Question

我有一个带有美国NBER衰退的简单数据集，目前被编码为虚拟变量。我想依序标记每次衰退。例如，在下表中，我希望经济衰退列读取“经济衰退1”，“无经济衰退”，“经济衰退2”等，从而对每个经济衰退进行分类。

Date    Recession 
1949-06-30 1
1949-09-30 1
1949-12-31 1
1950-03-31 0
1950-06-30 0
1953-09-30 1
1953-12-31 1

Answer 1

您可以使用rle来计数1的连续运行，并重复（rep）相应次数（lengths）

foo <- with(rle(input$Recession), rep(cumsum(values) * values, lengths))
ifelse(foo == 0, "No Recession", paste("Recession", foo))

输入：

structure(list(Date = c("1949-06-30", "1949-09-30", "1949-12-31", 
"1950-03-31", "1950-06-30", "1953-09-30", "1953-12-31"), Recession = c(1L, 
1L, 1L, 0L, 0L, 1L, 1L)), row.names = c(NA, -7L), class = "data.frame")

Answer 2

这是一种整洁的方法：

使用lag来确定衰退状态是否发生改变
使用&和cumsum来确定它是否从无衰退变为衰退
用if_else将所有应为“衰退”的行替换为“无衰退”

library(tidyverse)
df <- read_table2(
"Date  Recession
1949-06-30 1
1949-09-30 1
1949-12-31 1
1950-03-31 0
1950-06-30 0
1953-09-30 1
1953-12-31 1"
)

df %>%
  mutate(
    changed = Recession != lag(Recession, default = Recession[1]),
    to_recession = str_c("Recession ", cumsum(changed & as.logical(Recession)) + 1),
    Recession_Num = if_else(Recession == 1, to_recession, "No Recession")
    ) %>%
  select(-changed, -to_recession)
#> # A tibble: 7 x 3
#>   Date       Recession Recession_Num
#>   <date>         <int> <chr>        
#> 1 1949-06-30         1 Recession 1  
#> 2 1949-09-30         1 Recession 1  
#> 3 1949-12-31         1 Recession 1  
#> 4 1950-03-31         0 No Recession 
#> 5 1950-06-30         0 No Recession 
#> 6 1953-09-30         1 Recession 2  
#> 7 1953-12-31         1 Recession 2

^{由reprex package（v0.2.1）于2018-10-30创建}

Answer 3

这是cumsum的把戏。

x <- c(1, 1, 1, 0, 0, 1, 1)
i <- cumsum(c(1, diff(x) != 0) & as.logical(x))
ifelse(x == 0, "No Recession", paste("Recession", i))
#[1] "Recession 1"  "Recession 1"  "Recession 1"  "No Recession"
#[5] "No Recession" "Recession 2"  "Recession 2"

Answer 4

Date <- as.Date(c('1949-06-30', '1949-09-30', '1949-12-31', '1950-03-31', '1950-06-30', '1953-09-30', '1953-12-31'), 
                format = '%Y-%m-%d')
Recession <- c(1,1,1,0,0,1,1)

df <- data.frame(Date, Recession)

find_seq_1s <- function(x) {
        count <-  0
        in_seq <-  FALSE
        output <-  NULL
        for(i in x) {
                if(i == 1 && in_seq == FALSE) {
                        count <- count + 1
                        in_seq <-  TRUE
                        output <-  c(output, paste('Recession', as.character(count)))
                } else if(i == 1 && in_seq == TRUE) {
                        output <-  c(output, paste('Recession', as.character(count)))
                } else {
                        in_seq <-  FALSE
                        output <-  c(output, 'No Recession')
                }
        }
        return(output)
}

df$Rec_Seq <- find_seq_1s(df$Recession)

Answer 5

unlist(lapply(1:nrow(df), FUN = function(x) ifelse(df$recession[x]==1, paste("Recession", x), "No Recession")))

连续值的累积和

5 个答案: