I have the following problem, I have a tibble with mutliple character columns.
I tried to provide an MRE below:
library(tidyverse)
df <- tibble(food = c("pizza, bread, apple","joghurt, cereal, banana"),
food2 = c("bread, sausage, strawberry", "joghurt, oat, bacon"),
food3 = c("ice cream, bread, milkshake", "melon, cake, joghurt")
)
df %>%
# rowwise() %>%
mutate(allcolumns = map2(
str_split(food, ", "),
str_split(food2, ", "),
# str_split(food3, ", "),
intersect
) %>% unlist()
) -> df_new
My goal would be to get the common words for all columns. Words are separated by , in the columns. In the MRE I am able to find the intersect between two columns, however I couldnt get a solution for this issue. I experimented with Reduce but was not able to get it.
As an EDIT: I would also like to append it as a new row to the existing tibble
We may use map to loop over the columns, do the str_split and then reduce to get the intersect for elementwise intersect
library(dplyr)
library(purrr)
library(stringr)
df %>%
purrr::map(str_split, ", ") %>%
transpose %>%
purrr::map_chr(reduce, intersect) %>%
mutate(df, Intersect = .)
-output
# A tibble: 2 x 4
food food2 food3 Intersect
<chr> <chr> <chr> <chr>
1 pizza, bread, apple bread, sausage, strawberry ice cream, bread, milkshake bread
2 joghurt, cereal, banana joghurt, oat, bacon melon, cake, joghurt joghurt
or may also use pmap
df %>%
mutate(Intersect = pmap(across(everything(), str_split, ", "),
~ list(...) %>%
reduce(intersect)))
Related
Given a dataframe of types and values like so:
topic
keyword
cheese
cheddar
meat
beef
meat
chicken
cheese
swiss
bread
focaccia
bread
sourdough
cheese
gouda
My aim is to make a set of dynamic regexs based on the type, but I don't know how to make the variable names from the types. I can do this individually like so:
fn_get_topic_regex <- function(targettopic,df)
{
filter_df <- df |>
filter(topic == targettopic)
regex <- paste(filter_df$keyword, collapse = "|")
}
and do things like:
cheese_regex <- fn_get_topic_regex("cheese",df)
But what I'd like to be able to do is build all these regexes automatically without having to define each one.
The intended output would be something like:
cheese_regex: "cheddar|swiss|gouda"
bread_regex: "focaccia|sourdough"
meat_regex: "beef|chicken"
Where the start of the variable name is the distinct topic.
What's the best way to do that without defining each regex individually by hand?
You can use dplyr's group_by() and summarise()
df %>%
group_by(topic) %>%
summarise(regex = paste(keyword, collapse = "|"))
# A tibble: 3 × 2
topic regex
<chr> <chr>
1 bread focaccia|sourdough
2 cheese cheddar|swiss|gouda
3 meat beef|chicken
Or you can apply your function to every unique value in df$topic:
map_chr(unique(df$topic) %>% setNames(paste0(., "_regex")),
fn_get_topic_regex, df = df)
cheese_regex meat_regex bread_regex
"cheddar|swiss|gouda" "beef|chicken" "focaccia|sourdough"
Just remember to add return(regex) to the end of your function, or not to assign the last line to a variable at all. I would even put everything in a single pipe chain:
fn_get_topic_regex <- function(targettopic,df)
{
df |>
filter(topic == targettopic) |>
pull(keyword) |>
paste(collapse = "|")
}
Here is a base R solution with your intended output in a named list.
df <- structure(list(topic = c("cheese", "meat", "meat", "cheese", "bread", "bread", "cheese"),
keyword = c("cheddar", "beef", "chicken", "swiss", "focaccia", "sourdough", "gouda")),
class = "data.frame", row.names = c(NA, -7L))
#split into a list per topic
topics <- split(df, df$topic)
#collapse the keyword column
topics <- lapply(topics, function(t) {
paste(t$keyword, collapse = "|")
})
#rename
names(topics)<- paste0(names(topics), "_regex")
topics
$bread_regex
[1] "focaccia|sourdough"
$cheese_regex
[1] "cheddar|swiss|gouda"
$meat_regex
[1] "beef|chicken"
We could do something like this:
after grouping we could use summarise together with paste and collapse to get our regex s
Then, when the regex is needed we could refer to it by indexing like the example below:
library(dplyr)
library(stringr) #str_detect
my_regex <- df %>%
group_by(topic) %>%
summarise(regex = paste(keyword, collapse = "|"))
df %>%
mutate(new_col = ifelse(str_detect(keyword, my_regex$regex[1]), "it is bread", "it is not bread"))
A tibble: 3 × 2
topic regex
<chr> <chr>
1 bread focaccia|sourdough
2 cheese cheddar|swiss|gouda
3 meat beef|chicken
> df %>%
+ mutate(new_col = ifelse(str_detect(keyword, my_regex$regex[1]), "it is bread", "it is not bread"))
topic keyword new_col
1 cheese cheddar it is not bread
2 meat beef it is not bread
3 meat chicken it is not bread
4 cheese swiss it is not bread
5 bread focaccia it is bread
6 bread sourdough it is bread
7 cheese gouda it is not bread
Suppose I have diary entries from 5 people and I want to determine if they mention any food-related key words. I want an output of the key word with a window of one word before and after to provide context before determining if they are food-related.
The search should be case-insensitive, and it's ok if the key word is embedded in another word. E.g., If a key word is "rice", I want to output to include "price".
Assume I have the following data:
foods <- c('corn', 'hot dog', 'ham', 'rice')
df <- data.frame(id = 1:5,
diary = c('I ate rice and corn today',
'Sue ate my corn.',
'He just hammed it up',
'Corny jokes are my fave',
'What is the price of milk'))
The output I'm looking for is:
|ID|Output |
|--|--------------------------------|
|1 |"ate rice and", "and corn today"|
|2 |"my corn" |
|3 |"just hammed it" |
|4 |"Corny jokes" |
|5 |"the price of" |
I've used strings::stri_detect but the output includes the entire diary entry.
I've used strings::stri_extract but I can't find a way to include one word before and after the key word.
The following solution works when the same food appears multiple times in the same phrase. It is based on the splitting of each phrase into its individual words.
library(tidyverse)
extract3 <- function(txt, word)
{
str_split(txt, "\\W") %>%
unlist() %>%
{. ->> w} %>%
map(~ str_extract(.x,regex(paste0("(.)*",word,"(.)*"),ignore_case=T))) %>%
unlist() %>%
is.na() %>%
`!` %>%
which() %>%
map_chr(~ paste(
w[unique(c(max(c(.x-1,1)),.x,min(c(.x+1,length(w)))))], collapse = " ")) %>%
paste(collapse = ", ")
}
df_out <- tibble()
for (i in 1:nrow(df))
for (j in 1:length(foods))
df_out <- rbind(df_out,
tibble(
id=df$id[i],diary=df$diary[i], output=extract3(df$diary[i],foods[j])))
df_out %>%
filter(output != "") %>%
group_by(id) %>%
mutate(output=paste(output,collapse = ", ")) %>%
ungroup() %>%
distinct()
EDITED (WITHOUT FOR CYCLES)
library(tidyverse)
extract3 <- function(txt, word)
{
str_split(txt, "\\W") %>%
unlist() %>%
{. ->> w} %>%
map(~ str_extract(.x,regex(paste0("(.)*",word,"(.)*"),ignore_case=T))) %>%
unlist() %>%
is.na() %>%
`!` %>%
which() %>%
map_chr(~ paste(
w[unique(c(max(c(.x-1,1)),.x,min(c(.x+1,length(w)))))], collapse = " ")) %>%
paste(collapse = ", ") %>%
str_trim()
}
map_dfr(
1:nrow(df),
function(id) map_dfr(1:length(foods), ~ tibble(
id = df$id[id],
diary = df$diary[id],
output = extract3(df$diary[id], foods[.])))) %>%
filter(output != "") %>%
group_by(id) %>%
mutate(output = paste(output,collapse = ", ")) %>%
ungroup() %>%
distinct()
We can collapse the regex and extract the words ("\w+") that preceed or follow the collapsed pattern. The regex() function allows the argument ignore_case = TRUE, which is very useful for case-insensitive matching. We may have to include optional word boundaries arount the collapsed pattern, so both rice and price, ham or hammed are included.
I made some small changes to the data to make it more illustrative.
I posted two answers.
One will exclude matches inside larger words, such as "hammed" or "price", so non-food matches will return empty strings.
The other is more inclusive.
library(dplyr)
library(stringr)
df %>% mutate(Output = str_extract_all (diary,
regex(paste0("\\w+\\s+(",
paste("\\b",foods, "\\b", collapse = "|", sep=''),
")\\s+\\w+"),
ignore_case=TRUE)))
output 1
id diary Output
1 1 I ate rice and corn today ate rice and
2 2 Sue ate my corn.
3 3 He just hammed it up
4 4 Corny jokes are my fave
5 5 What is the price of milk
6 6 I like to eat ham sandwiches eat ham sandwiches
solution 2
df %>% mutate(Output = str_extract_all (diary,
regex(paste0("\\w+\\s+(",
paste("\\b\\w*",foods, "\\w*\\b", collapse = "|", sep=''),
")\\s+\\w+"),
ignore_case=TRUE)))
id diary Output
1 1 I ate rice and corn today ate rice and
2 2 Sue ate my corn.
3 3 He just hammed it up just hammed it
4 4 Corny jokes are my fave
5 5 What is the price of milk the price of
6 6 I like to eat ham sandwiches eat ham sandwiches
data
foods <- c('corn', 'hot dog', 'ham', 'rice')
df <- data.frame(id = 1:6,
diary = c('I ate rice and corn today',
'Sue ate my corn.',
'He just hammed it up',
'Corny jokes are my fave',
'What is the price of milk',
'I like to eat ham sandwiches'))
FINAL EDIT
I figured out the problem with "corn", and handled the multiple matches issue.
We have to do a nested loop. First loop through all entries in "diary"(outer loop). Then, in the inner loop, loop through all "foods", and call "str_extract_all", with the appropriate regex. The initial regex required a food word be preceded or followed by another word, so foods at sentence boundaries were not matched. I included a ? quantifier (0 or 1 matches) around the surrounding words (\\w+\\s+) so it all works smoothly. The only issue left is the order of the matches in multiple matches, it is still odd. But I think the solution is fine now.
df %>% mutate(output=map(df$diary,
~map(foods, \(x) str_extract_all(.x,
regex(paste0("(\\w+\\s+)?(",
paste("\\b\\w*", x, "\\w*\\b", collapse = "|", sep=''),
")(\\s+\\w+)?"),
ignore_case=TRUE))))%>%
map(unlist))
id diary output
1 1 I ate rice and corn today and corn today, ate rice and
2 2 Sue ate my corn. my corn
3 3 He just hammed it up just hammed it
4 4 Corny jokes are my fave Corny jokes
5 5 What is the price of milk the price of
6 6 I like to eat ham sandwiches eat ham sandwiches
Not entirely sure whether that's 100% helpful but worth a try:
First, define your keywords as a case-insensitive alternation pattern:
patt <- paste0("(?i)(", paste0(foods, collapse = "|"), ")")
Then extract the word on the left, the keyword itself called node, and the word on the right using stringr's function str_extract_all:
library(stringr)
df1 <- data.frame(
left = unlist(str_extract_all(gsub("[.,!?]", "", df$diary), paste0("(?i)(\\S+|^)(?=\\s?", patt, ")"))),
node = unlist(str_extract_all(gsub("[.,!?]", "", df$diary), patt)),
right = unlist(str_extract_all(gsub("[.,!?]", "", df$diary), paste0("(?<=\\s?", patt, "\\s?)(\\S+|$)")))
)
Result:
df1
left node right
1 ate rice and
2 and corn today
3 my corn
4 just ham med
5 Corn y
6 p rice of
While this is not exactly the expected output it may still serve your purpose iff that purpose is to check whether a match is indeed a keyword. In lines 5 and 6, for example, the view provided by df1 immediately makes it clear that these are not keyword matches.
EDIT:
This solution preserves the idvalues:
library(tidyverse)
library(purrr)
extract_ <- function(df_row){
df1 <- data.frame(
id = df_row$id,
left = unlist(str_extract_all(gsub("[.,!?]", "", df_row$diary), paste0("(?i)(\\S+|^)(?=\\s?", patt, ")"))),
node = unlist(str_extract_all(gsub("[.,!?]", "", df_row$diary), patt)),
right = unlist(str_extract_all(gsub("[.,!?]", "", df_row$diary), paste0("(?<=\\s?", patt, "\\s?)(\\S+|$)")))
)
}
df %>%
group_split(id) %>% # splits data frame into list of bins, i.e. by id
map_dfr(.x, .f = ~ extract_(.x)) # now we iterate over bins with our function
id left node right
1 1 ate rice and
2 1 and corn today
3 2 my corn
4 3 just ham med
5 4 Corn y
6 5 p rice of
I'm trying to find an effective way to extract words from an text column in a dataset. The approach I'm using is
library(dplyr)
library(stringr)
Text = c("A little bird told me about the dog", "A pig in a poke", "As busy as a bee")
data = as.data.frame(Text)
keywords <- paste0(c("bird", "dog", "pig","wolf","cat", "bee", "turtle"), collapse = "|")
data %>% mutate(Word = str_extract(Text, keywords))
It's just an example but I have more than 2000 possible words to extract from each row. I don't know yet another approach to use, but the fact I will have a big regex will make things slow or doesn't matter the size of the regex? I think it will not appear more than one of these words in each row, but there is a way to make multiple columns automatically if more than one word appear in each row?
We can use str_extract_all to return a list, convert the list elements to a named list or tibble and use unnest_wider
library(purrr)
library(stringr)
library(tidyr)
library(dplyr)
data %>%
mutate(Words = str_extract_all(Text, keywords),
Words = map(Words, ~ as.list(unique(.x)) %>%
set_names(str_c('col', seq_along(.))))) %>%
unnest_wider(Words)
# A tibble: 3 x 3
# Text col1 col2
# <fct> <chr> <chr>
#1 A little bird told me about the dog bird dog
#2 A pig in a poke pig <NA>
#3 As busy as a bee bee <NA>
Try intersect with keywords as an array
data <- data.frame(Text = Text, Word = sapply(Text, function(v) intersect(unlist(strsplit(v,split = " ")),keywords),USE.NAMES = F))
# Sample Data Frame
df <- data.frame(Column_A
=c("1011 Red Cat",
"Mouse 2011 is in the House 3001", "Yellow on Blue Dog walked around Park"))
I've a column of manually inputted data which I'm trying to clean.
Column_A
1|1011 Red Cat |
2|Mouse 2011 is in the House 3001 |
2|Yellow on Blue Dog walked around Park|
I want to separate each characteristic into it's own column, but still maintain Column A to pull out other characteristics later.
Colour Code Column_A
1|Red |1001 |Cat
2|NA |2001 3001 |Mouse is in the House
3|Yellow on Blue |NA |Dog walked around Park
To date, I've been re-ordering them with gsub and capturing groups, then using Tidyr::extract to separate them.
library(dplyr)
library(tidyr)
library(stringr)
df1 <- df %>%
# Reorders the Colours
mutate(Column_A = gsub("(.*?)?(Yellow|Blue|Red)(.*)?", "\\2 \\1\\3",
Column_A, perl = TRUE)) %>%
# Removes Whitespaces
mutate(Column_A =str_squish(Column_A)) %>%
# Extracts the Colours
extract(Column_A, c("Colour", "Column_A"), "(Red|Yellow|Blue)?(.*)") %>%
# Repeats the Prececding Steps for Codes
mutate(Column_A = gsub("(.*?)?(\\b\\d{1,}\\b)(.*)?", "\\2 \\1\\3",
Column_A, perl = TRUE)) %>%
mutate(Column_A =str_squish(Column_A)) %>%
extract(Column_A, c("Code", "Column_A"), "(\\b\\d{1,}\\b)?(.*)") %>%
mutate(Column_A = str_squish(Column_A))
Which Results in this:
Colour Code Column_A
|Red |1011 |Cat
|Yellow |NA |on Blue Dog walked around Park
|NA |1011 |Mouse is in the House 1001
This works fine for the first row, but not the proceeding space and word separated ones, which I've subsequently been extracting and uniting. What's a more elegant way of doing this?
Here's a solution with a combination of stringr and gsub, using a list of colours supplied in R:
library(dplyr)
library(stringr)
# list of colours from R colors()
cols <- as.character(colors())
apply(df,
1,
function(x)
tibble(
# Exctract CSV of colours
Color = cols[cols %in% str_split(tolower(x), " ", simplify = T)] %>%
paste0(collapse = ","),
# Extract CSV of sequential lists of digits
Code = str_extract_all(x, regex("\\d+"), simplify = T) %>%
paste0(collapse = ","),
# Remove colours and digits from Column_A
Column_A = gsub(paste0("(\\d+|",
paste0(cols, collapse = "|"),
")"), "", x, ignore.case = T) %>% trimws())) %>%
bind_rows()
# A tibble: 3 x 3
Color Code Column_A
<chr> <chr> <chr>
1 red 1011 Cat
2 "" 2011,3001 Mouse is in the House
3 blue,yellow "" on Dog walked around Park
Using tidyverse we can do
library(tidyverse)
colors <- paste0(c("Red", "Yellow", "Blue"), collapse = "|")
df %>%
mutate(Color = str_extract(Column_A,
paste0("(", colors, ").*(", colors, ")|(", colors, ")")),
Code = str_extract_all(Column_A, "\\d+", ),
Column_A = pmap_chr(list(Color, Code, Column_A), function(x, y, z)
trimws(gsub(paste0("\\b", c(x, y), "\\b", collapse = "|"), "", z))),
Code = map_chr(Code, paste, collapse = " "))
# Column_A Color Code
#1 Cat Red 1011
#2 Mouse is in the House <NA> 2011 3001
#3 Dog walked around Park Yellow on Blue
We first extract text between two colors using str_extract. You can include all the possible colors which can occur in the data in colors. We use paste0 to construct the regex. For this example it would be
paste0("(", colors, ").*(", colors, ")|(", colors, ")")
#[1] "(Red|Yellow|Blue).*(Red|Yellow|Blue)|(Red|Yellow|Blue)"
meaning extract text between and including colors or extract only colors.
For Code part as we can have multiple Code values, we use str_extract_all and get all the numbers from the column. This part is initially stored in a list.
For Column_A values we remove everything which was selected in Code and Color adding word boundaries using gsub and the remaining part is saved.
As we had stored Code in list previously, we convert them to one string by collapsing them. This returns empty strings for values that do not match. You can convert them back to NA by adding Code = replace(Code, Code == "", NA)) in the chain if needed.
I have a data frame, and for various reasons I need to keep one of the elements as a factor and, maintaining the order of the levels, replace periods in the levels with spaces. Here's an example
library(tidyverse) library(stringr)
sandwich <- c("bread", "mustard.sauce", "tuna.fish", "lettuce", "bread")
data_frame(sandwich_str = sandwich) %>%
mutate(sandwich_factor = factor(sandwich)) %>%
mutate(sandwich2 = factor(sandwich_factor,
levels = str_replace_all(levels(sandwich_factor), "\\.", " "))) %>%
mutate(sandwich3 = str_replace_all(sandwich_str, "\\.", " "))
print(sandwich_df)
# A tibble: 5 x 4
sandwich_str, sandwich_factor, sandwich2, sandwich3
<chr> <fctr>, <fctr> <chr>,
1 bread bread bread bread
2 mustard.sauce mustard.sauce <NA> mustard sauce
3 tuna.fish tuna.fish <NA> tuna fish
4 lettuce lettuce lettuce lettuce
5 bread bread bread bread
So in this data frame:
sandwich_str is an element of characters
sandwich_factor is an element of factors
in sandwich2 I tried replacing all of the periods in the levels of sandwich_factor. For whatever reason, this returns NA whenever there are periods.
in sandwich3 I take the more simple approach of just replacing all of the periods in strings with spaces. This works substantially better.
So I'm wondering what isn't working in my attempt at sandwich2. I'd like it to look more like sandwich3. Any advice?
Does this suit?
library(tidyverse)
library(stringr)
# Data --------------------------------------------------------------------
sandwich <-
c("bread", "mustard.sauce", "tuna.fish", "lettuce", "bread")
df <-
data_frame(sandwich_str = sandwich)
# Convert periods to spaces -----------------------------------------------
df$sandwich_str <-
df$sandwich_str %>%
as.character() %>%
str_replace("\\."," ") %>%
as.factor()
# Print output ------------------------------------------------------------
df %>%
print()
Credit to #aosmith for posting this answer as a comment. I'll post it here as an answer so I can accept and close this.
The problem was that factor levels are defined with the flag labels rather than levels. So the correct way for me to have written this previously would be:
library(tidyverse) library(stringr)
sandwich <- c("bread", "mustard.sauce", "tuna.fish", "lettuce", "bread")
data_frame(sandwich_str = sandwich) %>%
mutate(sandwich_factor = factor(sandwich)) %>%
mutate(sandwich2 = factor(sandwich_factor,
labels = str_replace_all(levels(sandwich_factor), "\\.", " "))) %>%
mutate(sandwich3 = str_replace_all(sandwich_str, "\\.", " "))
print(sandwich_df)