在数据帧中执行倍数pivot_longer

如何解决在数据帧中执行倍数pivot_longer

这是我的数据，每行对应一个巢，每行中有3个小鸡CHICK_RING1，CHICK_RING2，CHICK_RING3及其年龄和身体状况。

data = wrapr::build_frame(
  "YEAR","Nest","Clutch","RINGMALE","RINGFEMALE","CHICK_RING_1","CHICK_RING_2","CHICK_RING_3","LastAge1","LastAge2","LastAge3","RealMotherAge","RealFatherAge","BODYCOND1","BODYCOND2","BODYCOND3" |
    "2002","02-125","3","A72","C15","K67","K75","K82","90","68","9","14","7.070707","9.086538","8.622449"  |
    "2008","08-155","S09","R30","R40","75","72","73","15","20","7.075472","6.984925","7.511962"  |
    "2006","06-267","O30","O37","O59","70","66","13","18","9.227273","8.232323","9.44186"   |
    "1999","99-925","A39","A76","I00","I15","I73","69","10","7.989691","7.882883","8.043478"  |
    "2011","11-0305","A66","P48","W25","W30","W46","4","22","7.675676","10.4186","7.352941"  )

我想执行一个pivot_longer，以获取一个具有15行的数据帧，并在每行中添加小鸡ID及其身体状况，小鸡年龄（lastAge），父母的年龄，年份，巢和关键。

一种解决方案是为每个小鸡创建3个数据集，但更喜欢使用tidyverse。

谢谢

解决方法

您应该旋转两次：

library(tidyverse)
data %>% 
  pivot_longer(cols = all_of(ends_with(c("1","2","3"))),names_to = c("stat","CHICK_ID"),names_pattern = "(.*)(.)") %>% 
  pivot_wider(names_from = "stat") %>% 
  mutate(across(-c("Nest","RINGMALE","RINGFEMALE","CHICK_RING_"),as.numeric))

names_pattern是一个正则表达式，它将匹配两组，每个组在括号之间。 (.)将匹配任何单个字符。 (.*)将匹配尽可能多的字符，同时仍然允许末尾的(.)捕获最后一个字符。

第一个pivot_longer会将您的大部分数据转换为字符。您可以使用mutate(across(...))来迫使他们返回。