Question

我有以下几点：

library(tidyverse)
dput(test2)

structure(list(Cat2 = structure(c(3L, 1L, 1L, 1L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("A", 
"B", "NADA"), class = "factor"), start = c(0, 94, 95, 98, 98, 
98, 101, 104, 105, 106, 108, 108, 112, 112, 114, 114, 117, 119, 
119, 120, 120, 123, 123, 125, 128, 128, 131, 131, 134, 134, 137, 
137, 139, 141, 141, 153, 155, 156), end = c(94, 95, 98, 98, 98, 
101, 104, 105, 106, 108, 108, 112, 112, 114, 114, 117, 119, 119, 
120, 120, 123, 123, 125, 128, 128, 131, 131, 134, 134, 137, 137, 
139, 141, 141, 153, 155, 156, 157)), .Names = c("Cat2", "start", 
"end"), row.names = c(NA, -38L), class = c("tbl_df", "tbl", "data.frame"
))

如何转换此时间序列以便

我得到一行代表Cat2中状态的整个持续时间。现在，我有3行，状态A从95到98
如果A行和B行的开头和结尾重叠，我会得到一个新状态（A，B）
如果持续时间超过6

预期输出

NADA 0 94
A 94 98
A,B 98 98 
B 98 114
NADA 114 119
B 119 141
NADA 141 153 ###because of long B duration
B 153 157

Answer 1

这是一种不太方便的方式：

ab <- c('A', 'B')
df %>%
    mutate(
        Cat2 = ifelse(end - start + 1 > 6, 'NADA', as.character(Cat2)),  # change to NADA if > 6
        Cat2 = ifelse(
            start == lag(end) & Cat2 != lag(Cat2) & Cat2 %in% ab & lag(Cat2) %in% ab,
            'A,B', Cat2),  # change state to A,B if start and end of A and B overlap
        g = cumsum(Cat2 != lag(Cat2, default = Cat2[1]))  # group intervals to be merged
    ) %>% 
    group_by(g) %>%
    summarize(
        Cat2 = Cat2[1],
        start = start[1],
        end = end[n()]
    ) %>% select(-g)  # remove grouping variable

# # A tibble: 8 × 3
#    Cat2 start   end
#   <chr> <dbl> <dbl>
# 1  NADA     0    94
# 2     A    94    98
# 3   A,B    98    98
# 4     B    98   114
# 5  NADA   114   119
# 6     B   119   141
# 7  NADA   141   153
# 8     B   153   157

转换长格式时间序列和多个重叠状态

1 个答案: