【问题标题】:How to delete rows for leading and trailing NAs by group in R如何在R中按组删除前导和尾随NA的行
【发布时间】:2019-10-26 10:52:39
【问题描述】:

我需要删除包含 NA 的行,但前提是它们在前导(尾随),即在出现变量之前(之后)的任何数据。这非常类似于: How to find (not replace) leading NAs, gaps, and final NAs in data.table columns by category 和: How delete leading and trailing rows by condition in R?

但是,我需要按变量“ID”分组执行此过程。我将在后面的步骤中估算介于两者之间的 NA 数据。

这同样适用于尾随的 NA。

初始 data.frame 如下所示:

df1<-data.frame(ID=(rep(c("C1001","C1008","C1009","C1012"),each=17)),
  Year=(rep(c(1996:2012),4)),x1=(floor(runif(68,20,75))),x2= 
  (floor(runif(68,1,100))))

#Introduce leading / tailing NAs

df1[1:5,3]<-NA
df1[18:23,4]<-NA
df1[35:42,4]<-NA
df1[49:51,3]<-NA
df1[66:68,3]<-NA


#introduce "gap"- NAs
set.seed(123)
df1$x1[rbinom(68,1,0.1)==1]<-NA
df1$x2[rbinom(68,1,0.1)==1]<-NA

输出很长。这是为了在“间隙”-NA 和“领先/尾随” NA 之间做出适当的区分

head(df1,10)

      ID Year x1 x2
1  C1001 1996 NA 40
2  C1001 1997 NA 88
3  C1001 1998 NA 37
4  C1001 1999 NA 29
5  C1001 2000 NA 17
6  C1001 2001 42 18
7  C1001 2002 20 48
8  C1001 2003 30 26
9  C1001 2004 66 22
10 C1001 2005 32 67

输出应按 ID 组删除前导 NA(参见上面的第 1:5 行)。或以下示例中的第 18:23 行:

df1[18:28,]

      ID Year x1 x2
18 C1008 1996 33 NA
19 C1008 1997 26 NA
20 C1008 1998 NA NA
21 C1008 1999 51 NA
22 C1008 2000 31 NA
23 C1008 2001 44 NA
24 C1008 2002 NA 56
25 C1008 2003 47 70
26 C1008 2004 39 91
27 C1008 2005 55 62
28 C1008 2006 40 43

最终的输出应该是这样的(当然取决于抛出的随机 NA!):

      ID Year x1 x2
6  C1001 2001 42 18
7  C1001 2002 20 48
8  C1001 2003 30 26
9  C1001 2004 66 22
10 C1001 2005 32 67
11 C1001 2006 NA  5
12 C1001 2007 24 70
13 C1001 2008 33 35
14 C1001 2009 60 41
15 C1001 2010 66 82
16 C1001 2011 47 91
17 C1001 2012 41 28
24 C1008 2002 NA 56
25 C1008 2003 47 70
26 C1008 2004 39 91
27 C1008 2005 55 62
28 C1008 2006 40 43
29 C1008 2007 39 54
30 C1008 2008 49  6
31 C1008 2009 NA 26
32 C1008 2010 NA 40
33 C1008 2011 42 20
34 C1008 2012 34 83
44 C1009 2005 51 96
45 C1009 2006 66 96
46 C1009 2007 37 NA
47 C1009 2008 58 26
48 C1009 2009 34 22
52 C1012 1996 51 78
53 C1012 1997 70 17
54 C1012 1998 69 41
55 C1012 1999 35 47
56 C1012 2000 37 86
57 C1012 2001 74 92
58 C1012 2002 54 NA
59 C1012 2003 71 67
60 C1012 2004 45 95
61 C1012 2005 42 52
62 C1012 2006 56 58
63 C1012 2007 28 34
64 C1012 2008 51 35
65 C1012 2009 33  2

非常感谢!

【问题讨论】:

    标签: r group-by na delete-row


    【解决方案1】:

    这是一种使用filter_at() 的方法,该方法使用cumsum() 标识前导NA 值,以及具有相同想法但向量反转的尾随值。

    library(dplyr)
    
    df1 %>%
      group_by(ID) %>%
      filter_at(vars(-ID, -Year), all_vars(pmin(cumsum(!is.na(.)), rev(cumsum(!is.na(rev(.))))) != 0))
    
    # A tibble: 42 x 4
    # Groups:   ID [4]
       ID     Year    x1    x2
       <fct> <int> <dbl> <dbl>
     1 C1001  2001    42    18
     2 C1001  2002    20    48
     3 C1001  2003    30    26
     4 C1001  2004    66    22
     5 C1001  2005    32    67
     6 C1001  2006    NA     5
     7 C1001  2007    24    70
     8 C1001  2008    33    35
     9 C1001  2009    60    41
    10 C1001  2010    66    82
    # ... with 32 more rows
    

    【讨论】:

    • Tahnks H 1!适用于超过 2 列 - 很棒。
    【解决方案2】:

    这是一个 解决方案,它依赖于 rleid 仅删除领先的 NA:

    library(data.table)
    dt <- as.data.table(df1)
    
    dt[,
       .SD[!(rleid(x1) %in% c(1, max(rleid(x1))) & is.na(x1)) &
             !(rleid(x2) %in% c(1, max(rleid(x2))) & is.na(x2))],
       by = ID
       ]
    

    要自动处理多列,假设它们都以x 开头,您可以这样做:

    dt[dt[, Reduce('&',
                   lapply(.SD, function(x) !(rleid(x) %in% c(1, max(rleid(x1))) & is.na(x)))),
          by = ID,
          .SDcols = grep('x', names(dt))]$V1
       ]
    # or using .SD as before
    
    dt[,
       .SD[Reduce('&', lapply(.SD, function(x) !(rleid(x) %in% c(1, max(rleid(x1))) & is.na(x)))),
           .SDcols = grep('x', names(dt))],
       by = ID
       ]
    

    或与 相同的想法:

    library(dplyr)
    library(data.table)
    
    df1%>%
      group_by(ID)%>%
      filter_at(vars(starts_with('x')), all_vars(!(is.na(.) & rleid(.) %in% c(1, max(rleid(.))))))
    

    结果为 42 行:

    # A tibble: 42 x 4
    # Groups:   ID [4]
       ID     Year    x1    x2
       <fct> <int> <dbl> <dbl>
     1 C1001  2001    25    54
     2 C1001  2002    28    50
     3 C1001  2003    35    94
     4 C1001  2004    52    34
     5 C1001  2005    60    47
     6 C1001  2006    NA     9
     7 C1001  2007    67    86
     8 C1001  2008    58    40
     9 C1001  2009    61    73
    10 C1001  2010    28    18
    # ... with 32 more rows
    

    【讨论】:

    • 谢谢科尔!我更喜欢不添加任何额外列的解决方案,因为原始数据集大约有 50 个。这让我想到了我的后续问题:您知道将上述内容应用于除前两列之外的所有列的方法吗?我有一个列名列表,也许会有所帮助?
    • @Juan 查看修改后的data.table 解决方案的编辑。请注意,这为每个示例提供了 42 行。使用 rowSums() 会导致问题,特别是对于第 24 行,因为在 x1x2 之间有一串 NA。
    • @Juan 你能澄清一下预期的输出是 42 行还是 41 行?另一个答案在这个答案中提供了 41 对 42。
    • 研究了一下,我相信 42 更好。但我认为没有 100% 正确的答案。
    【解决方案3】:

    另一个使用的选项:

    f4 <- function(DT) {
        setindex(DT, ID)
        DT[, rn := .I]
        uid <- DT[,.(ID=unique(ID), V=TRUE)]
        rows <- rbindlist(lapply(cols, function(x) {
            merge(
                DT[, V := !is.na(get(x))][uid, on=c("ID", "V"), mult="first", .(ID, S=rn)],
                DT[uid, on=c("ID", "V"), mult="last", .(ID, E=rn)],
                by="ID")
        }))[, .(S=max(S), E=min(E)) , ID]
        DT[rows, on=.(ID, rn>=S, rn<=E), .SD]
    }
    f4(df1)
    

    输出:

           ID Year V1 V2 rn     V
     1: C1001 2001 45 70  6  TRUE
     2: C1001 2002 74 78  6  TRUE
     3: C1001 2003 48  9  6  TRUE
     4: C1001 2004 27 32  6  TRUE
     5: C1001 2005 39  3  6  TRUE
     6: C1001 2006 NA 89  6  TRUE
     7: C1001 2007 22  2  6  TRUE
     8: C1001 2008 56 12  6  TRUE
     9: C1001 2009 29 34  6  TRUE
    10: C1001 2010 30 53  6  TRUE
    11: C1001 2011 61 46  6  TRUE
    12: C1001 2012 23 42  6  TRUE
    13: C1008 2002 NA 95 24  TRUE
    14: C1008 2003 71 64 24  TRUE
    15: C1008 2004 41 92 24  TRUE
    16: C1008 2005 45 28 24  TRUE
    17: C1008 2006 74 59 24  TRUE
    18: C1008 2007 45 16 24  TRUE
    19: C1008 2008 57 64 24  TRUE
    20: C1008 2009 NA 35 24  TRUE
    21: C1008 2010 NA  2 24  TRUE
    22: C1008 2011 32 27 24  TRUE
    23: C1008 2012 69 41 24  TRUE
    24: C1009 2005 30 24 44  TRUE
    25: C1009 2006 43 49 44  TRUE
    26: C1009 2007 50 NA 44 FALSE
    27: C1009 2008 28 72 44  TRUE
    28: C1009 2009 43 20 44  TRUE
    29: C1012 1996 36 73 52  TRUE
    30: C1012 1997 52  4 52  TRUE
    31: C1012 1998 67 14 52  TRUE
    32: C1012 1999 39 59 52  TRUE
    33: C1012 2000 56 12 52  TRUE
    34: C1012 2001 25 92 52  TRUE
    35: C1012 2002 26 NA 52 FALSE
    36: C1012 2003 73 11 52  TRUE
    37: C1012 2004 39 50 52  TRUE
    38: C1012 2005 65 89 52  TRUE
    39: C1012 2006 70 21 52  TRUE
    40: C1012 2007 54 86 52  TRUE
    41: C1012 2008 37 70 52  TRUE
    42: C1012 2009 66 22 52  TRUE
           ID Year V1 V2 rn     V
    

    数据:

    library(data.table)
    df1 <- data.frame(ID=(rep(c("C1001","C1008","C1009","C1012"),each=17)),
        Year=(rep(c(1996:2012),4)), V1=(floor(runif(68,20,75))), V2=
            (floor(runif(68,1,100))))
    df1[1:5,3]<-NA
    df1[18:23,4]<-NA
    df1[35:42,4]<-NA
    df1[49:51,3]<-NA
    df1[66:68,3]<-NA
    set.seed(123)
    df1$V1[rbinom(68,1,0.1)==1]<-NA
    df1$V2[rbinom(68,1,0.1)==1]<-NA
    setDT(df1)[, rn := .I]
    cols <- paste0("V", 1:5)
    

    多行多组数据的时序码:

    set.seed(0L)
    if ((BIGDATA <- TRUE)) {
        nr <- 1e7
        nc <- 5
        nid <- 1e5
        dat <- data.table(ID=sample(nid, nr, TRUE),
            as.data.table(matrix(sample(c(1, NA), nr*nc, TRUE), ncol=nc)),
            key="ID")
        cols <- paste0("V", 1:5)
    } else {
        df1 <- data.frame(ID=(rep(c("C1001","C1008","C1009","C1012"),each=17)),
            Year=(rep(c(1996:2012),4)), V1=(floor(runif(68,20,75))), V2=
                (floor(runif(68,1,100))))
        df1[1:5,3]<-NA
        df1[18:23,4]<-NA
        df1[35:42,4]<-NA
        df1[49:51,3]<-NA
        df1[66:68,3]<-NA
        set.seed(123)
        df1$V1[rbinom(68,1,0.1)==1]<-NA
        df1$V2[rbinom(68,1,0.1)==1]<-NA
        dat <- setDT(df1)[, rn := .I]
        cols <- paste0("V", 1:2)
    }
    
    DT0 <- copy(dat)
    DT1 <- copy(dat)
    DT2 <- copy(dat)
    DT3 <- copy(dat)
    DT4 <- copy(dat)
    
    f0 <- function(DT) {
        DT[DT[, Reduce('&',
            lapply(.SD, function(x) {
                r <- rleid(x)
                !(r %in% c(1, max(r)) & is.na(x))
            })),
            ID,
            .SDcols=cols]$V1]
    }
    
    f1 <- function(DT) {
        DT[, c("rn", "rid") := .(.I, rowid(ID))][.N:1L, rev_rid := rowid(ID)]
    
        for (x in cols) {
            idx <- DT[is.na(get(x)) & ID %in% DT[is.na(get(x)) & (rid==1L | rev_rid==1L), ID],
                if (rid[1L]==1L || rev_rid[.N]==1L) rn,
                cumsum(c(0L, diff(rn) > 1L))]$V1
            DT <- DT[!rn %in% idx]
        }
    
        DT
    }
    
    f2 <- function(DT) {
        DT[, c("rn", "rid") := .(.I, rowid(ID))][.N:1L, rev_rid := rowid(ID)]
    
        for (x in cols) {
            DT <- DT[!rn %in% DT[is.na(get(x)),
                if (rid[1L]==1L || rev_rid[.N]==1L) rn,
                cumsum(c(0L, diff(rn) > 1L))]$V1]
        }
    
        DT
    }
    
    f3 <- function(DT) {
        DT[, rn := .I]
        rows <- DT[, transpose(lapply(.SD, function(x) c(rn[match(TRUE, !is.na(x))],
                rev(rn)[match(TRUE, !is.na(rev(x)))]))),
            ID, .SDcols=cols][, .(S=max(V1), E=min(V2)) , ID]
        DT[rows, on=.(ID, rn>=S, rn<=E), .SD]
    }
    
    f4 <- function(DT) {
        setindex(DT, ID)
        DT[, rn := .I]
        uid <- DT[,.(ID=unique(ID), V=TRUE)]
        rows <- rbindlist(lapply(cols, function(x) {
            merge(
                DT[, V := !is.na(get(x))][uid, on=c("ID", "V"), mult="first", .(ID, S=rn)],
                DT[uid, on=c("ID", "V"), mult="last", .(ID, E=rn)],
                by="ID")
        }))[, .(S=max(S), E=min(E)) , ID]
        DT[rows, on=.(ID, rn>=S, rn<=E), .SD]
    }
    
    microbenchmark::microbenchmark(f0(DT0), f1(DT1), f2(DT2), f3(DT3), f4(DT4), times=3L)
    

    时间安排:

    Unit: seconds
        expr       min        lq      mean    median        uq       max neval
     f0(DT0)  8.874985  8.950951  8.993281  9.026917  9.052429  9.077942     3
     f1(DT1) 16.249656 16.337013 16.657910 16.424370 16.862038 17.299706     3
     f2(DT2) 18.225748 18.284212 18.391198 18.342676 18.473922 18.605169     3
     f3(DT3) 10.361079 10.612313 10.698897 10.863548 10.867806 10.872063     3
     f4(DT4)  3.106936  3.137846  3.173174  3.168755  3.206293  3.243830     3
    

    另一个行数相同但组数少得多的测试:

    set.seed(0L)
    nr <- 1e7
    nc <- 5
    nid <- 1e2
    dat <- data.table(ID=sample(nid, nr, TRUE),
        as.data.table(matrix(sample(c(1, NA), nr*nc, TRUE), ncol=nc)),
        key="ID")
    cols <- paste0("V", 1:5)
    DT0 <- copy(dat)
    DT3 <- copy(dat)
    microbenchmark::microbenchmark(f0(DT0), f3(DT3), f4(DT4), times=3L)
    

    时间安排:

    Unit: seconds
        expr      min       lq     mean   median       uq      max neval
     f0(DT0) 2.317905 2.331944 2.358256 2.345983 2.378431 2.410880     3
     f3(DT3) 2.108385 2.123889 2.132315 2.139392 2.144280 2.149168     3
     f4(DT4) 2.050805 2.079687 2.101211 2.108569 2.126414 2.144258     3
    

    【讨论】:

    • @Cole 我不知道。只是发现 non equi join 比 foverlaps 更具可读性
    • 我同意并且出于同样的原因通常远离foverlaps()。我查了一下,不相信foverlaps()可以用。
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