连接并计算每行中的元素数？

Question

df <- structure(list(ID = c("1", "2", "3", "4", "5", "6"), `ID without mask` = c(NA_character_, 
 NA_character_, NA_character_, NA_character_, NA_character_, NA_character_
), `Other Years` = c("2011", "2015", "2015", "2006, 2006, 2005, 2005, 2007", 
 "2014, 2011", "2007"), `Cut off Year` = c("2011", "2015", "2015", "2005", 
 "2011", "2007"), `2005` = c(NA, NA, NA, "30", "18", NA), `2006` = c(NA_character_, 
 NA_character_, NA_character_, NA_character_, NA_character_, NA_character_
 ), `2007` = c("15", NA, "18", NA, "30, 18", NA), `2008` = c("16", 
 NA, NA, "30, 27", "18, 30", NA), `2009` = c("15", NA, NA, "20", 
 "30, 18", NA), `2010` = c(NA, NA, NA, "30, 20", NA, NA), `2011` = c(NA_character_, 
 NA_character_, NA_character_, NA_character_, NA_character_, NA_character_
 ), `2012` = c(NA, NA, NA, "20, 30", NA, "26"), `2013` = c("15", 
 NA, "19", NA, NA, NA), `2014` = c(NA, NA, "18", NA, NA, NA), 
`2015` = c(NA, NA, "18", NA, "18", NA), `2016` = c(NA_character_, 
 NA_character_, NA_character_, NA_character_, NA_character_, 
 NA_character_)), .Names = c("ID", "ID without mask", 
 "Other Years", "Cut off Year", "2005", "2006", "2007", "2008", 
 "2009", "2010", "2011", "2012", "2013", "2014", "2015", "2016"
 ), row.names = c(NA, 6L), class = "data.frame")

鉴于上述数据框。我希望 R 将连续的每个元素（多年来）连接在一起并计数，然后将其输出到新列中。

基于“截止年份”列，我希望 R 将截止年份之前的元素连接在一个列中，并将截止年份之后的元素连接在第二列中（包括截止年份）。

所以对于截止年份为 2011 的第一行，2007、2008 和 2009 年分别有 15、16、15，所以总共是 3，所以 R 应该在新列中输出数字 3 . 2011 年之后，只有 2013 年的元素有条目，因此“之后”列将只有数字 1。

“30、27”等元素算作二等

这是所需的输出：

structure(list(ID = c("1", "2", "3", "4", "5", "6"), `ID without mask` = c(NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_
), `Other Years` = c("2011", "2015", "2015", "2006, 2006, 2005, 2005, 2007", 
"2014, 2011", "2007"), `Cut off Year` = c("2011", "2015", "2015", "2005", 
"2011", "2007"), `2005` = c(NA, NA, NA, "30", "18", NA), `2006` = c(NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_
), `2007` = c("15", NA, "18", NA, "30, 18", NA), `2008` = c("16", 
NA, NA, "30, 27", "18, 30", NA), `2009` = c("15", NA, NA, "20", 
"30, 18", NA), `2010` = c(NA, NA, NA, "30, 20", NA, NA), `2011` = c(NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_
), `2012` = c(NA, NA, NA, "20, 30", NA, "26"), `2013` = c("15", 
NA, "19", NA, NA, NA), `2014` = c(NA, NA, "18", NA, NA, NA), 
    `2015` = c(NA, NA, "18", NA, "18", NA), `2016` = c(NA_character_, 
    NA_character_, NA_character_, NA_character_, NA_character_, 
    NA_character_), `Before cut` = c("3", "0", "3", "0", "7", "0"), `After cut` = c("1", "0", "1", "8", "1", "1")), .Names = c("ID", "Collab Years Patents", 
"Collab Years Publications", "Cut off Year", "2005", "2006", "2007", "2008", 
"2009", "2010", "2011", "2012", "2013", "2014", "2015", "2016", "Before cut", "After cut"
), row.names = c(NA, 6L), class = "data.frame")

Answer 1

我们转换成一个不错的长格式，计算值，然后加入原始格式：

library(tidyr)
library(dplyr)
library(stringr)

gather(df, key = "Year", value = "value", `2005`:`2016`) %>%
    mutate(val_count = str_count(value, pattern = ",") + 1) %>%
    group_by(ID) %>%
    summarize(Before = sum(val_count * (Year < `Cut off Year`), na.rm = TRUE),
              After = sum(val_count * (Year >= `Cut off Year`), na.rm = TRUE)) %>%
    right_join(df) %>%
    select(1:3)
# Joining, by = "ID"
# A tibble: 6 x 3
     ID Before After
  <chr>  <dbl> <dbl>
1     1      3     1
2     2      0     0
3     3      3     1
4     4      0     8
5     5      7     1
6     6      0     1

我使用select(1:3) 仅显示结果的相关部分 - 省略该行以获取所有其他列。为了获得值的数量，我们将逗号的数量加 1。