Filtering for multiple strings within the same column in r - r

Question

My large data set (Groceries) has a column in it containing character data (Fruits) all of which is lower case and all of which contains no punctuation.
It looks a bit like this:
# Groceries Data Frame
Id Groceries$Fruits
1 apple orange banana lemon grapefruit
2 grapes tomato passion fruit
3 strawberry orange kiwi
4 lemon orange passion fruit grapefruit lime
5 lemon orange passion fruit grapefruit lime peach
...
I'm trying to select all the rows (of which there are 3,320) from the Fruits column that contain 5 specific fruits (orange, lime, lemon, grapefruit & passion fruit). Initially I'm only interested in the rows that contain all 5 of these fruits and no additional Fruits. Thus, the only row out of these 5 that should be filtered/subsetted would be row 4. The fruits do not have to be in any particular order.
The data is actually answers to a test, so eventually I'm interested in determining who got 0/5 fruits, who got 1/5, 2/5 and so on...
I've tried 2 methods so far, both to no avail.
Firstly I tried using grep(), but no rows were stored in the resulting data frame.
# 1st attempt with grep()
Correct fruits <- Groceries[grep("orange, lemon, lime, passion fruit,
grapefruit", Groceries$Fruits), ]
And then I tried using filter(), but the selected rows don't contain just the 5 Fruits I'm seeking out, it selects all rows that contain any of the 5 fruits.
# 2nd attempt with filter
library(dplyr)
library(stringr)
CorrectFruits <- c("lemon", "orange", "passion fruit", "grapefruit",
"lime")
filter <- Groceries %>%
select(Id, Fruits) %>%
filter(str_detect(tolower(Fruits), pattern = CorrectFruits))
The result I'm after initially is a new DF containing all the columns in the Groceries table, but only the rows of those people who got all 5 of the chosen fruits correct.
Next, it would be cool to select the opposite; everyone who didn't get all 5 correct.
Finally, I'd love to be able to subset those who got a specific proportion correct. I.e. row 1 got 3 correct, row 2 only got 1 correct and row 3 only got 1 correct.
Any help would be greatly appreciated!
Here's an example of what some of the columns look like:
# Groceries
Id Age Nationality Colour question Fruits question
1 26-35 Canadian Red apple orange banana lemon grapefruit
2 26-35 US Blue grapes tomato passion fruit
3 46-55 Canadian Red strawberry orange kiwi
4 55+ US Red lemon orange passion fruit grapefruit lime
5 36-45 British Green lemon orange passion fruit grapefruit lime peach

Answer 1

Might need more clarification on what you intend on doing with answers that have all 5 fruits with some extra, but this should help you out. I substituted all instances of "passion fruit" with "passionfruit" to make it easier:
df$Fruits <- gsub("passion fruit", "passionfruit", df$Fruits)
CorrectFruits <- c("lemon", "orange", "passionfruit", "grapefruit",
"lime")
df$Count <- str_count(df$Fruits, paste(CorrectFruits, collapse = '|'))
df$Count <- ifelse((df$Count == 5 & str_count(df$Fruits, '\\w+') > 5), 0, df$Count)
which gives
ID Fruits Count
1 apple orange banana lemon grapefruit 3
2 grapes tomato passionfruit 1
3 strawberry orange kiwi 1
4 lemon orange passionfruit grapefruit lime 5
5 lemon orange passionfruit grapefruit lime peach 0
First line does the passionfruit substitution, and then str_count counts all occurrences of correct fruits in df$Fruit. Finally, if all 5 fruits are correct but there are extras, Count resets to 0.

Answer 2

Here is my answer after seeing others' genius solutions.
ID <- c(1:5)
Age <- c("26-35", "26-35", "46-55", "55+", "56-45")
Nationality <- c("Canadian", "US", "Canadian", "US", "British")
Color <- c("Correct", "Incorrect", "Incorrect", "Correct", "Correect")
Fruits <- c("pineapple",
"apple",
"apple orange kiwi fifth",
"orange apple pineapple kiwi fifth",
"pineapple orange apple fifth kiwi"
)
df <- data.frame(ID, Age, Nationality, Color, Fruits)
df
heds1's reponse looks great. However, you want to be careful using string exacts such as grepl because it could return compound words. For example, consider the word pineapple; it contains pine and apple. Notice here that searching for apple returns pineapples.
filter(df, grepl("apple", Fruits))
ID Age Nationality Color Fruits
1 1 26-35 Canadian Correct pineapple
2 2 26-35 US Incorrect apple
3 3 46-55 Canadian Incorrect apple orange kiwi fifth
4 4 55+ US Correct orange apple pineapple kiwi fifth
5 5 56-45 British Correect pineapple orange apple fifth kiwi
The answer provided by sumshyftw is awesome. And I love that I am learning something from sumshyftw. But to demonstrate my point that unrestrained string search could mess your count:
CorrectFruits <- c("apple")
df$Count <- str_count(df$Fruits, paste(CorrectFruits, collapse = '|'))
df$Count <- ifelse((df$Count == 5 & str_count(df$Fruits, '\\w+') > 5), 0, df$Count)
df
ID Age Nationality Color Fruits Count
1 1 26-35 Canadian Correct pineapple 1
2 2 26-35 US Incorrect apple 1
3 3 46-55 Canadian Incorrect apple orange kiwi fifth 1
4 4 55+ US Correct orange apple pineapple kiwi fifth 2
5 5 56-45 British Correect pineapple orange apple fifth kiwi 2
Notice that it counted the pineapple as a correct answer despite that the only correct fruit is an apple. To overcome this, you want to wrap your words with \\b.
CorrectFruits <- c("\\bapple\\b")
df$Count <- str_count(df$Fruits, paste(CorrectFruits, collapse = '|'))
df$Count <- ifelse((df$Count == 5 & str_count(df$Fruits, '\\w+') > 5), 0, df$Count)
df
ID Age Nationality Color Fruits Count
1 1 26-35 Canadian Correct pineapple 0
2 2 26-35 US Incorrect apple 1
3 3 46-55 Canadian Incorrect apple orange kiwi fifth 1
4 4 55+ US Correct orange apple pineapple kiwi fifth 1
5 5 56-45 British Correect pineapple orange apple fifth kiwi 1
R no longer counts pineapple as an apple.
But for the record, sumshyftw deserves the credit for working out the hard part in my example:
CorrectFruits <- c("\\bapple\\b", "\\borange\\b", "\\bpineapple\\b", "\\bfifth\\b", "\\bkiwi\\b")
df$Count <- str_count(df$Fruits, paste(CorrectFruits, collapse = '|'))
df$Count <- ifelse((df$Count == 5 & str_count(df$Fruits, '\\w+') > 5), 0, df$Count)
df
ID Age Nationality Color Fruits Count
1 1 26-35 Canadian Correct pineapple 1
2 2 26-35 US Incorrect apple 1
3 3 46-55 Canadian Incorrect apple orange kiwi fifth 4
4 4 55+ US Correct orange apple pineapple kiwi fifth 5
5 5 56-45 British Correect pineapple orange apple fifth kiwi 5
To show only those with all five fruits:
df2 <- filter(df, df$Count == 5)
df2
ID Age Nationality Color Fruits Count
1 4 55+ US Correct orange apple pineapple kiwi fifth 5
2 5 56-45 British Correect pineapple orange apple fifth kiwi 5

Answer 3

Here's one way using grepl with a target list of keywords.
df <- structure(list(v1 = structure(1:4, .Label = c("row1", "row2",
"row3", "row4"), class = "factor"), v2 = structure(c(2L, 4L,
1L, 3L), .Label = c("another invalid row", "apple banana mandarin orange pear",
"banana apple mandarin pear orange", "not a valid row"), class = "factor")), class = "data.frame", row.names = c(NA,
-4L))
targets <- c("banana", "apple", "orange", "pear", "mandarin")
bool_df <- as.data.frame(sapply(targets, grepl, df$v2))
match_rows <- which(rowSums(bool_df) == 5)
df <- df[match_rows,]
You can then change the criteria in the match_rows vector by changing the 5 to, for example 4 for four fruit matches, etc.