I have few raster images that represents bi-monthly data. I want to convert bi-monthly into monthly data taking averages of the two images.
There are total of 23 image (single band or single layer)If i stack the image using stack() from list.files, for some reason it reads 46 layers, but if i open all raster images using raster function individually and then stack it reads 23 layers only.
I open image individually and then stacked it, but while i convert the bi-monthly julian days it cannot read correctly after 4th month.
library(raster)
setwd("F:/LANDSAT-NDVI/testAverage")
x1<-raster("landsatNDVISC05SLC2000001.tif")
x2<-raster("landsatNDVISC05SLC2000017.tif")
x3<-raster("landsatNDVISC05SLC2000033.tif")
x4<-raster("landsatNDVISC05SLC2000049.tif")
x5<-raster("landsatNDVISC05SLC2000065.tif")
x6<-raster("landsatNDVISC05SLC2000081.tif")
x7<-raster("landsatNDVISC05SLC2000097.tif")
x8<-raster("landsatNDVISC05SLC2000113.tif")
x9<-raster("landsatNDVISC05SLC2000129.tif")
x10<-raster("landsatNDVISC05SLC2000145.tif")
x11<-raster("landsatNDVISC05SLC2000161.tif")
x12<-raster("landsatNDVISC05SLC2000177.tif")
x13<-raster("landsatNDVISC05SLC2000193.tif")
x14<-raster("landsatNDVISC05SLC2000209.tif")
x15<-raster("landsatNDVISC05SLC2000225.tif")
x16<-raster("landsatNDVISC05SLC2000241.tif")
x17<-raster("landsatNDVISC05SLC2000257.tif")
x18<-raster("landsatNDVISC05SLC2000273.tif")
x19<-raster("landsatNDVISC05SLC2000289.tif")
x20<-raster("landsatNDVISC05SLC2000305.tif")
x21<-raster("landsatNDVISC05SLC2000321.tif")
x22<-raster("landsatNDVISC05SLC2000337.tif")
x23<-raster("landsatNDVISC05SLC2000353.tif")
data<stack(x1,x2,x3,x4,x5,x6,x7,x8,x9,x10,x11,x12,x13,x14,x15,x16,x17,x18,x19,x20,x21,x22,x23)
julday <-c("landsatNDVISC05SLC2000001.tif","landsatNDVISC05SLC2000017.tif","landsatNDVISC05SLC2000033.tif",
"landsatNDVISC05SLC2000049.tif","landsatNDVISC05SLC2000065.tif","landsatNDVISC05SLC2000081.tif",
"landsatNDVISC05SLC2000097.tif","landsatNDVISC05SLC2000113.tif","landsatNDVISC05SLC2000129.tif",
"landsatNDVISC05SLC2000145.tif","landsatNDVISC05SLC2000161.tif","landsatNDVISC05SLC2000177.tif",
"landsatNDVISC05SLC2000193.tif","landsatNDVISC05SLC2000209.tif","landsatNDVISC05SLC2000225.tif",
"landsatNDVISC05SLC2000241.tif","landsatNDVISC05SLC2000257.tif","landsatNDVISC05SLC2000273.tif",
"landsatNDVISC05SLC2000289.tif","landsatNDVISC05SLC2000305.tif","landsatNDVISC05SLC2000321.tif",
"landsatNDVISC05SLC2000337.tif","landsatNDVISC05SLC2000353.tif")
julday <- as.numeric(substr(julday, 24,26)) #24 to 26th digit in the file name represents Julian days#
dates <- as.Date(julday, origin=as.Date("2000-01-01"))
combinddat <- setZ(data, dates)
monthly <- zApply(combinddat, by = format(dates,"%Y-%m"), fun = mean, na.rm = T)
The dates produced using that data is wrong; result is as follows
> dates
[1] "2000-01-02" "2000-01-18" "2000-02-03" "2000-02-19" "2000-03-06"
[6] "2000-03-22" "2000-04-07" "2000-01-14" "2000-01-30" "2000-02-15"
[11] "2000-03-02" "2000-03-18" "2000-04-03" "2000-01-10" "2000-01-26"
[16] "2000-02-11" "2000-02-27" "2000-03-14" "2000-03-30" "2000-01-06"
[21] "2000-01-22" "2000-02-07" "2000-02-23"
But i want the dates to be 12 months based on my julian days.
This does not answer your question but you probably could improve your code a lot with:
setwd("F:/LANDSAT-NDVI/testAverage")
library(raster)
f <- list.files(pattern="\\.tif$")
f <- sort(f)
data <- stack(f)
Related
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$Close)]; tmin
## [1] "2021-03-31"
data.ZIM[tmin]
## Open High Low Close
## 2021-03-31 24.75 24.99 24.15 24.34
Here I have a data frame which looks like following way,with the first column "POSIXct" and second "latitude"
> head(b)
sample_time latitude
3813442 2015-05-21 19:02:41 39.92770
3813483 2015-05-21 19:03:16 39.92770
3813485 2015-05-21 19:14:30 39.92433
3813515 2015-05-21 19:14:59 39.92469
3813550 2015-05-21 19:15:30 39.92520
3813585 2015-05-21 19:16:00 39.92585
Now,I want to plot latitude vs sample_time, with x axis representing 24 hours timestamp within a single day and group latitude by different days.
Any help will be appreciated!Many thanks.
First, you need to define "day", as opposed to the full time. Then you need to figure out what you mean by "group" ... let's just say you want to aggregate and take the daily mean. Third, you need to make the plot.
b$day <- round.Date(b[,"sample_time"], units="days")
b_agg <- aggregate(list(sample_time=b[,"sample_time"]), by=list(day=b[,"day"]), FUN=mean)
plot(b_agg)
Edit:
Just an additional thought, if you didn't want to aggregate, you could skip the second step, and change the third to plot(b[,"day"], b[,"latitude"]. Alternatively, you may even want something like boxplot(latitude~day, data=b).
Given a dataset of months, how do I calculate the "average" month, taking into account that months are circular?
months = c(1,1,1,2,3,5,7,9,11,12,12,12)
mean(months)
## [1] 6.333333
In this dummy example, the mean should be in January or December. I see that there are packages for circular statistics, but I'm not sure whether they suit my needs here.
I think
months <- c(1,1,1,2,3,5,7,9,11,12,12,12)
library("CircStats")
conv <- 2*pi/12 ## months -> radians
Now convert from months to radians, compute the circular mean, and convert back to months. I'm subtracting 1 here assuming that January is at "0 radians"/12 o'clock ...
(res1 <- circ.mean(conv*(months-1))/conv)
The result is -0.3457. You might want:
(res1 + 12) %% 12
which gives 11.65, i.e. partway through December (since we are still on the 0=January, 11=December scale)
I think this is right but haven't checked it too carefully.
For what it's worth, the CircStats::circ.mean function is very simple -- it might not be worth the overhead of loading the package if this is all you need:
function (x)
{
sinr <- sum(sin(x))
cosr <- sum(cos(x))
circmean <- atan2(sinr, cosr)
circmean
}
Incorporating #A.Webb's clever alternative from the comments:
m <- mean(exp(conv*(months-1)*1i))
12+Arg(m)/conv%%12 ## 'direction', i.e. average month
Mod(m) ## 'intensity'
How to select a value from time series corresponding needed date?
I create a monthly time series object with command:
producers.price <- ts(producers.price, start=2012+0/12, frequency=12)
Then I try to do next:
value <- producers.price[as.Date("01.2015", "%m.%Y")]
But this doesn't make that I want and value is equal
[1] NA
Instead of 10396.8212805739 if producers.price is:
producers.price <- structure(c(7481.52109434237, 6393.18959031561, 6416.63065650718,
5672.08354710121, 7606.24186413516, 5201.59247092013, 6488.18361474813,
8376.39182893415, 9199.50916585545, 8261.87133079494, 8293.8195347453,
8233.13630279516, 7883.17272003961, 7537.21001580393, 6566.60260432381,
7119.99345843556, 8086.40101607729, 9125.11104610046, 10134.0228610828,
10834.5732454454, 9410.35031874371, 9559.36933274129, 9952.38679679724,
10390.3628690951, 11134.8432864557, 11652.0075507499, 12626.9616107684,
12140.6698452193, 11336.8315981684, 10526.0309052316, 10632.1492109584,
8341.26367412737, 9338.95688558448, 9732.80173656971, 10724.5525831506,
11272.2273444623, 10396.8212805739, 10626.8428853062, 11701.0802817581,
NA), .Tsp = c(2012, 2015.25, 12), class = "ts")
So, I had/have a similar problem and was looking all over to solve it. My solution is not as great as I'd have wanted it to be, but it works. I tried it out with your data and it seems to give the right result.
Explanation
Turns out in R time series data is really stored as a sequence, starting at 1, and not with yout T. Eg. If you have a time series that starts in 1950 and ends in 1960 with each data at one year interval, the Y at 1950 will be ts[1] and Y at 1960 will be ts[11].
Based on this logic you will need to subtract the date from the start of the data and add 1 to get the value at that point.
This code in R gives you the result you expect.
producers.price[((as.yearmon("2015-01")- as.yearmon("2012-01"))*12)+1]
If you need help in the time calculations, check this answer
You will need the zoo and lubridate packages
Get the difference between dates in terms of weeks, months, quarters, and years
Hope it helps :)
1) window.ts
The window.ts function is used to subset a "ts" time series by a time window. The window command produces a time series with one data point and the [[1]] makes it a straight numeric value:
window(producers.price, start = 2015 + 0/12, end = 2015 + 0/12)[[1]]
## [1] 10396.82
2) zoo We can alternately convert it to zoo and subscript it by a yearmon class variable and then use [[1]] or coredata to convert it to a plain number or we can use window.zoo much as we did with window.ts :
library(zoo)
as.zoo(producers.price)[as.yearmon("2015-01")][[1]]
## [1] 10396.82
coredata(as.zoo(producers.price)[as.yearmon("2015-01")])
## [1] 10396.82
window(as.zoo(producers.price), 2015 + 0/12 )[[1]]
## [1] 10396.82
coredata(window(as.zoo(producers.price), 2015 + 0/12 ))
## [1] 10396.82
3) xts The four lines in (2) also work if library(zoo) is replaced with library(xts) and as.zoo is replaced with as.xts.
Looking for a simple command, one line and no library needed?
You might try this.
as.numeric(window(producers.price, 2015.1, 2015.2))
I know that I can get a date from an Excel based number (days since 1899-12-30) in the following way:
as.Date(41000, origin = "1899-12-30")
which will give me "2012-04-01". I want however the opposite. As a user I would like to input a date as a string and get the number of days since "1899-12-30".
Something along the lines
as.integer(as.Date('2014-03-01', origin="1899-12-30"))
which I hoped would result in 41000 and not in the R based days since 1970-01-01 which is 15431.
Maybe this is silly as I realize that I can add the days manually by writing something like:
as.integer(as.Date('2012-04-01')) + 25569
I just wondered if there is a function which does this?
I think you want difftime as in:
difftime(as.Date('2012-04-01'), as.Date("1899-12-30"))
## Time difference of 41000 days
Do it by hand, simpler and safer:
d0 <- as.Date('1899-12-30')
d1 <- as.Date('2014-10-28')
as.integer(d1 - d0)
##[1] 41940 # This is interpreted by Excel as '2014-10-28'
Of course, you can write a function to convert a R date to an Excel one:
convert_to_excel_date <- function(d) {
# Converts a R date value to an Excel date value
#
# Parameters:
# d: a R date object
d0 <- as.Date('1899-12-30')
return(as.integer(d - d0))
}
# Example:
# convert_to_excel_date(as.Date('2018-10-28'))