How to create quarterly data in R - r

Question

I have daily scores and their corresponding dates as seen below and currently struggling to convert them to quarterly. However, the years are first of all not chronological and am quite confused as to how to deal with the two situations.
see sample.
data
dates score
1 July 1, 2019 Monday 8
2 October 25, 2015 Sunday -3
3 June 17, 2020 Wednesday -5
4 January 17, 2018 Wednesday -1
5 April 15, 2019 Monday 6
6 October 30, 2019 Wednesday 10
7 March 6, 2017 Monday -2
8 November 19, 2018 Monday 3
9 June 11, 2020 Thursday 5
10 October 11, 2017 Wednesday -13
11 December 3, 2017 Sunday -8
12 November 14, 2018 Wednesday -6
13 August 22, 2017 Tuesday 8
14 December 13, 2017 Wednesday 5
15 January 22, 2016 Friday 5`
dates <- sapply(date, function(x)
trimws(grep(paste(month.name, collapse = '|'), x, value = TRUE)));
sort(as.Date(dates,'%B %d, %Y %A'))

Answer 1

This is a job for lubridate. You can parse your date column with lubridate::parse_date_time() and extract the quarter they fall in with lubridate::quarter():
library("tibble")
library("dplyr")
library("lubridate")
tbl <- tribble(~date, ~score,
"July 1, 2019 Monday", 8,
"October 25, 2015 Sunday", -3,
"June 17, 2020 Wednesday", -5,
"January 17, 2018 Wednesday", -1,
"April 15, 2019 Monday", 6,
"October 30, 2019 Wednesday", 10,
"March 6, 2017 Monday", -2,
"November 19, 2018 Monday", 3,
"June 11, 2020 Thursday", 5,
"October 11, 2017 Wednesday", -13,
"December 3, 2017 Sunday", -8,
"November 14, 2018 Wednesday", -6,
"August 22, 2017 Tuesday", 8,
"December 13, 2017 Wednesday", 5,
"January 22, 2016 Friday", 5)
tbl %>%
mutate(date = parse_date_time(date, "B d, Y")) %>%
mutate(quarter = quarter(date, with_year = TRUE))
#> # A tibble: 15 x 3
#> date score quarter
#> <dttm> <dbl> <dbl>
#> 1 2019-07-01 00:00:00 8 2019.3
#> 2 2015-10-25 00:00:00 -3 2015.4
#> 3 2020-06-17 00:00:00 -5 2020.2
#> 4 2018-01-17 00:00:00 -1 2018.1
#> 5 2019-04-15 00:00:00 6 2019.2
#> 6 2019-10-30 00:00:00 10 2019.4
#> 7 2017-03-06 00:00:00 -2 2017.1
#> 8 2018-11-19 00:00:00 3 2018.4
#> 9 2020-06-11 00:00:00 5 2020.2
#> 10 2017-10-11 00:00:00 -13 2017.4
#> 11 2017-12-03 00:00:00 -8 2017.4
#> 12 2018-11-14 00:00:00 -6 2018.4
#> 13 2017-08-22 00:00:00 8 2017.3
#> 14 2017-12-13 00:00:00 5 2017.4
#> 15 2016-01-22 00:00:00 5 2016.1

Answer 2

If you are trying to change dates column to Date class you can use as.Date.
df$new_date <- as.Date(trimws(df$dates), '%B %d, %Y')
Or this should also work with lubridate's mdy :
df$new_date <- lubridate::mdy(df$dates)

Answer 3

Once the data has been converted to date values per Ronak Shah's answer, we can use lubridate::quarter() to generate year and quarter values.
textData <- " dates|score
July 1, 2019 Monday| 8
October 25, 2015 Sunday| -3
June 17, 2020 Wednesday| -5
January 17, 2018 Wednesday| -1
April 15, 2019 Monday| 6
October 30, 2019 Wednesday| 10
March 6, 2017 Monday| -2
November 19, 2018 Monday| 3
June 11, 2020 Thursday| 5
October 11, 2017 Wednesday| -13
December 3, 2017 Sunday| -8
November 14, 2018 Wednesday| -6
August 22, 2017 Tuesday| 8
December 13, 2017 Wednesday| 5
January 22, 2016 Friday| 5
"
df <- read.csv(text=textData,
header=TRUE,
sep="|")
library(lubridate)
df$dt_quarter <- quarter(mdy(df$dates),with_year = TRUE,
fiscal_start = 1)
head(df)
We include with_year = TRUE and fiscal_start = 1 arguments to illustrate that one can change the output to include / exclude the year information, as well as change the start month for the year from the default of 1.
...and the output:
> head(df)
dates score dt_quarter
1 July 1, 2019 Monday 8 2019.3
2 October 25, 2015 Sunday -3 2015.4
3 June 17, 2020 Wednesday -5 2020.2
4 January 17, 2018 Wednesday -1 2018.1
5 April 15, 2019 Monday 6 2019.2
6 October 30, 2019 Wednesday 10 2019.4

Answer 4

The yearqtr class represents a year and quarter as the year plus 0 for Q1, 0.25 for Q2, 0.5 for Q3 and 0.75 for Q4. If date is defined as a yearqtr object as below then as.integer(date) is the year and cycle(date) is the quarter: 1, 2, 3 or 4. Note that junk at the end of the date field is ignored by as.yearqtr so we only need to specify month, day and year percent codes.
If you want a Date object instead of a yearqtr object then uncomment one of the commented out lines.
data is defined reproducibly in the Note at the end. (In the future please use dput to display your input data to prevent ambiguity as discussed in the information at the top of the r tag page.)
library(zoo)
date <- as.yearqtr(data$date, "%B %d, %Y")
# uncomment one of these lines if you want a Date object instead of yearqtr object
# date <- as.Date(date) # first day of quarter
# date <- as.Date(date, frac = 1) # last day of quarter
data.frame(date, score = data$score)[order(date), ]
giving the following sorted data frame assuming that we do not uncomment any of the commented out lines above.
date score
2 2015 Q4 -3
15 2016 Q1 5
7 2017 Q1 -2
...snip...
Time series
If this is supposed to be a time series with a single aggregated score per quarter then we can get a zoo series like this where data is the original data defined in the Note below.
library(zoo)
to_ym <- function(x) as.yearqtr(x, "%B %d, %Y")
z <- read.zoo(data, FUN = to_ym, aggregate = "mean")
z
## 2015 Q4 2016 Q1 2017 Q1 2017 Q3 2017 Q4 2018 Q1 2018 Q4 2019 Q2
## -3.000000 5.000000 -2.000000 8.000000 -5.333333 -1.000000 -1.500000 6.000000
## 2019 Q3 2019 Q4 2020 Q2
## 8.000000 10.000000 0.000000
or as a ts object like this:
as.ts(z)
## Qtr1 Qtr2 Qtr3 Qtr4
## 2015 -3.000000
## 2016 5.000000 NA NA NA
## 2017 -2.000000 NA 8.000000 -5.333333
## 2018 -1.000000 NA NA -1.500000
## 2019 NA 6.000000 8.000000 10.000000
## 2020 NA 0.000000
Note
The input data in reproducible form:
data <- structure(list(dates = c("July 1, 2019 Monday", "October 25, 2015 Sunday",
"June 17, 2020 Wednesday", "January 17, 2018 Wednesday", "April 15, 2019 Monday",
"October 30, 2019 Wednesday", "March 6, 2017 Monday", "November 19, 2018 Monday",
"June 11, 2020 Thursday", "October 11, 2017 Wednesday", "December 3, 2017 Sunday",
"November 14, 2018 Wednesday", "August 22, 2017 Tuesday", "December 13, 2017 Wednesday",
"January 22, 2016 Friday"), score = c(8L, -3L, -5L, -1L, 6L,
10L, -2L, 3L, 5L, -13L, -8L, -6L, 8L, 5L, 5L)), class = "data.frame", row.names = c(NA,
-15L))
Update
Have updated this answer several times so be sure you are looking at the latest version.