R：根據其他列創建多個新列

Question

假設我有一個看起來像這樣的數據框架

dd <- read.table(header = TRUE, text = "ID week1_t week1_a  week2_t week2_a
  1      12      22       17       4   
  1      15      32       18       5   
  1      24      12       29       6   
  2      45      11       19       8   
  2      23      33       20      10")

是否有一種直接的方法來創建week1_d列，week2_d列，等等每周，這是基於week1_t和week1_a之間的差異？ 或者我是否必須手動構建“差異”列？

預期輸出如下：

dd <- read.table(header = TRUE, text = "ID week1_t week1_a  week2_t week2_a week1_d week2_d 
  1      12      22       17       4       10       -13                 
  1      15      32       18       5       17       -13   
  1      24      12       29       6       -12      -23 
  2      45      11       19       8       -34      -11
  2      23      33       20      10       10       -10      ")

實際上，有大約30周，所以我試圖避免手動這樣做。 我正在考慮for循環每周的運行，並且grepping匹配week +（循環索引）的列。 有沒有更好的方法呢？

Answer 1

從“整潔的數據”角度來看，您的問題是您在列名中編碼（多個！）數據：周數和字母代表的數字。 我會轉換為長格式，其中week是一列，定義d = a - t ，並且（如果需要）轉換回寬格式。 但是我可能會把它保留為長格式，因為如果你想做任何其他操作，它們可能更容易在長數據上實現（更多的操作，建模，繪圖......）。

library(tidyr)
library(dplyr)

long = dd %>% 
    mutate(real_id = 1:n()) %>%
    gather(key = key, value = value, starts_with("week")) %>%
    separate(key, into = c("week", "letter")) %>% 
    spread(key = letter, value = value) %>%
    mutate(d = a - t)

head(long)
#   ID real_id  week  a  t   d
# 1  1       1 week1 22 12  10
# 2  1       1 week2  4 17 -13
# 3  1       2 week1 32 15  17
# 4  1       2 week2  5 18 -13
# 5  1       3 week1 12 24 -12
# 6  1       3 week2  6 29 -23

wide = gather(long, key = letter, value = value, a, t, d) %>%
    mutate(key = paste(week, letter, sep = "_")) %>%
    select(-week, -letter) %>%
    spread(key = key, value = value)

wide
#   ID real_id week1_a week1_d week1_t week2_a week2_d week2_t
# 1  1       1      22      10      12       4     -13      17
# 2  1       2      32      17      15       5     -13      18
# 3  1       3      12     -12      24       6     -23      29
# 4  2       4      11     -34      45       8     -11      19
# 5  2       5      33      10      23      10     -10      20

Answer 2

我們split帶有sub的后綴移除到list ，將'week'列（ dd[-1] ）除以數據集的names ，得到兩列之間的差異並分配list元素以在'dd'中創建新列。

lst <-  lapply(split.default(dd[-1], 
           sub("_.*", "", names(dd)[-1])), function(x) x[2]-x[1])
dd[paste0("week_", seq_along(lst), "d")] <- lapply(lst, unlist, use.names=FALSE)
dd
#    ID week1_t week1_a week2_t week2_a week1_d week2_d
#1  1      12      22      17       4      10     -13
#2  1      15      32      18       5      17     -13
#3  1      24      12      29       6     -12     -23
#4  2      45      11      19       8     -34     -11
#5  2      23      33      20      10      10     -10

---

如果列是交替的，即'week1_t'后跟'week1_a'，則'week2_t'，然后是'week2_a'，等等。

Un1 <- unique(sub("_.*", "", names(dd)[-1]))
i1 <-  c(TRUE, FALSE)
dd[paste0(Un1, "_d")] <-  dd[-1][!i1]- dd[-1][i1]
dd
#  ID week1_t week1_a week2_t week2_a week1_d week2_d
#1  1      12      22      17       4      10     -13
#2  1      15      32      18       5      17     -13
#3  1      24      12      29       6     -12     -23
#4  2      45      11      19       8     -34     -11
#5  2      23      33      20      10      10     -10