Let's say I have the following dataframe:
my_basket = data.frame(ITEM_GROUP = c("Fruit","Fruit","Fruit","Fruit","Fruit","Vegetable","Vegetable","Vegetable","Vegetable","Dairy","Dairy","Dairy","Dairy","Dairy"),
ITEM_NAME = c("Apple","Banana","Orange","Mango","Papaya","Carrot","Potato","Brinjal","Raddish","Milk","Curd","Cheese","Milk","Paneer"),
Price = c(100,80,80,90,65,70,60,70,25,60,40,35,50,NA),
Tax = c(2,4,5,6,2,3,5,1,3,4,5,6,4,NA))
This then yields:
> my_basket
ITEM_GROUP ITEM_NAME Price Tax
1 Fruit Apple 100 2
2 Fruit Banana 80 4
3 Fruit Orange 80 5
4 Fruit Mango 90 6
5 Fruit Papaya 65 2
6 Vegetable Carrot 70 3
7 Vegetable Potato 60 5
8 Vegetable Brinjal 70 1
9 Vegetable Raddish 25 3
10 Dairy Milk 60 4
11 Dairy Curd 40 5
12 Dairy Cheese 35 6
13 Dairy Milk 50 4
14 Dairy Paneer NA NA
What I now would like to do, is make a list of fruits I want to keep and then filter those, so:
fruitlist = c("Apple", "Banana")
How would I go about using tidyverse to filter the data in my data.frame to only keep the fruits in my fruitlist, but also all my Vegetables and Dairy? Normally I'd do:
my_basket %<>% filter(ITEM_NAME %in% fruitlist)
But then I'd also lose all the vegetables and dairy, which is not what I want. I've been trying to make something work with case_when but can't seem to make it work. There must be something obvious I'm missing here.
EDIT: Seconds after posting my question I finally realised:
my_basket %<>% filter(ITEM_NAME %in% fruitlist | ITEM_GROUP != "Fruit")
That solves it. I think if I'd have to filter multiple groups like this, piping the filter command repeatedly would work too.
You could use grepl with a regex alternation:
fruitlist <- c("Apple", "Banana")
regex <- paste0("^(?:", paste0(fruitlist, collapse="|"), ")$")
my_basket %<>% filter(grepl(regex, ITEM_NAME))
Related
I am trying to impute missing values in my dataset by matching against values in another dataset.
This is my data:
df1 %>% head()
<V1> <V2>
1 apple NA
2 cheese NA
3 butter NA
df2 %>% head()
<V1> <V2>
1 apple jacks
2 cheese whiz
3 butter scotch
4 apple turnover
5 cheese sliders
6 butter chicken
7 apple sauce
8 cheese doodles
9 butter milk
This is what I want df1 to look like:
<V1> <V2>
1 apple jacks, turnover, sauce
2 cheese whiz, sliders, doodles
3 butter scotch, chicken, milk
This is my code:
df1$V2[is.na(df1$V2)] <- df2$V2[match(df1$V1,df2$V1)][which(is.na(df1$V2))]
This code works fine, however it only pulls the first missing value and ignores the rest.
Another solution just using base R
aggregate(DF2$V2, list(DF2$V1), c, simplify=F)
Group.1 x
1 apple jacks, turnover, sauce
2 butter scotch, chicken, milk
3 cheese whiz, sliders, doodles
I don't think you even need to import the df1 in this case can do it all based on df2
df1 <- df2 %>% group_by(`<V1>`) %>% summarise(`<V2>`=paste0(`<V2>`, collapse = ", "))
This is my data frame
ID=c(1,2,3,4,5,6,7,8,9,10,11,12)
favFruit=c('apple','lemon','pear',
'apple','apple','pear',
'apple','lemon','pear',
'pear','pear','pear')
surveyDate = ('1/1/2005','1/1/2005','1/1/2005',
'2/1/2005','2/1/2005','2/1/2005',
'3/1/2005','3/1/2005','3/1/2005',
'4/1/2005','4/1/2005','4/1/2005')
df<-data.frame(ID,favFruit, surveyDate)
I need to aggregate it so I can plot a line graph in R for count of favFruit by date split by favFruit but I am unable to create an aggregate table. My data has 45000 rows so a manual solution is not possible.
surveyYear favFruit count
1/1/2005 apple 1
1/1/2005 lemon 1
1/1/2005 pear 1
2/1/2005 apple 2
2/1/2005 lemon 0
2/1/2005 pear 1
... etc
I tried this but R printed an error
df2 <- aggregate(df, favFruit, FUN = sum)
and I tried this, another error
df2 <- aggregate(df, date ~ favFruit, sum)
I checked for solutions online but their data generally included a column of quantities which I dont have and the solutions were overly complex. Is there an easy way to do this? Thanx in advance. Thank you to whoever suggested the link as a possible duplicate but it has has date and number of rows. But my question needs number of rows by date and favFruit (one more column) 1
Update:
Ronak Shah's solution worked. Thanx!
The solution provided by Ronak is very good.
In case you prefer to keep the zero counts in your dataframe.
You could use table function:
data.frame(with(df, table(favFruit, surveyDate)))
Output:
favFruit surveyDate Freq
1 apple 1/1/2005 1
2 lemon 1/1/2005 1
3 pear 1/1/2005 1
4 apple 2/1/2005 2
5 lemon 2/1/2005 0
6 pear 2/1/2005 1
7 apple 3/1/2005 1
8 lemon 3/1/2005 1
9 pear 3/1/2005 1
10 apple 4/1/2005 0
11 lemon 4/1/2005 0
12 pear 4/1/2005 3
I have a data frame named data1.
> data1 <- data.frame(name = c("apple","apple","pine","pine",
"apple","apple", "pine","pine","banana","banana"),
characters = c("red","green","yellow","brown",
"big","sweet","delicious","medium","soft", "long"))
> data1
name characters
1 apple red
2 apple green
3 pine yellow
4 pine brown
5 apple big
6 apple sweet
7 pine delicious
8 pine medium
9 banana soft
10 banana long
And I want to change the same values of the name variable according to the values of characters column.
Just like data2:
> data2
name characters
1 colapple red
2 colapple green
3 colorpine yellow
4 colorpine brown
5 othapple big
6 othapple sweet
7 despine delicious
8 despine medium
9 banana soft
10 banana long
In fact, data1 is very big. And I need to change the same values in the data1$name into special values. So I need a general way to realize it. I try to use If statement to do it, but there are some errors. How can I do it?
Like I said in the comment to the question, I am not seeing the relation between the columns, aren't the prefixes changing by groups of the 1st column?
If so, the code below will do what the question asks for. It creates an index k with a standard R cumsum trick. Then pastes the prefixes indexed by the index k and column data1$name.
pref <- c("col", "color", "oth", "des")
k <- cumsum(c(1, abs(diff(data1$name == "apple")) > 0))
data2 <- data.frame(name = paste0(pref[k], data1$name),
characters = data1$characters)
data2
# name characters
#1 colapple red
#2 colapple green
#3 colapple white
#4 colorpine yellow
#5 colorpine brown
#6 colorpine black
#7 othapple big
#8 othapple sweet
#9 othapple small
#10 despine delicious
#11 despine medium
#12 despine ache
Edit
With the new data set posted after the answer and following the discussion in comments, here is a solution with setNames and match.
pref3 <- c(rep("col", 2), rep("color", 2), rep("oth", 2), rep("des", 2), rep("", 2))
pref3 <- setNames(pref3, data3$characters)
k <- match(data3$characters, names(pref3))
data3$name <- paste0(pref3[k], data3$name)
Data
data1 <- data.frame(name = c("apple","apple","apple", "pine","pine","pine",
"apple","apple","apple", "pine","pine","pine"),
characters = c("red","green","white","yellow","brown","black",
"big","sweet","small","delicious","medium","ache"))
data3 <- data.frame(name = c("apple","apple","pine","pine",
"apple","apple", "pine","pine","banana","banana"),
characters = c("red","green","yellow","brown",
"big","sweet","delicious","medium","soft", "long"))
I have a dataframe "test" as below. I want to exclude all the rows of that person, if this person has "apple" in the "fruit" column, using R language.
I wrote:
filter(test, name != test$name[test$fruit=="apple"])
original "test" data frame
actual result
expected result
Any help is appreciated! Thanks!
> test
name fruit
1 kevin apple
2 kevin pear
3 kevin peach
4 jack apple
5 jack pear
6 jack peach
7 jack kiwi
8 caleb grapefruit
9 caleb kiwi
10 caleb pear
11 justin pineapple
12 justin grape
13 justin watermelon
14 justin kiwi
First, we find the all the 'name' which have 'apple' as a fruit.
df=unique(test$name[test$fruit=="apple"])
> df
[1] kevin jack
Levels: caleb jack justin kevin
Now we need to remove rows from rows from test where name is same as those in df, i.e 'kevin' or 'jack'.
test1= test[ (!(test$name %in% df)),]
> test1
name fruit
8 caleb grapefruit
9 caleb kiwi
10 caleb pear
11 justin pineapple
12 justin grape
13 justin watermelon
14 justin kiwi
Ofcourse we can write this in a single line :
test2=test[(!(test$name %in% (unique(test$name[test$fruit=="apple"])))),]
> test2
name fruit
8 caleb grapefruit
9 caleb kiwi
10 caleb pear
11 justin pineapple
12 justin grape
13 justin watermelon
14 justin kiwi
You can do this in multiple ways.
In base R :
subset(test, !ave(fruit == 'apple', name, FUN = any))
# name fruit
#4 Justin pineapple
#5 Justin grape
Using dplyr
test %>% group_by(name) %>% filter(!any(fruit == 'apple'))
Or data.table
setDT(test)[, .SD[!any(fruit == 'apple')], name]
Another option in base R without grouping could be
subset(test, !name %in% unique(name[fruit == "apple"]))
data
test <- data.frame(name = c('Jack', 'Jack', 'Jack', 'Justin', 'Justin'),
fruit =c('pineapple', 'apple', 'grape', 'pineapple', 'grape'))
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
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$Fruit. Finally, if all 5 fruits are correct but there are extras, Count resets to 0.
Here is my answer after seeing others' genius solutions.
ID <- c(1:5)
Age <- c("26-35", "26-35", "46-55", "55+", "56-45")
Nationality <- c("Canadian", "US", "Canadian", "US", "British")
Color <- c("Correct", "Incorrect", "Incorrect", "Correct", "Correect")
Fruits <- c("pineapple",
"apple",
"apple orange kiwi fifth",
"orange apple pineapple kiwi fifth",
"pineapple orange apple fifth kiwi"
)
df <- data.frame(ID, Age, Nationality, Color, Fruits)
df
heds1's reponse looks great. However, you want to be careful using string exacts such as grepl because it could return compound words. For example, consider the word pineapple; it contains pine and apple. Notice here that searching for apple returns pineapples.
filter(df, grepl("apple", Fruits))
ID Age Nationality Color Fruits
1 1 26-35 Canadian Correct pineapple
2 2 26-35 US Incorrect apple
3 3 46-55 Canadian Incorrect apple orange kiwi fifth
4 4 55+ US Correct orange apple pineapple kiwi fifth
5 5 56-45 British Correect pineapple orange apple fifth kiwi
The answer provided by sumshyftw is awesome. And I love that I am learning something from sumshyftw. But to demonstrate my point that unrestrained string search could mess your count:
CorrectFruits <- c("apple")
df$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
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.