Question

我想用该列的滞后值替换一列（“股票”）中的缺失值，并用其他列来进行其他处理。由于以下循环在我的原始数据集中需要花费大量时间，因此还有另一种方法可以不循环吗？

    for (i in 1990:1993){
  df <- df %>% 
    group_by(fuel) %>%
    mutate(stock=ifelse(i==year & year>1991,lag(stock)+formation+lag(sales),stock))%>% 
    mutate(sales=ifelse(i==year & year>1991, stock-lag(stock),sales))
}

数据集示例：

df <- structure(list(year = c(1990L, 1991L, 1992L, 1993L, 1990L, 1991L, 
1992L, 1993L), fuel = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 
2L), .Label = c("a", "b"), class = "factor"), stock = c(10, 11, 
NA, NA, 10, 11, NA, NA), sales = c(NA, 1, NA, NA, NA, 1, NA, 
NA), formation = c(0.3, 0.4, 0.5, 0.3, 0.7, 0.4, 0.5, 0.7)), class = c("grouped_df", 
"tbl_df", "tbl", "data.frame"), row.names = c(NA, -8L), vars = "fuel", labels = structure(list(
    fuel = structure(1:2, .Label = c("a", "b"), class = "factor")), class = "data.frame", row.names = c(NA, 
-2L), vars = "fuel", drop = TRUE), indices = list(0:3, 4:7), drop = TRUE, group_sizes = c(4L, 
4L), biggest_group_size = 4L)

Answer 1

这是您要找的吗？在这种情况下，TRUE基本上类似于ELSE语句。我相信您也可以使用类似的方法，但是结果是相同的。

df <- structure(list(year = c(
  1990L, 1991L, 1992L, 1993L, 1990L, 1991L,
  1992L, 1993L
), fuel = structure(c(
  1L, 1L, 1L, 1L, 2L, 2L, 2L,
  2L
), .Label = c("a", "b"), class = "factor"), stock = c(
  10, 11,
  NA, NA, 10, 11, NA, NA
), sales = c(
  NA, 1, NA, NA, NA, 1, NA,
  NA
), formation = c(0.3, 0.4, 0.5, 0.3, 0.7, 0.4, 0.5, 0.7)), class = c(
  "grouped_df",
  "tbl_df", "tbl", "data.frame"
), row.names = c(NA, -8L), vars = "fuel", labels = structure(list(
  fuel = structure(1:2, .Label = c("a", "b"), class = "factor")
), class = "data.frame", row.names = c(
  NA,
  -2L
), vars = "fuel", drop = TRUE), indices = list(0:3, 4:7), drop = TRUE, group_sizes = c(
  4L,
  4L
), biggest_group_size = 4L)

library(dplyr)
#> 
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#> 
#>     filter, lag
#> The following objects are masked from 'package:base':
#> 
#>     intersect, setdiff, setequal, union

cat("Before")
#> Before
print(df)
#> # A tibble: 8 x 5
#> # Groups:   fuel [2]
#>    year fuel  stock sales formation
#>   <int> <fct> <dbl> <dbl>     <dbl>
#> 1  1990 a        10    NA       0.3
#> 2  1991 a        11     1       0.4
#> 3  1992 a        NA    NA       0.5
#> 4  1993 a        NA    NA       0.3
#> 5  1990 b        10    NA       0.7
#> 6  1991 b        11     1       0.4
#> 7  1992 b        NA    NA       0.5
#> 8  1993 b        NA    NA       0.7

df <- df %>%
  mutate(
    stock = case_when(
      year == 1991 ~ stock,
      TRUE ~ lag(stock) + formation + lag(sales)
    ),
    sales = case_when(
      year == 1991 ~ sales,
      TRUE ~ stock - lag(stock)
    )
  )

cat("After")
#> After
print(df)
#> # A tibble: 8 x 5
#> # Groups:   fuel [2]
#>    year fuel  stock sales formation
#>   <int> <fct> <dbl> <dbl>     <dbl>
#> 1  1990 a      NA    NA         0.3
#> 2  1991 a      11     1         0.4
#> 3  1992 a      12.5   1.5       0.5
#> 4  1993 a      NA    NA         0.3
#> 5  1990 b      NA    NA         0.7
#> 6  1991 b      11     1         0.4
#> 7  1992 b      12.5   1.5       0.5
#> 8  1993 b      NA    NA         0.7

^{由reprex package（v0.2.1）于2019-01-04创建}

Answer 2

这是使用临时列的代数解决方案。也许有一种更优雅的方式，但我认为这可行。

df2 <- df %>%
  # Make temporary columns for stock and sales.
  mutate(stock_temp     = stock,
         sales_temp     = sales) %>%

  # For each fuel type, fill down over missing values in those columns
  group_by(fuel) %>%
  fill(stock_temp, sales_temp) %>%

  # Add temp column using formation when stock is missing, and collect the
  #  cumulative total of these. Add that to sales_temp to get stock.
  #  The missing sales values are defined in OP as change in stock.
  mutate(formation_temp = if_else(is.na(stock), 
                                  formation, 0),
         cuml_form      = cumsum(formation_temp),
         add            = if_else(is.na(stock), 
                                  lag(sales_temp) + cuml_form, 0),
         stock = if_else(is.na(stock), stock_temp + add, stock),
         sales = if_else(is.na(sales), stock - lag(stock), sales)) %>%
  select(year:formation)


> df2
## A tibble: 8 x 5
## Groups:   fuel [2]
#   year fuel  stock  sales formation
#  <int> <fct> <dbl>  <dbl>     <dbl>
#1  1990 a      10   NA           0.3
#2  1991 a      11    1           0.4
#3  1992 a      12.5  1.5         0.5
#4  1993 a      12.8  0.3         0.3
#5  1990 b      10   NA           0.7
#6  1991 b      11    1           0.4
#7  1992 b      12.5  1.5         0.5
#8  1993 b      13.2  0.700       0.7

在不使用循环的情况下替换行中R中的值

2 个答案: