Is it possible to extract data points from moving average data?

In other words, if a set of data only has simple moving averages of the previous 30 points, is it possible to extract the original data points?

If so, how?

**Answer**

+1 to fabee’s answer, which is complete. Just a note to translate it into R, based on the packages that I’ve found to do the operations at hand. In my case, I had data that is NOAA temperature forecasts on a three-month basis: Jan-Feb-Mar, Feb-Mar-Apr, Mar-Apr-May, etc, and I wanted to break it out into (approximate) monthly values, assuming that each three-month period’s temperature is essentially an average.

```
library (Matrix)
library (matrixcalc)
# Feb-Mar-Apr through Nov-Dec-Jan temperature forecasts:
qtemps <- c(46.0, 56.4, 65.8, 73.4, 77.4, 76.2, 69.5, 60.1, 49.5, 41.2)
# Thus I need a 10x12 matrix, which is a band matrix but with the first
# and last rows removed so that each row contains 3 1's, for three months.
# Yeah, the as.matrix and all is a bit obfuscated, but the results of
# band are not what svd.inverse wants.
a <- as.matrix (band (matrix (1, nrow=12, ncol=12), -1, 1)[-c(1, 12),])
ai <- svd.inverse (a)
mtemps <- t(qtemps) %*% t(ai) * 3
```

Which works great for me. Thanks @fabee.

EDIT: OK, back-translating my R to Python, I get:

```
from numpy import *
from numpy.linalg import *
qtemps = transpose ([[46.0, 56.4, 65.8, 73.4, 77.4, 76.2, 69.5, 60.1, 49.5, 41.2]])
a = tril (ones ((12, 12)), 2) - tril (ones ((12, 12)), -1)
a = a[0:10,:]
ai = pinv (a)
mtemps = dot (ai, qtemps) * 3
```

(Which took a lot longer to debug than the R version. First because I’m not as familiar with Python as with R, but also because R is much more usable interactively.)

**Attribution***Source : Link , Question Author : Community , Answer Author : Wayne*