插补中间缺失值

Question

我有一个组级别的数据。 data 如下所示。

我的实际数据是“Value”，所需数据是“Expected_Value”。

我尝试了以下代码：

setDT(file_to_share)[,Expected_Value := na.locf(na.locf(Value, na.rm=FALSE), fromLast=TRUE),by = c("Group_A",   "Group_B")]

但在这段代码中，插补是在整个缺失值上完成的。如果缺失值在值之间，我想计算缺失值。而缺失值将是先前可用值的复制。

如果有人可以指导我如何做，那将是一个很大的帮助。

注意：我尝试使用data.table 和zoo 进行计算。但任何其他方法也可以。

Answer 1

OP 已请求仅填充每个组内其他值之间的 NA 值。这意味着在应用zoo::na.locf() 时，在每个组的开头或结尾跳过任何NA 值序列。

使用data.table，这可以通过识别要跳过的行的索引和一种反连接来完成：

library(data.table)
setDT(DT)[!DT[, {
  na_grp <- rleid(is.na(Value))
  .I[na_grp %in% c(1L, max(na_grp))]
}, by = .(Group_A, GROUP_B)]$V1, Value := zoo::na.locf(Value)][]

    Group_A    GROUP_B     Date Value Expected_Value
 1: GROUP_1  Group_1_1 1/1/2017    NA             NA
 2: GROUP_1  Group_1_1 1/2/2017    NA             NA
 3: GROUP_1  Group_1_1 1/3/2017    34             34
 4: GROUP_1  Group_1_1 1/4/2017    20             20
 5: GROUP_1  Group_1_1 1/5/2017    20             20
 6: GROUP_1  Group_1_1 1/6/2017    20             20
 7: GROUP_1  Group_1_1 1/7/2017    38             38
 8: GROUP_1  Group_1_2 1/1/2017    35             35
 9: GROUP_1  Group_1_2 1/2/2017    28             28
10: GROUP_1  Group_1_2 1/3/2017    20             28
11: GROUP_1  Group_1_2 1/4/2017    32             32
12: GROUP_1  Group_1_2 1/5/2017    39             39
13: GROUP_1  Group_1_2 1/6/2017    28             28
14: GROUP_1  Group_1_2 1/7/2017    NA             NA
15: GROUP_2 Group_1_11 1/1/2017    NA             NA
16: GROUP_2 Group_1_11 1/2/2017    NA             NA
17: GROUP_2 Group_1_11 1/3/2017    40             40
18: GROUP_2 Group_1_11 1/4/2017    32             32
19: GROUP_2 Group_1_11 1/5/2017    20             20
20: GROUP_2 Group_1_11 1/6/2017    NA             NA
21: GROUP_2 Group_1_11 1/7/2017    NA             NA
22: GROUP_2 Group_1_21 1/1/2017    NA             NA
23: GROUP_2 Group_1_21 1/2/2017    32             32
24: GROUP_2 Group_1_21 1/3/2017    36             36
25: GROUP_2 Group_1_21 1/4/2017    36             36
26: GROUP_2 Group_1_21 1/5/2017    28             28
27: GROUP_2 Group_1_21 1/6/2017    33             33
28: GROUP_2 Group_1_21 1/7/2017    40             40
29: GROUP_3 Group_1_13 1/1/2017    NA             NA
30: GROUP_3 Group_1_13 1/2/2017    NA             NA
31: GROUP_3 Group_1_13 1/3/2017    NA             NA
32: GROUP_3 Group_1_13 1/4/2017    29             29
33: GROUP_3 Group_1_13 1/5/2017    31             31
34: GROUP_3 Group_1_13 1/6/2017    31             31
35: GROUP_3 Group_1_13 1/7/2017    34             34
36: GROUP_3 Group_1_23 1/1/2017    26             26
37: GROUP_3 Group_1_23 1/2/2017    33             33
38: GROUP_3 Group_1_23 1/3/2017    27             27
39: GROUP_3 Group_1_23 1/4/2017    23             23
40: GROUP_3 Group_1_23 1/5/2017    25             25
41: GROUP_3 Group_1_23 1/6/2017    41             41
42: GROUP_3 Group_1_23 1/7/2017    25             25
    Group_A    GROUP_B     Date Value Expected_Value

说明

• 对于每个组，NA/非NA 值的条纹已编号
• 每组中的第一个和最后一个条纹被选中，索引从特殊符号.I 中检索。 （由于Value 将在原地更新，无论第一条或最后一条是否包含NA 都无关紧要；它们无论如何都不会被更新。）
• 找到的索引
  DT[, {na_grp <- rleid(is.na(Value)); .I[na_grp %in% c(1L, max(na_grp))]}, by = .(Group_A, GROUP_B)]$V1
  被排除在外，因此zoo::na.locf(Value) 仅适用于每组的“内部”条纹。

数据

DT <- structure(list(Group_A = c("GROUP_1", "GROUP_1", "GROUP_1", "GROUP_1", 
"GROUP_1", "GROUP_1", "GROUP_1", "GROUP_1", "GROUP_1", "GROUP_1", 
"GROUP_1", "GROUP_1", "GROUP_1", "GROUP_1", "GROUP_2", "GROUP_2", 
"GROUP_2", "GROUP_2", "GROUP_2", "GROUP_2", "GROUP_2", "GROUP_2", 
"GROUP_2", "GROUP_2", "GROUP_2", "GROUP_2", "GROUP_2", "GROUP_2", 
"GROUP_3", "GROUP_3", "GROUP_3", "GROUP_3", "GROUP_3", "GROUP_3", 
"GROUP_3", "GROUP_3", "GROUP_3", "GROUP_3", "GROUP_3", "GROUP_3", 
"GROUP_3", "GROUP_3"), GROUP_B = c("Group_1_1", "Group_1_1", 
"Group_1_1", "Group_1_1", "Group_1_1", "Group_1_1", "Group_1_1", 
"Group_1_2", "Group_1_2", "Group_1_2", "Group_1_2", "Group_1_2", 
"Group_1_2", "Group_1_2", "Group_1_11", "Group_1_11", "Group_1_11", 
"Group_1_11", "Group_1_11", "Group_1_11", "Group_1_11", "Group_1_21", 
"Group_1_21", "Group_1_21", "Group_1_21", "Group_1_21", "Group_1_21", 
"Group_1_21", "Group_1_13", "Group_1_13", "Group_1_13", "Group_1_13", 
"Group_1_13", "Group_1_13", "Group_1_13", "Group_1_23", "Group_1_23", 
"Group_1_23", "Group_1_23", "Group_1_23", "Group_1_23", "Group_1_23"
), Date = c("1/1/2017", "1/2/2017", "1/3/2017", "1/4/2017", "1/5/2017", 
"1/6/2017", "1/7/2017", "1/1/2017", "1/2/2017", "1/3/2017", "1/4/2017", 
"1/5/2017", "1/6/2017", "1/7/2017", "1/1/2017", "1/2/2017", "1/3/2017", 
"1/4/2017", "1/5/2017", "1/6/2017", "1/7/2017", "1/1/2017", "1/2/2017", 
"1/3/2017", "1/4/2017", "1/5/2017", "1/6/2017", "1/7/2017", "1/1/2017", 
"1/2/2017", "1/3/2017", "1/4/2017", "1/5/2017", "1/6/2017", "1/7/2017", 
"1/1/2017", "1/2/2017", "1/3/2017", "1/4/2017", "1/5/2017", "1/6/2017", 
"1/7/2017"), Value = c(NA, NA, 34L, 20L, NA, NA, 38L, 35L, 28L, 
NA, 32L, 39L, 28L, NA, NA, NA, 40L, 32L, 20L, NA, NA, NA, 32L, 
36L, NA, 28L, 33L, 40L, NA, NA, NA, 29L, 31L, NA, 34L, 26L, 33L, 
27L, 23L, 25L, 41L, 25L), Expected_Value = c(NA, NA, 34L, 20L, 
20L, 20L, 38L, 35L, 28L, 28L, 32L, 39L, 28L, NA, NA, NA, 40L, 
32L, 20L, NA, NA, NA, 32L, 36L, 36L, 28L, 33L, 40L, NA, NA, NA, 
29L, 31L, 31L, 34L, 26L, 33L, 27L, 23L, 25L, 41L, 25L)), .Names = c("Group_A", 
"GROUP_B", "Date", "Value", "Expected_Value"), row.names = c(NA, 
-42L), class = "data.frame")

Answer 2

即使您使用data.table 解决方案，这里也有一个使用tidyverse 方法的解决方案。 （如果时间允许，我可能会尝试翻译成data.table）。

这个想法是创建一个分组变量来捕获您的周数和fill 分组 GroupA、groupB 和 week 下的值（这里称为 grp）。我们还从last 创建了Value 到fill 的副本（tidyr 术语是.direction = 'up'）。然后，我们使用NA 值的累积总和创建另一个分组变量，并在假设新组大小（Group_A、GROUP_B、grp 和 @ 987654336@) 是 1，它的 value1 是 NA。这给出了预期的结果。

library(tidyverse)

df2 <- df1 %>% 
  mutate(Date = as.POSIXct(Date, format = '%m/%d/%Y')) %>% 
  mutate(value1 = Value) %>%
  group_by(Group_A, GROUP_B, grp = cumsum(format(Date, '%d')=='01'))%>% 
  fill(Value) %>% 
  fill(value1, .direction = 'up') %>% 
  mutate(grp1 = cumsum(is.na(Value))) %>% 
  group_by(Group_A, GROUP_B, grp, grp1) %>% 
  mutate(new = n(), Value = replace(Value, new == 1 | is.na(value1), NA)) %>%
  ungroup() %>%
  select(-c(value1, grp, grp1, new))

给出，

# A tibble: 42 × 5
   Group_A   GROUP_B       Date Value Expected_Value
     <chr>     <chr>     <dttm> <int>          <int>
1  GROUP_1 Group_1_1 2017-01-01    NA             NA
2  GROUP_1 Group_1_1 2017-01-02    NA             NA
3  GROUP_1 Group_1_1 2017-01-03    34             34
4  GROUP_1 Group_1_1 2017-01-04    20             20
5  GROUP_1 Group_1_1 2017-01-05    20             20
6  GROUP_1 Group_1_1 2017-01-06    20             20
7  GROUP_1 Group_1_1 2017-01-07    38             38
8  GROUP_1 Group_1_2 2017-01-01    35             35
9  GROUP_1 Group_1_2 2017-01-02    28             28
10 GROUP_1 Group_1_2 2017-01-03    28             28
# ... with 32 more rows

#Where,

identical(df2$Value, df2$Expected_Value)
#[1] TRUE