Both maxMATX and maxZIM return no observation, which I am very confused about.
Here is the code
library(tseries)
\#teries have all the Financial Data , hence we need to load it
data.ZIM\<- get.hist.quote("ZIM")
data.MATX\<- get.hist.quote("MATX")
data.ZIM\<-data.ZIM\[Sys.Date()-0:364\]
data.MATX\<-data.MATX\[Sys.Date()-0:364\]
head(data.ZIM)
head(data.MATX)
min(data.ZIM$Close)
max(data.ZIM$Close)
minZIM=data.ZIM\[data.ZIM$Close==24.34\]
maxZIM=data.ZIM\[data.ZIM$Close==88.62\]
data.ZIM\[data.ZIM$Close==88.62\]
minZIM
maxZIM
min(data.MATX$Close)
max(data.MATX$Close)
minMATX=data.MATX\[data.MATX$Close==60.07,\]
maxMATX=data.MATX\[data.MATX$Close==121.47,\]
minMATX
maxMATX
I was trying to extract the data from Tseries and I have faced difficulty when trying to print the row (or specifically I was trying to find the date of which the 52 weeks low and high was happening ).
Use which.min and which.max to find indexes of minimum and maximum close and use those to look up the time.
library(tseries)
data.ZIM <- get.hist.quote("ZIM", start = Sys.Date() - 364)
tmin <- time(data.ZIM)[which.min(data.ZIM$Close)]; tmin
## [1] "2021-03-31"
data.ZIM[tmin]
## Open High Low Close
## 2021-03-31 24.75 24.99 24.15 24.34
I have a dataset with a level of radiation per hour. I need to get the average level of radiation from the previous 6 hours. So for point c I need: mean(data$radiation[(c-7):(c-1)])
This would be a solution to my problem, if the dataset where to be complete (it is not, sometimes a few hours are missing) and I have no idea how to automate it without a for-loop (which I would like to avoid as there are 199056 entries)
I have the data in a data frame with radiation and time in a POSIXct format:
GLOBAL_radiation POSTIME
1383116 98 2016-06-10 18:00:00
1383118 55 2016-06-10 19:00:00
1383125 26 2016-06-10 20:00:00
1383130 6 2016-06-10 21:00:00
1383137 0 2016-06-10 22:00:00
1383142 0 2016-06-10 23:00:00
I've been cracking my brain on this for a while now, I do hope a function exists for this that I'm unaware of. Thanks in advance.
I'm not quite sure if this meet your needs, but I give it a try:
library(dplyr)
# define start value for date, which is assumed to be the
# last value in the time-vector
start <- dat$POSTIME[nrow(dat)]
# compute difference of all time points in relation
# to latest time point in data set
dat$hours <- as.vector(difftime(start, dat$POSTIME, units = "hours"))
# create a "grouping" vector, where all 6-hours-span-timepoints
# are grouped together
dat$grp <- as.integer(dat$hours / 6)
# group by 6-hours-span and compute mean for each
# 6-hours time-period
dat %>% group_by(grp) %>% summarise(mean(STRALING))
Using the zoo package (and help from SO) I have created a time series from the following:
z <- read.zoo("D:\\Futures Data\\BNVol3.csv", sep = ",", header = TRUE, index = 1:2,
tz="", format = "%d/%m/%Y %H:%M")
This holds data in the following format:(Intra-day from 07:00 to 20.50)
2012-10-01 14:50:00 2012-10-01 15:00:00 2012-10-01 15:10:00 2012-10-01 15:20:00
8638 9014 9402 9505
I want to "deseasonalize" the intra-day component of this data so that 1 day is considered a complete seasonal cycle. (I am using the day component because not all days will run from 07.00 to 20.50 due to bank holidays etc, but running from 07.00 to 20.50 is usually the standard. I assume that if i used the 84 intra-day points as 1 seasonal cycle then as some point the deseasonalizing will begin to get thrown off track)
I have tried to use the decompose method but this has not worked.
x <- Decompose(z)
Not sure "zoo" and decompose method are compatible but I thought "zoo" and "ts" were designed to be. Is there another way to do this?
Thanks in advance for any help.
I have a time series of continuous data measured at 10 minute intervals for a period of five months. For simplicity's sake, the data is available in two columns as follows:
Timestamp Temp.Diff
2/14/2011 19:00 -0.385
2/14/2011 19:10 -0.535
2/14/2011 19:20 -0.484
2/14/2011 19:30 -0.409
2/14/2011 19:40 -0.385
2/14/2011 19:50 -0.215
... And it goes on for the next five months. I have parsed the Timestamp column using as.POSIXct.
I want to select rows with certain times of the day, (e.g. from 12 noon to 3 PM), I would like either like to exclude the other hours of the day, OR just extract those 3 hours but still have the data flow sequentially (i.e. in a time series).
You seem to know the basic idea, but are just missing the details. As you mentioned, we just transform the Timestamps into POSIX objects then subset.
lubridate Solution
The easiest way is probably with lubridate. First load the package:
library(lubridate)
Next convert the timestamp:
##*m*onth *d*ay *y*ear _ *h*our *m*inute
d = mdy_hm(dd$Timestamp)
Then we select what we want. In this case, I want any dates after 7:30pm (regardless of day):
dd[hour(d) == 19 & minute(d) > 30 | hour(d) >= 20,]
Base R solution
First create an upper limit:
lower = strptime("2/14/2011 19:30","%m/%d/%Y %H:%M")
Next transform the Timestamps in POSIX objects:
d = strptime(dd$Timestamp, "%m/%d/%Y %H:%M")
Finally, a bit of dataframe subsetting:
dd[format(d,"%H:%M") > format(lower,"%H:%M"),]
Thanks to plannapus for this last part
Data for the above example:
dd = read.table(textConnection('Timestamp Temp.Diff
"2/14/2011 19:00" -0.385
"2/14/2011 19:10" -0.535
"2/14/2011 19:20" -0.484
"2/14/2011 19:30" -0.409
"2/14/2011 19:40" -0.385
"2/14/2011 19:50" -0.215'), header=TRUE)
You can do this with easily with the time-based subsetting in the xts package. Assuming your data.frame is named Data:
library(xts)
x <- xts(Data$Temp.Diff, Data$Timestamp)
y <- x["T12:00/T15:00"]
# you need the leading zero if the hour is a single digit
z <- x["T09:00/T12:00"]
I have a year's worth of hourly data in a data frame in R:
> str(df.MHwind_load) # compactly displays structure of data frame
'data.frame': 8760 obs. of 6 variables:
$ Date : Factor w/ 365 levels "2010-04-01","2010-04-02",..: 1 1 1 1 1 1 1 1 1 1 ...
$ Time..HRs. : int 1 2 3 4 5 6 7 8 9 10 ...
$ Hour.of.Year : int 1 2 3 4 5 6 7 8 9 10 ...
$ Wind.MW : int 375 492 483 476 486 512 421 396 456 453 ...
$ MSEDCL.Demand: int 13293 13140 12806 12891 13113 13802 14186 14104 14117 14462 ...
$ Net.Load : int 12918 12648 12323 12415 12627 13290 13765 13708 13661 14009 ...
While preserving the hourly structure, I would like to know how to extract
a particular month/group of months
the first day/first week etc of each month
all mondays, all tuesdays etc of the year
I have tried using "cut" without result and after looking online think that "lubridate" might be able to do so but haven't found suitable examples. I'd greatly appreciate help on this issue.
Edit: a sample of data in the data frame is below:
Date Hour.of.Year Wind.MW datetime
1 2010-04-01 1 375 2010-04-01 00:00:00
2 2010-04-01 2 492 2010-04-01 01:00:00
3 2010-04-01 3 483 2010-04-01 02:00:00
4 2010-04-01 4 476 2010-04-01 03:00:00
5 2010-04-01 5 486 2010-04-01 04:00:00
6 2010-04-01 6 512 2010-04-01 05:00:00
7 2010-04-01 7 421 2010-04-01 06:00:00
8 2010-04-01 8 396 2010-04-01 07:00:00
9 2010-04-01 9 456 2010-04-01 08:00:00
10 2010-04-01 10 453 2010-04-01 09:00:00
.. .. ... .......... ........
8758 2011-03-31 8758 302 2011-03-31 21:00:00
8759 2011-03-31 8759 378 2011-03-31 22:00:00
8760 2011-03-31 8760 356 2011-03-31 23:00:00
EDIT: Additional time-based operations I would like to perform on the same dataset
1. Perform hour-by-hour averaging for all data points i.e average of all values in the first hour of each day in the year. The output will be an "hourly profile" of the entire year (24 time points)
2. Perform the same for each week and each month i.e obtain 52 and 12 hourly profiles respectively
3. Do seasonal averages, for example for June to September
Convert the date to the format which lubridate understands and then use the functions month, mday, wday respectively.
Suppose you have a data.frame with the time stored in column Date, then the answer for your questions would be:
###dummy data.frame
df <- data.frame(Date=c("2012-01-01","2012-02-15","2012-03-01","2012-04-01"),a=1:4)
##1. Select rows for particular month
subset(df,month(Date)==1)
##2a. Select the first day of each month
subset(df,mday(Date)==1)
##2b. Select the first week of each month
##get the week numbers which have the first day of the month
wkd <- subset(week(df$Date),mday(df$Date)==1)
##select the weeks with particular numbers
subset(df,week(Date) %in% wkd)
##3. Select all mondays
subset(df,wday(Date)==1)
First switch to a Date representation: as.Date(df.MHwind_load$Date)
Then call weekdays on the date vector to get a new factor labelled with day of week
Then call months on the date vector to get a new factor labelled with name of month
Optionally create a years variable (see below).
Now subset the data frame using the relevant combination of these.
Step 2. gets an answer to your task 3. Steps 3. and 4. get you to task 1. Task 2 might require a line or two of R. Or just select rows corresponding to, say, all the Mondays in a month and call unique, or its alter-ego duplicated on the results.
To get you going...
newdf <- df.MHwind_load ## build an augmented data set
newdf$d <- as.Date(newdf$Date)
newdf$month <- months(newdf$d)
newdf$day <- weekdays(newdf$d)
## for some reason R has no years function. Here's one
years <- function(x){ format(as.Date(x), format = "%Y") }
newdf$year <- years(newdf$d)
# get observations from January to March of every year
subset(newdf, month %*% in c('January', 'February', 'March'))
# get all Monday observations
subset(newdf, day == 'Monday')
# get all Mondays in 1999
subset(newdf, day == 'Monday' & year == '1999')
# slightly fancier: _first_ Monday of each month
# get the first weeks
first.week.of.month <- !duplicated(cbind(newdf$month, newdf$day))
# now pull out the mondays
subset(newdf, first.monday.of.month & day=='Monday')
Since you're not asking about the time (hourly) part of your data, it is best to then store your data as a Date object. Otherwise, you might be interested in chron, which also has some convenience functions like you'll see below.
With respect to Conjugate Prior's answer, you should store your date data as a Date object. Since your data already follows the default format ('yyyy-mm-dd') you can just call as.Date on it. Otherwise, you would have to specify your string format. I would also use as.character on your factor to make sure you don't get errors inline. I know I've ran into problems with factors-into-Dates for that reason (possibly corrected in current version).
df.MHwind_load <- transform(df.MHwind_load, Date = as.Date(as.character(Date)))
Now you would do well to create wrapper functions that extract the information you desire. You could use transform like I did above to simply add those columns that represent months, days, years, etc, and then subset on them logically. Alternatively, you might do something like this:
getMonth <- function(x, mo) { # This function assumes w/in single year vector
isMonth <- month(x) %in% mo # Boolean of matching months
return(x[which(isMonth)] # Return vector of matching months
} # end function
Or, in short form
getMonth <- function(x, mo) x[month(x) %in% mo]
This is just a tradeoff between storing that information (transform frame) or having it processed when desired (use accessor methods).
A more complicated process is your need for, say, the first day of a month. This is not entirely difficult, though. Below is a function that will return all of those values, but it is rather simple to just subset a sorted vector of values for a given month and take their first one.
getFirstDay <- function(x, mo) {
isMonth <- months(x) %in% mo
x <- sort(x[isMonth]) # Look at only those in the desired month.
# Sort them by date. We only want the first day.
nFirsts <- rle(as.numeric(x))$len[1] # Returns length of 1st days
return(x[seq(nFirsts)])
} # end function
The easier alternative would be
getFirstDayOnly <- function(x, mo) {sort(x[months(x) %in% mo])[1]}
I haven't prototyped these, as you didn't provide any data samples, but this is the sort of approach that can help you get the information you desire. It is up to you to figure out how to put these into your work flow. For instance, say you want to get the first day for each month of a given year (assuming we're only looking at one year; you can create wrappers or pre-process your vector to a single year beforehand).
# Return a vector of first days for each month
df <- transform(df, date = as.Date(as.character(date)))
sapply(unique(months(df$date)), # Iterate through months in Dates
function(month) {getFirstDayOnly(df$date, month)})
The above could also be designed as a separate convenience function that uses the other accessor function. In this way, you create a series of direct but concise methods for getting pieces of the information you want. Then you simply pull them together to create very simple and easy to interpret functions that you can use in your scripts to get you precise what you desire in the most efficient manner.
You should be able to use the above examples to figure out how to prototype other wrappers for accessing the date information you require. If you need help on those, feel free to ask in a comment.