# Cummulate and concatenate

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Excel BI’s Excel Challenge #313 — solved in R

### Defining the Puzzle:

Today we need to transform table into its cummulative (someway) version. Every column have to include content of column lagged by 2.

Generate the result table.

Here, Tn = T(n-2) & Tn where & is concatenation operator.

### Loading Data from Excel:

Lets start loading data and libraries:

library(tidyverse) library(readxl) library(data.table) input = read_excel(“Scan3.xlsx”, range = “A2:G5”, col_names = c(“X1”, “X2”, “X3”, “X4”, “X5”, “X6”, “X7”)) test = read_excel(“Scan3.xlsx”, range = “I2:O5”, col_names = c(“X1”, “X2”, “X3”, “X4”, “X5”, “X6”, “X7”))

### Approach 1: Tidyverse with purrr

process_columns <- function(df) { num_cols <- ncol(df) if (num_cols < 3) { return(df) } for (i in 3:num_cols) { df <- df %>% mutate(across(all_of(names(df)[i]), ~ paste0(df[[i — 2]], .))) } return(df) } result = process_columns(input)

### Approach 2: Base R

process_columns_base_R <- function(df) { num_cols <- ncol(df) if (num_cols < 3) { return(df) } for (i in 3:num_cols) { df[[i]] <- paste0(df[[i — 2]], df[[i]]) } return(df) } result_base_R = process_columns_base_R(input)

### Approach 3: Data.table

process_columns_data_table <- function(df) { setDT(df) num_cols <- ncol(df) if (num_cols < 3) { return(df) } cols_to_modify <- names(df)[3:num_cols] for (i in cols_to_modify) { df[, (i) := paste0(df[[which(names(df) == i) — 2]], df[[i]])] } return(as_tibble(as.data.frame(df))) } result_dt = process_columns_data_table(input)

### Validating Our Solutions:

identical(result, test) # [1] TRUE identical(result_base_R, test) # [1] TRUE identical(result_dt, test) # [1] TRUE

If you like my publications or have your own ways to solve those puzzles in R, Python or whatever tool you choose, let me know.

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