如何删除R中按组引导和跟踪NAs的行
如何删除R中按组引导和跟踪NAs的行
提问于 2019-10-26 10:52:39
我需要删除包含NA的行，但前提是它们正在引导(尾随)，即在变量出现之前(之后)出现任何数据。这非常类似于：如何按类别在data.table列中查找(而不是替换)引导NAs、间隙和最终NAs和：如何在R中按条件删除前导行和尾行？
但是，我需要按照变量"ID“分组这个过程。我将在稍后的步骤中将NAs的数据计算在两者之间。
这同样适用于尾随的NAs。
初始的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输出相当长。这是为了尽可能地区分"gap"-NAs和“引导/跟踪”
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输出应该按照ID组去除前面的NAs (请参见上面的第1:5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最后的输出应该如下所示(当然，这取决于随机抛出的NAs！)：
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谢谢一堆人！
回答 3
Stack Overflow用户
回答已采纳
发布于 2019-10-26 12:19:39
这里有一种使用filter_at()的方法，它用cumsum()标识领先的NA值，并用相同的思想标识尾随的值，但是向量相反。
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 rowsStack Overflow用户
发布于 2019-10-26 11:28:13
下面是一个data.table解决方案，它依赖于rleid只删除前面的NAs：
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
]若要自动使用多列执行此操作，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
]或与德普利有相同的想法
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(.))))))结果共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 rowsStack Overflow用户
发布于 2019-11-04 07:55:25
使用data.table的另一个选项
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)产出：
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$V2[rbinom(68,1,0.1)==1]<-NA
setDT(df1)[, rn := .I]
cols <- paste0("V", 1:5)具有大量行和大量组的数据的定时代码：
set.seed(0L)
if ((BIGDATA <- TRUE)) {
nr <- 1e7
nc <- 5
nid <- 1e5
dat <- data.table(ID=sample(nid, nr, TRUE),
as.data.table(matrix(sample(c(1, NA), nr*nc, TRUE), ncol=nc)),
key="ID")
cols <- paste0("V", 1:5)
} else {
df1 <- data.frame(ID=(rep(c("C1001","C1008","C1009","C1012"),each=17)),
Year=(rep(c(1996:2012),4)), V1=(floor(runif(68,20,75))), V2=
(floor(runif(68,1,100))))
df1[1:5,3]<-NA
df1[18:23,4]<-NA
df1[35:42,4]<-NA
df1[49:51,3]<-NA
df1[66:68,3]<-NA
set.seed(123)
df1$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)计时：
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另一个具有相同行数但组数要小得多的测试：
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)计时：
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/58570110
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