Importing multiple excel sheets into one dataframe adding the sheet name as variable-CodePudding

I am importing multiple excel sheets into one dataframe using the rio package.

WIOD_EA_EmRelEnergy <- import_list("EA 2016/Emission-relevant Energy Accounts_total.xlsx",
                                   setclass = "tbl", rbind = TRUE)

This line of code does already exactly what I want. It adds a column at the end called "file" indicating the the number of the sheet (data from the first sheet takes the value 1 and so on).

However, I am trying that column to take the name of the sheet rather than a number. The names of the sheets are country codes ("AUS", "AUT", etc.). Thus, the data of the first sheet should not take the value 1 but rather "AUS".

This should only be a small problem but i simply do not find the solution.

CodePudding user response：

No experience with the rio-package. My workflow usually goes like:

library(data.table)
library(readxl)
files.to.read <- list.files( whatever arguments you need )
L <- lapply(files.to.read, readxl::read_excel, whatever arguments you need)
names(L) <- basename(files.to.read)
final <- data.table::rbindlist(L, use.names = TRUE, fill = TRUE, id = "file")

CodePudding user response：

You could use openxlsx::getSheetNames to label the numeric file column as factor.

library(rio)
transform(import_list(xlsx_file, rbind=TRUE, rbind_label='file'), 
          file=factor(file, labels=openxlsx::getSheetNames(xlsx_file)))
#     mpg cyl  disp  hp drat    wt  qsec vs am gear carb    file             
# 1  21.0   6 160.0 110 3.90 2.620 16.46  0  1    4    4 mtcars1
# 2  21.0   6 160.0 110 3.90 2.875 17.02  0  1    4    4 mtcars1
# 3  22.8   4 108.0  93 3.85 2.320 18.61  1  1    4    1 mtcars1
# 4  21.4   6 258.0 110 3.08 3.215 19.44  1  0    3    1 mtcars1
# 5  18.7   8 360.0 175 3.15 3.440 17.02  0  0    3    2 mtcars1
# 6  18.1   6 225.0 105 2.76 3.460 20.22  1  0    3    1 mtcars1
# 7  14.3   8 360.0 245 3.21 3.570 15.84  0  0    3    4 mtcars1
# 8  24.4   4 146.7  62 3.69 3.190 20.00  1  0    4    2 mtcars1
# 9  22.8   4 140.8  95 3.92 3.150 22.90  1  0    4    2 mtcars1
# 10 19.2   6 167.6 123 3.92 3.440 18.30  1  0    4    4 mtcars1
# 11 17.8   6 167.6 123 3.92 3.440 18.90  1  0    4    4 mtcars2
# 12 16.4   8 275.8 180 3.07 4.070 17.40  0  0    3    3 mtcars2
# 13 17.3   8 275.8 180 3.07 3.730 17.60  0  0    3    3 mtcars2
# 14 15.2   8 275.8 180 3.07 3.780 18.00  0  0    3    3 mtcars2
# 15 10.4   8 472.0 205 2.93 5.250 17.98  0  0    3    4 mtcars2
# 16 10.4   8 460.0 215 3.00 5.424 17.82  0  0    3    4 mtcars2
# 17 14.7   8 440.0 230 3.23 5.345 17.42  0  0    3    4 mtcars2
# 18 32.4   4  78.7  66 4.08 2.200 19.47  1  1    4    1 mtcars2
# 19 30.4   4  75.7  52 4.93 1.615 18.52  1  1    4    2 mtcars2
# 20 33.9   4  71.1  65 4.22 1.835 19.90  1  1    4    1 mtcars2
# 21 21.5   4 120.1  97 3.70 2.465 20.01  1  0    3    1 mtcars3
# 22 15.5   8 318.0 150 2.76 3.520 16.87  0  0    3    2 mtcars3
# 23 15.2   8 304.0 150 3.15 3.435 17.30  0  0    3    2 mtcars3
# 24 13.3   8 350.0 245 3.73 3.840 15.41  0  0    3    4 mtcars3
# 25 19.2   8 400.0 175 3.08 3.845 17.05  0  0    3    2 mtcars3
# 26 27.3   4  79.0  66 4.08 1.935 18.90  1  1    4    1 mtcars3
# 27 26.0   4 120.3  91 4.43 2.140 16.70  0  1    5    2 mtcars3
# 28 30.4   4  95.1 113 3.77 1.513 16.90  1  1    5    2 mtcars3
# 29 15.8   8 351.0 264 4.22 3.170 14.50  0  1    5    4 mtcars3
# 30 19.7   6 145.0 175 3.62 2.770 15.50  0  1    5    6 mtcars3
# 31 15.0   8 301.0 335 3.54 3.570 14.60  0  1    5    8 mtcars3
# 32 21.4   4 121.0 109 4.11 2.780 18.60  1  1    4    2 mtcars3

Data:

export(list(mtcars1 = mtcars[1:10,], 
            mtcars2 = mtcars[11:20,],
            mtcars3 = mtcars[21:32,]),
       xlsx_file <- tempfile(fileext = ".xlsx")
)

CodePudding user response：

I would personally take the best of both, rio's import_list that prevents you from applying over the sheets and take the advantage that without the rbind = T it stores each list with the sheet as name. Not sure why they decided to swap from sheet name to sheet index when rbind = T, so then use rbindlist from data.table with use.names = T.

rbindlist(import_list("mtcars.xlsx"), use.names = T, fill = T, id = "file")

export(list(
  mtcars1 = mtcars[1:10,],
  mtcars2 = mtcars[11:20,],
  mtcars3 = mtcars[21:32,]), "mtcars.xlsx")

CodePudding user response：

Simply use bind_rows() in dplyr and set the arg .id = "sheet", then data in each sheet will be row-bind together and a new column named what you set in .id is added to record the sheet names which the data come from.

dplyr::bind_rows(
  import_list("path/to/file/test.xlsx", setclass = "tbl"),
  .id = "sheet"
)

Test

Write out an excel file with 2 sheets named AUS and AUT:

openxlsx::write.xlsx(
  list(AUS = data.frame(x = 1:2, y = 3:4),
       AUT = data.frame(x = 5:6, y = 7:8)),
  file = "test.xlsx"
)

Then

dplyr::bind_rows(
  rio::import_list("test.xlsx", setclass = "tbl"),
  .id = "sheet"
)

# # A tibble: 4 × 3                                                                 
#   sheet     x     y
#   <chr> <dbl> <dbl>
# 1 AUS       1     3
# 2 AUS       2     4
# 3 AUT       5     7
# 4 AUT       6     8

CodePudding user response：

You should ideally provide some sample data.

This should do the trick though:

library(readxl)
library(tidyverse)
library(rio)

# Sample data
list("Sheet 1" = tibble(n = 1:3),
     "Sheet 2" = tibble(n = 101:103)) %>% 
  writexl::write_xlsx("Sheet name.xlsx")

# What you want
purrr::map2_df(.x = rio::import_list("Sheet name.xlsx"),
               .y = readxl::excel_sheets("Sheet name.xlsx"),
               ~{
                 .x %>% 
                   dplyr::mutate(`Sheet name` = .y)
               })