如何计算字符的天数持久性

Question

假设我们有以下数据：

type <- paste("type", c(1,1,1,2,3,1,2,2,3,3,3,3,1,1))
dates <- seq(as.Date("2000/1/1"), by = "days", length.out = length(type)) 
mydataframe <- data.frame(type, dates)

我在其他 posts 中看到 rle 可能会完成这项工作，但我想获得一个数据框，其中对于每种类型，我的平均持续时间为几天。类似于：

> print(persistance)
  type1 type2 type3
1     2   1.5   2.5

有人知道怎么做吗？谢谢！

Answer 1

data.table

library(data.table)
runs <- setDT(rle(as.character(mydataframe$type)))
runs[, mean(lengths), values]

#    values  V1
# 1: type 1 2.0
# 2: type 2 1.5
# 3: type 3 2.5

tidyverse 和 magrittr

library(tidyverse)
library(magrittr)

rle(as.character(mydataframe$type)) %$% 
  tibble(lengths, values) %>% 
  group_by(values) %>% 
  summarise_all(mean)

# # A tibble: 3 x 2
#   values lengths
#   <chr>    <dbl>
# 1 type 1    2.00
# 2 type 2    1.50
# 3 type 3    2.50

dplyr

library(dplyr)
rle(as.character(mydataframe$type)) %>% 
  unclass %>%
  as.data.frame %>% 
  group_by(values) %>% 
  summarise_all(mean)

Answer 2

另一种（分组）解决方案：

type <- paste("type", c(1,1,1,2,3,1,2,2,3,3,3,3,1,1))
dates <- seq(as.Date("2000/1/1"), by = "days", length.out = length(type)) 
mydataframe <- data.frame(type, dates)

library(dplyr)

mydataframe %>%
  count(type, group = cumsum(type != lag(type, default = first(type)))) %>%
  group_by(type) %>%
  summarise(Avg = mean(n))

# # A tibble: 3 x 2
#     type     Avg
#    <fct>  <dbl>
# 1 type 1   2  
# 2 type 2   1.5
# 3 type 3   2.5

Answer 3

您可以使用基本 R 函数 rle 和 aggregate 来执行此操作。

# set up the data as in your question
type <- paste("type", c(1,1,1,2,3,1,2,2,3,3,3,3,1,1))
dates <- seq(as.Date("2000/1/1"), by = "days", length.out = length(type)) 
mydataframe <- data.frame(type, dates)

# calculate the length of the run using rle 
runs <- rle(as.character(mydataframe$type))
# calculate the average length of the run
aggregate(runs[[1]], by = runs[2], FUN = mean)

请注意，这假设您的日期列中的日期确实是连续的。如果您在日期上有差距并且想将其视为单独的运行，则必须稍微更改公式才能真正使用 dates 列中的日期。

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