Question

假设我有一个看起来像这样的数据框

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")

是否有一种简单的方法可以创建一个week1_d列，一个week2_d列，依此类推每周，这是基于week1_t和week1_a之间的差异？或者我是否必须手动构建“差异”列？

预期输出如下：

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      ")

实际上，大约有30周，所以我试图避免手动执行此操作。我正在考虑for循环每周的运行，并且grepping匹配week +（循环索引）的列。有没有更好的方法呢？

Answer 1

从“整洁数据”的角度来看，您的问题是您在列名中编码（多个！）数据：周数和字母代表的数字。我会转换为长格式，其中week是一列，定义d = a - t，并且（如果需要）转换回宽格式。但是我可能会把它保留为长格式，因为如果你想做任何其他操作，它们可能更容易在长数据上实现（更多操作，建模，绘图......）。

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

我们split周＆＃39;在将dd[-1]后缀移除到names后，数据集的sub列（list），获取两列之间的差异并指定list }元素在＆＃39; dd＆＃39;。

中创建新列

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＆quot; week1_t＆＃39;然后是＆＃39; week1_a＆＃39;，然后＆＃39; week2_t＆＃39;，然后是＆＃39; week2_a＆＃39;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

R：根据其他列创建多个新列

2 个答案: