I have a messy dataframe:
df <- data.frame(name = c('Chicken','Chicken1','ChiCKen','Chicke_N',
'Eg_g','EGG','egg'))
levels(as.factor(df$name))
[1] "Chicke_N" "Chicken" "ChiCKen" "Chicken1" "Eg_g" "egg" "EGG"
What would be the most efficient way of cleaning up these names, making them either 'chicken' or 'egg'?
I thought something like this would work:
df <- df%>% mutate(name = ifelse(name %in% c('Chicken','Chicken1','ChiCKen','Chicke_N'),'chicken','egg'))
I was wrong.
This changes to lower case and removes all punctuation and (contiguous) numbers, so it works on the given examples.
data.frame(name = c('Chicken','Chicken1','ChiCKen','Chicke_N',
'Eg_g','EGG','egg')) %>%
mutate(name = tolower(name) %>% stringr::str_remove("[[:punct:]]|\\d+"))
For a versatile approach, you might consider joining by stringdistance.
Make sure to read the helpfiles on the different methods for computing stringdistance (i.e. osa, lv, dl, hamming, lcs, qgram, cosine, jaccard, jw and soundex).
df.valid <- data.frame(name = c("chicken", "egg"))
library(tidyverse)
library(fuzzyjoin)
df %>% stringdist_left_join(df.valid, ignore_case = TRUE, max_dist = 5)
# name.x name.y
# 1 Chicken chicken
# 2 Chicken1 chicken
# 3 ChiCKen chicken
# 4 Chicke_N chicken
# 5 Eg_g egg
# 6 EGG egg
# 7 egg egg
Another alternative is to use approximate matching by using agrepl.
df |>
mutate(cleaned = ifelse(agrepl("chicken", df$name, ignore.case = TRUE), "chicken", "egg"))
name cleaned
1 Chicken chicken
2 Chicken1 chicken
3 ChiCKen chicken
4 Chicke_N chicken
5 Eg_g egg
6 EGG egg
7 egg egg
An alternative is to use lookup tables:
df <- data.frame(name = c('Chicken','Chicken1','ChiCKen','Chicke_N',
'Eg_g','EGG','egg'))
lkup <- data.frame(name = levels(as.factor(df$name)),
clean = c('chicken','chicken','chicken','chicken',
'egg','egg','egg'))
inner_join(df,lkup,by='name') %>% mutate(name = clean) %>% select(name)
This method is more verbose, but in some cases the name might not be coercible, for example when trying to match 'bird' to chicken..
Related
Given a dataframe of types and values like so:
topic
keyword
cheese
cheddar
meat
beef
meat
chicken
cheese
swiss
bread
focaccia
bread
sourdough
cheese
gouda
My aim is to make a set of dynamic regexs based on the type, but I don't know how to make the variable names from the types. I can do this individually like so:
fn_get_topic_regex <- function(targettopic,df)
{
filter_df <- df |>
filter(topic == targettopic)
regex <- paste(filter_df$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
My dataframe looks like this:
V1
c("cheese","bread","sugar","cream","milk","butter")
c("milk","butter","apples","cream","bread")
c("butter","milk","toffee")
c("cream","milk","butter","sugar")
I am trying to count the number of times each element appears and sum in a new column. I would like to end up with something like this:
V2 V3
cheese 1
bread 2
sugar 2
cream 3
milk 4
butter 4
apples 1
toffee 1
I have tried using the following code
counts <- unlist(V1, use.names = FALSE)
counts <- table(counts)
But for some reason the counts are wrong and values are being skipped.
If I understand you correctly and your data is organized as provided below, then we could do it this way:
Using separate_rows will allow to bring all your strings in one row.
remove c and empty rows
Use fct_inorder from forcats package (it is in tidyverse) to keep the order as provided
then apply count with the name argument:
library(tidyverse)
df %>%
separate_rows(V1) %>%
filter(!(V1 == "c" | V1 == "")) %>%
mutate(V1 = fct_inorder(V1)) %>%
count(V1, name ="V3")
V1 V3
<fct> <int>
1 cheese 1
2 bread 2
3 sugar 2
4 cream 3
5 milk 4
6 butter 4
7 apples 1
8 toffee 1
df <- structure(list(V1 = c("c(\"cheese\",\"bread\",\"sugar\",\"cream\",\"milk\",\"butter\")",
"c(\"milk\",\"butter\",\"apples\",\"cream\",\"bread\")", "c(\"butter\",\"milk\",\"toffee\")",
"c(\"cream\",\"milk\",\"butter\",\"sugar\")")), class = c("tbl_df",
"tbl", "data.frame"), row.names = c(NA, -4L))
A couple of little issues with the question. Found it hard to reproduce exactly so took some liberties with the DF and present a couple of options that might help:
Option 1 - data in one column
library(tidyverse)
df <- data.frame(V1 = c("cheese","bread","sugar","cream","milk","butter",
"milk","butter","apples","cream","bread",
"butter","milk","toffee",
"cream","milk","butter","sugar"))
df <- df %>% dplyr::group_by(V1) %>%
summarise(
V3 = n()
)
Option 2 - data in columns - added NAs so it made a DF
library(tidyverse)
df <- data.frame(c("cheese","bread","sugar","cream","milk","butter"),
c("milk","butter","apples","cream","bread",NA),
c("butter","milk","toffee",NA,NA,NA),
c("cream","milk","butter","sugar",NA,NA))
df <- data.frame(V1=unlist(df)) %>%
select(V1) %>%
drop_na() %>%
group_by(V1) %>%
summarise(V3 = n())
hope this helps!
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 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)))
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)