I struggle mightily with dates in R and could do this pretty easily in SPSS, but I would love to stay within R for my project.
I have a date column in my data frame and want to remove the year completely in order to leave the month and day. Here is a peak at my original data.
> head(ds$date)
[1] "2003-10-09" "2003-10-11" "2003-10-13" "2003-10-15" "2003-10-18" "2003-10-20"
> class((ds$date))
[1] "Date"
I "want" it to be.
> head(ds$date)
[1] "10-09" "10-11" "10-13" "10-15" "10-18" "10-20"
> class((ds$date))
[1] "Date"
If possible, I would love to set the first date to be October 1st instead of January 1st.
Any help you can provide will be greatly appreciated.
EDIT: I felt like I should add some context. I want to plot an NHL player's performance over the course of a season which starts in October and ends in April. To add to this, I would like to facet the plots by each season which is a separate column in my data frame. Because I want to compare cumulative performance over the course of the season, I believe that I need to remove the year portion, but maybe I don't; as I indicated, I struggle with dates in R. What I am looking to accomplish is a plot that compares cumulative performance over relative dates by season and have the x-axis start in October and end in April.
> d = as.Date("2003-10-09", format="%Y-%m-%d")
> format(d, "%m-%d")
[1] "10-09"
Is this what you are looking for?
library(ggplot2)
## make up data for two seasons a and b
a = as.Date("2010/10/1")
b = as.Date("2011/10/1")
a.date <- seq(a, by='1 week', length=28)
b.date <- seq(b, by='1 week', length=28)
## make up some score data
a.score <- abs(trunc(rnorm(28, mean = 10, sd = 5)))
b.score <- abs(trunc(rnorm(28, mean = 10, sd = 5)))
## create a data frame
df <- data.frame(a.date, b.date, a.score, b.score)
df
## Since I am using ggplot I better create a "long formated" data frame
df.molt <- melt(df, measure.vars = c("a.score", "b.score"))
levels(df.molt$variable) <- c("First season", "Second season")
df.molt
Then, I am using ggplot2 for plotting the data:
## plot it
ggplot(aes(y = value, x = a.date), data = df.molt) + geom_point() +
geom_line() + facet_wrap(~variable, ncol = 1) +
scale_x_date("Date", format = "%m-%d")
If you want to modify the x-axis (e.g., display format), then you'll probably be interested in scale_date.
You have to remember Date is a numeric format, representing the number of days passed since the "origin" of the internal date counting :
> str(Date)
Class 'Date' num [1:10] 14245 14360 14475 14590 14705 ...
This is the same as in EXCEL, if you want a reference. Hence the solution with format as perfectly valid.
Now if you want to set the first date of a year as October 1st, you can construct some year index like this :
redefine.year <- function(x,start="10-1"){
year <- as.numeric(strftime(x,"%Y"))
yearstart <- as.Date(paste(year,start,sep="-"))
year + (x >= yearstart) - min(year) + 1
}
Testing code :
Start <- as.Date("2009-1-1")
Stop <- as.Date("2011-11-1")
Date <- seq(Start,Stop,length.out=10)
data.frame( Date=as.character(Date),
year=redefine.year(Date))
gives
Date year
1 2009-01-01 1
2 2009-04-25 1
3 2009-08-18 1
4 2009-12-11 2
5 2010-04-05 2
6 2010-07-29 2
7 2010-11-21 3
8 2011-03-16 3
9 2011-07-09 3
10 2011-11-01 4
Related
I am trying to plot a time series from an excel file in R Studio. It has a single column named 'Dates'. This column contains datetime data of customer visits in the form 2/15/2014 6:17:22 AM. The datetime was originally in char format and I converted it into a Large POSIXct value using lubridate:
tsData <- mdy_hms(fullUsage$Dates)
Which gives me a value:
POSIXct[1:25,354], format: "2018-04-13 10:18:14" "2018-04-14 13:27:11" .....
I then tried converting it into a time series object using the code below:
require(xts)
visitTimes.ts <- xts(tsData, start = 1, order.by=as.POSIXct(tsData))
plot(visitTimes.ts)
ts_plot(visitTimes.ts)
ts_info(visitTimes.ts)
Im not 100% sure but it looks like the plot is coming out using the sum count of visits. I believe my problem may be in correctly indexing my data using the dates. I apologize in advance if this is a simple issue to deal with I am still learning R. I have included the screenshot of my plot.
yes you are right, you need to provide both the date column (x axis) and the value (y axis)
here's a simple example:
v1 <- data.frame(Date = mdy_hms(c("1-1-2020-00-00-00", "1-2-2020-00-00-00", "1-3-2020-00-00-00")), Value = c(1, 3, 6))
v2 <- xts(v1["Value"], order.by = v1[, "Date"])
plot(v2)
first argument of xts takes the x values, on the order.by i leave the actual ts object
You need to count the number of events in each time period and plot these values on the y axis. You didn't provide enough data for a reproducible example, so I have created a small example. We'll use the tidyverse packages dplyr and lubridate to help us out here:
library(lubridate)
library(dplyr)
library(ggplot2)
set.seed(69)
fullUsage <- data.frame(Dates = as.POSIXct("2020-01-01") +
minutes(round(cumsum(rexp(10000, 1/25))))
)
head(fullUsage)
#> Dates
#> 1 2020-01-01 00:02:00
#> 2 2020-01-01 00:15:00
#> 3 2020-01-01 00:22:00
#> 4 2020-01-01 00:29:00
#> 5 2020-01-01 01:13:00
#> 6 2020-01-01 01:27:00
First of all, we will create columns that show the hour of day and the month that events occurred:
fullUsage$hours <- hour(fullUsage$Dates)
fullUsage$month <- floor_date(fullUsage$Dates, "month")
Now we can effectively just count the number of events per month and plot this number for each month:
fullUsage %>%
group_by(month) %>%
summarise(n = length(hours)) %>%
ggplot(aes(month, n)) +
geom_col()
And we can do the same for the hour of day:
fullUsage %>%
group_by(hours) %>%
summarise(n = length(hours)) %>%
ggplot(aes(hours, n)) +
geom_col() +
scale_x_continuous(breaks = 0:23) +
labs(y = "Hour of day")
Created on 2020-08-05 by the reprex package (v0.3.0)
I have the following data:
Date,Rain
1979_8_9_0,8.775
1979_8_9_6,8.775
1979_8_9_12,8.775
1979_8_9_18,8.775
1979_8_10_0,0
1979_8_10_6,0
1979_8_10_12,0
1979_8_10_18,0
1979_8_11_0,8.025
1979_8_12_12,0
1979_8_12_18,0
1979_8_13_0,8.025
[1] The data is six hourly but some dates have incomplete 6 hourly data. For example, August 11 1979 has only one value at 00H. I would like to get the daily accumulated from this kind of data using R. Any suggestion on how to do this easily in R?
I'll appreciate any help.
You can transform your data to dates very easily with:
dat$Date <- as.Date(strptime(dat$Date, '%Y_%m_%d_%H'))
After that you should aggregate with:
aggregate(Rain ~ Date, dat, sum)
The result:
Date Rain
1 1979-08-09 35.100
2 1979-08-10 0.000
3 1979-08-11 8.025
4 1979-08-12 0.000
5 1979-08-13 8.025
Based on the comment of Henrik, you can also transform to dates with:
dat$Date <- as.Date(dat$Date, '%Y_%m_%d')
# split the "date" variable into new, separate variable
splitDate <- stringr::str_split_fixed(string = df$Date, pattern = "_", n = 4)
df$Day <- splitDate[,3]
# split data by Day, loop over each split and add rain variable
unlist(lapply(split(df$Rain, df$Day), sum))
Here my time period range:
start_day = as.Date('1974-01-01', format = '%Y-%m-%d')
end_day = as.Date('2014-12-21', format = '%Y-%m-%d')
df = as.data.frame(seq(from = start_day, to = end_day, by = 'day'))
colnames(df) = 'date'
I need to created 10,000 data.frames with different fake years of 365days each one. This means that each of the 10,000 data.frames needs to have different start and end of year.
In total df has got 14,965 days which, divided by 365 days = 41 years. In other words, df needs to be grouped 10,000 times differently by 41 years (of 365 days each one).
The start of each year has to be random, so it can be 1974-10-03, 1974-08-30, 1976-01-03, etc... and the remaining dates at the end df need to be recycled with the starting one.
The grouped fake years need to appear in a 3rd col of the data.frames.
I would put all the data.frames into a list but I don't know how to create the function which generates 10,000 different year's start dates and subsequently group each data.frame with a 365 days window 41 times.
Can anyone help me?
#gringer gave a good answer but it solved only 90% of the problem:
dates.df <- data.frame(replicate(10000, seq(sample(df$date, 1),
length.out=365, by="day"),
simplify=FALSE))
colnames(dates.df) <- 1:10000
What I need is 10,000 columns with 14,965 rows made by dates taken from df which need to be eventually recycled when reaching the end of df.
I tried to change length.out = 14965 but R does not recycle the dates.
Another option could be to change length.out = 1 and eventually add the remaining df rows for each column by maintaining the same order:
dates.df <- data.frame(replicate(10000, seq(sample(df$date, 1),
length.out=1, by="day"),
simplify=FALSE))
colnames(dates.df) <- 1:10000
How can I add the remaining df rows to each col?
The seq method also works if the to argument is unspecified, so it can be used to generate a specific number of days starting at a particular date:
> seq(from=df$date[20], length.out=10, by="day")
[1] "1974-01-20" "1974-01-21" "1974-01-22" "1974-01-23" "1974-01-24"
[6] "1974-01-25" "1974-01-26" "1974-01-27" "1974-01-28" "1974-01-29"
When used in combination with replicate and sample, I think this will give what you want in a list:
> replicate(2,seq(sample(df$date, 1), length.out=10, by="day"), simplify=FALSE)
[[1]]
[1] "1985-07-24" "1985-07-25" "1985-07-26" "1985-07-27" "1985-07-28"
[6] "1985-07-29" "1985-07-30" "1985-07-31" "1985-08-01" "1985-08-02"
[[2]]
[1] "2012-10-13" "2012-10-14" "2012-10-15" "2012-10-16" "2012-10-17"
[6] "2012-10-18" "2012-10-19" "2012-10-20" "2012-10-21" "2012-10-22"
Without the simplify=FALSE argument, it produces an array of integers (i.e. R's internal representation of dates), which is a bit trickier to convert back to dates. A slightly more convoluted way to do this is and produce Date output is to use data.frame on the unsimplified replicate result. Here's an example that will produce a 10,000-column data frame with 365 dates in each column (takes about 5s to generate on my computer):
dates.df <- data.frame(replicate(10000, seq(sample(df$date, 1),
length.out=365, by="day"),
simplify=FALSE));
colnames(dates.df) <- 1:10000;
> dates.df[1:5,1:5];
1 2 3 4 5
1 1988-09-06 1996-05-30 1987-07-09 1974-01-15 1992-03-07
2 1988-09-07 1996-05-31 1987-07-10 1974-01-16 1992-03-08
3 1988-09-08 1996-06-01 1987-07-11 1974-01-17 1992-03-09
4 1988-09-09 1996-06-02 1987-07-12 1974-01-18 1992-03-10
5 1988-09-10 1996-06-03 1987-07-13 1974-01-19 1992-03-11
To get the date wraparound working, a slight modification can be made to the original data frame, pasting a copy of itself on the end:
df <- as.data.frame(c(seq(from = start_day, to = end_day, by = 'day'),
seq(from = start_day, to = end_day, by = 'day')));
colnames(df) <- "date";
This is easier to code for downstream; the alternative being a double seq for each result column with additional calculations for the start/end and if statements to deal with boundary cases.
Now instead of doing date arithmetic, the result columns subset from the original data frame (where the arithmetic is already done). Starting with one date in the first half of the frame and choosing the next 14965 values. I'm using nrow(df)/2 instead for a more generic code:
dates.df <-
as.data.frame(lapply(sample.int(nrow(df)/2, 10000),
function(startPos){
df$date[startPos:(startPos+nrow(df)/2-1)];
}));
colnames(dates.df) <- 1:10000;
>dates.df[c(1:5,(nrow(dates.df)-5):nrow(dates.df)),1:5];
1 2 3 4 5
1 1988-10-21 1999-10-18 2009-04-06 2009-01-08 1988-12-28
2 1988-10-22 1999-10-19 2009-04-07 2009-01-09 1988-12-29
3 1988-10-23 1999-10-20 2009-04-08 2009-01-10 1988-12-30
4 1988-10-24 1999-10-21 2009-04-09 2009-01-11 1988-12-31
5 1988-10-25 1999-10-22 2009-04-10 2009-01-12 1989-01-01
14960 1988-10-15 1999-10-12 2009-03-31 2009-01-02 1988-12-22
14961 1988-10-16 1999-10-13 2009-04-01 2009-01-03 1988-12-23
14962 1988-10-17 1999-10-14 2009-04-02 2009-01-04 1988-12-24
14963 1988-10-18 1999-10-15 2009-04-03 2009-01-05 1988-12-25
14964 1988-10-19 1999-10-16 2009-04-04 2009-01-06 1988-12-26
14965 1988-10-20 1999-10-17 2009-04-05 2009-01-07 1988-12-27
This takes a bit less time now, presumably because the date values have been pre-caclulated.
Try this one, using subsetting instead:
start_day = as.Date('1974-01-01', format = '%Y-%m-%d')
end_day = as.Date('2014-12-21', format = '%Y-%m-%d')
date_vec <- seq.Date(from=start_day, to=end_day, by="day")
Now, I create a vector long enough so that I can use easy subsetting later on:
date_vec2 <- rep(date_vec,2)
Now, create the random start dates for 100 instances (replace this with 10000 for your application):
random_starts <- sample(1:14965, 100)
Now, create a list of dates by simply subsetting date_vec2 with your desired length:
dates <- lapply(random_starts, function(x) date_vec2[x:(x+14964)])
date_df <- data.frame(dates)
names(date_df) <- 1:100
date_df[1:5,1:5]
1 2 3 4 5
1 1997-05-05 2011-12-10 1978-11-11 1980-09-16 1989-07-24
2 1997-05-06 2011-12-11 1978-11-12 1980-09-17 1989-07-25
3 1997-05-07 2011-12-12 1978-11-13 1980-09-18 1989-07-26
4 1997-05-08 2011-12-13 1978-11-14 1980-09-19 1989-07-27
5 1997-05-09 2011-12-14 1978-11-15 1980-09-20 1989-07-28
With ggplot2, I would like to create a multiplot (facet_grid) where each plot is the weekly count values for the month.
My data are like this :
day_group count
1 2012-04-29 140
2 2012-05-06 12595
3 2012-05-13 12506
4 2012-05-20 14857
I have created for this dataset two others colums the Month and the Week based on day_group :
day_group count Month Week
1 2012-04-29 140 Apr 17
2 2012-05-06 12595 May 18
3 2012-05-13 12506 May 19
4 2012-05-20 14857 May 2
Now I would like for each Month to create a barplot where I have the sum of the count values aggregated by week. So for example for a year I would have 12 plots with 4 bars (one per week).
Below is what I use to generate the plot :
ggplot(data = count_by_day, aes(x=day_group, y=count)) +
stat_summary(fun.y="sum", geom = "bar") +
scale_x_date(date_breaks = "1 month", date_labels = "%B") +
facet_grid(facets = Month ~ ., scales="free", margins = FALSE)
So far, my plot looks like this
https://dl.dropboxusercontent.com/u/96280295/Rplot.png
As you can see the x axes is not as I'm looking for. Instead of showing only week 1, 2, 3 and 4, it displays all the month.
Do you know what I must change to get what I'm looking for ?
Thanks for your help
Okay, now that I see what you want, I wrote a small program to illustrate it. The key to your order of month problem is making month a factor with the levels in the right order:
library(dplyr)
library(ggplot2)
#initialization
set.seed(1234)
sday <- as.Date("2012-01-01")
eday <- as.Date("2012-07-31")
# List of the first day of the months
mfdays <- seq(sday,length.out=12,by="1 month")
# list of months - this is key to keeping the order straight
mlabs <- months(mfdays)
# list of first weeks of the months
mfweek <- trunc((mfdays-sday)/7)
names(mfweek) <- mlabs
# Generate a bunch of event-days, and then months, then week numbs in our range
n <- 1000
edf <-data.frame(date=sample(seq(sday,eday,by=1),n,T))
edf$month <- factor(months(edf$date),levels=mlabs) # use the factor in the right order
edf$week <- 1 + as.integer(((edf$date-sday)/7) - mfweek[edf$month])
# Now summarize with dplyr
ndf <- group_by(edf,month,week) %>% summarize( count = n() )
ggplot(ndf) + geom_bar(aes(x=week,y=count),stat="identity") + facet_wrap(~month,nrow=1)
Yielding:
(As an aside, I am kind of proud I did this without lubridate ...)
I think you have to do this but I am not sure I understand your question:
ggplot(data = count_by_day, aes(x=Week, y=count, group= Month, color=Month))
I need to convert date (m/d/y format) into 3 separate columns on which I hope to run an algorithm.(I'm trying to convert my dates into Julian Day Numbers). Saw this suggestion for another user for separating data out into multiple columns using Oracle. I'm using R and am throughly stuck about how to code this appropriately. Would A1,A2...represent my new column headings, and what would the format difference be with the "update set" section?
update <tablename> set A1 = substr(ORIG, 1, 4),
A2 = substr(ORIG, 5, 6),
A3 = substr(ORIG, 11, 6),
A4 = substr(ORIG, 17, 5);
I'm trying hard to improve my skills in R but cannot figure this one...any help is much appreciated. Thanks in advance... :)
I use the format() method for Date objects to pull apart dates in R. Using Dirk's datetext, here is how I would go about breaking up a date into its constituent parts:
datetxt <- c("2010-01-02", "2010-02-03", "2010-09-10")
datetxt <- as.Date(datetxt)
df <- data.frame(date = datetxt,
year = as.numeric(format(datetxt, format = "%Y")),
month = as.numeric(format(datetxt, format = "%m")),
day = as.numeric(format(datetxt, format = "%d")))
Which gives:
> df
date year month day
1 2010-01-02 2010 1 2
2 2010-02-03 2010 2 3
3 2010-09-10 2010 9 10
Note what several others have said; you can get the Julian dates without splitting out the various date components. I added this answer to show how you could do the breaking apart if you needed it for something else.
Given a text variable x, like this:
> x
[1] "10/3/2001"
then:
> as.Date(x,"%m/%d/%Y")
[1] "2001-10-03"
converts it to a date object. Then, if you need it:
> julian(as.Date(x,"%m/%d/%Y"))
[1] 11598
attr(,"origin")
[1] "1970-01-01"
gives you a Julian date (relative to 1970-01-01).
Don't try the substring thing...
See help(as.Date) for more.
Quick ones:
Julian date converters already exist in base R, see eg help(julian).
One approach may be to parse the date as a POSIXlt and to then read off the components. Other date / time classes and packages will work too but there is something to be said for base R.
Parsing dates as string is almost always a bad approach.
Here is an example:
datetxt <- c("2010-01-02", "2010-02-03", "2010-09-10")
dates <- as.Date(datetxt) ## you could examine these as well
plt <- as.POSIXlt(dates) ## now as POSIXlt types
plt[["year"]] + 1900 ## years are with offset 1900
#[1] 2010 2010 2010
plt[["mon"]] + 1 ## and months are on the 0 .. 11 intervasl
#[1] 1 2 9
plt[["mday"]]
#[1] 2 3 10
df <- data.frame(year=plt[["year"]] + 1900,
month=plt[["mon"]] + 1, day=plt[["mday"]])
df
# year month day
#1 2010 1 2
#2 2010 2 3
#3 2010 9 10
And of course
julian(dates)
#[1] 14611 14643 14862
#attr(,"origin")
#[1] "1970-01-01"
To convert date (m/d/y format) into 3 separate columns,consider the df,
df <- data.frame(date = c("01-02-18", "02-20-18", "03-23-18"))
df
date
1 01-02-18
2 02-20-18
3 03-23-18
Convert to date format
df$date <- as.Date(df$date, format="%m-%d-%y")
df
date
1 2018-01-02
2 2018-02-20
3 2018-03-23
To get three seperate columns with year, month and date,
library(lubridate)
df$year <- year(ymd(df$date))
df$month <- month(ymd(df$date))
df$day <- day(ymd(df$date))
df
date year month day
1 2018-01-02 2018 1 2
2 2018-02-20 2018 2 20
3 2018-03-23 2018 3 23
Hope this helps.
Hi Gavin: another way [using your idea] is:
The data-frame we will use is oilstocks which contains a variety of variables related to the changes over time of the oil and gas stocks.
The variables are:
colnames(stocks)
"bpV" "bpO" "bpC" "bpMN" "bpMX" "emdate" "emV" "emO" "emC"
"emMN" "emMN.1" "chdate" "chV" "cbO" "chC" "chMN" "chMX"
One of the first things to do is change the emdate field, which is an integer vector, into a date vector.
realdate<-as.Date(emdate,format="%m/%d/%Y")
Next we want to split emdate column into three separate columns representing month, day and year using the idea supplied by you.
> dfdate <- data.frame(date=realdate)
year=as.numeric (format(realdate,"%Y"))
month=as.numeric (format(realdate,"%m"))
day=as.numeric (format(realdate,"%d"))
ls() will include the individual vectors, day, month, year and dfdate.
Now merge the dfdate, day, month, year into the original data-frame [stocks].
ostocks<-cbind(dfdate,day,month,year,stocks)
colnames(ostocks)
"date" "day" "month" "year" "bpV" "bpO" "bpC" "bpMN" "bpMX" "emdate" "emV" "emO" "emC" "emMN" "emMX" "chdate" "chV"
"cbO" "chC" "chMN" "chMX"
Similar results and I also have date, day, month, year as separate vectors outside of the df.