Handling time with zoo in R - r

I'm trying to load time series in R with the 'zoo' library.
The observations I have varying precision. Some have the day/month/year, others only month and year, and others year:
02/10/1915
1917
07/1917
07/1918
30/08/2018
Subsequently, I need to aggregate the rows by year, year and month.
The basic R as.Date function doesn't handle that.
How can I model this data with zoo?
Thanks,
Mulone

We use the test data formed from the index data in the question followed by a number:
# test data
Lines <- "02/10/1915 1
1917 2
07/1917 3
07/1918 4
30/08/2018 5"
yearly aggregation
library(zoo)
to.year <- function(x) as.numeric(sub(".*/", "", as.character(x)))
read.zoo(text = Lines, FUN = to.year, aggregate = mean)
The last line returns:
1915 1917 1918 2018
1.0 2.5 4.0 5.0
year/month aggregation
Since year/month aggregation of data with no months makes no sense we first drop the year only data and aggregate the rest:
DF <- read.table(text = Lines, as.is = TRUE)
# remove year-only records. DF.ym has at least year and month.
yr <- suppressWarnings(as.numeric(DF[[1]]))
DF.ym <- DF[is.na(yr), ]
# remove day, if present, and convert to yearmon.
to.yearmon <- function(x) as.yearmon( sub("\\d{1,2}/(\\d{1,2}/)", "\\1", x), "%m/%Y" )
read.zoo(DF.ym, FUN = to.yearmon, aggregate = mean)
The last line gives:
Oct 1915 Jul 1917 Jul 1918 Aug 2018
1 3 4 5
UPDATE: simplifications

Related

How to aggregate using water years (oct 1 2008- sept 31 2009)

I have data measuring precipitation daily using R. My dates are in format 2008-01-01 and range for 10 years. I am trying to aggregate from 2008-10-01 to 2009-09-31 but I am not sure how. Is there a way in aggregate to set a start date of aggregation and group.
My current code is
data<- aggregate(data$total_snow_cm, by=list(data$year), FUN = 'sum')
but this output gives me a sum total of the snowfall for each year from jan - dec but I want it to include oct / 08 to sept / 09.
Assuming your data are in long format, I'd do something like this:
library(tidyverse)
#make sure R knows your dates are dates - you mention they're 'yyyy-mm-dd', so
yourdataframe <- yourdataframe %>%
mutate(yourcolumnforprecipdate = ymd(yourcolumnforprecipdate)
#in this script or another, define a water year function
water_year <- function(date) {
ifelse(month(date) < 10, year(date), year(date)+1)}
#new wateryear column for your data, using your new function
yourdataframe <- yourdataframe %>%
mutate(wateryear = water_year(yourcolumnforprecipdate)
#now group by water year (and location if there's more than one)
#and sum and create new data.frame
wy_sums <- yourdataframe %>% group_by(locationcolumn, wateryear) %>%
summarize(wy_totalprecip = sum(dailyprecip))
For more info, read up on the tidyverse 's great sublibrary called lubridate -
where the ymd() function is from. There are others like ymd_hms(). mutate() is from the tidyverse's dplyr libary. Both libraries are extremely useful!
I'd like to give the actual answer to the question, where the aggregate() way was asked.
You may use with() to wrap the data specification around aggregate(). In the with() you can define date intervals as you can with numbers.
df1.agg <- with(df1[as.Date("2008-10-01") <= df1$year & df1$year <= as.Date("2009-09-30"), ],
aggregate(total_snow_cm, by=list(year), FUN=sum))
Another way is to use aggregate()'s formula interface, where data and, hence, also the interval can be specified inside the aggregate() call.
df1.agg <- aggregate(total_snow_cm ~ year,
data=df1[as.Date("2008-10-01") <= df1$year &
df1$year <= as.Date("2009-09-30"), ], FUN=sum)
Result
head(df1.agg)
# year total_snow_cm
# 1 2008-10-01 171
# 2 2008-10-02 226
# 3 2008-10-03 182
# 4 2008-10-04 129
# 5 2008-10-05 135
# 6 2008-10-06 222
Data
set.seed(42)
df1 <- data.frame(total_snow_cm=sample(120:240, 4018, replace=TRUE),
year=seq(as.Date("2000-01-01"),as.Date("2010-12-31"), by="day"))

convert irregular 6hourly data to daily accumulated using R

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))

hydrological year time series

Currently I am working on a river discharge data analysis. I have the daily discharge record from 1935 to now. I want to extract the annual maximum discharge for each hydrolocial year (start from 01/11 to next year 31/10). However, I found that the hydroTSM package can only deal with the natural year. I tried to use the "zoo" package, but I found it's difficult to compute, as each year have different days. Does anyone have some idea? Thanks.
the data looks like:
01-11-1935 663
02-11-1935 596
03-11-1935 450
04-11-1935 381
05-11-1935 354
06-11-1935 312
my code:
mydata<-read.table("discharge")
colnames(mydata) <- c("date","discharge")
library(zoo)
z<-zooreg(mydata[,2],start=as.Date("1935-11-1"))
mydta$date <- as.POSIXct(dat$date)
q.month<-daily2monthly(z,FUN=max,na.rm = TRUE,date.fmt = "%Y-%m-%d",out.fmt="numeric")
q.month.plain=coredata(q.month)
z.month<-zooreg(q.month.plain,start=1,frequency=12)
With dates stored in a vector of class Date, you can just use cut() and tapply(), like this:
## Example data
df <- data.frame(date = seq(as.Date("1935-01-01"), length = 100, by = "week"),
flow = (runif(n = 100, min = 0, max = 1000)))
## Use vector of November 1st dates to cut data into hydro-years
breaks <- seq(as.Date("1934-11-01"), length=4, by="year")
df$hydroYear <- cut(df$date, breaks, labels=1935:1937)
## Find the maximum flow in each hydro-year
with(df, tapply(flow, hydroYear, max))
# 1935 1936 1937
# 984.7327 951.0440 727.4210
## Note: whenever using `cut()`, I take care to double-check that
## I've got the cuts exactly right
cut(as.Date(c("1935-10-31", "1935-11-01")), breaks, labels=1935:1937)
# [1] 1935 1936
# Levels: 1935 1936 1937
Here is a one-liner to do that.
First convert the dates to "yearmon" class. This class represents a year month as the sum of a year as the integer part and a month as the fractional part (Jan = 0, Feb = 1/12, etc.). Add 2/12 to shift November to January and then truncate to give just the years. Aggregate over those. Although the test data we used starts at the beginning of the hydro year this solution works even if the data does not start on the beginning of the hydro year.
# test data
library(zoo)
z <- zooreg(1:1000, as.Date("2000-11-01")) # test input
aggregate(z, as.integer(as.yearmon(time(z)) + 2/12), max)
This gives:
2001 2002 2003
365 730 1000
Try the xts package, which works together with zoo:
require(zoo)
require(xts)
dates = seq(Sys.Date(), by = 'day', length = 365 * 3)
y = cumsum(rnorm(365 * 3))
serie = zoo(y, dates)
# if you need to specify `start` and `end`
# serie = window(serie, start = "2015-06-01")
# xts function
apply.yearly(serie, FUN = max)

How do I calculate a monthly rate of change from a daily time series in R?

I'm beginning to get my feet wet with R, and I'm brand new to time series concepts. Can anyone point me in the right direction to calculate a monthly % change, based on a daily data point? I want the change between the first and last data points of each month. For example:
tseries data:
1/1/2000 10.00
...
1/31/2000 10.10
2/1/2000 10.20
...
2/28/2000 11.00
I'm looking for a return data frame of the form:
1/31/2000 .01
2/28/2000 .0784
Ideally, I'd be able to calculate from the endpoint of the prior month to the endpoint of current month, but I'm supposing partitioning by month is easier as a starting point. I'm looking at packages zoo and xts, but am still stuck. Any takers? Thanks...
Here's one way to do it using plyr and ddply.
I use ddply sequentially, first to get the first and last rows of each month, and again to calculate the monthlyReturn.
(Perhaps using xts or zoo might be easier, I am not sure.)
#Using plyr and the data in df
df$Date <- as.POSIXlt(as.Date(df$Date, "%m/%d/%Y"))
df$Month <- (df$Date$mon + 1) #0 = January
sdf <- df[,-1] #drop the Date Column, ddply doesn't like it
library("plyr")
#this function is called with 2 row data frames
monthlyReturn<- function(df) {
(df$Value[2] - df$Value[1])/(df$Value[1])
}
adf <- ddply(sdf, .(Month), function(x) x[c(1, nrow(x)), ]) #get first and last values for each Month
mon.returns <- ddply(adf, .(Month), monthlyReturn)
Here's the data I used to test it out:
> df
Date Value
1 1/1/2000 10.0
2 1/31/2000 10.1
3 2/1/2000 10.2
4 2/28/2000 11.0
5 3/1/2000 10.0
6 3/31/2000 24.1
7 5/10/2000 510.0
8 5/22/2000 522.0
9 6/04/2000 604.0
10 7/03/2000 10.1
11 7/30/2000 7.2
12 12/28/2000 11.0
13 12/30/2000 3.0
> mon.returns
Month V1
1 1 0.01000000
2 2 0.07843137
3 3 1.41000000
4 5 0.02352941
5 6 0.00000000
6 7 -0.28712871
7 12 -0.72727273
Hope that helps.
Here is another way to do this(using the quantmod package):
This calculates the monthly return from the daily price of AAPL.
*library(quantmod) # load the quantmod package
getSymbols("AAPL") # download daily price for stock AAPL
monthlyReturn = periodReturn(AAPL,period="monthly")
monthlyReturn2014 = periodReturn(AAPL,period="monthly",subset='2014:') # for 2014*
This is a pretty old thread, but for reference, here comes a data.table solution using same data as #Ram:
structure(list(Date = structure(c(10957, 10987, 10988, 11015, 11017, 11047, 11087, 11099, 11112, 11141, 11168, 11319, 11321), class = "Date"), Value = c(10, 10.1, 10.2, 11, 10, 24.1, 510, 522, 604, 10.1, 7.2, 11, 3)), .Names = c("Date", "Value"), row.names = c(NA, -13L), class = c("data.table", "data.frame"), .internal.selfref = <pointer: 0x00000000001b0788>)
It's essentially a one-liner that uses the data.table::month function:
library(data.table)
setDT(df)[ , diff(Value) / Value[1], by= .(month(Date))]
This will produce the change, relative to the first recorded day in each month. If the change relative to the last day is preferred, then the expression in the middle should be changed to diff(Value) / Vale[2].
1) no packages Try this:
DF <- read.table(text = Lines)
fmt <- "%m/%d/%Y"
ym <- format(as.Date(DF$V1, format = fmt), "%Y-%m")
ret <- function(x) diff(range(x))/x[1]
ag <- aggregate(V2 ~ ym, DF, ret)
giving:
> ag
ym V2
1 2000-01 0.01000000
2 2000-02 0.07843137
We could convert this to "ts" class, if desired. Assuming no missing months:
ts(ag$V2, start = 2000, freq = 12)
giving:
Jan Feb
2000 0.01000000 0.07843137
2) It's a bit easier if you use the zoo or xts time series packages. fmt and ret are from above:
library(zoo)
z <- read.zoo(text = Lines, format = fmt)
z.ret <- aggregate(z, as.yearmon, ret)
giving:
> z.ret
Jan 2000 Feb 2000
0.01000000 0.07843137
If you already have a data.frame DF then the read.zoo statement could be replaced with z <- read.zoo(DF, format = fmt) or omit the format arg if the first column is of "Date" class.
If "ts" class were desired then use as.ts(z.ret)
Note: The input Lines is:
Lines <- "1/1/2000 10.00
1/31/2000 10.10
2/1/2000 10.20
2/28/2000 11.00"
The ROC function in the TTR package will do this. You can use to.monthly or endpoints() (From daily time series to weekly time series in R xts object) first if you will only be looking at monthly behaviour.
library(TTR)
# data.monthly <- to.monthly( data, indexAt='periodEnd' ) # if OHLC data
# OR
data.monthly <- data[ endpoints(data, on="months", k=1), ]
data.roc <- ROC(data.monthly, n = 1, type = "discrete")

Aggregating weekly (7 day) data to monthly in R

I have data measured over a 7 day period. Part of the data looks as follows:
start wk end wk X1
2/1/2004 2/7/2004 89
2/8/2004 2/14/2004 65
2/15/2004 2/21/2004 64
2/22/2004 2/28/2004 95
2/29/2004 3/6/2004 79
3/7/2004 3/13/2004 79
I want to convert this weekly (7 day) data into monthly data using weighted averages of X1. Notice that some of the 7 day X1 data will overlap from one month to the other (X1=79 for the period 2/29 to 3/6 of 2004).
Specifically I would obtain the February 2004 monthly data (say, Y1) the following way
(7*89 + 7*65 + 7*64 + 7*95 + 1*79)/29 = 78.27
Does R have a function that will properly do this? (to.monthly in the xts library DOES NOT do what I need) If, not what is the best way to do this in R?
Convert the data to daily data and then aggregate:
Lines <- "start end X1
2/1/2004 2/7/2004 89
2/8/2004 2/14/2004 65
2/15/2004 2/21/2004 64
2/22/2004 2/28/2004 95
2/29/2004 3/6/2004 79
3/7/2004 3/13/2004 79
"
library(zoo)
# read data into data frame DF
DF <- read.table(text = Lines, header = TRUE)
# convert date columns to "Date" class
fmt <- "%m/%d/%Y"
DF <- transform(DF, start = as.Date(start, fmt), end = as.Date(end, fmt))
# convert to daily zoo series
to.day <- function(i) with(DF, zoo(X1[i], seq(start[i], end[i], "day")))
z.day <- do.call(c, lapply(1:nrow(DF), to.day))
# aggregate by month
aggregate(z.day, as.yearmon, mean)
The last line gives:
Feb 2004 Mar 2004
78.27586 79.00000
If you are willing to get rid of "end week" from your DF, apply.monthly will work like a charm.
DF.xts <- xts(DF$X1, order.by=DF$start_wk)
DF.xts.monthly <- apply.monthly(DF.xts, "sum")
Then you can always recreate end dates if you absolutely need them by adding 30.

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