| Title: | Quickly Get Datetime Data Ready for Analysis |
|---|---|
| Description: | Transforms datetime data into a format ready for analysis. It offers two core functionalities; aggregating data to a higher level interval (thicken) and imputing records where observations were absent (pad). |
| Authors: | Edwin Thoen [aut, cre] |
| Maintainer: | Edwin Thoen <[email protected]> |
| License: | MIT + file LICENSE |
| Version: | 0.6.3 |
| Built: | 2024-11-25 16:21:21 UTC |
| Source: | CRAN |
After thickening all the values are either
shifted to the first or the last value of their interval.
This function creates a vector from x, with the values shifted to
the (approximate) center of the interval. This can give a more accurate
picture of the aggregated data when plotting.
center_interval(x, shift = c("up", "down"), interval = NULL)center_interval(x, shift = c("up", "down"), interval = NULL)
x |
A vector of class |
shift |
"up" or "down". |
interval |
The interval to be used for centering. If |
The interval will be translated to number of days when
x is of class Date, or number of seconds when x is of
class POSIXt. For months and quarters this will be the average
length of the interval. The translated units divided by two
will be added by or subtracted from each value of x.
Vector of the same class as x, with the values shifted to the
(approximate) center.
library(dplyr) library(ggplot2) plot_set <- emergency %>% thicken("hour", "h") %>% count(h) %>% head(24) ggplot(plot_set, aes(h, n)) + geom_col() plot_set %>% mutate(h_center = center_interval(h)) %>% ggplot(aes(h_center, n)) + geom_col()library(dplyr) library(ggplot2) plot_set <- emergency %>% thicken("hour", "h") %>% count(h) %>% head(24) ggplot(plot_set, aes(h, n)) + geom_col() plot_set %>% mutate(h_center = center_interval(h)) %>% ggplot(aes(h_center, n)) + geom_col()
Find the closest instance of the requested weekday to min(x).
Helper function for thicken with the interval "week", when the user
desires the start day of the weeks to be different from Sundays.
closest_weekday(x, wday = 1, direction = c("down", "up"))closest_weekday(x, wday = 1, direction = c("down", "up"))
x |
A vector of class |
wday |
Integer in the range 0-6 specifying the desired weekday start (0 = Sun, 1 = Mon, 2 = Tue, 3 = Wed, 4 = Thu, 5 = Fri, 6 = Sat). |
direction |
The first desired weekday before ("down") or after ("up")
the first day in |
Object of class Date, reflecting the closest desired weekday
to x.
closest_weekday(coffee$time_stamp) closest_weekday(coffee$time_stamp, 5) closest_weekday(coffee$time_stamp, 1, direction = "up") closest_weekday(coffee$time_stamp, 5, direction = "up")closest_weekday(coffee$time_stamp) closest_weekday(coffee$time_stamp, 5) closest_weekday(coffee$time_stamp, 1, direction = "up") closest_weekday(coffee$time_stamp, 5, direction = "up")
Made-up data set for demonstrating padr.
coffeecoffee
A data frame with 4 rows and 2 variables:
YYYY-MM-DD HH:MM:SS
Amount spent on coffee
The emergency calls coming in at Montgomery County, PA since 2015-12-10. Data set was created at 2016-10-17 16:15:40 CEST from the API and contains events until 2016-10-17 09:47:03 EST. From the original set the columns desc and e are not included.
emergencyemergency
A data frame with 120450 rows and 6 variables:
Latitude from Google maps, based on the address
Longitude from Google maps, based on the address
Zipcode from Google, when possible
Title, emergency category
YYYY-MM-DD HH:MM:SS
Township
For each specified column in x replace the missing values by a
function of the nonmissing values.
fill_by_function(x, ..., fun = mean)fill_by_function(x, ..., fun = mean)
x |
A data frame. |
... |
The unquoted column names of the variables that should be filled. |
fun |
The function to apply on the nonmissing values. |
x with the altered columns.
library(dplyr) # for the pipe operator x <- seq(as.Date('2016-01-01'), by = 'day', length.out = 366) x <- x[sample(1:366, 200)] %>% sort x_df <- data_frame(x = x, y1 = runif(200, 10, 20) %>% round, y2 = runif(200, 1, 50) %>% round) x_df %>% pad %>% fill_by_function(y1, y2) x_df %>% pad %>% fill_by_function(y1, y2, fun = median)library(dplyr) # for the pipe operator x <- seq(as.Date('2016-01-01'), by = 'day', length.out = 366) x <- x[sample(1:366, 200)] %>% sort x_df <- data_frame(x = x, y1 = runif(200, 10, 20) %>% round, y2 = runif(200, 1, 50) %>% round) x_df %>% pad %>% fill_by_function(y1, y2) x_df %>% pad %>% fill_by_function(y1, y2, fun = median)
For each specified column in x replace the missing values by the most
prevalent nonmissing value.
fill_by_prevalent(x, ...)fill_by_prevalent(x, ...)
x |
A data frame. |
... |
The unquoted column names of the variables that should be filled. |
x with the altered columns.
library(dplyr) # for the pipe operator x <- seq(as.Date('2016-01-01'), by = 'day', length.out = 366) x <- x[sample(1:366, 200)] %>% sort x_df <- data_frame(x = x, y1 = rep(letters[1:3], c(80, 70, 50)) %>% sample, y2 = rep(letters[2:5], c(60, 80, 40, 20)) %>% sample) x_df %>% pad %>% fill_by_prevalent(y1, y2)library(dplyr) # for the pipe operator x <- seq(as.Date('2016-01-01'), by = 'day', length.out = 366) x <- x[sample(1:366, 200)] %>% sort x_df <- data_frame(x = x, y1 = rep(letters[1:3], c(80, 70, 50)) %>% sample, y2 = rep(letters[2:5], c(60, 80, 40, 20)) %>% sample) x_df %>% pad %>% fill_by_prevalent(y1, y2)
Replace all missing values in the specified columns by the same value.
fill_by_value(x, ..., value = 0)fill_by_value(x, ..., value = 0)
x |
A data frame. |
... |
The unquoted column names of the variables that should be filled. |
value |
The value to replace the missing values by. |
x with the altered columns.
library(dplyr) # for the pipe operator x <- seq(as.Date('2016-01-01'), by = 'day', length.out = 366) x <- x[sample(1:366, 200)] %>% sort x_df <- data_frame(x = x, y1 = runif(200, 10, 20) %>% round, y2 = runif(200, 1, 50) %>% round, y3 = runif(200, 20, 40) %>% round, y4 = sample(letters[1:5], 200, replace = TRUE)) x_padded <- x_df %>% pad x_padded %>% fill_by_value(y1) x_df %>% pad %>% fill_by_value(y1, y2, value = 42)library(dplyr) # for the pipe operator x <- seq(as.Date('2016-01-01'), by = 'day', length.out = 366) x <- x[sample(1:366, 200)] %>% sort x_df <- data_frame(x = x, y1 = runif(200, 10, 20) %>% round, y2 = runif(200, 1, 50) %>% round, y3 = runif(200, 20, 40) %>% round, y4 = sample(letters[1:5], 200, replace = TRUE)) x_padded <- x_df %>% pad x_padded %>% fill_by_value(y1) x_df %>% pad %>% fill_by_value(y1, y2, value = 42)
After applying thicken all the observations of a period are mapped
to a single time point. This function will convert a datetime variable to
a character vector that reflects the period, instead of a single time point.
strftime is used to format the start and the end of the interval.
format_interval( x, start_format = "%Y-%m-%d", end_format = start_format, sep = " ", end_offset = 0, units_to_last = NULL )format_interval( x, start_format = "%Y-%m-%d", end_format = start_format, sep = " ", end_offset = 0, units_to_last = NULL )
x |
A vector of class |
start_format |
String to format the start values of each period, to be used
in |
end_format |
String to format the end values of each period, to be used
in |
sep |
Character string that separates the |
end_offset |
Units in days if |
units_to_last |
To determine the formatting of the last value in |
The end of the periods will be determined by the next unique value
in x. It does so without regarding the interval of x. If a specific
interval is desired, thicken and / or pad should first be
applied to create an equally spaced datetime variable.
A character vector showing the interval.
library(dplyr) library(ggplot2) plot_set <- emergency %>% head(500) %>% thicken("hour", "h") %>% count(h) # this will show the data on the full hour ggplot(plot_set, aes(h, n)) + geom_col() # adding a character to indicate the hours of the interval. plot_set %>% mutate(h_int = format_interval(h, "%H", sep = "-"))library(dplyr) library(ggplot2) plot_set <- emergency %>% head(500) %>% thicken("hour", "h") %>% count(h) # this will show the data on the full hour ggplot(plot_set, aes(h, n)) + geom_col() # adding a character to indicate the hours of the interval. plot_set %>% mutate(h_int = format_interval(h, "%H", sep = "-"))
The interval is the highest datetime unit that can explain all instances of a
variable of class Date, class POSIXct, or class POSIXct.
This function will determine what the interval of the variable is.
get_interval(x)get_interval(x)
x |
A variable of class of class |
See vignette("padr") for more information on intervals.
A character string indicating the interval of x.
x_month <- seq(as.Date('2016-01-01'), as.Date('2016-05-01'), by = 'month') get_interval(x_month) x_sec <- seq(as.POSIXct('2016-01-01 00:00:00'), length.out = 100, by = 'sec') get_interval(x_sec) get_interval(x_sec[seq(0, length(x_sec), by = 5)])x_month <- seq(as.Date('2016-01-01'), as.Date('2016-05-01'), by = 'month') get_interval(x_month) x_sec <- seq(as.POSIXct('2016-01-01 00:00:00'), length.out = 100, by = 'sec') get_interval(x_sec) get_interval(x_sec[seq(0, length(x_sec), by = 5)])
pad will fill the gaps in incomplete datetime variables, by figuring out
what the interval of the data is and what instances are missing. It will insert
a record for each of the missing time points. For all
other variables in the data frame a missing value will be inserted at the padded rows.
pad( x, interval = NULL, start_val = NULL, end_val = NULL, by = NULL, group = NULL, break_above = 1 )pad( x, interval = NULL, start_val = NULL, end_val = NULL, by = NULL, group = NULL, break_above = 1 )
x |
A data frame containing at least one variable of class |
interval |
The interval of the returned datetime variable.
Any character string that would be accepted by |
start_val |
An object of class |
end_val |
An object of class |
by |
Only needs to be specified when |
group |
Optional character vector that specifies the grouping
variable(s). Padding will take place within the different groups. When
interval is not specified, it will be determined applying |
break_above |
Numeric value that indicates the number of rows in millions above which the function will break. Safety net for situations where the interval is different than expected and padding yields a very large dataframe, possibly overflowing memory. |
The interval of a datetime variable is the time unit at which the
observations occur. The eight intervals in padr are from high to low
year, quarter, month, week, day,
hour, min, and sec. Since padr v.0.3.0 the
interval is no longer limited to be of a single unit.
(Intervals like 5 minutes, 6 hours, 10 days are possible). pad will
figure out the interval of the input variable and the step size, and will
fill the gaps for the instances that would be expected from the interval and
step size, but are missing in the input data.
Note that when start_val and/or end_val are specified, they are
concatenated with the datetime variable before the interval is determined.
Rows with missing values in the datetime variables will be retained. However, they will be moved to the end of the returned data frame.
The data frame x with the datetime variable padded. All
non-grouping variables in the data frame will have missing values at the rows
that are padded. The result will always be sorted on the datetime variable.
If group is not NULL result is sorted on grouping variable(s)
first, then on the datetime variable.
simple_df <- data.frame(day = as.Date(c('2016-04-01', '2016-04-03')), some_value = c(3,4)) pad(simple_df) pad(simple_df, interval = "day") library(dplyr) # for the pipe operator month <- seq(as.Date('2016-04-01'), as.Date('2017-04-01'), by = 'month')[c(1, 4, 5, 7, 9, 10, 13)] month_df <- data.frame(month = month, y = runif(length(month), 10, 20) %>% round) # forward fill the padded values with tidyr's fill month_df %>% pad %>% tidyr::fill(y) # or fill all y with 0 month_df %>% pad %>% fill_by_value(y) # padding a data.frame on group level day_var <- seq(as.Date('2016-01-01'), length.out = 12, by = 'month') x_df_grp <- data.frame(grp1 = rep(LETTERS[1:3], each =4), grp2 = letters[1:2], y = runif(12, 10, 20) %>% round(0), date = sample(day_var, 12, TRUE)) %>% arrange(grp1, grp2, date) # pad by one grouping var x_df_grp %>% pad(group = 'grp1') # pad by two groups vars x_df_grp %>% pad(group = c('grp1', 'grp2'), interval = "month") # Using group argument the interval is determined over all the observations, # ignoring the groups. x <- data.frame(dt_var = as.Date(c("2017-01-01", "2017-03-01", "2017-05-01", "2017-01-01", "2017-02-01", "2017-04-01")), id = rep(1:2, each = 3), val = round(rnorm(6))) pad(x, group = "id") # applying pad with do, interval is determined individualle for each group x %>% group_by(id) %>% do(pad(.))simple_df <- data.frame(day = as.Date(c('2016-04-01', '2016-04-03')), some_value = c(3,4)) pad(simple_df) pad(simple_df, interval = "day") library(dplyr) # for the pipe operator month <- seq(as.Date('2016-04-01'), as.Date('2017-04-01'), by = 'month')[c(1, 4, 5, 7, 9, 10, 13)] month_df <- data.frame(month = month, y = runif(length(month), 10, 20) %>% round) # forward fill the padded values with tidyr's fill month_df %>% pad %>% tidyr::fill(y) # or fill all y with 0 month_df %>% pad %>% fill_by_value(y) # padding a data.frame on group level day_var <- seq(as.Date('2016-01-01'), length.out = 12, by = 'month') x_df_grp <- data.frame(grp1 = rep(LETTERS[1:3], each =4), grp2 = letters[1:2], y = runif(12, 10, 20) %>% round(0), date = sample(day_var, 12, TRUE)) %>% arrange(grp1, grp2, date) # pad by one grouping var x_df_grp %>% pad(group = 'grp1') # pad by two groups vars x_df_grp %>% pad(group = c('grp1', 'grp2'), interval = "month") # Using group argument the interval is determined over all the observations, # ignoring the groups. x <- data.frame(dt_var = as.Date(c("2017-01-01", "2017-03-01", "2017-05-01", "2017-01-01", "2017-02-01", "2017-04-01")), id = rep(1:2, each = 3), val = round(rnorm(6))) pad(x, group = "id") # applying pad with do, interval is determined individualle for each group x %>% group_by(id) %>% do(pad(.))
Pad the datetime variable after thicken_cust is applied, using the same
spanning.
pad_cust(x, spanned, by = NULL, group = NULL, drop_last_spanned = TRUE)pad_cust(x, spanned, by = NULL, group = NULL, drop_last_spanned = TRUE)
x |
A data frame containing at least one datetime variable of
class |
spanned |
A datetime vector to which the the datetime variable in
|
by |
Only needs to be specified when |
group |
Optional character vector that specifies the grouping variable(s). Padding will take place within the different group values. |
drop_last_spanned |
Logical, indicating whether to drop the last value
from |
The data frame x with the datetime column padded.
library(dplyr) # analysis of traffic accidents in traffic jam hours and other hours. accidents <- emergency %>% filter(title == "Traffic: VEHICLE ACCIDENT -") spanning <- span_time("20151210 16", "20161017 17", tz = "EST") %>% subset_span(list(hour = c(6, 9, 16, 19))) thicken_cust(accidents, spanning, "period") %>% count(period) %>% pad_cust(spanning)library(dplyr) # analysis of traffic accidents in traffic jam hours and other hours. accidents <- emergency %>% filter(title == "Traffic: VEHICLE ACCIDENT -") spanning <- span_time("20151210 16", "20161017 17", tz = "EST") %>% subset_span(list(hour = c(6, 9, 16, 19))) thicken_cust(accidents, spanning, "period") %>% count(period) %>% pad_cust(spanning)
pad_int fills the gaps in incomplete integer variables. It will insert
a record for each of the missing value. For all
other variables in the data frame a missing value will be inserted at the
padded rows.
pad_int(x, by, start_val = NULL, end_val = NULL, group = NULL, step = 1)pad_int(x, by, start_val = NULL, end_val = NULL, group = NULL, step = 1)
x |
A data frame. |
by |
The column to be padded. |
start_val |
The first value of the returned variable. If NULL it will use the lowest value of the input variable. |
end_val |
The last value of the returned variable. If NULL it will use the highest value of the input variable. |
group |
Optional character vector that specifies the grouping variable(s). Padding will take place within the different group values. |
step |
The step size of the returned variable. |
The data frame x with the specified variable padded. All
non-grouping variables in the data frame will have missing values at the rows
that are padded.
int_df <- data.frame(x = c(2005, 2007, 2008, 2011), val = c(3, 2, 6, 3)) pad_int(int_df, 'x') pad_int(int_df, 'x', start_val = 2006, end_val = 2013) int_df2 <- data.frame(x = c(2005, 2015), val = c(3, 4)) pad_int(int_df2, 'x', step = 2) pad_int(int_df2, 'x', step = 5) int_df3 <- data.frame(x = c(2005, 2006, 2008, 2006, 2007, 2009), g = rep(LETTERS[1:2], each = 3), val = c(6, 6, 3, 5, 4, 3)) pad_int(int_df3, 'x', group = 'g') pad_int(int_df3, 'x', group = 'g', start_val = 2005, end_val = 2009)int_df <- data.frame(x = c(2005, 2007, 2008, 2011), val = c(3, 2, 6, 3)) pad_int(int_df, 'x') pad_int(int_df, 'x', start_val = 2006, end_val = 2013) int_df2 <- data.frame(x = c(2005, 2015), val = c(3, 4)) pad_int(int_df2, 'x', step = 2) pad_int(int_df2, 'x', step = 5) int_df3 <- data.frame(x = c(2005, 2006, 2008, 2006, 2007, 2009), g = rep(LETTERS[1:2], each = 3), val = c(6, 6, 3, 5, 4, 3)) pad_int(int_df3, 'x', group = 'g') pad_int(int_df3, 'x', group = 'g', start_val = 2005, end_val = 2009)
Span a vector of specified interval around a variable of class Date,
POSIXct, or POSIXlt..
span_around(x, interval, start_shift = NULL, end_shift = start_shift)span_around(x, interval, start_shift = NULL, end_shift = start_shift)
x |
A vector of class |
interval |
Character, specifying the desired interval. |
start_shift |
Character, indicating the time to shift back from the first observation. |
end_shift |
Character, indicating the time to shift forward from the last observation. |
Note that use of the start_shift and end_shift
arguments change the entire spanning when they are not in line with
the interval. It is not checked for.
A datetime vector, with the first observation smaller or equal than
min(x) and the last observation larger or equal than max(x).
Spaces between points are equal to interval.
span_around(coffee$time_stamp, "hour") span_around(coffee$time_stamp, "hour", end_shift = "2 hour") span_around(coffee$time_stamp, "2 day") span_around(coffee$time_stamp, "2 day", start_shift = "2 day") span_around(emergency$time_stamp, "week") span_around(emergency$time_stamp, "2 month")span_around(coffee$time_stamp, "hour") span_around(coffee$time_stamp, "hour", end_shift = "2 hour") span_around(coffee$time_stamp, "2 day") span_around(coffee$time_stamp, "2 day", start_shift = "2 day") span_around(emergency$time_stamp, "week") span_around(emergency$time_stamp, "2 month")
seq.Date.Quickly create a sequence of dates from minimal specifications.
span_date(from, to = NULL, len_out = NULL, by = NULL)span_date(from, to = NULL, len_out = NULL, by = NULL)
from |
Integer or character of length 4 (yyyy), 6 (yyyymm), or 8 (yyymmdd). Indicating the start value of the sequence. |
to |
Integer or character of length 4 (yyyy), 6 (yyyymm), or 8 (yyymmdd). Optional. |
len_out |
The desired length of the sequence. Optional. |
by |
The desired interval. Optional. |
Minimal specification of dates, sets unspecified date parts to default values. These are 01 for both month and day.
In addition to from, either to or len_out must be specified.
If by is not specified, span_date will set the interval to the
highest of the specified date parts in either from or to.
For example, if they are 2011 and 2015 it will be "year", if they are 2011
and 201501 it will be "month".
An object of class Date.
# using "to" argument span_date(2011, 2015) span_date(201101, 201501) span_date(2011, 2015, by = "month") span_date(2011, 201501) span_date(20111225, 2012) # using "len_out" argument span_date(2011, len_out = 4) span_date(201101, len_out = 4) span_date(20110101, len_out = 4) span_date(20110101, len_out = 4, by = "month")# using "to" argument span_date(2011, 2015) span_date(201101, 201501) span_date(2011, 2015, by = "month") span_date(2011, 201501) span_date(20111225, 2012) # using "len_out" argument span_date(2011, len_out = 4) span_date(201101, len_out = 4) span_date(20110101, len_out = 4) span_date(20110101, len_out = 4, by = "month")
seq.POSIXct.Quickly create a sequence of datetimes from minimal specifications.
span_time(from, to = NULL, len_out = NULL, by = NULL, tz = "UTC")span_time(from, to = NULL, len_out = NULL, by = NULL, tz = "UTC")
from |
Integer or character of length 4 (yyyy), 6 (yyyymm), or 8 ( yyymmdd). Character of length 11 (yyyymmdd hh), 13 (yyyymmdd hhmm), or 15 ( yyyymmdd hhmmss). Indicating the start value of the sequence. |
to |
Integer or character of length 4 (yyyy), 6 (yyyymm), or 8 ( yyymmdd). Character of length 11 (yyyymmdd hh), 13 (yyyymmdd hhmm), or 15 ( yyyymmdd hhmmss). Indicating the end value of the sequence. Optional. |
len_out |
The desired length of the sequence. Optional. |
by |
The desired interval. Optional. |
tz |
The desired timezone. |
Minimal specification of datetimes, sets unspecified date parts to default values. These are 01 for both month and day and 00 for hour, minute, and second.
In addition to from, either to or length must be specified.
If the by is not specified, span_time will set the interval to
the highest of the specified datetime parts in either from or
to. For example, if they are "20160103 01" and "20160108 05" it will
be "hour", if they are "2011" and "20110101 021823" it will be "second".
An object of class POSIXct.
# using to span_time(2011, 2013) span_time("2011", "2013") span_time(2011, 201301) span_time(2011, 20130101) span_time(2011, "20110101 0023") span_time(2011, "20110101 002300") # using len_out span_time(2011, len_out = 3) span_time("2011", len_out = 3) span_time(2011, len_out = 10, by = "month") span_time(2011, len_out = 10, by = "day") span_time(2011, len_out = 10, by = "hour") span_time("20110101 00", len_out = 10) span_time("20110101 002300", len_out = 10)# using to span_time(2011, 2013) span_time("2011", "2013") span_time(2011, 201301) span_time(2011, 20130101) span_time(2011, "20110101 0023") span_time(2011, "20110101 002300") # using len_out span_time(2011, len_out = 3) span_time("2011", len_out = 3) span_time(2011, len_out = 10, by = "month") span_time(2011, len_out = 10, by = "day") span_time(2011, len_out = 10, by = "hour") span_time("20110101 00", len_out = 10) span_time("20110101 002300", len_out = 10)
Take a Date, POSIXct, or POSIXlt vector and subset it by
a pattern of date and/or time parts.
subset_span(spanned, pattern_list)subset_span(spanned, pattern_list)
spanned |
A vector of class |
pattern_list |
A list with the desired pattern for each of the following datetime parts: year, mon, mday, wday, hour, min, sec. |
For subsetting weekdays, they run from 0 (Sunday) to 6 (Saturday).
Vector of the same class as spanned, containing all the data points in
spanned that meets the requirements in pattern_list.
date_span <- span_date(20170701, len_out = 100) subset_span(date_span, list(wday = 1:5)) time_span <- span_time("20170101 00", 201702) subset_span(time_span, list(hour = 7:17)) subset_span(time_span, list(hour = c(10, 16), mday = seq(5, 30, 5)))date_span <- span_date(20170701, len_out = 100) subset_span(date_span, list(wday = 1:5)) time_span <- span_time("20170101 00", 201702) subset_span(time_span, list(hour = 7:17)) subset_span(time_span, list(hour = c(10, 16), mday = seq(5, 30, 5)))
Take the datetime variable in a data frame and map this to a variable of a higher interval. The mapping is added to the data frame in a new variable.
thicken( x, interval, colname = NULL, rounding = c("down", "up"), by = NULL, start_val = NULL, drop = FALSE, ties_to_earlier = FALSE )thicken( x, interval, colname = NULL, rounding = c("down", "up"), by = NULL, start_val = NULL, drop = FALSE, ties_to_earlier = FALSE )
x |
A data frame containing at least one datetime variable of
class |
interval |
The interval of the added datetime variable.
Any character string that would be accepted by |
colname |
The column name of the added variable. If |
rounding |
Should a value in the input datetime variable be mapped to
the closest value that is lower ( |
by |
Only needs to be specified when |
start_val |
By default the first instance of |
drop |
Should the original datetime variable be dropped from the
returned data frame? Defaults to |
ties_to_earlier |
By default when the original datetime observations is
tied with a value in the added datetime variable, it is assigned to the
current value when rounding is down or to the next value when rounding
is up. When |
When the datetime variable contains missing values, they are left in place in the dataframe. The added column with the new datetime variable, will have a missing values for these rows as well.
See vignette("padr") for more information on thicken.
See vignette("padr_implementation") for detailed information on
daylight savings time, different timezones, and the implementation of
thicken.
The data frame x with the variable added to it.
x_hour <- seq(lubridate::ymd_hms('20160302 000000'), by = 'hour', length.out = 200) some_df <- data.frame(x_hour = x_hour) thicken(some_df, 'week') thicken(some_df, 'month') thicken(some_df, 'day', start_val = lubridate::ymd_hms('20160301 120000')) library(dplyr) x_df <- data.frame( x = seq(lubridate::ymd(20130101), by = 'day', length.out = 1000) %>% sample(500), y = runif(500, 10, 50) %>% round) %>% arrange(x) # get the max per month x_df %>% thicken('month') %>% group_by(x_month) %>% summarise(y_max = max(y)) # get the average per week, but you want your week to start on Mondays # instead of Sundays x_df %>% thicken('week', start_val = closest_weekday(x_df$x, 2)) %>% group_by(x_week) %>% summarise(y_avg = mean(y)) # rounding up instead of down x <- data.frame(dt = lubridate::ymd_hms('20171021 160000', '20171021 163100')) thicken(x, interval = "hour", rounding = "up") thicken(x, interval = "hour", rounding = "up", ties_to_earlier = TRUE)x_hour <- seq(lubridate::ymd_hms('20160302 000000'), by = 'hour', length.out = 200) some_df <- data.frame(x_hour = x_hour) thicken(some_df, 'week') thicken(some_df, 'month') thicken(some_df, 'day', start_val = lubridate::ymd_hms('20160301 120000')) library(dplyr) x_df <- data.frame( x = seq(lubridate::ymd(20130101), by = 'day', length.out = 1000) %>% sample(500), y = runif(500, 10, 50) %>% round) %>% arrange(x) # get the max per month x_df %>% thicken('month') %>% group_by(x_month) %>% summarise(y_max = max(y)) # get the average per week, but you want your week to start on Mondays # instead of Sundays x_df %>% thicken('week', start_val = closest_weekday(x_df$x, 2)) %>% group_by(x_week) %>% summarise(y_avg = mean(y)) # rounding up instead of down x <- data.frame(dt = lubridate::ymd_hms('20171021 160000', '20171021 163100')) thicken(x, interval = "hour", rounding = "up") thicken(x, interval = "hour", rounding = "up", ties_to_earlier = TRUE)
Like thicken, it will find the datetime variable in x
and add a variable of a higher periodicity to it. However, the variable to
which to map the observation is provided by the user. This enables mapping to
time points that are unequally spaced.
thicken_cust(x, spanned, colname, by = NULL, drop = FALSE)thicken_cust(x, spanned, colname, by = NULL, drop = FALSE)
x |
A data frame containing at least one datetime variable of
class |
spanned |
A datetime vector to which the the datetime variable in
|
colname |
Character, the column name of the added variable. |
by |
Only needs to be specified when |
drop |
Should the original datetime variable be dropped from the
returned data frame? Defaults to |
Only rounding down is available for custom thickening.
The data frame x with the variable added to it.
library(dplyr) # analysis of traffic accidents in traffic jam hours and other hours. accidents <- emergency %>% filter(title == "Traffic: VEHICLE ACCIDENT -") spanning <- span_time("20151210 16", "20161017 17", tz = "EST") %>% subset_span(list(hour = c(6, 9, 16, 19))) thicken_cust(accidents, spanning, "period") %>% count(period) %>% pad_cust(spanning)library(dplyr) # analysis of traffic accidents in traffic jam hours and other hours. accidents <- emergency %>% filter(title == "Traffic: VEHICLE ACCIDENT -") spanning <- span_time("20151210 16", "20161017 17", tz = "EST") %>% subset_span(list(hour = c(6, 9, 16, 19))) thicken_cust(accidents, spanning, "period") %>% count(period) %>% pad_cust(spanning)