r - 使用 grep 函数在 R 中正确转换数据

Question

我正在尝试根据特定列(即“向上”和“向下”)中的值 reshape 我的数据。 Up 和 Down 在数据框中的顺序不同，因此我很难将数据“转换”为正确的形状。

我试过使用转换函数来移动数据，但我无法得到以一致(又名准确)的方式工作的答案。

这是我的输入:

input = structure(list(X = 1:6, Report = c("Sales.csv", "Sales.csv", 
"Sales.csv", "Sales.csv", "Sales.csv", "Sales.csv"), Shock = c("Currencies.USD_Up", 
"Currencies.USD_Down", "Currencies.AUD_Up", "Currencies.AUD_Down", 
"Currencies.EUR_Down", "Currencies.EUR_Up"), Result = c(-519375.9816, 
-7388851.423, -42950.77683, -667.367063, -12819532.15, -138054.0061
), FX = c("USD", "USD", "AUD", "AUD", "EUR", "EUR")), class = "data.frame", row.names = c(NA, 
-6L))

这是我的首选输出:

output = structure(list(X = 1:3, Report = c("Sales.csv", "Sales.csv", 
"Sales.csv"), Shock = c("Currencies.USD", "Currencies.AUD", "Currencies.EUR"
), Currency = c("USD", "AUD", "EUR"), Up = c(-519375.9816, -42950.77683, 
-138054.0061), Down = c(-7388851.423, -667.367063, -12819532.15
)), class = "data.frame", row.names = c(NA, -3L))

因为输入中的 EUR 数据顺序不同，我似乎无法正确调整数据形状。我试过使用 grep 函数来订购它，但我做不到。谁能提出更好的方法？

Answer 1

这是一种 tidyverse 方法:

library(dplyr)
library(tidyr)
library(tibble)
input %>% 
  as_tibble() %>% 
  separate(Shock, c("Shock", "tmp"), sep = "_") %>% 
  rename(Currency = FX) %>% 
  select(-X) %>% 
  spread(tmp, Result) %>% 
  mutate(X = row_number()) %>% 
  select(X, Report, Shock, Currency, Up, Down)