Mutate each day and each hour using tidyverse functions in R
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Currently I have code returns each a tibble of events that occur each day using the following:
online_toy_purchases %>%
mutate(interval = lubridate::date(date)) %>%
group_by(interval) %>%
summarise(count = n())
This currently returns the following:
# A tibble: 31 x 2
interval count
2018-12-01 500
2018-12-02 300
2018-12-03 400
2018-12-04 200
2018-12-05 600
...
2018-12-31 100
I would like my code to group by each hour and each day for a more granular view of the data, which would return the following:
# A tibble: 744 x 2
interval count
2018-12-01 01:00:00 50
2018-12-01 02:00:00 60
2018-12-01 03:00:00 20
2018-12-01 04:00:00 80
...
2018-12-31 24:00:00 10
online_toy_purchases is a tibble that contains, among other features, the ID of the transaction and a timestamp containing the date and the hour, minute and second of the purchase (i.e -> "2018-12-01 01:20:58")
r datetime time-series tidyverse tidyr
asked Jan 3 at 22:06
SepaSepa
123
add a comment |
Currently I have code returns each a tibble of events that occur each day using the following:
online_toy_purchases %>%
mutate(interval = lubridate::date(date)) %>%
group_by(interval) %>%
summarise(count = n())
This currently returns the following:
# A tibble: 31 x 2
interval count
2018-12-01 500
2018-12-02 300
2018-12-03 400
2018-12-04 200
2018-12-05 600
...
2018-12-31 100
I would like my code to group by each hour and each day for a more granular view of the data, which would return the following:
# A tibble: 744 x 2
interval count
2018-12-01 01:00:00 50
2018-12-01 02:00:00 60
2018-12-01 03:00:00 20
2018-12-01 04:00:00 80
...
2018-12-31 24:00:00 10
online_toy_purchases is a tibble that contains, among other features, the ID of the transaction and a timestamp containing the date and the hour, minute and second of the purchase (i.e -> "2018-12-01 01:20:58")
r datetime time-series tidyverse tidyr
asked Jan 3 at 22:06
SepaSepa
123
3
It's easier to help you if you include a simple reproducible example with sample input and desired output that can be used to test and verify possible solutions.
– MrFlick
Jan 3 at 22:07
1
What if you just didgroup_by(interval, lubridate::hour(date))? Unable to test because there is no reproducible example.
– MrFlick
Jan 3 at 22:08
This returns a tibble with interval, 'lubridate::hour(date)', count as features, with the middle feature displaying the hours. This is really close to what I want, but wouldn't be suitable for plotting. Working on getting some reproducible data to this post.
– Sepa
Jan 3 at 22:19
add a comment |
Currently I have code returns each a tibble of events that occur each day using the following:
online_toy_purchases %>%
mutate(interval = lubridate::date(date)) %>%
group_by(interval) %>%
summarise(count = n())
This currently returns the following:
# A tibble: 31 x 2
interval count
2018-12-01 500
2018-12-02 300
2018-12-03 400
2018-12-04 200
2018-12-05 600
...
2018-12-31 100
I would like my code to group by each hour and each day for a more granular view of the data, which would return the following:
# A tibble: 744 x 2
interval count
2018-12-01 01:00:00 50
2018-12-01 02:00:00 60
2018-12-01 03:00:00 20
2018-12-01 04:00:00 80
...
2018-12-31 24:00:00 10
online_toy_purchases is a tibble that contains, among other features, the ID of the transaction and a timestamp containing the date and the hour, minute and second of the purchase (i.e -> "2018-12-01 01:20:58")
r datetime time-series tidyverse tidyr
asked Jan 3 at 22:06
SepaSepa
123
Currently I have code returns each a tibble of events that occur each day using the following:
online_toy_purchases %>%
mutate(interval = lubridate::date(date)) %>%
group_by(interval) %>%
summarise(count = n())
This currently returns the following:
# A tibble: 31 x 2
interval count
2018-12-01 500
2018-12-02 300
2018-12-03 400
2018-12-04 200
2018-12-05 600
...
2018-12-31 100
I would like my code to group by each hour and each day for a more granular view of the data, which would return the following:
# A tibble: 744 x 2
interval count
2018-12-01 01:00:00 50
2018-12-01 02:00:00 60
2018-12-01 03:00:00 20
2018-12-01 04:00:00 80
...
2018-12-31 24:00:00 10
online_toy_purchases is a tibble that contains, among other features, the ID of the transaction and a timestamp containing the date and the hour, minute and second of the purchase (i.e -> "2018-12-01 01:20:58")
r datetime time-series tidyverse tidyr
r datetime time-series tidyverse tidyr
asked Jan 3 at 22:06
SepaSepa
123
asked Jan 3 at 22:06
SepaSepa
123
asked Jan 3 at 22:06
SepaSepa
123
asked Jan 3 at 22:06
SepaSepa
123
asked Jan 3 at 22:06
SepaSepa
123
123
3
It's easier to help you if you include a simple reproducible example with sample input and desired output that can be used to test and verify possible solutions.
– MrFlick
Jan 3 at 22:07
1
What if you just didgroup_by(interval, lubridate::hour(date))? Unable to test because there is no reproducible example.
– MrFlick
Jan 3 at 22:08
This returns a tibble with interval, 'lubridate::hour(date)', count as features, with the middle feature displaying the hours. This is really close to what I want, but wouldn't be suitable for plotting. Working on getting some reproducible data to this post.
– Sepa
Jan 3 at 22:19
add a comment |
3
It's easier to help you if you include a simple reproducible example with sample input and desired output that can be used to test and verify possible solutions.
– MrFlick
Jan 3 at 22:07
1
What if you just didgroup_by(interval, lubridate::hour(date))? Unable to test because there is no reproducible example.
– MrFlick
Jan 3 at 22:08
This returns a tibble with interval, 'lubridate::hour(date)', count as features, with the middle feature displaying the hours. This is really close to what I want, but wouldn't be suitable for plotting. Working on getting some reproducible data to this post.
– Sepa
Jan 3 at 22:19
3
3
It's easier to help you if you include a simple reproducible example with sample input and desired output that can be used to test and verify possible solutions.
– MrFlick
Jan 3 at 22:07
It's easier to help you if you include a simple reproducible example with sample input and desired output that can be used to test and verify possible solutions.
– MrFlick
Jan 3 at 22:07
1
1
What if you just did
group_by(interval, lubridate::hour(date))? Unable to test because there is no reproducible example.– MrFlick
Jan 3 at 22:08
What if you just did
group_by(interval, lubridate::hour(date))? Unable to test because there is no reproducible example.– MrFlick
Jan 3 at 22:08
This returns a tibble with interval, 'lubridate::hour(date)', count as features, with the middle feature displaying the hours. This is really close to what I want, but wouldn't be suitable for plotting. Working on getting some reproducible data to this post.
– Sepa
Jan 3 at 22:19
This returns a tibble with interval, 'lubridate::hour(date)', count as features, with the middle feature displaying the hours. This is really close to what I want, but wouldn't be suitable for plotting. Working on getting some reproducible data to this post.
– Sepa
Jan 3 at 22:19
add a comment |
1 Answer
1
active
oldest
votes
This will count the number of rows within each hour of the data.
library(tidyverse)
online_toy_purchases %>%
# assuming that "date" is formatted as a datetime variable already
count(time = lubridate::floor_date(date, "1 hour")) %>%
# additional step using padr::pad to add missing hours and
# tidyr::replace_na to make NAs into zeroes
padr::pad() %>%
replace_na(list(n=0))
For visualization and further analysis, it will be helpful to have rows recording periods with no data. You might alternatively accomplish something similar by converting to a tsibble.
answered Jan 3 at 22:16
Jon SpringJon Spring
7,9532929
Thanks! This gets me very close. How would I use padr or tsibble to return a "0" value for unmentioned hours? Those exist in this data.
– Sepa
Jan 3 at 22:25
updated answer to include padr step.
– Jon Spring
Jan 3 at 23:50
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
votes
active
oldest
votes
This will count the number of rows within each hour of the data.
library(tidyverse)
online_toy_purchases %>%
# assuming that "date" is formatted as a datetime variable already
count(time = lubridate::floor_date(date, "1 hour")) %>%
# additional step using padr::pad to add missing hours and
# tidyr::replace_na to make NAs into zeroes
padr::pad() %>%
replace_na(list(n=0))
For visualization and further analysis, it will be helpful to have rows recording periods with no data. You might alternatively accomplish something similar by converting to a tsibble.
answered Jan 3 at 22:16
Jon SpringJon Spring
7,9532929
Thanks! This gets me very close. How would I use padr or tsibble to return a "0" value for unmentioned hours? Those exist in this data.
– Sepa
Jan 3 at 22:25
updated answer to include padr step.
– Jon Spring
Jan 3 at 23:50
add a comment |
This will count the number of rows within each hour of the data.
library(tidyverse)
online_toy_purchases %>%
# assuming that "date" is formatted as a datetime variable already
count(time = lubridate::floor_date(date, "1 hour")) %>%
# additional step using padr::pad to add missing hours and
# tidyr::replace_na to make NAs into zeroes
padr::pad() %>%
replace_na(list(n=0))
For visualization and further analysis, it will be helpful to have rows recording periods with no data. You might alternatively accomplish something similar by converting to a tsibble.
answered Jan 3 at 22:16
Jon SpringJon Spring
7,9532929
Thanks! This gets me very close. How would I use padr or tsibble to return a "0" value for unmentioned hours? Those exist in this data.
– Sepa
Jan 3 at 22:25
updated answer to include padr step.
– Jon Spring
Jan 3 at 23:50
add a comment |
This will count the number of rows within each hour of the data.
library(tidyverse)
online_toy_purchases %>%
# assuming that "date" is formatted as a datetime variable already
count(time = lubridate::floor_date(date, "1 hour")) %>%
# additional step using padr::pad to add missing hours and
# tidyr::replace_na to make NAs into zeroes
padr::pad() %>%
replace_na(list(n=0))
For visualization and further analysis, it will be helpful to have rows recording periods with no data. You might alternatively accomplish something similar by converting to a tsibble.
answered Jan 3 at 22:16
Jon SpringJon Spring
7,9532929
This will count the number of rows within each hour of the data.
library(tidyverse)
online_toy_purchases %>%
# assuming that "date" is formatted as a datetime variable already
count(time = lubridate::floor_date(date, "1 hour")) %>%
# additional step using padr::pad to add missing hours and
# tidyr::replace_na to make NAs into zeroes
padr::pad() %>%
replace_na(list(n=0))
For visualization and further analysis, it will be helpful to have rows recording periods with no data. You might alternatively accomplish something similar by converting to a tsibble.
answered Jan 3 at 22:16
Jon SpringJon Spring
7,9532929
edited Jan 3 at 23:48
answered Jan 3 at 22:16
Jon SpringJon Spring
7,9532929
answered Jan 3 at 22:16
Jon SpringJon Spring
7,9532929
answered Jan 3 at 22:16
Jon SpringJon Spring
7,9532929
7,9532929
Thanks! This gets me very close. How would I use padr or tsibble to return a "0" value for unmentioned hours? Those exist in this data.
– Sepa
Jan 3 at 22:25
updated answer to include padr step.
– Jon Spring
Jan 3 at 23:50
add a comment |
Thanks! This gets me very close. How would I use padr or tsibble to return a "0" value for unmentioned hours? Those exist in this data.
– Sepa
Jan 3 at 22:25
updated answer to include padr step.
– Jon Spring
Jan 3 at 23:50
Thanks! This gets me very close. How would I use padr or tsibble to return a "0" value for unmentioned hours? Those exist in this data.
– Sepa
Jan 3 at 22:25
Thanks! This gets me very close. How would I use padr or tsibble to return a "0" value for unmentioned hours? Those exist in this data.
– Sepa
Jan 3 at 22:25
updated answer to include padr step.
– Jon Spring
Jan 3 at 23:50
updated answer to include padr step.
– Jon Spring
Jan 3 at 23:50
add a comment |
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3
It's easier to help you if you include a simple reproducible example with sample input and desired output that can be used to test and verify possible solutions.
– MrFlick
Jan 3 at 22:07
1
What if you just did
group_by(interval, lubridate::hour(date))? Unable to test because there is no reproducible example.– MrFlick
Jan 3 at 22:08
This returns a tibble with interval, 'lubridate::hour(date)', count as features, with the middle feature displaying the hours. This is really close to what I want, but wouldn't be suitable for plotting. Working on getting some reproducible data to this post.
– Sepa
Jan 3 at 22:19