This question already has an answer here:
Subsetting winter (Dez, Jan, Feb) from daily time series (zoo)
(1 answer)
Closed 7 years ago.
My zoo object
I Would like to create a subset of values (these are discharge values) containing only December flow values.
Thank you!
We can extract the 'months' from the index with format, get a logical index by comparing with 'Dec', and use that to subset the zoo object.
z1[format(index(z1), '%b')=='Dec']
#1938-12-03 1938-12-10 1938-12-17 1938-12-24 1938-12-31
# 49 50 51 52 53
If we convert to xts object, .indexmon from the xts package can be also used. The .indexmon starts from 0, so December is 11.
library(xts)
z1[.indexmon(as.xts(z1))==11]
Other options from the comments are using grep on the index to get the numeric index and subset (from #Pierre Lafortune)
z1[grep("-12-",index(z1))]
Or with subset/month option (from # G. Grothendieck)
subset(z1, months(time(z1)) == "December")
data
library(zoo)
z1 <- zoo(1:100, order.by = seq(as.Date('1938-01-01'),
length.out=100, by = '1 week'))
Related
I am working with data collected monthly. In my dataset, there are some months where no data was collected and thus, there is no entry in my data. I have previously used bfastts for similar occurrences when data was collected daily, so that I may have NA values in my data. How may I do the same for monthly data, using bfastts or some other function?
eg. below if needed
2006-06-01 2.260121
2006-07-01 2.306800
2006-08-01 2.246624
2006-09-01 1.724565
2006-11-01 1.630561
2007-05-01 2.228918
2007-06-01 2.228918
2007-07-01 2.22891
I wish to have NA fields for December to March.
The question did not specify what class of object is desired but here are three. zoo supports an irregularly spaced index so it does not need to insert NA's but ts does not and converting from zoo to ts automatically inserts NA's. Convert the ts object back to zoo again or to a data frame to get a zoo or data frame object with NA's.
The zoo and data frame objects use yearmon class for the index which internally represents year/month as year + fraction where fraction is 0, 1/12, ..., 11/12 for Jan, Feb, ..., Dec and displays in meaningful form. as.Date can be used to convert yearmon objects to Date objects although in this case yearmon probably makes more sense since it directly represents year and month without day.
If you want to go in the other direction and remove NA's use na.omit(z_na) or na.omit(DF_na) .
library(zoo)
# zoo object - no NA's
z <- read.zoo(DF, FUN = as.yearmon)
# ts object with NA's
tt <- as.ts(z)
# zoo object with NA's
z_na <- as.zoo(tt)
# data.frame with NA's
DF_na <- fortify.zoo(tt)
Note
Lines <- "2006-06-01 2.260121
2006-07-01 2.306800
2006-08-01 2.246624
2006-09-01 1.724565
2006-11-01 1.630561
2007-05-01 2.228918
2007-06-01 2.228918
2007-07-01 2.22891"
DF <- read.table(text = Lines)
I have a data frame which is a time series of meteorological measurement with monthly resolution from 1961 till 2018. I am interested in the variable that measures the monthly average temperature since I need the multi-annual average temperature for the summers.
To do this I must filter from the "DateVaraible" column the fifth and sixth digit, which are the month.
The values in time column are formatted like this
"19610701". So I need the 07(Juli) after 1961.
I start coding for 1 month for other purposes, so I did not try anything worth to mention. I guess that .grepl could do the work, but I do not know how the "matching" operator works.
So I started with this code that works.
summersmonth<- Df[DateVariable %like% "19610101" I DateVariable %like% "19610201"]
I am expecting a code like this
summermonths <- Df[DateVariable %like% "**06**" I DateVariable%like% "**07**..]
So that all entries with month digit from 06 to 09 are saved in the new dataframe summermonths.
Thanks in advance for any reply or feedback regarding my question.
Update
Thank to your answers I got the first part, which is to convert the variable in a as.date with the format "month"(Class=char)
Now I need to select months from Juni to September .
A horrible way to get the result I wanted is to do several subset and a rbind afterward.
Sommer1<-subset(Df, MonthVar == "Mai")
Sommer2<-subset(Df, MonthVar == "Juli")
Sommer3<-subset(Df, MonthVar == "September")
SummerTotal<-rbind(Sommer1,Sommer2,Sommer3)
I would be very glad to see this written in a tidy way.
Update 2 - Solution
Here is the tidy way, as here Using multiple criteria in subset function and logical operators
Veg_Seas<-subset(Df, subset = MonthVar %in% c("Mai","Juni","Juli","August","September"))
You can convert your date variable as date (format) and take the month:
allmonths <- month(as.Date(Df$DateVariable, format="%Y%m%d"))
Note that of your column has been originally imported as factor you need to convert it to character first:
allmonths <- month(as.Date(as.character(Df$DateVariable), format="%Y%m%d"))
Then you can check whether it is a summermonth:
summersmonth <- Df[allmonths %in% 6:9, ]
Example:
as.Date("20190702", format="%Y%m%d")
[1] "2019-07-02"
month(as.Date("20190702", format="%Y%m%d"))
[1] 7
We can use anydate from anytime to convert to Date class and then extract the month
library(anytime)
month(anydate(as.character(Df$DateVariable)))
This question already has answers here:
Convert four digit year values to class Date
(5 answers)
Closed 5 years ago.
OK, this should be really simple but I'm not 'getting it.' I have a data frame with a column "Year" that I want to convert to a time series, but the format is tripping me up. How do I convert the "Year" value to a date, with the actual date being the end of each respective year (e.g. 2015 -> December 31st 2015)?
Year Production
1 1900 38400000
2 1901 43400000
3 1902 49000000
4 1903 44100000
5 1904 49800000
Goal is to get this to a time series data frame. (e.g. xts)
It is not quite the same as a previous question that converted a vector of years to dates. "Convert four digit year values to date type". Goal is to index the data by date, converting it to xts or similar object.
Edited:
This was the final solution:
df <- xts(x = df_original, order.by = as.Date(paste0(df_original[,1], "-12-31")))
whereby the "[,1]" indicates the first column of the original data frame.
If you want each full date to be 31 December, you could use paste along with as.Date to cast to a date:
df$date <- as.Date(paste0(df$Year, "-12-31"))
In addition to Tim Biegeleisen's answer, I will just add another way
df$final_date <- as.Date(ISOdate(df$Year, 12, 31))
I'm creating an xts object with a weekly (7 day) frequency to use in forecasting. However, even when using the frequency=7 argument in the xts call, the resulting xts object has a frequency of 1.
Here's an example with random data:
> values <- rnorm(364, 10)
> days <- seq.Date(from=as.Date("2014-01-01"), to=as.Date("2014-12-30"), by='days')
> x <- xts(values, order.by=days, frequency=7)
> frequency(x)
[1] 1
I have also tried, after using the above code, frequency(x) <- 7. However, this changes the class of x to only zooreg and zoo, losing the xts class and messing with the time stamp formats.
Does xts automatically choose a frequency based on analyzing the data in some way? If so, how can you override this to set a specific frequency for forecasting purposes (in this case, passing a seasonal time series to ets from the forecast package)?
I understand that xts may not allow frequencies that don't make sense, but a frequency of 7 with daily time stamps seems pretty logical.
Consecutive Date class dates always have a frequency of 1 since consecutive dates are 1 apart. Use ts or zooreg to get a frequency of 7:
tt <- ts(values, frequency = 7)
library(zoo)
zr <- as.zooreg(tt)
# or
zr <- zooreg(values, frequency = 7)
These will create a series whose times are 1, 1+1/7, 1+2/7, ...
If we have some index values of zr
zrdates <- index(zr)[5:12]
we can recover the dates from zrdates like this:
days[match(zrdates, index(zr))]
As pointed out in the comments xts does not support this type of series.
I have a dataframe in R where one of the columns is a timestamp such as 2014-03-08 00:20:34
I would like to convert all of these timestamps in the dataframe to the day of the year. For example, March 29, 2014 would be 88.
This will do the conversion if 'df' is my dataframe and 'updated' is my timestamp column:
strftime(df$updated, format = "%j")
How can I apply this across the entire dataframe?
Thanks,
as.numeric(as.Date("2014-03-08 00:20:34")-as.Date("2014-01-01"))+1
#[1] 67
So just use an assignment to whatever name column you want and replace df$updated for the character value. The other way would be with POSIXlt objects which are really multi-element named lists:
as.POSIXlt("2014-03-08 00:20:34")$yday
[1] 66
as.POSIXlt("2014-03-08 00:20:34")$yday +1
[1] 67 #b ecause `yday`s` are zero referenced unlike the rest of R