  7.36
#3:   NA   NA  9.5 11.3 16.6 12.2   NA   NA 69.6   NA   NA 12.4 10.8 10.5  8.8  9.9   NA  7.7 12.1   NA  8.87
#4:  9.1  8.7 29.9 23.1 18.3 23.5 21.5 23.0 18.2 28.8 39.9 16.4 16.9 23.4 18.8 31.9 26.2 22.4 29.2 25.2 15.67
#5: 14.7 17.5 21.1 19.4 20.0 14.5 14.1 12.6  9.9 12.6  6.4  9.6 18.5 14.3 26.2 10.7  6.4  6.9  7.1  9.0  6.85
#      25%   50%    75% 100%
#1:     NA    NA     NA   NA
#2:  9.375 10.75 11.725 15.6
#3:  9.800 11.05 12.250 69.6
#4: 18.275 23.05 26.850 39.9
#5:  9.450 13.35 17.750 26.2

Table <- structure(list(`2000` = c(NA, 11.7, NA, 9.1, 14.7), `2001` = c(NA, 
 10.7, NA, 8.7, 17.5), `2002` = c(NA, 10.8, 9.5, 29.9, 21.1), 
     `2003` = c(NA, 11.8, 11.3, 23.1, 19.4), `2004` = c(NA, 12.2, 
     16.6, 18.3, 20), `2005` = c(NA, 13.8, 12.2, 23.5, 14.5), 
     `2006` = c(NA, 7, NA, 21.5, 14.1), `2007` = c(NA, 10.2, NA, 
     23, 12.6), `2008` = c(NA, 11.2, 69.6, 18.2, 9.9), `2009` = c(NA, 
     6.8, NA, 28.8, 12.6), `2010` = c(NA, 7.4, NA, 39.9, 6.4), 
     `2011` = c(NA, 9.1, 12.4, 16.4, 9.6), `2012` = c(NA, 9.5, 
     10.8, 16.9, 18.5), `2013` = c(NA, 9.4, 10.5, 23.4, 14.3), 
     `2014` = c(NA, 9.3, 8.8, 18.8, 26.2), `2015` = c(NA, 15.6, 
     9.9, 31.9, 10.7), `2016` = c(NA, 11.3, NA, 26.2, 6.4), `2017` = c(NA, 
     13, 7.7, 22.4, 6.9), `2018` = c(NA, 10.9, 12.1, 29.2, 7.1
     ), `2019` = c(NA, 10.5, NA, 25.2, 9)), class = c("data.table", 
 "data.frame"), row.names = c(NA, -5L))