# Weather Prediction – How Far Can You Go? 2

Posted by **Ned Gulley**,

This summer my mother-in-law is renting a house on a lake in New Hampshire. Looking at the calendar, my wife said: "The ten-day forecast makes it look like it's going to be pretty hot up at the lake next week." This led to a more general discussion of the merits of ten-day forecasts.

It's funny how we can make decisions based on long-term predictions of weather even though we rarely go back and verify that the forecast was any good. Somehow the fact that the forecast exists at all gives it value. I'm left pondering this question: how much should we trust a ten-day prediction? As it happens, I have some data that can be useful here. For some time, I have been collecting some relevant data on Trendy: the ten day forecast for Natick, Massachusetts (hometown for MathWorks). So let's run some numbers.

Here's the trend: Ten Day Forecast Highs for Natick, MA.

Once a day this trend collects ten data points: today's high temperature and the predicted high temperature for the next nine days. In MATLAB, we'll be working with a matrix with one row for each day and ten columns.

Let's get the data into MATLAB so we can play around with it. I can retrieve (and so can you) the data from Trendy as a JSON object using the following call:

http://www.mathworks.com/matlabcentral/trendy/trends/1655/trend_data.json

In order to read this into MATLAB, I'm going to use Joe Hicklin's JSON parser.

```
url = 'http://www.mathworks.com/matlabcentral/trendy/trends/1655/trend_data.json';
json = urlread(url);
raw = JSON.parse(json);
```

t = zeros(length(raw),1); d = zeros(length(raw),10); for i = 1:length(raw) t(i) = raw{i}{1}; predictions = raw{i}{2}; for j = 1:10 d(i,j) = str2num(predictions{j}); end end firstTenRows = d(1:10,:)

firstTenRows = 44 51 59 66 46 48 53 50 51 54 50 58 63 49 48 52 48 46 57 54 59 61 49 47 53 49 48 43 41 48 62 49 48 54 49 39 39 44 47 46 49 48 54 50 40 38 39 47 51 54 47 55 50 39 40 48 52 53 53 53 54 50 40 39 48 53 54 52 52 50 49 40 38 50 55 54 56 56 53 49 40 39 50 56 54 52 56 54 47 43 39 50 55 55 55 59 58 40 41 46

Now I have a temperature prediction matrix that's structured like this.

I want to re-order this matrix so that each line shows the prediction trajectory for a single day in time. That means picking off the diagonals highlighted in the diagram above. So let's write some code that does this shift. I'm going to end up with two new matrices, d1 and d2

rowIndex = 1:10; colIndex = 10:-1:1; sz = size(d); len = (size(d,1)-10); d1 = zeros(len,10); d2 = zeros(len,10); t1 = zeros(len,1); for i = 1:len ind = sub2ind(sz,rowIndex+i-1,colIndex); trend = d(ind); d1(i,:) = trend; d2(i,:) = trend-trend(end); t1(i) = t(i+9); end firstTenRows = d1(1:10,:)

firstTenRows = 54 57 43 39 38 40 39 38 39 39 54 41 44 39 48 48 50 50 50 49 48 47 47 52 53 55 56 55 54 57 46 51 53 54 54 54 55 55 57 58 54 53 52 56 52 55 57 57 60 58 53 52 56 56 59 59 61 63 63 64 50 53 54 58 45 42 45 44 44 43 49 47 40 39 37 42 43 43 43 43 43 41 41 42 44 48 47 49 48 47 46 44 48 49 46 52 51 50 49 48

In d1, each row is the evolving temperature prediction for each day. So when we plot the first row of d1, we're getting the predictive arc for November 13th of last year.

i = 1; plot(-9:0,d1(i,:)) title(sprintf('Predicted Temperature for %s',datestr(t1(i),1))) xlabel('Time of Prediction (Offset in Days)') ylabel('Predicted Temperature (Deg. F)')

In d2, we just subtract from each row the last value in each row. Since this last value is the final (and presumably correct) temperature, this difference gives us the predictive error across the ten days. Here's the error for the November 13th prediction.

i = 1; plot(-9:0,d2(i,:)) title(sprintf('Error in Predicted Temperature for %s',datestr(t1(i),1))) xlabel('Time of Prediction (Offset in Days)') ylabel('Prediction Error (Deg. F)')

Notice that it shrinks to zero over time. That's good! Our predictions get more accurate as we approach the actual day in question. But the early predictions were off by as much as 18 degrees. Is that good or bad? You tell me.

Now let's look at all the days.

plot(-9:0,d2','Color',[0.5 0.5 1]) title('Error in Predicted Temperature') xlabel('Time of Prediction (Offset in Days)') ylabel('Prediction Error (Deg. F)')

It's hard to get a sense of the error distribution. So let's finish with a histogram of the absolute value of the error. Out of 240 measurements in this data set, the median error for a ten-day prediction is six degrees.

hist(abs(d2(:,1)),1:25) title('Histogram of Error in the Ten-Day Forecast') xlabel('Error (deg. F)') ylabel('Number of Samples')

That seems pretty good. Most of the time that error is going to be less than seven or so degrees Fahrenheit (or four degrees Celsius). I probably don't need to pack a sweater for the weekend at the lake.

Get the MATLAB code

Published with MATLAB® R2014a

### Note

Comments are closed.

## 2 CommentsOldest to Newest

**1**of 2

Hi Ned,

Very interesting project, I’m actually interested in doing this for my local university. I do have one question about this methodology though.. so correct me if I’m wrong. In your code, you’re pulling the forecasted temperatures for 10 days off of weather.com, and the first entry in your code is the forecasted maximum temperature.. not necessarily the actual maximum temperature for that day. With this in mind, isn’t your error analysis showing the error of forecasting the forecasted maximum temperature, rather than the error of forecasting the actual maximum temperature?

If this is the case, I would expect a more chaotic pattern in the error analysis due to atmospheric variability or local effects. Including in the maximum temperature from the Norwood Memorial Airport (http://w1.weather.gov/obhistory/KOWD.html) or Worcester Regional Airport (http://w1.weather.gov/obhistory/KORH.html) as your validation temperature (i.e., Day 0) would create a more robust error analysis, or so I imagine. What do you think about it??

**2**of 2

Hi John:

Thanks for the note. The first number in each column of collected data is the *actual observed high temperature* for that day. This gets flipped around to be the last column of the d2 matrix. The error analysis is showing how the forecasted high for a given day converges on the actual high for that day as you get closer to it.

## Recent Comments