在 R 中使用 lapply 迭代列时计算行值的百分比变化

Question

I have a dataframe with weekly values for a large number of variables. I want to iterate through each column and obtain the weekly change per row and variable expressed as percent.

Example:

a = c(2,3,1,9)
b = c(4,5,8,1)
sentiment = cbind(a,b) %>% 
as.data.frame()`



Outcome should be: 
     a  b  a_delta  b_delta 
     2  4     NA      NA
     3  5     0.5     0.3
     1  8    -0.7     0.6
     9  1     8.0    -0.8

In my current approach I use two steps: (1) create a weekly lag, (2) calculate the percentage difference between the lagged value and the value. There is no error message, but the calculation is still incorrect and I am not sure why. Any help would be much appreciated!

library(data.table) 

a = c(2,2.5,2,4)
b = c(4,5,8,1)
sentiment = cbind(a,b) %>% 
  as.data.frame()

setDT(sentiment)[, paste0(names(sentiment), "_delta") := lapply(.SD, function(x) shift(x, 1L, 
type="lag")/x -1)]

Answer 1

Here is a base R solution using sapply passed in a function to lapply that iterates over the columns of sentiment with the desired output column names using setNames .

sentiment <- data.frame(a = c(2,3,1,9), b = c(4,5,8,1))
calc_lag <- function(x) {
  c(NA, round(sapply(2:length(x), function(y) {
    (x[y] - x[y-1]) / x[y-1]
  }), 1))
}
cbind(sentiment, lapply(setNames(sentiment, paste0(colnames(sentiment), '_lag')), calc_lag))
#  a b a_lag b_lag
#1 2 4    NA    NA
#2 3 5   0.5   0.2
#3 1 8  -0.7   0.6
#4 9 1   8.0  -0.9

Answer 2

We can use diff

library(dplyr)
sentiment %>%
      mutate_all(list(delta = ~ round(c(NA, diff(.))/lag(.), 1)))

Or if we use the devel version of dplyr

sentiment %>% 
    mutate(across(everything(),  ~ round(c(NA, diff(.x))/lag(.x), 1), 
           names = "{col}_delta"))
#  a b a_delta b_delta
#1 2 4      NA      NA
#2 3 5     0.5     0.2
#3 1 8    -0.7     0.6
#4 9 1     8.0    -0.9