Introduction to unpivotr

In unpivotr: Unpivot Complex and Irregular Data Layouts

knitr::opts_chunk$set(
  collapse = TRUE,
  comment = "#>"
)
library(here)

# print every row of a data frame
library(knitr)
knit_print.data.frame = function(...) {
  print(..., n = Inf)
}
# register the method
registerS3method("knit_print", "data.frame", knit_print.data.frame)

Preface

This is based on a talk. You might want to watch the video or read the slides (see speaker notes by clicking the cog):

https://docs.google.com/presentation/d/1tVwn_-QVGZTflnF9APiPACNvyAKqujdl6JmxmrdDjok

1. Reading easy spreadsheets with {readxl}

It is easy to read a spreadsheet into R when it has:

• A rectangular shape
• One row of column headers
• No meaningful colour or other formatting
• Consistent data types in each column, e.g. all numbers or all text

Here is an example, a dataset of student test marks in different subjects.

knitr::include_graphics(here("vignettes/images/hp-tidy.png"))

To test whether a table will be easy to import, ask yourself "Is every row self-sufficient? Could I read only one row and understand all the data in it?" In this case, one row will tell you that Ron got two marks in potions in his second year.

Because this table is simple -- or 'tidy' -- it is easily imported into a data frame, using the {readxl} package.

library(readxl) # for read_excel()

hp_xlsx <- system.file("extdata/harry-potter.xlsx", package = "unpivotr")

tidy <- read_excel(hp_xlsx, sheet = "tidy")

tidy

Note that the row of column names in the spreadsheet has been used as column names of the data frame. Also the data type of each column is either dbl (double, which means a number), or chr (character) as appropriate.

2. Trying to read a hard spreadsheet with {readxl}

Here's is the same data but this time it is in a spreadsheet that the {readxl} package can't read so easily. Why not?

knitr::include_graphics(here("vignettes/images/hp-untidy.png"))

The Pupil and Year columns have been combined into one, so the names of the pupils aren't in the same rows as their marks, nor are they in the same columns. There is also a text value "10 - really?" amongst a column of numbers.

Is every row self-sufficient? Could you read only one row and understand all the data in it? No, because one row will only tell you the mark, subject and year, but not the name. Or else it will tell you the name, but not the mark, subject or year.

Here is what happens when the table is read with the {readxl} package.

untidy <- read_excel(hp_xlsx, sheet = "untidy")

untidy

What has gone wrong? The spreadsheet has broken the assumptions that the {readxl} package makes about data.

• A rectangular shape. The spreadsheet is not rectangular because the top-left cell is deliberately blank (is not 'missing' data), and so are the cells to the right of "Ron" and "Ginny".
• No meaningful colour or other formatting. The spreadsheet has meaningful formatting to distinguish between names in bold ("Ron", "Ginny") and years in plain type, ("1st year", "2nd year").
• Consistent data types in each column. The spreadsheet has mixed data types in the Herbology column, where some cells are numbers and one is text: "10 - really?".

The readxl package has done its best with a difficult file.

• It has dealt with the non-rectangular shape by filling the gaps, using ...1 to fill the cell in the top-left corner with a column header, and NA to fill the cells to the right of Ron and Ginny.
• It has dealt with the mixed data types in the Herbology column by treating everything as text, even the numbers, so that it can accommodate the text value "10 - really?".
• It hasn't dealt with the meaningful formatting (bold names) because it is blind to formatting -- {readxl} doesn't know anything about formatting except for data types.

Unfortunately {readxl} hasn't been able to make the data tidy. Each row still isn't self-sufficient. You couldn't read only one row and understand all the data in it.

Here is a final example of a spreadsheet that breaks the one remaining assumption: that there is a single row of column headers. This file has two rows of column headers.

knitr::include_graphics(here("vignettes/images/hp-pivoted.png"))

The rest of this tutorial will demonstrate how to use the {tidyxl} and {unpivotr} packages to import that spreadsheet.

3. Demonstration of {tidyxl} and {unpivotr}

Don't expect to understand yet how the following code works. It is here to show you what to expect later, and it is the entire code to import the spreadsheet above.

library(dplyr)
library(tidyr)
library(tidyxl)
library(unpivotr)

hp_xlsx <- system.file("extdata/harry-potter.xlsx", package = "unpivotr")

cells <- xlsx_cells(hp_xlsx, sheets = "pivoted")
formats <- xlsx_formats(hp_xlsx)

indent <- formats$local$alignment$indent

tidied <-
  cells %>%
  filter(!is_blank) %>%
  behead("up-left", "dormitory") %>%
  behead("up", "name") %>%
  behead_if(indent[local_format_id] == 0,
            direction = "left-up",
            name = "location") %>%
  behead("left", "subject") %>%
  select(address, dormitory, name, location, subject, mark = numeric) %>%
  arrange(dormitory, name, location, subject)

tidied

4. Explanation of tidyxl::xlsx_cells()

The first step to import a difficult spreadsheet is to read it with tidyxl::xlsx_cells().

What does tidyxl::xlsx_cells() do that is different from readxl::read_excel()? Instead of returning the data in a data frame, it returns individual cells in a data frame. Try matching each row of the output of xlsx_cells() to a cell in the spreadsheet.

cells <-
  xlsx_cells(hp_xlsx, sheets = "pivoted") %>%
  # Drop some columns to make it clearer what is going on
  select(row, col, is_blank, data_type, character, numeric, local_format_id)

cells

The first row of the output describes the cell B2 (row 1, column 2) of the spreadsheet, with the character value "Witch".

# A tibble: 47 x 7
     row   col is_blank data_type character numeric local_format_id
   <int> <int> <lgl>    <chr>     <chr>       <dbl>           <int>
 1     1     2 FALSE    character Witch          NA               2

Row 10 describes the cell C3 (row 3, column 3) of the spreadsheet, with the numeric value 11.

So what xlsx_cells() has done is give you a data frame that isn't data itself, but it describes the data in the spreadsheet. Each row describes one cell. This allows you to do some fancy tricks, like filter for all the numeric cells.

cells %>%
  filter(data_type == "numeric")

Or you could filter for a particular cell by its row and column position.

cells %>%
  filter(row == 2, col == 4)

And you can filter out all 'blank' cells. A cell is 'blank' if it has formatting but no value. Sometimes it's useful to have these, but usually you should discard them.

cells %>%
  filter(!is_blank)

That is all you need to know about the tidyxl package for now. Later you will be shown how to filter for cells by their formatting (e.g. bold cells, indented cells, or cells with coloured text).

5. Explanation of unpivotr::behead()

You've seen that tidyxl::xlsx_cells() reads a spreadsheet one cell at a time, so that you can filter for particular cells by their position, their value, their data type, etc. You could now write code to tidy up any spreadsheet.

The unpivotr package gives you some pre-packaged tools for tidying up a spreadsheet. The most important tool is behead(), which deals with one layer of header cells at a time.

Let's look again at the original spreadsheet. I have highlighted the first row of header cells.

knitr::include_graphics(here("vignettes/images/untidy-header-row-1.png"))

Use unpivotr::behead() to tag data cells with "Witch" or "Wizard", and then strip (or behead!) those header cells from the rest -- they are no longer required.

cells %>%
  filter(!is_blank) %>%
  behead("up-left", "dormitory")

Click through table to check that every cell belonging to the Witch header has been taggged "Witch" in the column dormitory, and the same for wizards Notice that the locations Castle and Grounds have also been tagged witch or wizard. Also, all the cells in row 1 have disappeared -- they have become values in the dormitory column.

What do the arguments to behead("up-left", "dormitory") mean? The second one, "dormitory" becomes the column name of the male/female tags. But the direction "up-left" is the most important one. It tells behead() which way to look for a header cell.

For example, starting from the cell C3 (row 3 column 3), behead() looks up and to the left to find the header "Witch". Starting from the cell D4 (row 4, column 4) it finds the header "Wizard". Starting from cells in the first column, there is no header cell in the "up-left" direction, so they are tagged with missing values. Don't worry about them -- they will come right later.

What if we try a different direction instead, "up-right" (up and to the right)? Again, compare the table with the spreadsheet

knitr::include_graphics(here("vignettes/images/untidy-header-row-1.png"))

cells %>%
  filter(!is_blank) %>%
  behead("up-right", "dormitory")

Check that Ginny has been tagged "Wizard", and so have her marks in cells below. Unpivotr doesn't know that this is wrong, it has just done what it was told. The behead() function isn't magic, it just enables you to tell unpivotr which data cells relate to which header cells.

6. Continuing unpivotr::behead()

Let's carry on with the second row of header cells (highlighted). This time the direction is simply "up" for directly up because there is a header in every column. Notice that we're building up a pipeline of transformations, one set of headers at a time.

knitr::include_graphics("images/untidy-header-row-2.png")

cells %>%
  filter(!is_blank) %>%
  behead("up-left", "dormitory") %>%
  behead("up", "name")

Click through the table to match it to the spreadsheet. The header cells in rows 1 and 2 have all disappeared to become values in the dormitory and name columns. The cell C3 (row 3, column 3) has been tagged "Witch" and "Ginny"

7. Handling meaningful formatting with unpivotr::behead_if()

Applying the same procedure to the headers in column A, which describe the location and subject, what are the directions?

Starting from a data cell, say, B7 (row 7, column 2), the location is"Grounds", which is to the left and then up, "left-up".

knitr::include_graphics("images/untidy-header-col-1.png")

But there is a complication. When unpivotr::behead() is travelling up the cells in column 1, how does it know to stop at "Grounds" and not overshoot to "Potions" or any of the cells further up? You must tell behead() to stop at the first cell that isn't indented. Alternatively, you could tell it to stop at the first cell that is bold.

Use unpivotr::behead_if() when there is a rule to identify a header cell. In this case the rule will be "when the cell has bold formatting".

A spreadsheet cell can have so many different formats that it would be unweildy for {tidyxl} to import them all at once. Instead, {tidyxl} imports a kind of lookup table of formatting, and each cell has a key into the lookup table, called local_format_id.

Here's how to look up the indented property of a cell.

formats <- xlsx_formats(hp_xlsx) # load the format lookup table from the file

indent <- formats$local$alignment$indent # find the 'indent' property

indent[cells$local_format_id] # look up the indent property of each cell

When you look up a format from inside behead_if(), you don't need to mention cell$, but you do have to name the other arguments to behead().

formats <- xlsx_formats(hp_xlsx) # load the format lookup table from the file
indent <- formats$local$alignment$indent # find the 'indent' property

cells %>%
  filter(!is_blank) %>%
  behead("up-left", "dormitory") %>%
  behead("up", "name") %>%
  behead_if(indent[local_format_id] == 0,
            direction = "left-up", # This argument has to be named now.
            name = "location")     # So does this one.

You can give more than one rule to behead_if() at once. They are applied together, so all the rules must evaluate to TRUE for a cell to be treated as a header cell. Here's an example applying the additional rule that a cell must be bold. The result in this case is the same.

formats <- xlsx_formats(hp_xlsx)
indent <- formats$local$alignment$indent
bold <- formats$local$font$bold # find the 'bold' property

cells %>%
  filter(!is_blank) %>%
  behead("up-left", "dormitory") %>%
  behead("up", "name") %>%
  behead_if(indent[local_format_id] == 0, # First rule
            bold[local_format_id],        # Second rule. Both must be TRUE
            direction = "left-up",
            name = "location")

Check that Hermione got 5 marks in a subject taken in Hogwarts grounds, by looking at cell B7 (row 7, column 2).

8. Finishing and cleaning up

Only one layer of headers remains: the subjects in column 1. The direction is directly "left".

knitr::include_graphics("images/untidy-header-col-2.png")

cells %>%
  filter(!is_blank) %>%
  behead("up-left", "dormitory") %>%
  behead("up", "name") %>%
  behead_if(indent[local_format_id] == 0,
            direction = "left-up",
            name = "location") %>%
  behead("left", "subject")

Check that Hermione got 5 marks in Herbology in particular, taken in Hogwarts grounds, by looking at cell B7 (row 7, column 2).

The final cleanup is straightforward; choose the columns to keep, using the standard tidyverse function dplyr::select(). At the same time you can rename the column numeric to mark, and the column character to other. What is the column other for? For the value "10 - really?".

cells %>%
  filter(!is_blank) %>%
  behead("up-left", "dormitory") %>%
  behead("up", "name") %>%
  behead_if(indent[local_format_id] == 0,
            direction = "left-up",
            name = "location") %>%
  behead("left", "subject") %>%
  select(dormitory, name, location, subject, mark = numeric, other = character)

It is up to you now what to do with the 'total' values for the castle and the grounds. If you don't want to keep them, it's easy enough to filter them out using !is.na(subject). That is done in the final code listing below.

library(dplyr)
library(tidyr)
library(tidyxl)
library(unpivotr)

hp_xlsx <- system.file("extdata/harry-potter.xlsx", package = "unpivotr")

cells <- xlsx_cells(hp_xlsx, sheet = "pivoted")
formats <- xlsx_formats(hp_xlsx)

indent <- formats$local$alignment$indent

tidied <-
  cells %>%
  filter(!is_blank) %>%
  behead("up-left", "dormitory") %>%
  behead("up", "name") %>%
  behead_if(indent[local_format_id] == 0,
            direction = "left-up",
            name = "location") %>%
  behead("left", "subject") %>%
  select(address, dormitory, name, location, subject, mark = numeric) %>%
  arrange(dormitory, name, location, subject)

tidied

Review

Well done for making it this far. If you have struggled to follow, that is normal -- it means you are learning. Try reading through a second or third time, and change parts of the code to see what happens.

unpivotr documentation built on Jan. 23, 2023, 5:40 p.m.