dplyr Rolling Conditional Counts
up vote
0
down vote
favorite
I have a data frame as follows:
df <- data.frame(
Item=c("A","A","A","A","A","B","B","B","B","B"),
Date=c("2018-1-1","2018-2-1","2018-3-1","2018-4-1","2018-5-1","2018-1-1","2018-2-1",
"2018-3-1","2018-4-1","2018-5-1"),
Value=rnorm(10))
I want to mutate a new column grouped by Item, to count the number of values higher than 0 within the window of 3 (or any other integer I specify).
I am familiar with tidyverse
, therefore, a dplyr
solution would be most welcome.
r dplyr
add a comment |
up vote
0
down vote
favorite
I have a data frame as follows:
df <- data.frame(
Item=c("A","A","A","A","A","B","B","B","B","B"),
Date=c("2018-1-1","2018-2-1","2018-3-1","2018-4-1","2018-5-1","2018-1-1","2018-2-1",
"2018-3-1","2018-4-1","2018-5-1"),
Value=rnorm(10))
I want to mutate a new column grouped by Item, to count the number of values higher than 0 within the window of 3 (or any other integer I specify).
I am familiar with tidyverse
, therefore, a dplyr
solution would be most welcome.
r dplyr
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a data frame as follows:
df <- data.frame(
Item=c("A","A","A","A","A","B","B","B","B","B"),
Date=c("2018-1-1","2018-2-1","2018-3-1","2018-4-1","2018-5-1","2018-1-1","2018-2-1",
"2018-3-1","2018-4-1","2018-5-1"),
Value=rnorm(10))
I want to mutate a new column grouped by Item, to count the number of values higher than 0 within the window of 3 (or any other integer I specify).
I am familiar with tidyverse
, therefore, a dplyr
solution would be most welcome.
r dplyr
I have a data frame as follows:
df <- data.frame(
Item=c("A","A","A","A","A","B","B","B","B","B"),
Date=c("2018-1-1","2018-2-1","2018-3-1","2018-4-1","2018-5-1","2018-1-1","2018-2-1",
"2018-3-1","2018-4-1","2018-5-1"),
Value=rnorm(10))
I want to mutate a new column grouped by Item, to count the number of values higher than 0 within the window of 3 (or any other integer I specify).
I am familiar with tidyverse
, therefore, a dplyr
solution would be most welcome.
r dplyr
r dplyr
edited Nov 20 at 13:47
Ronak Shah
30.7k103753
30.7k103753
asked Nov 20 at 12:51
Felix Zhao
4614
4614
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
0
down vote
accepted
Item Date Value
<fct> <date> <int>
1 A 2018-01-01 3
2 B 2018-01-01 2
3 B 2018-02-01 -5
4 A 2018-02-01 -3
5 A 2018-03-01 4
6 B 2018-03-01 -2
7 A 2018-04-01 5
8 B 2018-04-01 0
9 A 2018-05-01 1
10 B 2018-05-01 -4
Changed rnorm example for clarity, used sample(-5:5):
> df <- df %>% mutate(greater_than = (Value>0)*Value) %>%
group_by(Item) %>% arrange(Date) %>% mutate(greater_than =
zoo::rollapplyr(greater_than, 3, sum, partial = T))
df %>% arrange(Item) %>% head(10)
Should look like this:
1 A 2018-01-01 3 3
2 A 2018-02-01 -3 3
3 A 2018-03-01 4 7
4 A 2018-04-01 5 9
5 A 2018-05-01 1 10
6 B 2018-01-01 2 2
7 B 2018-02-01 -5 2
8 B 2018-03-01 -2 2
9 B 2018-04-01 0 0
10 B 2018-05-01 -4 0
add a comment |
up vote
3
down vote
Think zoo::
package if you want to roll anything.
df$new<-
zoo::rollsum( df$Value > 0, 3, fill = NA )
# Item Date Value new
#1 A 2018-1-1 0.5852699 NA
#2 A 2018-2-1 -0.7383377 1
#3 A 2018-3-1 -0.3157693 1
#4 A 2018-4-1 1.2475237 1
#5 A 2018-5-1 -1.5479757 1
#6 B 2018-1-1 -0.6913331 0
#7 B 2018-2-1 -0.2423809 0
#8 B 2018-3-1 -1.6363024 0
#9 B 2018-4-1 -0.3256263 1
#10 B 2018-5-1 0.3563144 NA
You have an option of the "window-position". Have a closer look at argument align = c("center", "left", "right")
.
So as a dplyr chain:
df %>% group_by(Item) %>% dplyr::mutate( new = zoo::rollsum( Value > 0, 3, fill = NA ))
Thanks Andre for your help. I tested with your method, it works!
– Felix Zhao
Nov 21 at 12:04
If your problem got solved please choose an answer.
– Andre Elrico
Nov 22 at 11:11
add a comment |
up vote
1
down vote
You could use the RcppRoll
package.
require(RcppRoll)
df$new <- df$new <- RcppRoll::roll_sum(df$Value > 0, 3, fill = NA)
Using Tidyverse:
df %>%
group_by(Item) %>%
dplyr::mutate(new = RcppRoll::roll_sum(Value > 0, 3, fill = NA))
Speedwise this is faster than the zoo
Package:
n <- 10000
df <- data.frame(
Item = sample(LETTERS, n, replace = TRUE),
Value = rnorm(n))
df_grouped <- df %>%
group_by(Item)
microbenchmark::microbenchmark(
RcppRoll = df_grouped <- df_grouped %>% dplyr::mutate(new_RcppRoll = RcppRoll::roll_sum(Value > 0, 3, fill = NA)),
zoo = df_grouped <- df_grouped %>% dplyr::mutate(new_zoo = zoo::rollsum( Value > 0, 3, fill = NA ))
)
Results in:
Unit: milliseconds
expr min lq mean median uq max neval
RcppRoll 2.509003 2.741993 2.929227 2.83913 2.983726 5.832962 100
zoo 11.172920 11.785113 13.288970 12.43320 13.607826 25.879754 100
And
all.equal(df_grouped$new_RcppRoll, df_grouped$new_zoo)
TRUE
Hi Rentrop, it took me to figure out the alignment clause and it is a really good package dealing with rolling. Thanks a lot for your help!
– Felix Zhao
Nov 21 at 12:06
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Item Date Value
<fct> <date> <int>
1 A 2018-01-01 3
2 B 2018-01-01 2
3 B 2018-02-01 -5
4 A 2018-02-01 -3
5 A 2018-03-01 4
6 B 2018-03-01 -2
7 A 2018-04-01 5
8 B 2018-04-01 0
9 A 2018-05-01 1
10 B 2018-05-01 -4
Changed rnorm example for clarity, used sample(-5:5):
> df <- df %>% mutate(greater_than = (Value>0)*Value) %>%
group_by(Item) %>% arrange(Date) %>% mutate(greater_than =
zoo::rollapplyr(greater_than, 3, sum, partial = T))
df %>% arrange(Item) %>% head(10)
Should look like this:
1 A 2018-01-01 3 3
2 A 2018-02-01 -3 3
3 A 2018-03-01 4 7
4 A 2018-04-01 5 9
5 A 2018-05-01 1 10
6 B 2018-01-01 2 2
7 B 2018-02-01 -5 2
8 B 2018-03-01 -2 2
9 B 2018-04-01 0 0
10 B 2018-05-01 -4 0
add a comment |
up vote
0
down vote
accepted
Item Date Value
<fct> <date> <int>
1 A 2018-01-01 3
2 B 2018-01-01 2
3 B 2018-02-01 -5
4 A 2018-02-01 -3
5 A 2018-03-01 4
6 B 2018-03-01 -2
7 A 2018-04-01 5
8 B 2018-04-01 0
9 A 2018-05-01 1
10 B 2018-05-01 -4
Changed rnorm example for clarity, used sample(-5:5):
> df <- df %>% mutate(greater_than = (Value>0)*Value) %>%
group_by(Item) %>% arrange(Date) %>% mutate(greater_than =
zoo::rollapplyr(greater_than, 3, sum, partial = T))
df %>% arrange(Item) %>% head(10)
Should look like this:
1 A 2018-01-01 3 3
2 A 2018-02-01 -3 3
3 A 2018-03-01 4 7
4 A 2018-04-01 5 9
5 A 2018-05-01 1 10
6 B 2018-01-01 2 2
7 B 2018-02-01 -5 2
8 B 2018-03-01 -2 2
9 B 2018-04-01 0 0
10 B 2018-05-01 -4 0
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Item Date Value
<fct> <date> <int>
1 A 2018-01-01 3
2 B 2018-01-01 2
3 B 2018-02-01 -5
4 A 2018-02-01 -3
5 A 2018-03-01 4
6 B 2018-03-01 -2
7 A 2018-04-01 5
8 B 2018-04-01 0
9 A 2018-05-01 1
10 B 2018-05-01 -4
Changed rnorm example for clarity, used sample(-5:5):
> df <- df %>% mutate(greater_than = (Value>0)*Value) %>%
group_by(Item) %>% arrange(Date) %>% mutate(greater_than =
zoo::rollapplyr(greater_than, 3, sum, partial = T))
df %>% arrange(Item) %>% head(10)
Should look like this:
1 A 2018-01-01 3 3
2 A 2018-02-01 -3 3
3 A 2018-03-01 4 7
4 A 2018-04-01 5 9
5 A 2018-05-01 1 10
6 B 2018-01-01 2 2
7 B 2018-02-01 -5 2
8 B 2018-03-01 -2 2
9 B 2018-04-01 0 0
10 B 2018-05-01 -4 0
Item Date Value
<fct> <date> <int>
1 A 2018-01-01 3
2 B 2018-01-01 2
3 B 2018-02-01 -5
4 A 2018-02-01 -3
5 A 2018-03-01 4
6 B 2018-03-01 -2
7 A 2018-04-01 5
8 B 2018-04-01 0
9 A 2018-05-01 1
10 B 2018-05-01 -4
Changed rnorm example for clarity, used sample(-5:5):
> df <- df %>% mutate(greater_than = (Value>0)*Value) %>%
group_by(Item) %>% arrange(Date) %>% mutate(greater_than =
zoo::rollapplyr(greater_than, 3, sum, partial = T))
df %>% arrange(Item) %>% head(10)
Should look like this:
1 A 2018-01-01 3 3
2 A 2018-02-01 -3 3
3 A 2018-03-01 4 7
4 A 2018-04-01 5 9
5 A 2018-05-01 1 10
6 B 2018-01-01 2 2
7 B 2018-02-01 -5 2
8 B 2018-03-01 -2 2
9 B 2018-04-01 0 0
10 B 2018-05-01 -4 0
answered Nov 20 at 13:37
Matheus Deister Veiga
16
16
add a comment |
add a comment |
up vote
3
down vote
Think zoo::
package if you want to roll anything.
df$new<-
zoo::rollsum( df$Value > 0, 3, fill = NA )
# Item Date Value new
#1 A 2018-1-1 0.5852699 NA
#2 A 2018-2-1 -0.7383377 1
#3 A 2018-3-1 -0.3157693 1
#4 A 2018-4-1 1.2475237 1
#5 A 2018-5-1 -1.5479757 1
#6 B 2018-1-1 -0.6913331 0
#7 B 2018-2-1 -0.2423809 0
#8 B 2018-3-1 -1.6363024 0
#9 B 2018-4-1 -0.3256263 1
#10 B 2018-5-1 0.3563144 NA
You have an option of the "window-position". Have a closer look at argument align = c("center", "left", "right")
.
So as a dplyr chain:
df %>% group_by(Item) %>% dplyr::mutate( new = zoo::rollsum( Value > 0, 3, fill = NA ))
Thanks Andre for your help. I tested with your method, it works!
– Felix Zhao
Nov 21 at 12:04
If your problem got solved please choose an answer.
– Andre Elrico
Nov 22 at 11:11
add a comment |
up vote
3
down vote
Think zoo::
package if you want to roll anything.
df$new<-
zoo::rollsum( df$Value > 0, 3, fill = NA )
# Item Date Value new
#1 A 2018-1-1 0.5852699 NA
#2 A 2018-2-1 -0.7383377 1
#3 A 2018-3-1 -0.3157693 1
#4 A 2018-4-1 1.2475237 1
#5 A 2018-5-1 -1.5479757 1
#6 B 2018-1-1 -0.6913331 0
#7 B 2018-2-1 -0.2423809 0
#8 B 2018-3-1 -1.6363024 0
#9 B 2018-4-1 -0.3256263 1
#10 B 2018-5-1 0.3563144 NA
You have an option of the "window-position". Have a closer look at argument align = c("center", "left", "right")
.
So as a dplyr chain:
df %>% group_by(Item) %>% dplyr::mutate( new = zoo::rollsum( Value > 0, 3, fill = NA ))
Thanks Andre for your help. I tested with your method, it works!
– Felix Zhao
Nov 21 at 12:04
If your problem got solved please choose an answer.
– Andre Elrico
Nov 22 at 11:11
add a comment |
up vote
3
down vote
up vote
3
down vote
Think zoo::
package if you want to roll anything.
df$new<-
zoo::rollsum( df$Value > 0, 3, fill = NA )
# Item Date Value new
#1 A 2018-1-1 0.5852699 NA
#2 A 2018-2-1 -0.7383377 1
#3 A 2018-3-1 -0.3157693 1
#4 A 2018-4-1 1.2475237 1
#5 A 2018-5-1 -1.5479757 1
#6 B 2018-1-1 -0.6913331 0
#7 B 2018-2-1 -0.2423809 0
#8 B 2018-3-1 -1.6363024 0
#9 B 2018-4-1 -0.3256263 1
#10 B 2018-5-1 0.3563144 NA
You have an option of the "window-position". Have a closer look at argument align = c("center", "left", "right")
.
So as a dplyr chain:
df %>% group_by(Item) %>% dplyr::mutate( new = zoo::rollsum( Value > 0, 3, fill = NA ))
Think zoo::
package if you want to roll anything.
df$new<-
zoo::rollsum( df$Value > 0, 3, fill = NA )
# Item Date Value new
#1 A 2018-1-1 0.5852699 NA
#2 A 2018-2-1 -0.7383377 1
#3 A 2018-3-1 -0.3157693 1
#4 A 2018-4-1 1.2475237 1
#5 A 2018-5-1 -1.5479757 1
#6 B 2018-1-1 -0.6913331 0
#7 B 2018-2-1 -0.2423809 0
#8 B 2018-3-1 -1.6363024 0
#9 B 2018-4-1 -0.3256263 1
#10 B 2018-5-1 0.3563144 NA
You have an option of the "window-position". Have a closer look at argument align = c("center", "left", "right")
.
So as a dplyr chain:
df %>% group_by(Item) %>% dplyr::mutate( new = zoo::rollsum( Value > 0, 3, fill = NA ))
edited Nov 20 at 13:16
answered Nov 20 at 13:03
Andre Elrico
5,56311027
5,56311027
Thanks Andre for your help. I tested with your method, it works!
– Felix Zhao
Nov 21 at 12:04
If your problem got solved please choose an answer.
– Andre Elrico
Nov 22 at 11:11
add a comment |
Thanks Andre for your help. I tested with your method, it works!
– Felix Zhao
Nov 21 at 12:04
If your problem got solved please choose an answer.
– Andre Elrico
Nov 22 at 11:11
Thanks Andre for your help. I tested with your method, it works!
– Felix Zhao
Nov 21 at 12:04
Thanks Andre for your help. I tested with your method, it works!
– Felix Zhao
Nov 21 at 12:04
If your problem got solved please choose an answer.
– Andre Elrico
Nov 22 at 11:11
If your problem got solved please choose an answer.
– Andre Elrico
Nov 22 at 11:11
add a comment |
up vote
1
down vote
You could use the RcppRoll
package.
require(RcppRoll)
df$new <- df$new <- RcppRoll::roll_sum(df$Value > 0, 3, fill = NA)
Using Tidyverse:
df %>%
group_by(Item) %>%
dplyr::mutate(new = RcppRoll::roll_sum(Value > 0, 3, fill = NA))
Speedwise this is faster than the zoo
Package:
n <- 10000
df <- data.frame(
Item = sample(LETTERS, n, replace = TRUE),
Value = rnorm(n))
df_grouped <- df %>%
group_by(Item)
microbenchmark::microbenchmark(
RcppRoll = df_grouped <- df_grouped %>% dplyr::mutate(new_RcppRoll = RcppRoll::roll_sum(Value > 0, 3, fill = NA)),
zoo = df_grouped <- df_grouped %>% dplyr::mutate(new_zoo = zoo::rollsum( Value > 0, 3, fill = NA ))
)
Results in:
Unit: milliseconds
expr min lq mean median uq max neval
RcppRoll 2.509003 2.741993 2.929227 2.83913 2.983726 5.832962 100
zoo 11.172920 11.785113 13.288970 12.43320 13.607826 25.879754 100
And
all.equal(df_grouped$new_RcppRoll, df_grouped$new_zoo)
TRUE
Hi Rentrop, it took me to figure out the alignment clause and it is a really good package dealing with rolling. Thanks a lot for your help!
– Felix Zhao
Nov 21 at 12:06
add a comment |
up vote
1
down vote
You could use the RcppRoll
package.
require(RcppRoll)
df$new <- df$new <- RcppRoll::roll_sum(df$Value > 0, 3, fill = NA)
Using Tidyverse:
df %>%
group_by(Item) %>%
dplyr::mutate(new = RcppRoll::roll_sum(Value > 0, 3, fill = NA))
Speedwise this is faster than the zoo
Package:
n <- 10000
df <- data.frame(
Item = sample(LETTERS, n, replace = TRUE),
Value = rnorm(n))
df_grouped <- df %>%
group_by(Item)
microbenchmark::microbenchmark(
RcppRoll = df_grouped <- df_grouped %>% dplyr::mutate(new_RcppRoll = RcppRoll::roll_sum(Value > 0, 3, fill = NA)),
zoo = df_grouped <- df_grouped %>% dplyr::mutate(new_zoo = zoo::rollsum( Value > 0, 3, fill = NA ))
)
Results in:
Unit: milliseconds
expr min lq mean median uq max neval
RcppRoll 2.509003 2.741993 2.929227 2.83913 2.983726 5.832962 100
zoo 11.172920 11.785113 13.288970 12.43320 13.607826 25.879754 100
And
all.equal(df_grouped$new_RcppRoll, df_grouped$new_zoo)
TRUE
Hi Rentrop, it took me to figure out the alignment clause and it is a really good package dealing with rolling. Thanks a lot for your help!
– Felix Zhao
Nov 21 at 12:06
add a comment |
up vote
1
down vote
up vote
1
down vote
You could use the RcppRoll
package.
require(RcppRoll)
df$new <- df$new <- RcppRoll::roll_sum(df$Value > 0, 3, fill = NA)
Using Tidyverse:
df %>%
group_by(Item) %>%
dplyr::mutate(new = RcppRoll::roll_sum(Value > 0, 3, fill = NA))
Speedwise this is faster than the zoo
Package:
n <- 10000
df <- data.frame(
Item = sample(LETTERS, n, replace = TRUE),
Value = rnorm(n))
df_grouped <- df %>%
group_by(Item)
microbenchmark::microbenchmark(
RcppRoll = df_grouped <- df_grouped %>% dplyr::mutate(new_RcppRoll = RcppRoll::roll_sum(Value > 0, 3, fill = NA)),
zoo = df_grouped <- df_grouped %>% dplyr::mutate(new_zoo = zoo::rollsum( Value > 0, 3, fill = NA ))
)
Results in:
Unit: milliseconds
expr min lq mean median uq max neval
RcppRoll 2.509003 2.741993 2.929227 2.83913 2.983726 5.832962 100
zoo 11.172920 11.785113 13.288970 12.43320 13.607826 25.879754 100
And
all.equal(df_grouped$new_RcppRoll, df_grouped$new_zoo)
TRUE
You could use the RcppRoll
package.
require(RcppRoll)
df$new <- df$new <- RcppRoll::roll_sum(df$Value > 0, 3, fill = NA)
Using Tidyverse:
df %>%
group_by(Item) %>%
dplyr::mutate(new = RcppRoll::roll_sum(Value > 0, 3, fill = NA))
Speedwise this is faster than the zoo
Package:
n <- 10000
df <- data.frame(
Item = sample(LETTERS, n, replace = TRUE),
Value = rnorm(n))
df_grouped <- df %>%
group_by(Item)
microbenchmark::microbenchmark(
RcppRoll = df_grouped <- df_grouped %>% dplyr::mutate(new_RcppRoll = RcppRoll::roll_sum(Value > 0, 3, fill = NA)),
zoo = df_grouped <- df_grouped %>% dplyr::mutate(new_zoo = zoo::rollsum( Value > 0, 3, fill = NA ))
)
Results in:
Unit: milliseconds
expr min lq mean median uq max neval
RcppRoll 2.509003 2.741993 2.929227 2.83913 2.983726 5.832962 100
zoo 11.172920 11.785113 13.288970 12.43320 13.607826 25.879754 100
And
all.equal(df_grouped$new_RcppRoll, df_grouped$new_zoo)
TRUE
answered Nov 20 at 13:33
Rentrop
13.9k33871
13.9k33871
Hi Rentrop, it took me to figure out the alignment clause and it is a really good package dealing with rolling. Thanks a lot for your help!
– Felix Zhao
Nov 21 at 12:06
add a comment |
Hi Rentrop, it took me to figure out the alignment clause and it is a really good package dealing with rolling. Thanks a lot for your help!
– Felix Zhao
Nov 21 at 12:06
Hi Rentrop, it took me to figure out the alignment clause and it is a really good package dealing with rolling. Thanks a lot for your help!
– Felix Zhao
Nov 21 at 12:06
Hi Rentrop, it took me to figure out the alignment clause and it is a really good package dealing with rolling. Thanks a lot for your help!
– Felix Zhao
Nov 21 at 12:06
add a comment |
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
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Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown