Data wrangling into long format in R - r

Question

I have a source dataset from a nature article. I was wondering how I could extract the values from rows 4 and 12 into a long data format with the relevant assigned group (i.e. Inefficient/Efficient).
This is the code I have used to get the data into R.
# load the required libraries
library(ggsignif)
library(readxl)
library(svglite)
library(tidyverse)
library(tidyr)
library(dplyr)
# The paper from which the figure is taken is Tasdogen et al. (2020)
# Metabolic heterogeneity confers differences in melanoma metastatic potential
# The figure is 2b and can be accessed at
# https://www.nature.com/articles/s41586-019-1847-2#MOESM3
# The link to the raw data used in the article is given below and directly improted for plotting
url <-'https://static-content.springer.com/esm/art%3A10.1038%2Fs41586-019-1847-2/MediaObjects/41586_2019_1847_MOESM3_ESM.xlsx'
#create a dataframe from the Excel data
temp <- tempfile()
download.file(url, temp, mode='wb')
myData <- read_excel(path = temp)
I cant figure out how to insert an image of the dataset but it should show up with the previous code. I need columns 2-31 for efficient and 2 to 37 for inefficient.
I hope that's enough information for people to understand want I'm talking about.

Answer 1

This data is really not structured well for a general read like that, but I'll try to make do:
### myData <- read_excel(...)
Data_wide<- myData[c(2:4,10:12), c(2:37)]
tmp <- as.data.frame(t(Data_wide))
head(tmp)
# V1 V2 V3 V4 V5 V6
# ...2 Efficient #1 0.47699999999999998 Inefficient #1 0.48499999999999999
# ...3 Efficient #2 0.376 Inefficient #2 0.47399999999999998
# ...4 Efficient #3 0.496 Inefficient #3 0.48799999999999999
# ...5 Efficient #4 0.32500000000000001 Inefficient #4 0.45600000000000002
# ...6 Efficient #5 8.8999999999999996E-2 Inefficient #5 0.53100000000000003
# ...7 Efficient #6 4.5999999999999999E-2 Inefficient #6 0.318
tmp <- rbind(tmp[,1:3], setNames(tmp[,4:6], names(tmp)[1:3]))
head(tmp)
# V1 V2 V3
# ...2 Efficient #1 0.47699999999999998
# ...3 Efficient #2 0.376
# ...4 Efficient #3 0.496
# ...5 Efficient #4 0.32500000000000001
# ...6 Efficient #5 8.8999999999999996E-2
# ...7 Efficient #6 4.5999999999999999E-2
tmp <- tmp[complete.cases(tmp),]
tmp$V3 <- as.numeric(tmp$V3)
rownames(tmp) <- NULL
head(tmp,3); tail(tmp,3)
# V1 V2 V3
# 1 Efficient #1 0.477
# 2 Efficient #2 0.376
# 3 Efficient #3 0.496
# V1 V2 V3
# 64 Inefficient #34 0.2451
# 65 Inefficient #35 0.2450
# 66 Inefficient #36 0.2529
With this structure, you can subset (remove V2, though I wonder why you feel it is not important) and rename (colnames(tmp) <- c(...)).

Answer 2

Although it might not be pretty, I believe this would be your solution using only readxl and tidyverse packages:
# Select first set of rows with group and value
set1 <-
myData %>%
filter(row_number() %in% c(2, 4))
# Select second set of rows with group and value
set2 <-
myData %>%
filter(row_number() %in% c(10, 12))
# Join both sets of data, so that all group labels are in one row and all values are in one row.
left_join(set1, set2, by = "Fractional enrichment of glucose m+6 in primary subcutaneous tumors after [U-13C]glucose infusion") %>%
#pivot the table to a long format with group lable and value labels in separate columns
pivot_longer(cols = !`Fractional enrichment of glucose m+6 in primary subcutaneous tumors after [U-13C]glucose infusion`) %>%
# pivot wider to a format with group lable and value labels in separate columns
pivot_wider(names_from = `Fractional enrichment of glucose m+6 in primary subcutaneous tumors after [U-13C]glucose infusion`, values_from = value) %>%
# Remove old column names/numbers
select(-name)
# A tibble: 72 x 2
Group `Glucose m+6`
<chr> <chr>
1 Inefficient 0.48499999999999999
2 Inefficient 0.47399999999999998
3 Inefficient 0.48799999999999999
4 Inefficient 0.45600000000000002
5 Inefficient 0.53100000000000003
6 Inefficient 0.318
7 Inefficient 0.26600000000000001
8 Inefficient 0.30399999999999999
9 Inefficient 0.309
10 Inefficient 0.33
# ... with 62 more rows

Answer 3

A clean way to address your problem is to use the libraries tidyxl and unpivotr.
They may seem rather complicated at first, but it's probably the cleanest way to handle excel files. I left some comments to help you go through it.
I suggest you to have a look at unpivotr vignettes.
# libraries
library(tidyverse)
library(tidyxl)
library(unpivotr)
# download data
url <-'https://static-content.springer.com/esm/art%3A10.1038%2Fs41586-019-1847-2/MediaObjects/41586_2019_1847_MOESM3_ESM.xlsx'
temp <- tempfile()
download.file(url, temp, mode='wb')
# read excel file
myData <- xlsx_cells(path = temp)
# select the sheet
figure1a <- myData %>% filter(sheet == "Figure 1 A")
# you can visualize data in an excel-like format with
# View(rectify(figure1a))
# since the sheet is composed by two tables
# get the top-left corner of each table (where in the first column you find Group)
corners <- figure1a %>% filter(character == "Group")
# partition the spreadsheet based on the corners you just got
# select the rows you will need
partitions <- figure1a %>% filter(row %in% c(3:5, 11:13)) %>% partition(corners)
# get the two partitions and edit them
# with purrr::map it will be easy
df <- partitions$cells %>%
# the first column for each partition shows the headers
map(behead, "left", "header") %>%
# the first row for each partition shows the Group: Efficient/Inefficient
map(behead, "up", "Group") %>%
# the second row for each partition shows the mouse id
# and bind the edited partitions together
map_dfr(behead, "up", "Mouse_ID") %>%
# select the columns we need
select(Group, Mouse_ID, Glucose_m6 = numeric)
# the final result
df
#> # A tibble: 66 x 3
#> Group Mouse_ID Glucose_m6
#> <chr> <chr> <dbl>
#> 1 Efficient #1 0.477
#> 2 Efficient #2 0.376
#> 3 Efficient #3 0.496
#> 4 Efficient #4 0.325
#> 5 Efficient #5 0.089
#> 6 Efficient #6 0.046
#> 7 Efficient #7 0.213
#> 8 Efficient #8 0.082
#> 9 Efficient #9 0.359
#> 10 Efficient #10 0.306
#> # ... with 56 more rows
Created on 2021-11-04 by the reprex package (v2.0.0)