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I have a database containing a list of events. Each event has an associated start date, and a date when the event ended or was completed, eg:
dataset <- tibble(
eventid = sample(1:100, 25, replace=TRUE),
start_date = sample(seq(as.Date('2011/01/01'), as.Date('2012/01/01'), by="day"), 25),
completed_date = sample(seq(as.Date('2012/01/01'), as.Date('2014/01/01'), by="day"), 25)
)
> dataset
# A tibble: 25 x 3
eventid start_date completed_date
<int> <date> <date>
1 57 2011-01-14 2013-01-07
2 97 2011-01-21 2011-03-03
3 58 2011-01-26 2011-02-05
4 25 2011-03-22 2013-07-20
5 8 2011-04-20 2012-07-16
6 81 2011-04-26 2013-03-04
7 42 2011-05-02 2012-01-16
8 77 2011-05-03 2012-08-14
9 78 2011-05-21 2013-09-26
10 49 2011-05-22 2013-01-04
# ... with 15 more rows
>
I am trying to produce a rolling "snapshot" of how many tasks were pending a different points in time, e.g. month by month. Expected result:
# A tibble: 25 x 2
month count
<date> <int>
1 2011-01-01 0
2 2011-02-01 3
3 2011-03-01 2
4 2011-04-01 2
5 2011-05-01 4
6 2011-06-01 8
I have attempted to group my variables using group_by(period=floor_date(start_date,"month")), but I'm a bit stuck and would appreciate a pointer in the right direction!
I would prefer a solution using dplyr if possible.
Thanks!
You can expand rows for each month included in the range of dates with map2 from purrr. map2 will iterate over multiple inputs simultaneously. In this case, it will iterate through the start and end dates at the same time.
In each iteration, if will create a monthly sequence using seq (or seq.Date) from start to end month (determined from floor_date). The result is nested for each row of data (since one row can have multiple months in the sequence). So, unnest is needed afterwards.
The transmute will add a new variable called month_year (and drop the old ones) and use substr to extract the year and month only (no day). This is the first through seventh character of the date.
Then, you can group_by the month-year and count up the number of pending projects for each month_year.
I included set.seed to reproduce from data below.
library(dplyr)
library(tidyr)
library(purrr)
library(lubridate)
dataset %>%
mutate(month = map2(floor_date(start_date, "month"),
floor_date(completed_date, "month"),
seq.Date,
by = "month")) %>%
unnest(month) %>%
transmute(month_year = substr(month, 1, 7)) %>%
group_by(month_year) %>%
summarise(count = n())
Output
month_year count
<chr> <int>
1 2011-01 1
2 2011-02 3
3 2011-03 9
4 2011-04 10
5 2011-05 13
6 2011-06 15
7 2011-07 16
8 2011-08 18
9 2011-09 19
10 2011-10 20
# … with 22 more rows
If you want to exclude the completed month (except when start month and completed month are the same, if that can exist), you can subtract 1 month from the sequence of months created. In this case, you can use pmax so that if both start and end months are the same, it will still count the month).
Here is the modified mutate with map2:
mutate(month = map2(floor_date(start_date, "month"),
pmax(floor_date(completed_date, "month") - 1, floor_date(start_date, "month")),
seq.Date,
by = "month"))
Data
set.seed(123)
dataset <- tibble(
eventid = sample(1:100, 25, replace=TRUE),
start_date = sample(seq(as.Date('2011/01/01'), as.Date('2012/01/01'), by="day"), 25),
completed_date = sample(seq(as.Date('2012/01/01'), as.Date('2014/01/01'), by="day"), 25)
)
I have an R dataframe df_demand with a date column (depdate) and a dependent variable column bookings. The duration is 365 days starting from 2017-11-02 and ending at 2018-11-01, sorted in ascending order.
We have booking data for only 279 days in the year.
dplyr::arrange(df_demand, depdate)
depdate bookings
1 2017-11-02 43
2 2017-11-03 27
3 2017-11-05 27
4 2017-11-06 22
5 2017-11-07 39
6 2017-11-08 48
.
.
279 2018-11-01 60
I want to introduce another column day_of_year in the following way:
depdate day_of_year bookings
1 2017-11-02 1 43
2 2017-11-03 2 27
3 2017-11-04 3 NA
4 2017-11-05 4 27
.
.
.
365 2018-11-01 365 60
I am trying to find the best possible way to do this.
In Python, I could use something like :
df_demand['day_of_year'] = df_demand['depdate'].sub(df_demand['depdate'].iat[0]).dt.days + 1
I wanted to know about an R equivalent of the same.
When I run
typeof(df_demand_2$depdate)
the output is
"double"
Am I missing something?
You can create a row for every date using the complete function from the tidyr package.
First, I'm creating a data frame with some sample data:
df <- data.frame(
depdate = as.Date(c('2017-11-02', '2017-11-03', '2017-11-05')),
bookings = c(43, 27, 27)
)
Next, I'm performing two operations. First, using tidyr::complete, I'm specifying all the dates I want in my analysis. I can do that using seq.Date, creating a sequence from the first to the last day.
Once that is done, the day_of_year column is simply equal to the row number.
df_complete <- tidyr::complete(df,
depdate = seq.Date(from = min(df$depdate), to = max(df$depdate), by = 1)
)
df_complete$day_of_year <- 1:nrow(df_complete)
> df_complete
#> # A tibble: 4 x 3
#> depdate bookings day_of_year
#> <date> <dbl> <int>
#> 1 2017-11-02 43 1
#> 2 2017-11-03 27 2
#> 3 2017-11-04 NA 3
#> 4 2017-11-05 27 4
An equivalent solution with the pipe operator from dplyr:
df %>%
complete(depdate = seq.Date(from = min(df$depdate), to = max(df$depdate), by = 1)) %>%
mutate(days_of_year = row_number())
I have a "date" vector, that contains dates in mm/dd/yyyy format:
head(Entered_Date,5)
[1] 1/5/1998 1/5/1998 1/5/1998 1/5/1998 1/5/1998
I am trying to plot a frequency variable against the date, but I want to group the dates that it is by month or year. As it is now, there is a frequency per day, but I want to plot the frequency by month or year. So instead of having a frequency of 1 for 1/5/1998, 1 for 1/7/1998, and 3 for 1/8/1998, I would like to display it as 5 for 1/1998. It is a relatively large data set, with dates from 1998 to present, and I would like to find some automated way to accomplish this.
> dput(head(Entered_Date))
structure(c(260L, 260L, 260L, 260L, 260L, 260L), .Label = c("1/1/1998",
"1/1/1999", "1/1/2001", "1/1/2002", "1/10/2000", "1/10/2001",
"1/10/2002", "1/10/2003", "1/10/2005", "1/10/2006", "1/10/2007",
"1/10/2008", "1/10/2011", "1/10/2012", "1/10/2013", "1/11/1999",
"1/11/2000", "1/11/2001", "1/11/2002", "1/11/2005", "1/11/2006",
"1/11/2008", "1/11/2010", "1/11/2011", "1/11/2012", "1/11/2013",
"1/12/1998", "1/12/1999", "1/12/2001", "1/12/2004", "1/12/2005", ...
The floor_date() function from the lubridate package does this nicely.
data %>%
group_by(month = lubridate::floor_date(date, "month")) %>%
summarize(summary_variable = sum(value))
Thanks to Roman Cheplyaka
https://ro-che.info/articles/2017-02-22-group_by_month_r
See more on how to use the function: https://lubridate.tidyverse.org/reference/round_date.html
Here is an example using dplyr. You simply use the corresponding date format string for month %m or year %Y in the format statement.
set.seed(123)
df <- data.frame(date = seq.Date(from =as.Date("01/01/1998", "%d/%m/%Y"),
to=as.Date("01/01/2000", "%d/%m/%Y"), by="day"),
value = sample(seq(5), 731, replace = TRUE))
head(df)
date value
1 1998-01-01 2
2 1998-01-02 4
3 1998-01-03 3
4 1998-01-04 5
5 1998-01-05 5
6 1998-01-06 1
library(dplyr)
df %>%
mutate(month = format(date, "%m"), year = format(date, "%Y")) %>%
group_by(month, year) %>%
summarise(total = sum(value))
Source: local data frame [25 x 3]
Groups: month [?]
month year total
(chr) (chr) (int)
1 01 1998 105
2 01 1999 91
3 01 2000 3
4 02 1998 74
5 02 1999 77
6 03 1998 96
7 03 1999 86
8 04 1998 91
9 04 1999 95
10 05 1998 93
.. ... ... ...
Just to add to #cdeterman answer, you can use lubridate along with dplyr to make this even easier:
df <- data.frame(date = seq.Date(from =as.Date("01/01/1998", "%d/%m/%Y"),
to=as.Date("01/01/2000", "%d/%m/%Y"), by="day"),
value = sample(seq(5), 731, replace = TRUE))
library(dplyr)
library(lubridate)
df %>%
mutate(month = month(date), year = year(date)) %>%
group_by(month, year) %>%
summarise(total = sum(value))
Maybe you just add a column in your data like this:
Year <- format(as.Date(Entered_Date, "%d/%m/%Y"), "%Y")
Dont need dplyr. Look at ?as.POSIXlt
df$date<-as.POSIXlt(df$date)
mon<-df$date$mon
yr<-df$date$year
monyr<-as.factor(paste(mon,yr,sep="/"))
df$date<-monyr
Don't need to use ggplot2 but its nice for this kind of thing.
c <- ggplot(df, aes(factor(date)))
c + geom_bar()
If you want to see the actual numbers
aggregate(. ~ date,data = df,FUN=length )
df2<-aggregate(. ~ date,data = df,FUN=length )
df2
date value
1 0/98 31
2 0/99 31
3 1/98 28
4 1/99 28
5 10/98 30
6 10/99 30
7 11/97 1
8 11/98 31
9 11/99 31
10 2/98 31
11 2/99 31
12 3/98 30
13 3/99 30
14 4/98 31
15 4/99 31
16 5/98 30
17 5/99 30
18 6/98 31
19 6/99 31
20 7/98 31
21 7/99 31
22 8/98 30
23 8/99 30
24 9/98 31
25 9/99 31
There is a super easy way using the cut() function:
list = as.Date(c("1998-5-2", "1993-4-16", "1998-5-10"))
cut(list, breaks = "month")
and you will get this:
[1] 1998-05-01 1993-04-01 1998-05-01
62 Levels: 1993-04-01 1993-05-01 1993-06-01 1993-07-01 1993-08-01 ... 1998-05-01
Another solution is slider::slide_period:
library(slider)
library(dplyr)
monthly_summary <- function(data) summarise(data, date = format(max(date), "%Y-%m"), value = sum(value))
slide_period_dfr(df, df$date, "month", monthly_summary)
date value
1 1998-01 92
2 1998-02 82
3 1998-03 113
4 1998-04 94
5 1998-05 92
6 1998-06 74
7 1998-07 89
8 1998-08 92
9 1998-09 91
10 1998-10 100
...
There is also group_by(month_yr = cut(date, breaks = "1 month") in base R, without needing to use lubridate or other packages.
I have data table here:
row V1 velocity
1 2009-04-06 95.9230769230769
2 2009-04-11 95.0985074626866
3 2009-04-17 95.8064935064935
4 2009-04-22 94.6357142857143
5 2009-04-27 95.3626865671642
6 2009-05-03 95.9101265822785
7 2009-05-08 95.826582278481
8 2009-05-14 94.5126582278481
9 2009-05-20 95.8371428571429
10 2009-05-25 94.6981481481481
11 2009-05-30 96.397619047619
12 2009-06-05 94.8132530120482
13 2009-06-10 96.4558139534884
14 2009-06-16 94.9627906976744
15 2009-06-21 95.2666666666667
16 2009-06-26 95.2919540229885
17 2009-07-01 95.4333333333333
18 2009-07-07 95.3375
19 2009-07-12 95.0534246575343
20 2009-07-18 96.0277777777778
21 2009-07-24 95.6885057471264
22 2009-07-29 93.9375
23 2009-08-03 95.2776315789474
24 2009-08-08 94.9089285714286
25 2009-08-13 96.8906976744186
26 2009-08-19 95.4487804878049
27 2009-08-24 97.2444444444444
28 2009-08-30 95.1174418604651
I want to write a r code to find a mean value of velocity by month. (There are May, June, July, and August.
What could I do?
Or jusr:
tapply(df$velocity, months(as.Date(df$V1)), mean)
April August Juli Juni Mai
95.36530 95.81465 95.24634 95.35810 95.53038
Here's how I would do it
Use lubridate to create a month variable to group by in dplyr and then get means.
library(lubridate)
library(dplyr)
df %>% group_by(month = month(df$V1)) %>% summarize(mean = mean(velocity))
month mean
1 4 95.36530
2 5 95.53038
3 6 95.35810
4 7 95.24634
5 8 95.81465
If you add label=T you get this:
df %>% group_by(month = month(df$V1,label=T)) %>% summarize(mean = mean(velocity))
month mean
1 Apr 95.36530
2 May 95.53038
3 Jun 95.35810
4 Jul 95.24634
5 Aug 95.81465
Let's say I have several years worth of data which look like the following
# load date package and set random seed
library(lubridate)
set.seed(42)
# create data.frame of dates and income
date <- seq(dmy("26-12-2010"), dmy("15-01-2011"), by = "days")
df <- data.frame(date = date,
wday = wday(date),
wday.name = wday(date, label = TRUE, abbr = TRUE),
income = round(runif(21, 0, 100)),
week = format(date, format="%Y-%U"),
stringsAsFactors = FALSE)
# date wday wday.name income week
# 1 2010-12-26 1 Sun 91 2010-52
# 2 2010-12-27 2 Mon 94 2010-52
# 3 2010-12-28 3 Tues 29 2010-52
# 4 2010-12-29 4 Wed 83 2010-52
# 5 2010-12-30 5 Thurs 64 2010-52
# 6 2010-12-31 6 Fri 52 2010-52
# 7 2011-01-01 7 Sat 74 2011-00
# 8 2011-01-02 1 Sun 13 2011-01
# 9 2011-01-03 2 Mon 66 2011-01
# 10 2011-01-04 3 Tues 71 2011-01
# 11 2011-01-05 4 Wed 46 2011-01
# 12 2011-01-06 5 Thurs 72 2011-01
# 13 2011-01-07 6 Fri 93 2011-01
# 14 2011-01-08 7 Sat 26 2011-01
# 15 2011-01-09 1 Sun 46 2011-02
# 16 2011-01-10 2 Mon 94 2011-02
# 17 2011-01-11 3 Tues 98 2011-02
# 18 2011-01-12 4 Wed 12 2011-02
# 19 2011-01-13 5 Thurs 47 2011-02
# 20 2011-01-14 6 Fri 56 2011-02
# 21 2011-01-15 7 Sat 90 2011-02
I would like to sum 'income' for each week (Sunday thru Saturday). Currently I do the following:
Weekending 2011-01-01 = sum(df$income[1:7]) = 487
Weekending 2011-01-08 = sum(df$income[8:14]) = 387
Weekending 2011-01-15 = sum(df$income[15:21]) = 443
However I would like a more robust approach which will automatically sum by week. I can't work out how to automatically subset the data into weeks. Any help would be much appreciated.
First use format to convert your dates to week numbers, then plyr::ddply() to calculate the summaries:
library(plyr)
df$week <- format(df$date, format="%Y-%U")
ddply(df, .(week), summarize, income=sum(income))
week income
1 2011-52 413
2 2012-01 435
3 2012-02 379
For more information on format.date, see ?strptime, particular the bit that defines %U as the week number.
EDIT:
Given the modified data and requirement, one way is to divide the date by 7 to get a numeric number indicating the week. (Or more precisely, divide by the number of seconds in a week to get the number of weeks since the epoch, which is 1970-01-01 by default.
In code:
df$week <- as.Date("1970-01-01")+7*trunc(as.numeric(df$date)/(3600*24*7))
library(plyr)
ddply(df, .(week), summarize, income=sum(income))
week income
1 2010-12-23 298
2 2010-12-30 392
3 2011-01-06 294
4 2011-01-13 152
I have not checked that the week boundaries are on Sunday. You will have to check this, and insert an appropriate offset into the formula.
This is now simple using dplyr. Also I would suggest using cut(breaks = "week") rather than format() to cut the dates into weeks.
library(dplyr)
df %>% group_by(week = cut(date, "week")) %>% mutate(weekly_income = sum(income))
I Googled "group week days into weeks R" and came across this SO question. You mention you have multiple years, so I think we need to keep up with both the week number and also the year, so I modified the answers there as so format(date, format = "%U%y")
In use it looks like this:
library(plyr) #for aggregating
df <- transform(df, weeknum = format(date, format = "%y%U"))
ddply(df, "weeknum", summarize, suminc = sum(income))
#----
weeknum suminc
1 1152 413
2 1201 435
3 1202 379
See ?strptime for all the format abbreviations.
Try rollapply from the zoo package:
rollapply(df$income, width=7, FUN = sum, by = 7)
# [1] 487 387 443
Or, use period.sum from the xts package:
period.sum(xts(df$income, order.by=df$date), which(df$wday %in% 7))
# [,1]
# 2011-01-01 487
# 2011-01-08 387
# 2011-01-15 443
Or, to get the output in the format you want:
data.frame(income = period.sum(xts(df$income, order.by=df$date),
which(df$wday %in% 7)),
week = df$week[which(df$wday %in% 7)])
# income week
# 2011-01-01 487 2011-00
# 2011-01-08 387 2011-01
# 2011-01-15 443 2011-02
Note that the first week shows as 2011-00 because that's how it is entered in your data. You could also use week = df$week[which(df$wday %in% 1)] which would match your output.
This solution is influenced by #Andrie and #Chase.
# load plyr
library(plyr)
# format weeks as per requirement (replace "00" with "52" and adjust corresponding year)
tmp <- list()
tmp$y <- format(df$date, format="%Y")
tmp$w <- format(df$date, format="%U")
tmp$y[tmp$w=="00"] <- as.character(as.numeric(tmp$y[tmp$w=="00"]) - 1)
tmp$w[tmp$w=="00"] <- "52"
df$week <- paste(tmp$y, tmp$w, sep = "-")
# get summary
df2 <- ddply(df, .(week), summarize, income=sum(income))
# include week ending date
tmp$week.ending <- lapply(df2$week, function(x) rev(df[df$week==x, "date"])[[1]])
df2$week.ending <- sapply(tmp$week.ending, as.character)
# week income week.ending
# 1 2010-52 487 2011-01-01
# 2 2011-01 387 2011-01-08
# 3 2011-02 443 2011-01-15
df.index = df['week'] #the the dt variable as index
df.resample('W').sum() #sum using resample
With dplyr:
df %>%
arrange(date) %>%
mutate(week = as.numeric(date - date[1])%/%7) %>%
group_by(week) %>%
summarise(weekincome= sum(income))
Instead of date[1] you can have any date from when you want to start your weekly study.