I'm trying to plot a line graph with R using the dataset that can be found here . I'm looking specifically at how to plot the number of cases in each region i.e. north east, north west etc against the period of time.
However, as the date is a period of a week rather than a standard date, how can I convert it to make the line graph actually possible? For example, right now it has the dates as 01/09/2020 - 07/09/2020. How can I use this for a line graph?
Sorry if my explanation isn't clear, here is a picture below.
I assume you're trying to plot a time series? You could just trim the dates to the beginning of the week and label the time axis as "Week beginning on date". You could do this with substr() in base r and keep the first 10 characters.
substr(data$column,1,10)
You may also want to format it as a date, easiest with the lubridate package, something like dmy() (day month year).
Here is the full code you would want:
library(tidyverse)
#Read in data
data <- read.csv("/Users/sabrinaxie/Downloads/covid19casesbysociodemographiccharacteristicengland1sep2020to10dec20213.csv")
#Modify data and remove extraneous top rows
data <- data %>%
rename(Period=Table.9..Weekly.estimates.of.age.standardised.COVID.19.case.rates..per.100.000.person.weeks..by.region..England..1.September.2020.to.6.December.20211.2.3) %>%
slice(3:n())
#Keep first 10 characters of Period column and assign to old column to replace
data$Period <- substr(data$Period,1,10)
#Parse as date
data$Period <- dmy(data$Period)
Related
I am trying to pull stock price data using tq_get in tidyquant, then want to plot the current price against the 52 week range. Here is an example of what I am looking to create.
Basically just a visual representation of where the stock is currently trading in relation to its 52 week range. Below is the code I have begun to load in the appropriate values for TSLA. First, I am wondering if it is possible to set the "from" and "to" dates so that they constantly update to be exactly one year ago and the current date, respectively? Second, is there a ggplot or another package that might be able to generate a similar plot? I've explored boxplots, but really I need something even more simple than that, as I really only need one axis. Thanks in advance!
X <- tq_get(c("^GSPC","TSLA"),get="stock.prices",from="2019-05-04", to="2020-05-04")
TSLA <- X %>% filter(symbol == "TSLA") %>% tk_xts()
chartSeries(TSLA)
TSLAlow <- min(TSLA$close)
TSLAlow
TSLAhigh <- max(TSLA$close)
TSLAhigh
TSLAclose <- tail(X$close, n=1)
TSLAclose
TSLArange <- tibble(TSLAlow, TSLAhigh, TSLAclose)
I'm very new to R, so this might seem straightforward. But I have a data frame with an original date column that has values that look like this: 4-02-91, 5-29-93 (i.e. m-d-y). I am trying to separate this column into 3, where months, days, and years are separate. Then I need to combine them again to this format 19910402, 19930529 - I need it this way in order to compare it to another dataset with similar dates.
Here is what I've been trying to do:
# Make DATE an actual date column
dataframe$DATE <- as.Date(used$DATE, format="%m-%d-%Y")
# This changes the original date column into something that looks like this: 1991-04-02, 1993-05-29
# Separate DATE into multiple columns
dataframe$year <- year(dataframe$DATE)
dataframe$month <- month(dataframe$DATE)
dataframe$day <- day(dataframe$DATE)
# Combine dates again to get string
dataframe$raster_date<-paste(dataframe$year, dataframe$month, dataframe$day, sep = "")
The last step looks great except where the months or days are single digits. It's coming out as 199142 and 1993529 instead of 19910402 and 19930529. How do I insert zeros when the month and day values are 1 digit?
Here, we can use sprintf instead of paste as the year, month, day from lubridate extracts those as numeric values and numeric class would drop the 0 padded as prefix. We add those prefix with 0s in sprintf
sprintf("%04d%02d%02d", dataframe$year, dataframe$month, dataframe$day)
There is a data frame like this:
The first two columns in the df describe the start date (month and year) and the end date (month and year). Column names describe every single month and year of a certain time period.
I need a function/loop that insterts "1" or "0" in each cell - "1" when the date from given column name is within the period described by the two first columns, and "0" if not.
I would appreciate any help.
You want to do two different things. (a) create a dummy variable and (b) see if a particular date is in an interval.
Making a dummy variable is the easiest one, in base R you can use ifelse. For example in the iris data frame:
iris$dummy <- ifelse(iris$Sepal.Width > 2.5, 1, 0)
Now working with dates is more complicated. In this answer we will use the library lubridate. First you need to convert all those dates to a format 'Month Year' to something that R can understand. For example for February you could do:
new_format_february_2016 <- interval(ymd('2016-02-01'), ymd('2016-03-01') - dseconds(1))
#[1] 2016-02-01 UTC--2016-02-29 23:59:59 UTC
This is February, the interval of time from the 1 of February to one second before the 1 of March. You can do the same with your start date column and you end date column.
To compare two intevals of time (so, to see if a particular month fall into your other intervals) you can do:
int_overlaps(new_format_february_2016, other_interval)
If this returns true, the two intervals (one particular month and another one) overlaps. This is not the same as one being inside another, but in your case it will work. Using this you can iterate over different columns and rows and build your dummy variable.
But before doing so, I would recommend to clean your data, as your current format is complicate to work with. To get all the power that vector types in R provides ideally you would want to have one row per observation and one variable per column. This does not seem to be the case with your data frame. Take a look to the chapter 'Tidy data' of 'R for Data Science' specially the spreading and gathering subsection:
Tidy data
Let's say I have a data frame with lots of values under these headers:
df <- data.frame(c("Tid", "Value"))
#Tid.format = %Y-%m-%d %H:%M
Then I turn that data frame over to zoo, because I want to handle it as a time series:
library("zoo")
df <- zoo(df$Value, df$Tid)
Now I want to produce a smooth scatter plot over which time of day each measurement was taken (i.e. discard date information and only keep time) which supposedly should be done something like this: https://stat.ethz.ch/pipermail/r-help/2009-March/191302.html
But it seems the time() function doesn't produce any time at all; instead it just produces a number sequence. Whatever I do from that link, I can't get a scatter plot of values over an average day. The data.frame code that actually does work (without using zoo time series) looks like this (i.e. extracting the hour from the time and converting it to numeric):
smoothScatter(data.frame(as.numeric(format(df$Tid,"%H")),df$Value)
Another thing I want to do is produce a density plot of how many measurements I have per hour. I have plotted on hours using a regular data.frame with no problems, so the data I have is fine. But when I try to do it using zoo then I either get errors or I get the wrong results when trying what I have found through Google.
I did manage to get something plotted through this line:
plot(density(as.numeric(trunc(time(df),"01:00:00"))))
But it is not correct. It seems again that it is just producing a sequence from 1 to 217, where I wanted it to be truncating any date information and just keep the time rounded off to hours.
I am able to plot this:
plot(density(df))
Which produces a density plot of the Values. But I want a density plot over how many values were recorded per hour of the day.
So, if someone could please help me sort this out, that would be great. In short, what I want to do is:
1) smoothScatter(x-axis: time of day (0-24), y-axis: value)
2) plot(density(x-axis: time of day (0-24)))
EDIT:
library("zoo")
df <- data.frame(Tid=strptime(c("2011-01-14 12:00:00","2011-01-31 07:00:00","2011-02-05 09:36:00","2011-02-27 10:19:00"),"%Y-%m-%d %H:%M"),Values=c(50,52,51,52))
df <- zoo(df$Values,df$Tid)
summary(df)
df.hr <- aggregate(df, trunc(df, "hours"), mean)
summary(df.hr)
png("temp.png")
plot(df.hr)
dev.off()
This code is some actual values that I have. I would have expected the plot of "df.hr" to be an hourly average, but instead I get some weird new index that is not time at all...
There are three problems with the aggregate statement in the question:
We wish to truncate the times not df.
trunc.POSIXt unfortunately returns a POSIXlt result so it needs to be converted back to POSIXct
It seems you did not intend to truncate to the hour in the first place but wanted to extract the hours.
To address the first two points the aggregate statement needs to be changed to:
tt <- as.POSIXct(trunc(time(df), "hours"))
aggregate(df, tt, mean)
but to address the last point it needs to be changed entirely to
tt <- as.POSIXlt(time(df))$hour
aggregate(df, tt, mean)
I have a time series dataset for several meteorological variables. The time data is logged in three separate columns:
Year (e.g. 2012)
Day of year (e.g. 261 representing 17-September in a Leap Year)
Hrs:Mins (e.g. 1610)
Is there a way I can merge the three columns to create a single timestamp in R? I'm not very familiar with how R deals with the Day of Year variable.
Thanks for any help with this!
It looks like the timeDate package can handle gregorian time frames. I haven't used it personally but it looks straightforward. There is a shift argument in some methods that allow you to set the offset from your data.
http://cran.r-project.org/web/packages/timeDate/timeDate.pdf
Because you mentioned it, I thought I'd show the actual code to merge together separate columns. When you have the values you need in separate columns you can use paste to bring them together and lubridate::mdy to parse them.
library(lubridate)
col.month <- "Jan"
col.year <- "2012"
col.day <- "23"
date <- mdy(paste(col.month, col.day, col.year, sep = "-"))
Lubridate is a great package, here's the official page: https://github.com/hadley/lubridate
And here is a nice set of examples: http://www.r-statistics.com/2012/03/do-more-with-dates-and-times-in-r-with-lubridate-1-1-0/
You should get quite far using ISOdatetime. This function takes vectors of year, day, hour, and minute as input and outputs an POSIXct object which represents time. You just have to split the third column into two separate hour minute columns and you can use the function.