Creating variable averaged over two strata
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Hi there,
I have a two long variables (+40,000 measures) of doubles x & y and i would like to calculate a third variable from them z. However X and Y are measured over time and at various depths in the ocean, not always at the same exact depths so in order to calculate my z variable i need to do some matlab wizardry?? (I have the date in every format under the sun and also separate variable for the year,month and day seperately)
I would like a code that calculates monthly averages at certain depths.
For example to go from a matrix of [date, x, y, depth] which contain various NaNs to one which has 1:20:4000 depth in the first column repeated for every month (total of 350 months in my data set) and then corresponding monthly average for x and y.
Thank you for any help with this in advance
1 个评论
John BG
2016-3-1
would it be possible to read a sample of the data?
采纳的回答
Mohammad Abouali
2016-3-1
编辑：Mohammad Abouali
2016-3-2
first convert your date so that you get year and month separately. If you have the date as a string you can use datevec() command as follows:
dateVector = datevec(dateStr);
year = dateVector(:,1);
month = dateVector(:,2);
then you can use grpstats(). I suggest to create a table first, but you can also use a regular array.
dataTBL = table();
dataTBL.Year= year;
dataTBL.month= month;
dataTBL.depth = depth;
dataTBL.x = x;
dataTBL.y = y;
monthlyMean = grpstats(dataTBL, {'year',month','depth'},{@(c) (nanmean(c(:)))})
If you upload a sample data we can be of more help.
2 个评论
aaron Harvey
2016-3-1
Mohammad Abouali
2016-3-2
编辑：Mohammad Abouali
2016-3-2
I am not sure what you are asking. If you are trying to multiply two vectors element by element you need to use .* instead of * (pay attention to the dot before *).
So it would be something like this
a=[1 2 3 NaN 2 3]'
b=[3 2 NaN 2 1 5]'
a.*b
ans =
3
4
NaN
NaN
2
15
If one of the element is NaN, then the multiplication results would be NaN.
Since you mentinoed your data has some NaN, earlier I suggested using nanmean instead of mean.
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