I have a data frame with a grouping variable ID, a factor F and a value V that looks something like this:
df <- data.frame(ID = c(rep(1, 3), rep(2, 3)),
F = factor(c("A","B","X","C","D","X")),
V = c(30, 32, 25, 31, 37, 24)
)
> df
ID F V
1 1 A 30
2 1 B 32
3 1 X 25
4 2 C 31
5 2 D 37
6 2 X 24
Now, I would like to add a new column New, which has the same value within each group (by ID) based on the value for V in the row where F==X using the tidyverse environment. Ideally, those rows would be removed afterwards so that the new data frame looks like this:
> df
ID F V New
1 1 A 30 25
2 1 B 32 25
3 2 C 31 24
4 2 D 37 24
I know that I have to use the group_by() function and probably also mutate(), but I couldn't quite manage to get my desired result.
df %>%
group_by(ID) %>%
mutate(New = V[F =='X']) %>%
filter(F != 'X')
# A tibble: 4 × 4
# Groups: ID [2]
ID F V New
<dbl> <fct> <dbl> <dbl>
1 1 A 30 25
2 1 B 32 25
3 2 C 31 24
4 2 D 37 24
library(dplyr)
df %>%
group_by(ID) %>% # grouping variables by ID
mutate(New = ifelse(F == "X",
V,
NA)) %>% # adding New column
summarise(New = max(New, na.rm = T)) %>% # Filtering rows with filled New column
right_join(df %>% filter(F != "X"), by = "ID") %>% # SQL-like join
select(ID, F, V, New) # reordering the columns to the desired order
And you get this output:
# A tibble: 4 × 4
ID F V New
<dbl> <fct> <dbl> <dbl>
1 1 A 30 25
2 1 B 32 25
3 2 C 31 24
4 2 D 37 24
Or even simplier:
df %>% filter(F == "X") %>% # filtering the rows with "X" in F column
right_join(df %>% filter(F != "X"), by = "ID") %>% joining to the same dataset without "X" rows
select(ID, F= F.y, V = V.y, New = V.x) #reordering and renaming of columns
Related
I have a data frame like below
d1<-c('a','b','c','d','e','f','g','h','i','j','k','l')
d2<-c(1,5,1,2,13,2,32,2,1,2,4,5)
df1<-data.frame(d1,d2)
Which looks like the data table in this picture
My goal is to filter the rows based on which value of d2 in every 3 rows is biggest. So it would look like this:
Thank you!
We may use rollmax from zoo to filter the rows
library(dplyr)
library(zoo)
df1 %>%
filter(d2 == na.locf0(rollmax(d2, k = 3, fill = NA)))
d1 d2
1 b 5
2 e 13
3 g 32
4 l 5
You can create a grouping variable that puts observations into groups of 3. I have first created a sequence from 1 to the total number of rows, incremented by 3. And then repeated each number of this sequence 3 times and subset the result to get a vector the same length of the data, incase the number of observations is not perfectly divisible by 3. Then simply filter rows based by the largest number of each group in d2 column.
library(dplyr)
df1 %>%
mutate(group = rep(seq(1, n(), by = 3), each = 3)[1:n()]) %>%
group_by(group) %>%
filter(d2 == max(d2))
# A tibble: 4 x 3
# Groups: group [4]
# d1 d2 group
# <chr> <dbl> <dbl>
# 1 b 5 1
# 2 e 13 4
# 3 g 32 7
# 4 l 5 10
Yet another solution:
library(tidyverse)
d1<-c('a','b','c','d','e','f','g','h','i','j','k','l')
d2<-c(1,5,1,2,13,2,32,2,1,2,4,5)
df1<-data.frame(d1,d2)
df1 %>%
mutate(id = rep(1:(n()/3), each=3)) %>%
group_by(id) %>%
slice_max(d2) %>%
ungroup %>% select(-id)
#> # A tibble: 4 × 2
#> d1 d2
#> <chr> <dbl>
#> 1 b 5
#> 2 e 13
#> 3 g 32
#> 4 l 5
I have a very simple case where I want to combine several data frames into one based on a common id elements of a particular data frame.
Example:
id <- c(1, 2, 3)
x <- c(10, 12, 14)
data1 <- data.frame(id, x)
id <- c(2, 3)
x <- c(20, 22)
data2 <- data.frame(id, x)
id <- c(1, 3)
x <- c(30, 32)
data3 <- data.frame(id, x)
Which gives us,
$data1
id x
1 1 10
2 2 12
3 3 14
$data2
id x
1 2 20
2 3 22
$data3
id x
1 1 30
2 3 32
Now, I want to combine all three data frames based on the id's of the data3. The expected output should look like
> comb
id x
1 1 10
2 1 NA
3 1 30
4 3 14
5 3 22
6 3 32
I am trying the following, but not getting the expected output.
library(dplyr)
library(tidyr)
combined <- bind_rows(data1, data2, data3, .id = "id") %>% arrange(id)
Any idea how to get the expected output?
Does this work:
library(dplyr)
library(tidyr)
data1 %>% full_join(data2, by = 'id') %>% full_join(data3, by = 'id') %>% arrange(id) %>% right_join(data3, by = 'id') %>%
pivot_longer(cols = -id) %>% select(-name) %>% distinct()
# A tibble: 6 x 2
id value
<dbl> <dbl>
1 1 10
2 1 NA
3 1 30
4 3 14
5 3 22
6 3 32
Combine the 3 dataframes in one list and use filter to select only the id's in 3rd dataframe.
library(dplyr)
library(tidyr)
bind_rows(data1, data2, data3, .id = "new_id") %>%
filter(id %in% id[new_id == 3]) %>%
complete(new_id, id)
# new_id id x
# <chr> <dbl> <dbl>
#1 1 1 10
#2 1 3 14
#3 2 1 NA
#4 2 3 22
#5 3 1 30
#6 3 3 32
A pure base R solution can also make it
lst <- list(data1, data2, data3)
reshape(
subset(
reshape(
do.call(rbind, Map(cbind, lst, grp = seq_along(lst))),
idvar = "id",
timevar = "grp",
direction = "wide"
),
id %in% lst[[3]]$id
),
idvar = "id",
varying = -1,
direction = "long"
)[c("id", "x")]
which gives
id x
1.1 1 10
3.1 3 14
1.2 1 NA
3.2 3 22
1.3 1 30
3.3 3 32
>
Using base R
do.call(rbind, unname(lapply(mget(ls(pattern = "^data\\d+$")), \(x) {
x1 <- subset(x, id %in% data3$id)
v1 <- setdiff(data3$id, x1$id)
if(length(v1) > 0) rbind(x1, cbind(id = v1, x = NA)) else x1
})))
-output
id x
1 1 10
3 3 14
2 3 22
11 1 NA
12 1 30
21 3 32
bind_rows(data1, data2, data3, .id = 'grp')%>%
complete(id, grp)%>%
select(-grp) %>%
filter(id%in%data3$id)
# A tibble: 6 x 2
id x
<dbl> <dbl>
1 1 10
2 1 NA
3 1 30
4 3 14
5 3 22
6 3 32
library(tidyverse)
df <- tibble(a = as.factor(1:20), b = c(50, 20, 13, rep(2, 10), rep(1, 7)))
How do I make dplyr look at this data frame df and collapse all these occurences of 2 into a single summed group, and collapse all the occurrences of 1 into a single summed group? And also keep the rest of the data frame.
Turn this:
# A tibble: 20 x 2
a b
<fct> <dbl>
1 1 50
2 2 20
3 3 13
4 4 2
5 5 2
6 6 2
7 7 2
8 8 2
9 9 2
10 10 2
11 11 2
12 12 2
13 13 2
14 14 1
15 15 1
16 16 1
17 17 1
18 18 1
19 19 1
20 20 1
into this:
# A tibble: 5 x 2
a b
<fct> <dbl>
1 1 50
2 2 20
3 3 13
4 grp2 20
5 grp1 7
[Edit] - I fixed the example data. Sorry about that.
We group by a manufactured sortkey to maintain sort order. We used the fact that b is in descending order in the input but if that is not the case in your actual data then replace sortkey = -b with the more general sortkey = data.table::rleid(b) or the longer sortkey = cumsum(coalesce(b != lag(b), FALSE)) .
We also convert b to the group names giving a new a. It wasn't clear which groups are to be converted to grp... form. Hard-coded 1 and 2? Any group with more than one row? Groups at the end with more than one row? At any rate it would be easy enough to change the condition in the if_else once that were clarified.
Finally perform the summation and then remove the sortkey.
df %>%
group_by(sortkey = -b, a = paste0(if_else(b %in% 1:2, "grp", ""), b)) %>%
summarize(b = sum(b)) %>%
ungroup %>%
select(-sortkey)
giving:
# A tibble: 5 x 2
a b
<chr> <int>
1 50 50
2 20 20
3 13 13
4 grp2 20
5 grp1 7
Here's a way. I have converted a from factor to character to make things easier. You can convert it back to factor if you want. Also your test data was a bit wrong.
df <- tibble(a = as.character(1:20), b = c(50, 20, 13, rep(2, 10), rep(1, 7)))
df %>%
mutate(
a = case_when(
b == 1 ~ "grp1",
b == 2 ~ "grp2",
TRUE ~ a
)
) %>%
group_by(a) %>%
summarise(b = sum(b))
# A tibble: 5 x 2
a b
<chr> <dbl>
1 1 50
2 2 20
3 3 13
4 grp1 7
5 grp2 20
This is an approach which gives you the desired names for groups & where you don't need to think in advance how many cases like that you would need (e.g. it would create grp3, grp4, ... depending on the number in b).
library(dplyr)
df %>%
mutate(
grp = as.numeric(lag(df$b) != df$b),
grp = cumsum(ifelse(is.na(grp), 0, grp))
) %>% group_by(grp) %>%
mutate(
a = ifelse(n() > 1, paste0("grp", b), a),
b = sum(b)
) %>% ungroup() %>% distinct(a, b)
Output:
a b
<chr> <dbl>
1 1 50
2 2 20
3 3 13
4 grp2 20
5 grp1 7
Note that the code could be also condensed but that leads to a certain lack of readability in my opinion:
df %>%
group_by(grp = cumsum(ifelse(is.na(as.numeric(lag(df$b) != df$b)), 0, as.numeric(lag(df$b) != df$b)))) %>%
mutate(
a = ifelse(n() > 1, paste0("grp", b), a),
b = sum(b)
) %>% ungroup() %>% distinct(a, b)
Here I have a data of three fields Dealer,Product,Freq.
My aim is to create a data which will contain top 2 sells for each dealer.
I have done it using data.table as bellow:
library(data.table)
library(dplyr)
dt <- data.table(Dealer = c("A","B","A","A","B","A"),
Product = c("a","b","b","c","d","d"),
Freq = c(10,12,23,24,23,12))
dt[,.SD[order(Freq, decreasing = T)][seq_along(Freq) < 3], by = Dealer]
How to do the similar thing using 'dplyr' package.
Here, I group by Dealer then find the top 2 values of Freq in each group.
dt %>% group_by(Dealer) %>% top_n(2, Freq) %>% ungroup
# # A tibble: 4 x 3
# Dealer Product Freq
# <fct> <fct> <dbl>
# 1 B b 12
# 2 A b 23
# 3 A c 24
# 4 B d 23
We can use slice or filter after doing the group_by and arrange (same methodology as in the OP's post)
library(dplyr)
dt %>%
group_by(Dealer) %>%
arrange(Dealer, desc(Freq)) %>%
slice(1:2)
# or with
# filter(row_number() < 3)
# A tibble: 4 x 3
# Groups: Dealer [2]
# Dealer Product Freq
# <chr> <chr> <dbl>
#1 A c 24
#2 A b 23
#3 B d 23
#4 B b 12
NOTE: In case of ties, this will get the output exactly the number of rows specified in the slice or filter
I have a data frame as follow where I would like to group the data by grp and index and use group a as a reference to perform some simple calculations. I would like to subtract the variable value from other group from the values of group a.
df <- data.frame(grp = rep(letters[1:3], each = 2),
index = rep(1:2, times = 3),
value = seq(10, 60, length.out = 6))
df
## grp index value
## 1 a 1 10
## 2 a 2 20
## 3 b 1 30
## 4 b 2 40
## 5 c 1 50
## 6 c 2 60
The desired outpout would be like:
## grp index value
## 1 b 1 20
## 2 b 2 20
## 3 c 1 40
## 4 c 2 40
My guess is it will be something close to:
group_by(df, grp, index) %>%
mutate(diff = value - value[grp == "a"])
Ideally I would like to do it using dplyr.
Regards, Philippe
We can filter for 'grp' that are not 'a' and then do the difference within mutate.
df %>%
filter(grp!="a") %>%
mutate(value = value- df$value[df$grp=="a"])
Or another option would be join
df %>%
filter(grp!="a") %>%
left_join(., subset(df, grp=="a", select=-1), by = "index") %>%
mutate(value = value.x- value.y) %>%
select(1, 2, 5)
# grp index value
#1 b 1 20
#2 b 2 20
#3 c 1 40
#4 c 2 40