I'd like to find consecutive month by client. I thought this is easy but
still can't find solutions..
My goal is to find months' consecutive purchases for each client. Any
My data
Client Month consecutive
A 1 1
A 1 2
A 2 3
A 5 1
A 6 2
A 8 1
B 8 1
In base R, we can use ave
df$consecutive <- with(df, ave(Month, Client, cumsum(c(TRUE, diff(Month) > 1)),
FUN = seq_along))
df
# Client Month consecutive
#1 A 1 1
#2 A 1 2
#3 A 2 3
#4 A 5 1
#5 A 6 2
#6 A 8 1
#7 B 8 1
In dplyr, we can create a new group with lag to compare the current month with the previous month and assign row_number() in each group.
library(dplyr)
df %>%
group_by(Client,group=cumsum(Month-lag(Month, default = first(Month)) > 1)) %>%
mutate(consecutive = row_number()) %>%
ungroup %>%
select(-group)
We can create a grouping variable based on the difference in adjacent 'Month' for each 'Client' and use that to create the sequence
library(dplyr)
df1 %>%
group_by(Client) %>%
group_by(grp =cumsum(c(TRUE, diff(Month) > 1)), add = TRUE) %>%
mutate(consec = row_number()) %>%
ungroup %>%
select(-grp)
# A tibble: 7 x 4
# Client Month consecutive consec
# <chr> <int> <int> <int>
#1 A 1 1 1
#2 A 1 2 2
#3 A 2 3 3
#4 A 5 1 1
#5 A 6 2 2
#6 A 8 1 1
#7 B 8 1 1
Or using data.table
library(data.table)
setDT(df1)[, grp := cumsum(c(TRUE, diff(Month) > 1)), Client
][, consec := seq_len(.N), .(Client, grp)
][, grp := NULL][]
data
df1 <- structure(list(Client = c("A", "A", "A", "A", "A", "A", "B"),
Month = c(1L, 1L, 2L, 5L, 6L, 8L, 8L), consecutive = c(1L,
2L, 3L, 1L, 2L, 1L, 1L)), class = "data.frame", row.names = c(NA,
-7L))
Related
I have data of names within an ID number along with a number of associated values. It looks something like this:
structure(list(id = c("a", "a", "b", "b"), name = c("bob", "jane",
"mark", "brittney"), number = c(1L, 2L, 1L, 2L), value = c(1L,
2L, 1L, 2L)), class = "data.frame", row.names = c(NA, -4L))
# id name number value
# 1 a bob 1 1
# 2 a jane 2 2
# 3 b mark 1 1
# 4 b brittney 2 2
I would like to create all the combinations of name, regardless of how many there are, and paste them together separated with commas, and sum their number and value within each id. The desired output from the example above is then:
structure(list(id = c("a", "a", "a", "b", "b", "b"), name = c("bob",
"jane", "bob, jane", "mark", "brittney", "mark, brittney"), number = c(1L,
2L, 3L, 1L, 2L, 3L), value = c(1L, 2L, 3L, 1L, 2L, 3L)), class = "data.frame", row.names = c(NA, -6L))
# id name number value
# 1 a bob 1 1
# 2 a jane 2 2
# 3 a bob, jane 3 3
# 4 b mark 1 1
# 5 b brittney 2 2
# 6 b mark, brittney 3 3
Thanks all!
You could use group_modify() + add_row():
library(dplyr)
df %>%
group_by(id) %>%
group_modify( ~ .x %>%
summarise(name = toString(name), across(c(number, value), sum)) %>%
add_row(.x, .)
) %>%
ungroup()
# # A tibble: 6 × 4
# id name number value
# <chr> <chr> <int> <int>
# 1 a bob 1 1
# 2 a jane 2 2
# 3 a bob, jane 3 3
# 4 b mark 1 1
# 5 b brittney 2 2
# 6 b mark, brittney 3 3
You can create pairwise indices using combn() and expand the data frame with these using slice(). Then just group by these row pairs and summarise. I'm assuming you want pairwise combinations but this can be adapted for larger sets if needed. Some code to handle groups < 2 is included but can be removed if these don't exist in your data.
library(dplyr)
library(purrr)
df1 %>%
group_by(id) %>%
slice(c(combn(seq(n()), min(n(), 2)))) %>%
mutate(id2 = (row_number()-1) %/% 2) %>%
group_by(id, id2) %>%
summarise(name = toString(name),
across(where(is.numeric), sum), .groups = "drop") %>%
select(-id2) %>%
bind_rows(df1 %>%
group_by(id) %>%
filter(n() > 1), .) %>%
arrange(id) %>%
ungroup()
# A tibble: 6 × 4
id name number value
<chr> <chr> <int> <int>
1 a bob 1 1
2 a jane 2 2
3 a bob, jane 3 3
4 b mark 1 1
5 b brittney 2 2
6 b mark, brittney 3 3
Edit:
To adapt for all possible combinations you can iterate over the values up to the max group size. Using edited data which has a couple of rows added to the first group:
map_df(seq(max(table(df2$id))), ~
df2 %>%
group_by(id) %>%
slice(c(combn(seq(n()), .x * (.x <= n())))) %>%
mutate(id2 = (row_number() - 1) %/% .x) %>%
group_by(id, id2) %>%
summarise(name = toString(name),
across(where(is.numeric), sum), .groups = "drop")
) %>%
select(-id2) %>%
arrange(id)
# A tibble: 18 × 4
id name number value
<chr> <chr> <int> <int>
1 a bob 1 1
2 a jane 2 2
3 a sophie 1 1
4 a jeremy 2 2
5 a bob, jane 3 3
6 a bob, sophie 2 2
7 a bob, jeremy 3 3
8 a jane, sophie 3 3
9 a jane, jeremy 4 4
10 a sophie, jeremy 3 3
11 a bob, jane, sophie 4 4
12 a bob, jane, jeremy 5 5
13 a bob, sophie, jeremy 4 4
14 a jane, sophie, jeremy 5 5
15 a bob, jane, sophie, jeremy 6 6
16 b mark 3 5
17 b brittney 4 6
18 b mark, brittney 7 11
Data for df2:
df2 <- structure(list(id = c("a", "a", "a", "a", "b", "b"), name = c("bob",
"jane", "sophie", "jeremy", "mark", "brittney"), number = c(1L,
2L, 1L, 2L, 3L, 4L), value = c(1L, 2L, 1L, 2L, 5L, 6L)), class = "data.frame", row.names = c(NA,
-6L))
A data.table option
setDT(df)[
,
lapply(
.SD,
function(x) {
unlist(
lapply(
seq_along(x),
combn,
x = x,
function(v) {
ifelse(all(is.character(v)), toString, sum)(v)
}
)
)
}
),
id
]
gives
id name number value
1: a bob 1 1
2: a jane 2 2
3: a bob, jane 3 3
4: b mark 1 1
5: b brittney 2 2
6: b mark, brittney 3 3
I have the something like the following:
person_ID visit date
1 2/25/2001
1 2/30/2001
1 4/2/2001
2 3/18/2004
3 9/22/2004
3 10/27/2004
3 5/15/2008
I want to add another column to see if the person has a reoccurring observation within 90 days, like:
person_ID visit date reoccurrence
1 2/25/2001 1
1 2/30/2001 1
1 4/2/2001 0
2 3/18/2004 0
3 9/22/2004 1
3 10/27/2004 0
3 5/15/2008 0
any help is appreciated, thank you!
If the second 'date' is not 2/30/2001, convert the 'visit_date' to Date class, grouped by 'person_id', get the difference between current and next 'visit_date' in 'day', check if it is less than 90, replace the NA with 0
library(dplyr)
library(lubridate)
library(tidyr)
df1 <- df1 %>%
mutate(visit_date = mdy(visit_date)) %>%
group_by(person_ID) %>%
mutate(reoccurrence = replace_na(+(difftime(lead(visit_date),
visit_date, units = 'day') < 90), 0)) %>%
ungroup
-output
# A tibble: 7 x 3
# person_ID visit_date reoccurrence
# <int> <date> <dbl>
#1 1 2001-02-25 1
#2 1 2001-02-28 1
#3 1 2001-04-02 0
#4 2 2004-03-18 0
#5 3 2004-09-22 1
#6 3 2004-10-27 0
#7 3 2008-05-15 0
Or using data.table
library(data.table)
setDT(df1)[, visit_date := as.IDate(visit_date, '%m/%d/%Y')
][, reoccurence := +(difftime(shift(visit_date, type = 'lead'),
visit_date, units = 'day') < 90))
][is.na(reoccurence), reoccurence := 0]
Or with base R
df1$visit_date <- as.Date(df1$visit_date, '%m/%d/%Y')
with(df1, ave(as.integer(visit_date), person_ID, FUN =
function(x) c(+(diff(x) < 90), 0)))
#[1] 1 1 0 0 1 0 0
data
df1 <- structure(list(person_ID = c(1L, 1L, 1L, 2L, 3L, 3L, 3L), visit_date = c("2/25/2001",
"2/28/2001", "4/2/2001", "3/18/2004", "9/22/2004", "10/27/2004",
"5/15/2008")), row.names = c(NA, -7L), class = "data.frame")
Base R variant:
reoccur <- function(x, lim=90) {
m <- outer(x, x, `-`)
m[upper.tri(m, diag=TRUE)] <- NA
colSums(!is.na(m) & m >= 0 & m <= lim) > 0
}
### make your dates *dates*
dat$visit <- as.Date(dat$visit, format="%m/%d/%Y")
### calculate if you have reoccurrences
ave(as.numeric(dat$visit), dat$person_ID, FUN=reoccur)
# [1] 1 1 0 0 1 0 0
Data:
dat <- structure(list(person_ID = c(1L, 1L, 1L, 2L, 3L, 3L, 3L), visit = c("2/25/2001", "2/27/2001", "4/2/2001", "3/18/2004", "9/22/2004", "10/27/2004", "5/15/2008")), class = "data.frame", row.names = c(NA, -7L))
(I changed "2/30/2001" to "2/27/2001" to get a real Date out of it.)
This question already has an answer here:
dplyr::first() to choose first non NA value
(1 answer)
Closed 2 years ago.
I understand we can use the dplyr function coalesce() to unite different columns, but is there such function to unite rows?
I am struggling with a confusing incomplete/doubled dataframe with duplicate rows for the same id, but with different columns filled. E.g.
id sex age source
12 M NA 1
12 NA 3 1
13 NA 2 2
13 NA NA NA
13 F 2 NA
and I am trying to achieve:
id sex age source
12 M 3 1
13 F 2 2
You can try:
library(dplyr)
#Data
df <- structure(list(id = c(12L, 12L, 13L, 13L, 13L), sex = structure(c(2L,
NA, NA, NA, 1L), .Label = c("F", "M"), class = "factor"), age = c(NA,
3L, 2L, NA, 2L), source = c(1L, 1L, 2L, NA, NA)), class = "data.frame", row.names = c(NA,
-5L))
df %>%
group_by(id) %>%
fill(everything(), .direction = "down") %>%
fill(everything(), .direction = "up") %>%
slice(1)
# A tibble: 2 x 4
# Groups: id [2]
id sex age source
<int> <fct> <int> <int>
1 12 M 3 1
2 13 F 2 2
As mentioned by #A5C1D2H2I1M1N2O1R2T1 you can select the first non-NA value in each group. This can be done using dplyr :
library(dplyr)
df %>% group_by(id) %>% summarise(across(.fns = ~na.omit(.)[1]))
# A tibble: 2 x 4
# id sex age source
# <int> <fct> <int> <int>
#1 12 M 3 1
#2 13 F 2 2
Base R :
aggregate(.~id, df, function(x) na.omit(x)[1], na.action = 'na.pass')
Or data.table :
library(data.table)
setDT(df)[, lapply(.SD, function(x) na.omit(x)[1]), id]
My data.frame df looks like this:
A 1
A 2
A 5
B 2
B 3
B 4
C 3
C 7
C 9
I want it to look like this:
A B C
1 2 3
2 3 7
5 4 9
I have tried spread() but probably not in the right way. Any ideas?
We can use unstack from base R
unstack(df1, col2 ~ col1)
# A B C
#1 1 2 3
#2 2 3 7
#3 5 4 9
Or with split
data.frame(split(df1$col2, df1$col1))
Or if we use spread or pivot_wider, make sure to create a sequence column
library(dplyr)
library(tidyr)
df1 %>%
group_by(col1) %>%
mutate(rn = row_number()) %>%
ungroup %>%
pivot_wider(names_from = col1, values_from = col2) %>%
# or use
# spread(col1, col2) %>%
select(-rn)
# A tibble: 3 x 3
# A B C
# <int> <int> <int>
#1 1 2 3
#2 2 3 7
#3 5 4 9
Or using dcast
library(data.table)
dcast(setDT(df1), rowid(col1) ~ col1)[, .(A, B, C)]
data
df1 <- structure(list(col1 = c("A", "A", "A", "B", "B", "B", "C", "C",
"C"), col2 = c(1L, 2L, 5L, 2L, 3L, 4L, 3L, 7L, 9L)),
class = "data.frame", row.names = c(NA,
-9L))
In data.table, we can use dcast :
library(data.table)
dcast(setDT(df), rowid(col1)~col1, value.var = 'col2')[, col1 := NULL][]
# A B C
#1: 1 2 3
#2: 2 3 7
#3: 5 4 9
I have a table that looks something like this:
ID Date Type
1 2019/03/12 A
1 2019/03/12 A
2 2019/01/07 A
2 2019/04/20 B
3 2019/02/09 C
4 2019/01/19 A
4 2019/01/23 A
I want to deduplicate this table by ID, but only if the span between the dates listed is greater than 7 days. If it is less than 7 days, then I want to keep the earliest date.
Want:
ID Date Type
1 2019/03/12 A
2 2019/01/07 A
2 2019/04/20 B
3 2019/02/09 C
4 2019/01/19 A
I'm just struggling with where to start conceptually.
An option would be to convert the 'Date' to Date class (ymd from lubridate is used here), then grouped by 'ID', filter the difference of 'Date' that is greater than or equal to 7
library(dplyr)
library(lubridate)
df1 %>%
mutate(Date = ymd(Date)) %>%
group_by(ID) %>%
filter(c(TRUE, diff(Date) >= 7))
# A tibble: 5 x 3
# Groups: ID [4]
# ID Date Type
# <int> <date> <chr>
#1 1 2019-03-12 A
#2 2 2019-01-07 A
#3 2 2019-04-20 B
#4 3 2019-02-09 C
#5 4 2019-01-19 A
data
df1 <- structure(list(ID = c(1L, 1L, 2L, 2L, 3L, 4L, 4L), Date = c("2019/03/12",
"2019/03/12", "2019/01/07", "2019/04/20", "2019/02/09", "2019/01/19",
"2019/01/23"), Type = c("A", "A", "A", "B", "C", "A", "A")),
class = "data.frame", row.names = c(NA,
-7L))