基于时间的序列分类

Question

我有一个数据集，其中包含一周内进行的各种测量。我想确定每周测量的开始和结束，并将它们分组如下。

如果开始日和结束日之间没有间隔，则：

 a.) 0 days (all week filled witth 0's)
 b.) 1 day (start=end); 
 c.) 2 days; 
 d.) 3 days; 
 e.) 4 days; 
 f.) 5 days; 
 g.) 6 days and 
 h.) 7 days.

如果测量之间存在差距，我想保留 id 和每周测量结构。

输出

样本数据：

    df<-structure(list(Id = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 
13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 
29, 30, 31, 32, 33, 34, 35, 36, 37, 38), A = c(0, 0, 0, 0, 41, 
0, 51, 0, 0, 41, 0, 0, 0, 0, 0, 43, 49, 0, 0, 29, 0, 48, 0, 0, 
0, 0, 0, 0, 52, 62, 47, 21, 0, 42, 0, 3, 0, 0), B = c(0, 0, 0, 
0, 0, 0, 51, 0, 0, 7, 45, 0, 46, 0, 44, 21, 51, 48, 0, 0, 47, 
42, 0, 0, 0, 0, 43, 0, 59, 56, 0, 57, 0, 46, 0, 44, 0, 0), C = c(0, 
0, 0, 0, 25, 0, 50, 0, 0, 0, 55, 0, 49, 0, 46, 17, 51, 41, 0, 
49, 51, 23, 0, 0, 0, 0, 38, 0, 57, 70, 46, 53, 0, 4, 0, 2, 0, 
0), D = c(0, 0, 0, 0, 42, 0, 63, 0, 0, 8, 0, 0, 0, 0, 47, 24, 
29, 0, 0, 0, 53, 35, 0, 0, 48, 0, 0, 0, 0, 14, 0, 60, 0, 53, 
0, 49, 0, 0), E = c(0, 0, 0, 0, 0, 0, 46, 0, 0, 48, 0, 0, 46, 
0, 43, 0, 0, 0, 0, 0, 46, 0, 0, 0, 48, 0, 26, 0, 0, 58, 46, 51, 
0, 40, 0, 48, 0, 0), F = c(0, 0, 0, 0, 0, 0, 0, 0, 55, 0, 0, 
0, 0, 0, 40, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 22, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0), G = c(0, 0, 0, 0, 0, 0, 0, 0, 60, 0, 0, 
0, 0, 0, 52, 0, 0, 0, 0, 0, 0, 0, 14, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0), Total = c(0, 0, 0, 0, 108, 0, 261, 0, 55, 
104, 100, 0, 141, 0, 220, 105, 180, 89, 0, 78, 197, 148, 0, 0, 
96, 0, 129, 0, 168, 260, 139, 242, 0, 185, 0, 146, 0, 0)), class = c("spec_tbl_df", 
"tbl_df", "tbl", "data.frame"), row.names = c(NA, -38L), spec = structure(list(
    cols = list(Id = structure(list(), class = c("collector_double", 
    "collector")), A = structure(list(), class = c("collector_double", 
    "collector")), B = structure(list(), class = c("collector_double", 
    "collector")), C = structure(list(), class = c("collector_double", 
    "collector")), D = structure(list(), class = c("collector_double", 
    "collector")), E = structure(list(), class = c("collector_double", 
    "collector")), F = structure(list(), class = c("collector_double", 
    "collector")), G = structure(list(), class = c("collector_double", 
    "collector")), Total = structure(list(), class = c("collector_double", 
    "collector"))), default = structure(list(), class = c("collector_guess", 
    "collector")), skip = 1L), class = "col_spec")) 

Answer 1

您可以通过测量来“编码”您的日子来解决这个问题。为此

1. 定义一个辅助函数，选择“0”表示无测量值，选择“v”表示测量值
2. 编码你的一周

要评估不间断的测量范围：

1. 删除任何前导或尾随 0。这说明测量不是从第一天开始或在最后一天结束 - 因为这是允许的。
2. 检查剩余代码是否有中断（即break）
3. 如果没有中断，长度，即nchars() 的字符数，给出了测量的天数。

为了展示正在发生的事情，我将中间代码存储在单独的列中。 完成后，您可以合并此列或删除中间列。

# helper function
check_value <- function(x){
   ifelse(x == 0, "0", "v")     # returns "0" for no measurement, else "v"
}


df %>% mutate(
# ------------ code your week = combinations of "0" or "v"
    combis = paste0(check_value(A), check_value(B), check_value(C), check_value(D), 
                    check_value(E), check_value(F))

# ------------ eliminate leading and trailing "0"
    , seqs = gsub(pattern = "(^0+)|(0+$)", "", combis)

# ------------ check for breaks
    , breaks = grepl(pattern = "0", x = seqs)

# ------------ if uninterrupted, count number of days
    , days = ifelse(breaks == FALSE, nchar(seqs), NA)

# ----------- construct comment summary
    , comment = case_when(
         breaks == TRUE & is.na(days) ~ "With"
        ,breaks == FALSE ~ paste0("Without-", days, " days")
        ,TRUE ~ as.character(NA))
    )

这会产生前 10 行：

# A tibble: 38 x 14
      Id     A     B     C     D     E     F     G Total combis seqs    breaks  days comment       
   <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <chr>  <chr>   <lgl>  <int> <chr>         
 1     1     0     0     0     0     0     0     0     0 000000 ""      FALSE      0 Without-0 days
 2     2     0     0     0     0     0     0     0     0 000000 ""      FALSE      0 Without-0 days
 3     3     0     0     0     0     0     0     0     0 000000 ""      FALSE      0 Without-0 days
 4     4     0     0     0     0     0     0     0     0 000000 ""      FALSE      0 Without-0 days
 5     5    41     0    25    42     0     0     0   108 v0vv00 "v0vv"  TRUE      NA With          
 6     6     0     0     0     0     0     0     0     0 000000 ""      FALSE      0 Without-0 days
 7     7    51    51    50    63    46     0     0   261 vvvvv0 "vvvvv" FALSE      5 Without-5 days
 8     8     0     0     0     0     0     0     0     0 000000 ""      FALSE      0 Without-0 days
 9     9     0     0     0     0     0    55    60    55 00000v "v"     FALSE      1 Without-1 days
10    10    41     7     0     8    48     0     0   104 vv0vv0 "vv0vv" TRUE      NA With