如何从 R 中的数据帧的开头和结尾删除 NA?

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【中文标题】如何从 R 中的数据帧的开头和结尾删除 NA?【英文标题】:How to remove NAs from the beginning and the end of a dataframe in R? 【发布时间】:2022-01-23 18:33:19 【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】:

【参考方案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】:

【参考方案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】:

【参考方案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】:

【参考方案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】:

【参考方案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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