R 动态data.frame中的累积和

Question

df <- data.frame('a'=c(1,2,3,4,5), 'M1'=c(1,20,3,4,50))

Now I want to create an additional 179 columns (M2 - M180) which is dependent on the sum of previous Mx columns. For instance, M2 should be the sum for all previous Mx values ie M1 in this case. And M3 should be the sum M1+M2 and so on all the way to M180.

The expected result for the first three Mx columsn should be:

Expected_res <- data.frame('a'=c(1,2,3,4,5), 'M1'=c(1,20,3,4,50), 'M2'=c(1,20,3,4,50), 'M3'=c(2,40,6,8,100))

Basically I want the excel equivalent to Sum($E1:E1) and then dragging it out for 180 columns, for each row of course.

Any help is appreciated.

Answer 1

You can use a for loop:

for(i in 2:180) {
  df[paste0('M', i)] <- rowSums(df[-1])
}

-1 ignores the 1st column ( a ) and takes sum of all other columns to create a new column.

Here's the output of first 5 columns.

df
#  a M1 M2  M3  M4  M5
#1 1  1  1   2   4   8
#2 2 20 20  40  80 160
#3 3  3  3   6  12  24
#4 4  4  4   8  16  32
#5 5 50 50 100 200 400

Answer 2

using tidyverse

library(tidyverse)
df <- data.frame('a'=c(1,2,3,4,5), 'M1'=c(1,20,3,4,50))
n <- 5
bind_cols(df, map_dfc(seq_len(n), ~transmute(df, !!paste0("M", .x + 1) := M1 * 2 ^ (.x - 1))))
#>   a M1 M2  M3  M4  M5  M6
#> 1 1  1  1   2   4   8  16
#> 2 2 20 20  40  80 160 320
#> 3 3  3  3   6  12  24  48
#> 4 4  4  4   8  16  32  64
#> 5 5 50 50 100 200 400 800

^{Created on 2021-02-02 by the reprex package (v1.0.0)}

using data.table

library(data.table)
df <- data.frame('a'=c(1,2,3,4,5), 'M1'=c(1,20,3,4,50))
setDT(df)
n <- 5
col_names <- paste0("M", seq_len(n) + 1)
df[, (col_names) := lapply(seq_len(n), function(x) M1 * 2 ^ (x - 1))]
df
#>    a M1 M2  M3  M4  M5  M6
#> 1: 1  1  1   2   4   8  16
#> 2: 2 20 20  40  80 160 320
#> 3: 3  3  3   6  12  24  48
#> 4: 4  4  4   8  16  32  64
#> 5: 5 50 50 100 200 400 800

^{Created on 2021-02-02 by the reprex package (v1.0.0)}