We wish to shift the values of a vector up or down by a given offset. Shifting allows us to access a vector’s previous or following values enabling us to compute changes.
We wish to shift the value of a vector down by one i.e. we wish to obtain the previous value.
In this example, we wish to compute the difference and rate of change between the value of each element of the column attr_1
and the value of its previous element. The data frame is sorted by the value of the column date
.
library(lubridate)
df_2 = df %>%
arrange(date) %>%
mutate(
previous = lag(attr_1),
delta = attr_1 - previous,
change = delta / previous)
Here is how this works:
lag()
from the dplyr
package to shift the values of the column attr_1
down and store the resulting vector in a new column previous
.lag()
shifts down by 1 position, i.e. an offset of 1.delta = attr_1 - previous
, we compute the difference between the current value of attr_1
and its previous value stored in previous
.change = delta / previous
, we compute the rate of change.date
column. See Sorting.Extension: Specify Offset Value
We wish to shift the value of a vector down by an arbitrary offset.
In this example, we wish to compute the difference and rate of change between the value of each element of the column attr_1
and its value one week prior (seven days earlier). The data frame is sorted by the value of the column date
.
library(lubridate)
df_2 = df %>%
arrange(date) %>%
mutate(
last_week = lag(attr_1, 7),
delta = attr_1 - last_week,
change = delta / last_week)
Here is how this works:
This code is similar to the code above with one exception: We pass the desired offset value to the argument n
of lag()
, which here is n=7
.
Extension: Fill NA
When we shift a vector, we inadvertently create missing values corresponding to the shift. By default, those missing values are encoded as NA
. We can specify an alternative value.
In this example, we wish to fill the missing values resulting from the shift with 0.
library(lubridate)
df_2 = df %>%
arrange(date) %>%
mutate(
previous = lag(attr_1, default = 0),
delta = attr_1 - previous)
Here is how this works:
default
of lag()
, which here is default=0
.NA
s after shifting because there may be other NA
s in the data that we do not necessarily wish to replace.Extension: Shift Per Group
We wish to shift the value of a column down by one i.e. we wish to obtain the previous value based on an ordered column per group, where the groups are specified by the column col_1
In this example, we wish to compute the difference between the value of each element of the column col_2
and the value of its previous element based on the column date
for each group in col_1
.
df_2 = df %>%
arrange(date) %>%
group_by(col_1) %>%
mutate(previous = lag(col_2))
Here is how this works:
This code is similar to the code above with one exception: We group the data frame by col_1
and then calculate lag. We sort the data frame by date
to calculate the lag based on date
column.
We wish to shift the value of a vector up by one i.e. we wish to obtain the next value.
In this example, we wish to compute the difference and rate of change between the value of each element of the column attr_1
and the value of its next element. The data frame is sorted by the value of the column date
.
library(lubridate)
df_2 = df %>%
arrange(date) %>%
mutate(
next_val = lead(attr_1),
delta = next_val - attr_1,
change = delta / attr_1)
Here is how this works:
lead()
instead of lag()
because we wish to shift up i.e. obtain the next value.