【问题标题】:How to remove NAs from the beginning and the end of a dataframe in R?如何从 R 中的数据帧的开头和结尾删除 NA?
【发布时间】:2023-03-20 06:03:01
【问题描述】:

我正在尝试使用zoo:na.approx 按组插入一些值。数据帧需要以非 NA 值开始和结束。有没有办法删除它们但保留“内部”NA?我不能使用基于其他变量的过滤器,因为插值是按组执行的,并且缺失值因组而异。

这是我的代码示例:

library(zoo)
library(lubridate)
library(dplyr)
set.seed(471)

db <- rep(seq(ymd("2021-12-20"), ymd("2021-12-30"), by = "days"),4) %>% merge(seq(1,4,1)) %>%
  mutate(z=rnorm(176))
db$z[db$z<0] <- NA

db %>% group_by(y) %>% mutate(aa=na.approx(z))

【问题讨论】:

    标签: r dplyr na


    【解决方案1】:

    rule=2 参数添加到na.approx 以在每个组的开头和结尾推断NAs,以便它们不是NA

    db %>%
      group_by(y) %>%
      mutate(aa=na.approx(z, rule = 2)) %>%
      ungroup
    

    或使用na.trim 删除每组开头和结尾的NA。

    db %>%
      group_by(y) %>%
      group_modify(~ na.trim(.)) %>%
      mutate(aa = na.approx(z)) %>%
      ungroup
    

    【讨论】:

      【解决方案2】:

      我将重点关注每组的前/后 3 行:

      db %>%
        group_by(y) %>%
        slice(c(1:3, n() - 2:0)) %>%
        print(n=99)
      # # A tibble: 24 x 3
      # # Groups:   y [4]
      #    x              y        z
      #    <date>     <dbl>    <dbl>
      #  1 2021-12-20     1 NA      
      #  2 2021-12-21     1  0.605  
      #  3 2021-12-22     1  0.185  
      #  4 2021-12-28     1  0.805  
      #  5 2021-12-29     1 NA      
      #  6 2021-12-30     1 NA      
      #  7 2021-12-20     2 NA      
      #  8 2021-12-21     2  0.402  
      #  9 2021-12-22     2 NA      
      # 10 2021-12-28     2 NA      
      # 11 2021-12-29     2  0.163  
      # 12 2021-12-30     2  0.796  
      # 13 2021-12-20     3  1.00   
      # 14 2021-12-21     3 NA      
      # 15 2021-12-22     3  0.733  
      # 16 2021-12-28     3  0.00858
      # 17 2021-12-29     3 NA      
      # 18 2021-12-30     3  0.179  
      # 19 2021-12-20     4 NA      
      # 20 2021-12-21     4  0.298  
      # 21 2021-12-22     4 NA      
      # 22 2021-12-28     4  0.355  
      # 23 2021-12-29     4  2.42   
      # 24 2021-12-30     4 NA      
      

      第 1 组和第 4 组在 NA 开始/结束,第 2 组在 NA 开始。

      试试这个:

      db %>%
        group_by(y) %>%
        filter(cumany(!is.na(z)) & rev(cumany(rev(!is.na(z))))) %>%
        slice(c(1:3, n() - 2:0)) %>%
        print(n=99)
      # # A tibble: 24 x 3
      # # Groups:   y [4]
      #    x              y        z
      #    <date>     <dbl>    <dbl>
      #  1 2021-12-21     1  0.605  
      #  2 2021-12-22     1  0.185  
      #  3 2021-12-23     1 NA      
      #  4 2021-12-26     1  0.871  
      #  5 2021-12-27     1 NA      
      #  6 2021-12-28     1  0.805  
      #  7 2021-12-21     2  0.402  
      #  8 2021-12-22     2 NA      
      #  9 2021-12-23     2  0.364  
      # 10 2021-12-28     2 NA      
      # 11 2021-12-29     2  0.163  
      # 12 2021-12-30     2  0.796  
      # 13 2021-12-20     3  1.00   
      # 14 2021-12-21     3 NA      
      # 15 2021-12-22     3  0.733  
      # 16 2021-12-28     3  0.00858
      # 17 2021-12-29     3 NA      
      # 18 2021-12-30     3  0.179  
      # 19 2021-12-21     4  0.298  
      # 20 2021-12-22     4 NA      
      # 21 2021-12-23     4  0.660  
      # 22 2021-12-27     4 NA      
      # 23 2021-12-28     4  0.355  
      # 24 2021-12-29     4  2.42   
      

      【讨论】:

        【解决方案3】:

        您可以先进行近似,然后删除NAs:

        db %>% 
          group_by(y) %>% 
          mutate(output = zoo::na.approx(z, na.rm = FALSE))
        

        输出:

        # A tibble: 176 x 4
        # Groups:   y [4]
           x              y      z   test
           <date>     <dbl>  <dbl>  <dbl>
         1 2021-12-20     1 NA     NA    
         2 2021-12-21     1  0.605  0.605
         3 2021-12-22     1  0.185  0.185
         4 2021-12-23     1 NA      0.455
         5 2021-12-24     1  0.725  0.725
         6 2021-12-25     1  1.51   1.51 
         7 2021-12-26     1 NA      1.41 
         8 2021-12-27     1  1.31   1.31 
         9 2021-12-28     1  1.07   1.07 
        10 2021-12-29     1  1.14   1.14 
        

        您可以部分地看到,na.approx 中的 na.rm = FALSE 参数保留每个组的顶部和底部 NA,同时计算组内的近似值。然后您可以过滤数据以删除新创建的列中的NA

        db %>% 
          group_by(y) %>% 
          mutate(output = zoo::na.approx(z, na.rm = F)) %>% 
          ungroup() %>% 
          filter(!is.na(output))
        

        【讨论】:

          【解决方案4】:

          您可以使用imputeTS::na_kalman,它也可以推断。

          r <- do.call(rbind, by(db, db$y, FUN=\(x) transform(x, aa=imputeTS::na_kalman(z))))
          
          tail(r[r$y == 1, ])
          #               x y           z          aa
          # 1.39 2021-12-25 1 0.020848035 0.020848035
          # 1.40 2021-12-26 1 0.017171691 0.017171691
          # 1.41 2021-12-27 1 0.007122718 0.007122718
          # 1.42 2021-12-28 1          NA 0.392535303
          # 1.43 2021-12-29 1 0.629796532 0.629796532
          # 1.44 2021-12-30 1          NA 0.258814648
          

          数据:

          db <- structure(list(x = structure(c(18981, 18982, 18983, 18984, 18985, 
          18986, 18987, 18988, 18989, 18990, 18991, 18981, 18982, 18983, 
          18984, 18985, 18986, 18987, 18988, 18989, 18990, 18991, 18981, 
          18982, 18983, 18984, 18985, 18986, 18987, 18988, 18989, 18990, 
          18991, 18981, 18982, 18983, 18984, 18985, 18986, 18987, 18988, 
          18989, 18990, 18991, 18981, 18982, 18983, 18984, 18985, 18986, 
          18987, 18988, 18989, 18990, 18991, 18981, 18982, 18983, 18984, 
          18985, 18986, 18987, 18988, 18989, 18990, 18991, 18981, 18982, 
          18983, 18984, 18985, 18986, 18987, 18988, 18989, 18990, 18991, 
          18981, 18982, 18983, 18984, 18985, 18986, 18987, 18988, 18989, 
          18990, 18991, 18981, 18982, 18983, 18984, 18985, 18986, 18987, 
          18988, 18989, 18990, 18991, 18981, 18982, 18983, 18984, 18985, 
          18986, 18987, 18988, 18989, 18990, 18991, 18981, 18982, 18983, 
          18984, 18985, 18986, 18987, 18988, 18989, 18990, 18991, 18981, 
          18982, 18983, 18984, 18985, 18986, 18987, 18988, 18989, 18990, 
          18991, 18981, 18982, 18983, 18984, 18985, 18986, 18987, 18988, 
          18989, 18990, 18991, 18981, 18982, 18983, 18984, 18985, 18986, 
          18987, 18988, 18989, 18990, 18991, 18981, 18982, 18983, 18984, 
          18985, 18986, 18987, 18988, 18989, 18990, 18991, 18981, 18982, 
          18983, 18984, 18985, 18986, 18987, 18988, 18989, 18990, 18991
          ), class = "Date"), y = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 
          1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 
          1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 
          2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 
          2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 
          3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 
          3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 
          4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 
          4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4), z = c(0.305344789017667, 
          0.256644623614096, NA, 1.31852719135355, 0.115506505762677, 0.732802091953865, 
          NA, 0.239925107412262, 0.685318244939073, 0.691973256906341, 
          1.32378575746467, NA, 0.384693043255873, 1.45895509632899, NA, 
          0.0599714441492927, NA, NA, NA, NA, NA, 0.71683339822062, NA, 
          3.27310516365819, 1.69204573033578, NA, 0.14017486940184, NA, 
          1.16261380170504, NA, NA, NA, 1.68438289810619, NA, NA, 1.31386940315565, 
          0.594623922245712, NA, 0.0208480351055444, 0.0171716909393243, 
          0.00712271758331095, NA, 0.629796532479193, NA, 0.244580018794366, 
          NA, 0.820911116824006, NA, NA, 0.557088403848106, 0.0130780982496676, 
          NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 1.28902764727033, 
          0.242390057597798, NA, 1.75609046517858, 0.921685169855448, 0.240269454747801, 
          NA, 0.133290865347424, 0.760944667549314, NA, 2.10865624982592, 
          0.201965354187563, NA, 0.372617511993437, 0.40925122336274, 0.598185767876918, 
          NA, NA, 1.51486434937749, NA, 0.365799492559624, 1.93980359376164, 
          NA, NA, NA, 1.39839171014837, NA, NA, 1.131273582479, 1.35134680218024, 
          NA, 1.02956577738351, 0.271873664141861, 0.777813782525466, NA, 
          NA, 0.286721974151372, 0.0305405702707527, NA, NA, 0.922064532313788, 
          NA, 0.211308210750866, NA, NA, 0.416086290075234, 0.744175318362445, 
          1.05570394997758, NA, 2.10096763825364, NA, NA, 0.945801512771798, 
          1.64923864766573, NA, 0.0338301608791077, 1.93867810865554, 0.611903344641826, 
          NA, NA, NA, 0.664664842786913, 0.992532329760494, 0.106067365628389, 
          NA, NA, 0.253237072580547, 1.39727781231248, 0.750659506338532, 
          NA, NA, 0.531677176826455, NA, 0.334496935245917, NA, 0.237217689673067, 
          NA, 0.729615340974382, 0.418007005399876, NA, NA, NA, 0.575142620388619, 
          2.27297683347494, NA, 1.0088509112411, NA, NA, NA, 1.07213691727514, 
          NA, 0.950964366873889, NA, NA, 1.37008596018781, NA, 0.581570283604887, 
          0.903895963902468, NA, 0.170520505104898, 0.664123540127705, 
          1.20066990898952, NA, 0.243496848502427, 0.679868588335254, NA, 
          2.09127742408436, 0.77948087799739, NA, 0.658167166169738, NA, 
          2.15919199233993, NA, 0.778191585042783)), row.names = c(NA, 
          -176L), class = "data.frame")
          

          【讨论】:

            【解决方案5】:

            另一种可能的解决方案:

            library(zoo)
            library(lubridate)
            library(dplyr)
            
            set.seed(471)
            
            db <- rep(seq(ymd("2021-12-20"), ymd("2021-12-30"), by = "days"),4) %>% merge(seq(1,4,1)) %>%
              mutate(z=rnorm(176))
            db$z[db$z<0] <- NA
            
            db %>% 
              group_by(y) %>% 
              mutate(aux = data.table::rleid(z)) %>% 
              filter(!((aux == 1 | aux == max(aux)) & is.na(z))) %>% 
              ungroup %>% select(-aux) %>% mutate(aa=na.approx(z))
            
            #> # A tibble: 170 × 4
            #>    x              y      z    aa
            #>    <date>     <dbl>  <dbl> <dbl>
            #>  1 2021-12-21     1  0.605 0.605
            #>  2 2021-12-22     1  0.185 0.185
            #>  3 2021-12-23     1 NA     0.455
            #>  4 2021-12-24     1  0.725 0.725
            #>  5 2021-12-25     1  1.51  1.51 
            #>  6 2021-12-26     1 NA     1.41 
            #>  7 2021-12-27     1  1.31  1.31 
            #>  8 2021-12-28     1  1.07  1.07 
            #>  9 2021-12-29     1  1.14  1.14 
            #> 10 2021-12-30     1 NA     0.585
            #> # … with 160 more rows
            

            【讨论】:

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