如果 value 等于 NA,则根据 R 中的份额进行估计

If value is equal to NA, then estimate based on share in R

考虑以下两个数据框:

df  <- data.frame(REGION   = c("REG01","REG02","REG03","REGSUM"),
                  INDUSTRY = c("INDU01","INDU01","INDU01","INDU01"),
                  VALUE    = c(NA,10,NA,30))

和:

df2 <- data.frame(REGION   = c("REG01","REG02","REG03","REGSUM"),
                  INDUSTRY = c("INDU01","INDU01","INDU01","INDU01"),
                  VALUE    = c(5,15,20,40))

我想做以下计算:如果 df 中的值等于 NA,那么我想根据 df2 中的份额进行估算。因为我知道 df 中的总和,所以我知道我必须在 df.

中的两个具有 NA 的元素之间分配值 df[REGSUM,INDU01] - df[REG02,INDU01] = 30 - 10 = 20

那么它应该将df2中相同的元素除以NA的元素之和:

df2_share[REG01,INDU01] = 5  / (5 + 20) = 0.2
df2_share[REG03,INDU01] = 20 / (5 + 20) = 0.8

此份额应用于估计 df1 中的 NA。所以我最终会得到以下数据框:

    REGION  INDUSTRY   VALUE
1   REG01   INDU01     0.2 * 20 = 4 
2   REG02   INDU01     10   
3   REG03   INDU01     0.8 * 20 = 16    
4   REGSUM  INDU01     30

我可以在 R 中做到这一点吗(我的数据框中有很多地区和行业)。

这是一个方法。
df 中既不是 NA 也不是 "REGSUM" 的值求和。使用此值计算分配给 NA 值的总数。然后获取 df2 中与 NA 条目对应的值并计算要分配的比例。

not_na_values <- sum(df$VALUE[df$REGION != "REGSUM"], na.rm = TRUE)
to_assign <- df$VALUE[df$REGION == "REGSUM"] - not_na_values

na <- is.na(df$VALUE)
numer <- df2$VALUE[na]
denom <- sum(numer)
df$VALUE[na] <- numer/denom * to_assign

df
#  REGION INDUSTRY VALUE
#1  REG01   INDU01     4
#2  REG02   INDU01    10
#3  REG03   INDU01    16
#4 REGSUM   INDU01    30

下面的函数将上面的代码概括为 data.frames 许多行业。它的工作原理是按行业拆分输入 data.frames 并 lapplying 之前的代码,作为每个拆分列表成员的函数编写。最后它重新组装这些子数据帧和 returns 给调用者。

assign_na_values <- function(x, y,
                             region_col = "REGION", 
                             industry_col = "INDUSTRY",
                             value_col = "VALUE", 
                             regsum = "REGSUM") {
  f <- function(x, y, region_col, value_col, regsum){
    i <- x[[region_col]] != regsum
    not_na_values <- sum(x[[value_col]][ i ], na.rm = TRUE)
    to_assign <- x[[value_col]][ !i ] - not_na_values
    
    na <- is.na(x[[value_col]])
    numer <- y[[value_col]][na]
    denom <- sum(numer)
    x[[value_col]][na] <- numer/denom * to_assign
    x
  }
  sp_x <- split(x, x[[industry_col]])
  sp_y <- split(y, y[[industry_col]])
  res <- lapply(seq_along(sp_x), function(i){
    f(sp_x[[i]], sp_y[[i]], region_col, value_col, regsum)
  })
  res <- do.call(rbind, res)
  row.names(res) <- NULL
  res
}

assign_na_values(df, df2)
#  REGION INDUSTRY VALUE
#1  REG01   INDU01     4
#2  REG02   INDU01    10
#3  REG03   INDU01    16
#4 REGSUM   INDU01    30
#5  REG01   INDU02    30
#6  REG02   INDU02     6
#7  REG03   INDU02     4
#8 REGSUM   INDU02    40

新测试数据

df <- data.frame(
  REGION = c("REG01","REG02","REG03","REGSUM","REG01","REG02","REG03","REGSUM"),    
  INDUSTRY = c("INDU01","INDU01","INDU01","INDU01","INDU02","INDU02","INDU02","INDU02"), 
  VALUE = c(NA,10,NA,30,30,NA,NA,40)
)

df2 <- data.frame(
  REGION = c("REG01","REG02","REG03","REGSUM","REG01","REG02","REG03","REGSUM"), 
  INDUSTRY = c("INDU01","INDU01","INDU01","INDU01","INDU02","INDU02","INDU02","INDU02"), 
  VALUE = c(5,15,20,40,10,30,20,60)
)

这是另一个基本的 R 解决方案

idx <- which(REGION == "REGSUM")
df <- transform(
  df,
  VALUE = replace(
    VALUE,
    is.na(VALUE),
    prop.table(df2$VALUE[is.na(VALUE)]) * (VALUE[idx] - sum(VALUE[-idx], na.rm = TRUE))
  )
)

这给出了

  REGION INDUSTRY VALUE
1  REG01   INDU01     4
2  REG02   INDU01    10
3  REG03   INDU01    16
4 REGSUM   INDU01    30

如果有多个'INDUSTRY',我们可以通过操作进行连接和使用组

library(dplyr)
df %>% 
  left_join(df2, by = c("REGION", "INDUSTRY")) %>% 
  group_by(INDUSTRY) %>%
  transmute(REGION, INDUSTRY, VALUE = case_when(is.na(VALUE.x) ~ 
    VALUE.y/sum(VALUE.y[is.na(VALUE.x)]) * (VALUE.x[n()] - 
         sum(VALUE.x[-n()], na.rm = TRUE)), TRUE ~ VALUE.x)) %>%
  ungroup

-输出

# A tibble: 4 x 3
#  REGION INDUSTRY VALUE
#  <chr>  <chr>    <dbl>
#1 REG01  INDU01       4
#2 REG02  INDU01      10
#3 REG03  INDU01      16
#4 REGSUM INDU01      30