如何记录全年的数值？

Question

我正在处理时间序列数据，如下所示：

     cname year govstruct
6091 China 1960         3
6092 China 1961         3
6093 China 1962         3
6094 China 1963         3
6095 China 1964         3
6096 China 1965         3
6097 China 1966         3
6098 China 1967         3
6099 China 1968         3
6100 China 1969         3
6101 China 1970         3
6102 China 1971         3
6103 China 1972         3
6104 China 1973         3
6105 China 1974         3
6106 China 1975         3
6107 China 1976         3
6108 China 1977         3
6109 China 1978         3
6110 China 1979         3
6111 China 1980         3
6112 China 1981         3
6113 China 1982         3
6114 China 1983         1
6115 China 1984         1
6116 China 1985         1
6117 China 1986         1
6118 China 1987         1
6119 China 1988         1
6120 China 1989         1
6121 China 1990         1
6122 China 1991         1
6123 China 1992         1
6124 China 1993         1
6125 China 1994         1
6126 China 1995         1
6127 China 1996         1
6128 China 1997         1
6129 China 1998         1
6130 China 1999         1
6131 China 2000         1
6132 China 2001         1
6133 China 2002         1
6134 China 2003         1
6135 China 2004         1
6136 China 2005         1
6137 China 2006         3
6138 China 2007         3
6139 China 2008         3
6140 China 2009         3
6141 China 2010         3
6142 China 2011         3
6143 China 2012         3

我想要构建一个数据集来记录govstruct所涵盖的日期范围。

我想要的是一个数据集，它记录了国家的名称、年份的范围和政府结构的价值。最后的数据集如下所示：

cname    years  govstruct
China 1960-1982    3
China 1983-2005    1
China 2006-2012    3

请注意，我将在世界各地转圈。因此，任何能够这样做的代码都将是非常感谢的。

非常感谢你的帮助。

Answer 1

这里有一个使用dplyr/data.table的选项，我们使用“cname”进行分组，使用“govstruct”的run-length-id进行分组，并在“年份”的range中使用paste对summarise进行分组。

library(dplyr)
library(stringr)
library(data.table)
df1 %>% 
    group_by(cname, grp = rleid(govstruct)) %>%
    summarise(govstructure = first(govstruct), 
         years = str_c(range(year), collapse="-"))  %>%
    ungroup %>%
    select(-grp)
# A tibble: 3 x 3
#  cname govstructure years    
#  <chr>        <int> <chr>    
#1 China            3 1960-1982
#2 China            1 1983-2005
#3 China            3 2006-2012

或者，我们也可以通过比较相邻元素来构造grp。

df1 %>%
   group_by(cname, grp = cumsum(c(TRUE, diff(govstruct) != 0))) %>%
   summarise(govstructure = first(govstruct), 
         years = str_c(range(year), collapse="-")) 

或者使用与dplyr中相同的方法使用dplyr。即按'govstruct‘的rleid和'cname’paste‘range’进行分组

library(data.table)
setDT(df1)[ , .(govstructure = first(govstruct),
      year = paste(range(year), collapse = "-")), 
      .(cname, grp = rleid(govstruct))][, grp := NULL][]
#  cname govstructure      year
#1: China            3 1960-1982
#2: China            1 1983-2005
#3: China            3 2006-2012

或使用base R的另一个选项

grp <- with(rle(df1$govstruct), rep(seq_along(values), lengths))
aggregate(year ~ cname + grp, data = df1, 
      FUN = function(x) paste(range(x), collapse="-"))

数据

df1 <- structure(list(cname = c("China", "China", "China", "China", 
"China", "China", "China", "China", "China", "China", "China", 
"China", "China", "China", "China", "China", "China", "China", 
"China", "China", "China", "China", "China", "China", "China", 
"China", "China", "China", "China", "China", "China", "China", 
"China", "China", "China", "China", "China", "China", "China", 
"China", "China", "China", "China", "China", "China", "China", 
"China", "China", "China", "China", "China", "China", "China"
), year = 1960:2012, govstruct = c(3L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 3L, 3L, 3L, 3L, 3L)), 
class = "data.frame", row.names = c(NA, 
-53L))

Answer 2

我们可以使用data.table来计算每个cname的first和last year值以及govstruct的游程编码值。

library(data.table)

setDT(df)[ , .(year = paste(first(year), last(year), sep = "-"), 
           govstruct = first(govstruct)), .(cname, rleid(govstruct))]


#   cname rleid      year govstruct
#1: China     1 1960-1982         3
#2: China     2 1983-2005         1
#3: China     3 2006-2012         3