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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")
user2952921
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    Some reproducible data would help you get an answer quicker and more verifiable. – cdeterman May 28 '15 at 15:05
  • It looks like your are subsetting values that do not exist. Values are prices, in this case your data seems to be a series of prices of class ts and so with attributes tsp. Why don't you create a data frame with dates in one column and prices in another? subsetting would be much easier in my opinion. – SabDeM May 28 '15 at 15:43
  • as.Date does not support dates that consist of only a month and a year, thus `as.Date("01.2015", "%m.%Y")` returns a NULL value. This post may help you: http://stackoverflow.com/questions/6242955/converting-year-and-month-to-a-date-in-r – C_Z_ May 28 '15 at 15:51
  • SabDeM, you absolutely right that this will be easier with dataframe. But for my program it seems better to tackle the task without dataframe. – user2952921 May 28 '15 at 15:52

3 Answers3

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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 :)

Community
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codeCruncher
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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.

G. Grothendieck
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Looking for a simple command, one line and no library needed? You might try this.

as.numeric(window(producers.price, 2015.1, 2015.2))
Henrique L.R.
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