In R, my dataframe ("sampledata") looks like this:
The timestamp column is POSIXct, format: "2018-10-01 00:03:23"
The state column is Factor w/ 3 levels "AVAILABLE", "MUST_NOT_RUN", "MUST_RUN"
There are 6 unique device_id. The timestamps for each device are not the same, meaning data was not always collected at the same minute for each device. In some cases, there are multiple records per minute for the same device.
I want to transform the data into a visualization that shows distribution of "state" across a "typical" day. Ideally, something like this:
I've tried to count each occurrence of "state" grouped by timestamp minutes but failed (Error: can't sum factors). I've been trying to use ggplot and geom_area for the visualization, but believe I need to restructure my data before it will work. Very new to R (obviously). Happy to read any tutorials or links provided as background and appreciate any help you can provide. Thanks!
Other information that may/may not be helpful:
There are a handful of columns in the dataframe not shown.
223,446 entries between 10/2/18 - 11/8/18.
You can take the hours from the timestamps and then compute proportions of your states by hour:
library(ggplot2)
library(plyr)
#get hours from timestamp
obj$hour <- as.POSIXlt(obj$timestamp)$hour
#get average state proportions per hour
plot_obj <- ddply(obj,.(hour), #take data.frame "obj" and group by "hour"
function(x) with(x,
data.frame(100*table(state)/length(state))))
ggplot(plot_obj, aes(x=hour,y=Freq,fill=state)) +
geom_area()
Related
I hope we're all doing great
I have several decades of daily rainfall data from several monitoring stations. The data all beings at separate dates. I have combined them into a single data frame with the date in the first column, with the rainfall depth in the second column. I want to sort the variable 'Total' by the variable: 'Date and time' (please see the links below)
ms1 <- read.csv('ms1.csv')
ms2 <- read.csv('ms2.csv')
etc.etc
df <- merge(ms1, ms2 etc. etc, by = "Date and Time")
The problem is that the range of dates would differ for each monitoring station (csv file). There may also missing dates in a range. Is there a way around this?
Would I have to create a separate vector with the greatest possible date range? Or would it automatically detect the earliest start date from the imported data.
for monitoring station 1 (ms1)
for monitoring station 2 (ms2)
Note: the data continues to the current date
There is a data frame like this:
The first two columns in the df describe the start date (month and year) and the end date (month and year). Column names describe every single month and year of a certain time period.
I need a function/loop that insterts "1" or "0" in each cell - "1" when the date from given column name is within the period described by the two first columns, and "0" if not.
I would appreciate any help.
You want to do two different things. (a) create a dummy variable and (b) see if a particular date is in an interval.
Making a dummy variable is the easiest one, in base R you can use ifelse. For example in the iris data frame:
iris$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
I have two data frames: rainfall data collected daily and nitrate concentrations of water samples collected irregularly, approximately once a month. I would like to create a vector of values for each nitrate concentration that is the sum of the previous 5 days' rainfall. Basically, I need to match the nitrate date with the rain date, sum the previous 5 days' rainfall, then print the sum with the nitrate data.
I think I need to either make a function, a for loop, or use tapply to do this, but I don't know how. I'm not an expert at any of those, though I've used them in simple cases. I've searched for similar posts, but none get at this exactly. This one deals with summing by factor groups. This one deals with summing each possible pair of rows. This one deals with summing by aggregate.
Here are 2 example data frames:
# rainfall df
mm<- c(0,0,0,0,5, 0,0,2,0,0, 10,0,0,0,0)
date<- c(1:15)
rain <- data.frame(cbind(mm, date))
# b/c sums of rainfall depend on correct chronological order, make sure the data are in order by date.
rain[ do.call(order, list(rain$date)),]
# nitrate df
nconc <- c(15, 12, 14, 20, 8.5) # nitrate concentration
ndate<- c(6,8,11,13,14)
nitrate <- data.frame(cbind(nconc, ndate))
I would like to have a way of finding the matching rainfall date for each nitrate measurement, such as:
match(nitrate$date[i] %in% rain$date)
(Note: Will match work with as.Date dates?) And then sum the preceding 5 days' rainfall (not including the measurement date), such as:
sum(rain$mm[j-6:j-1]
And prints the sum in a new column in nitrate
print(nitrate$mm_sum[i])
To make sure it's clear what result I'm looking for, here's how to do the calculation 'by hand'. The first nitrate concentration was collected on day 6, so the sum of rainfall on days 1-5 is 5mm.
Many thanks in advance.
You were more or less there!
nitrate$prev_five_rainfall = NA
for (i in 1:length(nitrate$ndate)) {
day = nitrate$ndate[i]
nitrate$prev_five_rainfall[i] = sum(rain$mm[(day-6):(day-1)])
}
Step by step explanation:
Initialize empty result column:
nitrate$prev_five_rainfall = NA
For each line in the nitrate df: (i = 1,2,3,4,5)
for (i in 1:length(nitrate$ndate)) {
Grab the day we want final result for:
day = nitrate$ndate[i]
Take the rainfull sum and it put in in the results column
nitrate$prev_five_rainfall[i] = sum(rain$mm[(day-6):(day-1)])
Close the for loop :)
}
Disclaimer: This answer is basic in that:
It will break if nitrate's ndate < 6
It will be incorrect if some dates are missing in the rain dataframe
It will be slow on larger data
As you get more experience with R, you might use data manipulation packages like dplyr or data.table for these types of manipulations.
#nelsonauner's answer does all the heavy lifting. But one thing to note, in my actual data my dates are not numerical like they are in the example above, they are dates listed as MM/DD/YYYY with the appropriate as.Date(nitrate$date, "%m/%d/%Y").
I found that the for loop above gave me all zeros for nitrate$prev_five_rainfall and I suspected it was a problem with the dates.
So I changed my dates in both data sets to numerical using the difference in number of days between a common start date and the recorded date, so that the for loop would look for a matching number of days in each data frame rather than a date. First, make a column of the start date using rep_len() and format it:
nitrate$startdate <- rep_len("01/01/1980", nrow(nitrate))
nitrate$startdate <- as.Date(all$startdate, "%m/%d/%Y")
Then, calculate the difference using difftime():
nitrate$diffdays <- as.numeric(difftime(nitrate$date, nitrate$startdate, units="days"))
Do the same for the rain data frame. Finally, the for loop looks like this:
nitrate$prev_five_rainfall = NA
for (i in 1:length(nitrate$diffdays)) {
day = nitrate$diffdays[i]
nitrate$prev_five_rainfall[i] = sum(rain$mm[(day-5):(day-1)]) # 5 days
}
Let's say I have a data frame with lots of values under these headers:
df <- data.frame(c("Tid", "Value"))
#Tid.format = %Y-%m-%d %H:%M
Then I turn that data frame over to zoo, because I want to handle it as a time series:
library("zoo")
df <- zoo(df$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.