Separate a shopping list into multiple columns - r

Question

I have a shopping list data like this:
df <- data.frame(id = 1:5, item = c("apple2milk5", "milk1", "juice3apple5", "egg10juice1", "egg8milk2"), stringsAsFactors = F)
# id item
# 1 1 apple2milk5
# 2 2 milk1
# 3 3 juice3apple5
# 4 4 egg10juice1
# 5 5 egg8milk2
I want to separate the variable item into multiple columns and record the number behind the goods. The problem I met is that the goods each person purchases are different so I cannot solve it using tidyr::separate() or other analogous functions. What I expect is:
# id apple milk juice egg
# 1 1 2 5 NA NA
# 2 2 NA 1 NA NA
# 3 3 5 NA 3 NA
# 4 4 NA NA 1 10
# 5 5 NA 2 NA 8
Note: The categories of goods in the market are unknown. So don't assume there are only 4 kinds of goods.
Thanks for any helps!

Answer 1

I just came up with a tidyverse solution which uses stringr::str_extract_all() to extract the quantities, sets their names as product names, and expands them to wide using tidyr::unnest_wider().
library(tidyverse)
df %>%
mutate(N = map2(str_extract_all(item, "\\d+"), str_extract_all(item, "\\D+"), set_names)) %>%
unnest_wider(N, transform = as.numeric)
# # A tibble: 5 × 6
# id item apple milk juice egg
# <int> <chr> <dbl> <dbl> <dbl> <dbl>
# 1 1 apple2milk5 2 5 NA NA
# 2 2 milk1 NA 1 NA NA
# 3 3 juice3apple5 5 NA 3 NA
# 4 4 egg10juice1 NA NA 1 10
# 5 5 egg8milk2 NA 2 NA 8

Answer 2

I'll add yet another answer. It only slightly differs from #ASuliman's but uses a bit of the newer tidyr and some cute regex to become a bit more straightforward.
The regex trick is that the pattern "(?<=\\d)\\B(?=[a-z])" will match the non-boundary (i.e. an empty location) between numbers and letters, allowing you to create rows for every "apple5" type of entry. Extract the letters into an item column and numbers into a count column. Using the new pivot_wider which replaces spread, you can convert those counts to numeric values as you reshape.
library(dplyr)
library(tidyr)
df %>%
separate_rows(item, sep = "(?<=\\d)\\B(?=[a-z])") %>%
extract(item, into = c("item", "count"), regex = "^([a-z]+)(\\d+)$") %>%
pivot_wider(names_from = item, values_from = count, values_fn = list(count = as.numeric))
#> # A tibble: 5 x 5
#> id apple milk juice egg
#> <int> <dbl> <dbl> <dbl> <dbl>
#> 1 1 2 5 NA NA
#> 2 2 NA 1 NA NA
#> 3 3 5 NA 3 NA
#> 4 4 NA NA 1 10
#> 5 5 NA 2 NA 8

Answer 3

Possibily something like this, and should work with any item/quantity.
It just assumes that the quantity follows the item.
Lets use a custom function which extracts item and quantity:
my_fun <- function(w) {
items <- stringr::str_split(w, "\\d+", simplify = T)
items <- items[items!=""] # dont now why but you get en empty spot each time
quantities <- stringr::str_split(w, "\\D+", simplify = T)
quantities <- quantities[quantities!=""]
d <- data.frame(item = items, quantity=quantities, stringsAsFactors = F)
return(d)
}
Example:
my_fun("apple2milk5")
# gives:
# item quantity
# 1 apple 2
# 2 milk 5
Now we can apply the function to each id, using nest and map:
library(dplyr)
df_result <- df %>%
nest(item) %>%
mutate(res = purrr::map(data, ~my_fun(.x))) %>%
unnest(res)
df_results
# # A tibble: 9 x 3
# id item quantity
# <int> <chr> <chr>
# 1 1 apple 2
# 2 1 milk 5
# 3 2 milk 1
# 4 3 juice 3
# 5 3 apple 5
# 6 4 egg 10
# 7 4 juice 1
# 8 5 egg 8
# 9 5 milk 2
Now we can use dcast() (probabily spread would work too):
data.table::dcast(df_result, id~item, value.var="quantity")
# id apple egg juice milk
# 1 1 2 <NA> <NA> 5
# 2 2 <NA> <NA> <NA> 1
# 3 3 5 <NA> 3 <NA>
# 4 4 <NA> 10 1 <NA>
# 5 5 <NA> 8 <NA> 2
Data:
df <- data.frame(id = 1:5, item = c("apple2milk5", "milk1", "juice3apple5", "egg10juice1", "egg8milk2"), stringsAsFactors = F)

Answer 4

tmp = lapply(strsplit(df$item, "(?<=\\d)(?=\\D)|(?<=\\D)(?=\\d)", perl = TRUE),
function(x) {
d = split(x, 0:1)
setNames(as.numeric(d[[2]]), d[[1]])
})
nm = unique(unlist(lapply(tmp, names)))
cbind(df, do.call(rbind, lapply(tmp, function(x) setNames(x[nm], nm))))
# id item apple milk juice egg
#1 1 apple2milk5 2 5 NA NA
#2 2 milk1 NA 1 NA NA
#3 3 juice3apple5 5 NA 3 NA
#4 4 egg10juice1 NA NA 1 10
#5 5 egg8milk2 NA 2 NA 8

Answer 5

Place a space before each numeric substring and a newline after it. Then read that data using read.table and unnest it. Finally use pivot_wider to convert from long to wide form.
library(dplyr)
library(tidyr)
df %>%
mutate(item = gsub("(\\d+)", " \\1\n", item)) %>%
rowwise %>%
mutate(item = list(read.table(text = item, as.is = TRUE))) %>%
ungroup %>%
unnest(item) %>%
pivot_wider(names_from = "V1", values_from = "V2")
giving:
# A tibble: 5 x 5
id apple milk juice egg
<int> <int> <int> <int> <int>
1 1 2 5 NA NA
2 2 NA 1 NA NA
3 3 5 NA 3 NA
4 4 NA NA 1 10
5 5 NA 2 NA 8
Variation
This is a variation of the above code that eliminates the unnest. We replace each numeric string by a space, that string, another space, the id and a newline. Then use read.table to read that in. Note the use of %$% rather than %>% before the read.table. Finally use pivot_wider to convert from long to wide form.
library(dplyr)
library(magrittr)
library(tidyr)
df %>%
rowwise %>%
mutate(item = gsub("(\\d+)", paste(" \\1", id, "\n"), item)) %$%
read.table(text = item, as.is = TRUE, col.names = c("nm", "no", "id")) %>%
ungroup %>%
pivot_wider(names_from = "nm", values_from = "no")

Answer 6

You can try
library(tidyverse)
library(stringi)
df %>%
mutate(item2 =gsub("[0-9]", " ", df$item)) %>%
mutate(item3 =gsub("[a-z]", " ", df$item)) %>%
mutate_at(vars(item2, item3), ~stringi::stri_extract_all_words(.) %>% map(paste, collapse=",")) %>%
separate_rows(item2, item3, sep = ",") %>%
spread(item2, item3)
id item apple egg juice milk
1 1 apple2milk5 2 <NA> <NA> 5
2 2 milk1 <NA> <NA> <NA> 1
3 3 juice3apple5 5 <NA> 3 <NA>
4 4 egg10juice1 <NA> 10 1 <NA>
5 5 egg8milk2 <NA> 8 <NA> 2

Answer 7

#replace any digit followed by a character "positive look-ahead assertion" by the digit plus a comma
library(dplyr)
library(tidyr)
df %>% mutate(item=gsub('(\\d+(?=\\D))','\\1,' ,item, perl = TRUE)) %>%
separate_rows(item, sep = ",") %>%
extract(item, into = c('prod','quan'), '(\\D+)(\\d+)') %>%
spread(prod, quan, fill=0)
id apple egg juice milk
1 1 2 0 0 5
2 2 0 0 0 1
3 3 5 0 3 0
4 4 0 10 1 0
5 5 0 8 0 2

Answer 8

This is a simple solution in base R and stringr:
goods <- unique(unlist(stringr::str_split(df$item, pattern = "[0-9]")))
goods <- goods[goods != ""]
df <- cbind(df$id, sapply(goods,
function(x) stringr::str_extract(df$item, pattern = paste0(x,"[0-9]*"))))
df <- as.data.frame(df)
df[-1] <- lapply(df[-1], function(x) as.numeric(stringr::str_extract(x, pattern = "[0-9]*$")))
names(df)[1] <- "id"
Output
id apple milk juice egg
1 1 2 5 NA NA
2 2 NA 1 NA NA
3 3 5 NA 3 NA
4 4 NA NA 1 10
5 5 NA 2 NA 8

Answer 9

Mostly base R with some input from stringr and data.table:
library(stringr)
library(data.table)
cbind(
id = df$id,
rbindlist(
lapply(df$item, function(x) as.list(setNames(str_extract_all(x, "[0-9]+")[[1]], strsplit(x, "[0-9]+")[[1]]))),
fill = TRUE
)
)
id apple milk juice egg
1: 1 2 5 <NA> <NA>
2: 2 <NA> 1 <NA> <NA>
3: 3 5 <NA> 3 <NA>
4: 4 <NA> <NA> 1 10
5: 5 <NA> 2 <NA> 8

Answer 10

A cleaner data.table solution with input from stringr:
df[,
.(it_count = str_extract_all(item, "[0-9]+")[[1]],
it_name = str_extract_all(item, "[^0-9]+")[[1]]),
by = id
][, dcast(.SD, id ~ it_name, value.var = "it_count")]
id apple egg juice milk
1: 1 2 <NA> <NA> 5
2: 2 <NA> <NA> <NA> 1
3: 3 5 <NA> 3 <NA>
4: 4 <NA> 10 1 <NA>
5: 5 <NA> 8 <NA> 2