Could you please help me for this matter?
I have 3 matrices, P (Power), T (Temperature) and H (Humidity)
every matrix has 31 columns (days) and there are 24 rows for every column
which are the data for the March of year 2000, i.e.
for example, the matrix P has 31 columns where every column represents
a day data for Power through 24 hours and the same idea is for T and H
I tried to write a MATLAB program that accomplish my goal but
It gave me errors.
My aim is:
In the MATLAB command window, the program should ask the user the following phrase:
Please enter the day number of March, 2000 from 1 to 31:
And I know it is as follows:
Name=input (Please enter the day number of March, 2000 from 1 to 31:)
Then, when, for example, number 5 is entered, the result shown is a matrix containing the following:
1st column: The day name or it can be represented by numbers
2nd column: simple numbers from 1 to 24 representing the hours for that day
3rd column: the 24 points of P of that day extracted from the original P
(the column number 5 of the original P)
4th column: the 24 points of T of that day extracted from the original T
(the column number 5 of the original T)
5th column: the 24 points of H of that day extracted from the original H
(the column number 5 of the original H)
Any help will be highly appreciated,
Regards
Here is what you ask for:
% some sample data
P = rand(24,31);
T = rand(24,31);
H = rand(24,31);
% input day number
daynum=input('Please enter the day number of March, 2000 from 1 to 31: ');
[r, c] = size(P);
% generate output matrix
OutputMatrix = zeros(r,5);
OutputMatrix(:,1) = repmat(weekday(datenum(2000,3,daynum)),r,1);
OutputMatrix(:,2) = 1:r;
OutputMatrix(:,3) = P(:,daynum);
OutputMatrix(:,4) = T(:,daynum);
OutputMatrix(:,5) = H(:,daynum);
disp(OutputMatrix)
The matrix can be generated in a one line, but this way is clearer.
Is it always for March 2000? :) Where do you get this information from?
Related
Suppose I have a date vector shown here by tt and a corresponding data series corresponding to aa. For example:
dd = datestr(datenum('2007-01-01 00:00','yyyy-mm-dd HH:MM'):1/24:...
datenum('2011-12-31 23:00','yyyy-mm-dd HH:MM'),...
'yyyy-mm-dd HH:MM');
tt = datevec(datenum(dd,'yyyy-mm-dd HH:MM'));
tt(1002,:) = [];
aa = rand(length(tt),1)
How is it possible to ensure that the hours and days are consistent among the years?
For example, I only want to keep times that are the same among years e.g.
2009-01-01 01:00
would be the same as
2010-01-01 01:00
ad so on.
If one year has a measurements at
2009-01-01 02:00
but yyyy-01-01 02:00
is not present in the other years, this time should re removed.
I would like the to return tt and aa where only those times that are consistent among the years are kept. how can this be done?
I was considering finding the indices for the unique years first as:
[~,~,iyears] = unique(tt(:,1),'rows');
and then find the indices for the unique month, day, and hour as:
[~,~,iid] = unique(tt(:,2:4),'rows');
but I am not sure how to combine these to give the desired output?
The solution below uses a loop to store data in an unitialized array, which can be inefficient, but unless your dataset is huge (with many, many years) it should do the job. The general idea is to break the dataset up into years. I am storing the resulting time-vectors in a cell array, because they probably won't have the same length. I then do a set-intersection of all the time vectors, to get a vector of common times. From there it's straight forward.
years = unique(tt(:,1), 'rows');
% Put the "sub-times" of each year into cell array
for ii = 1:length(years)
times_each_year{ii} = tt(tt(:,1)==years(ii),2:end);
end
% Do intersection of all "sub-times" sets
common_times = times_each_year{1};
for ii = 2:length(years)
common_times = intersect(common_times, times_each_year{ii},'rows');
end
% Find and delete the points that are not member of the "sub-times":
idx = ~ismember(tt(:,2:end),common_times,'rows');
deleted_points = datestr(tt(idx,:)); % for later review
tt(idx,:) = [];
However, note that the deleted_points vector contains more points than one might expect. That's because 2008 was a leap year, and all the points corresponding to Febr. 29th were deleted.
Another such oddity might await you if your data is "contaminated" by daylight savings time.
Code
a1 = str2num(datestr(tt,'mmddHHMM')); %// If in your data minutes are always 00, you can use 'mmddHH' instead and save some runtime
k1 = unique(a1);
gt1 = histc(a1,k1);
valid_rows = ismember(a1,k1(gt1==max(gt1)));
new_tt = tt(valid_rows,:); %// Desired tt output
new_aa = aa(valid_rows,:); %// Desired aa output
Explanation
To understand how it works, let's test out the code at a micro-level. Let's assume some small data that corresponds to tt -
data1 = [4 5 1 4 5 1 4 5 6]
data1 is the data collected over few sets and resembles tt that has data over few years with month, date, hour and minutes when these four parameters are conglomerated into a single parameter.
One can notice it would represent data from three sets/years with data as {4,5}, {1,4,5} and {1,4,5,6}. Our job is to found out all those values in data1 that is repeated across all the three years/sets of data. Thus, the final output must be {4,5}.
Let's see how this can be coded up.
Step 1: Get the unique values
unique_val = unique(data1)
We would have - [1 4 5 6]
Step 2: Get the count of unique values in the data
count_unique_val = histc(data1,unique_val)
Output is - [2 3 3 1]
Step 3: Get the indices from the unique values array where their counts are equal to the maximum of counts, indicating those are the unique values that are repeated across all the sets.
index1 = count_unique_val==max(count_unique_val)
Output comes out as - [0 1 1 0]
Step 4: Get those "consistent" unique values
consistent_val = unique_val(index1)
Gives us - [4 5], which is what we were looking for.
Step 5: Finally get the indices where the consistent data is present,
which can be used later on to select the rows with "consistent" data.
index_consistent_val = ismember(data1,consistent_val)
Output is - [1 1 0 1 1 0 1 1 0], which makes sense too.
Please note that in the original code a1 = str2num(datestr(tt,'mmddHHMM')); gets us the single parameter from the four parameters of month, date, hour and minutes as discussed in the comments earlier too.
good morning. I have a matrix (eout) with 14610 rows and 16 column.
14610 row represent one of each day of the period between 01-01-1960 and 31-12-2000.
What I need is a new matrix with 40 rows and 16 columns with a mean for each year.
something like a mean of 365 rows continuously.
The problem I have are the leap years each 4 years.
Suggestions to solve this?
For a start, to get the number of days for a given year:
function n = ndays(year)
tmp = repmat([1,1,0,0,0],numel(year),1);
n = datenum([year(:)+1,tmp])-datenum([year(:), tmp]);
end
With this, you could collect the rows e.g. with mat2cell:
rows_per_year = ndays(1960:2000);
chunks = mat2cell(yourInputMatrix, rows_per_year, size(yourInputMatrix,2));
means = cellfun(#(x) mean(x,1), chunks);
(The latter part comes untested..)
I have observed daily data that I need to compare to generated Monthly data so I need to get a mean of each month over the thirty year period.
My observed data set is currently in 365x31 with rows being each day (no leap years!) and the extra column being the month number (1-12).
the problem I am having is that I can only seem to get a script to get the mean of all years. ie. I cannot figure how to get the script to do it for each column separately. Example of the data is below:
1 12 14
1 -15 10
2 13 3
2 2 37
...all the way to 12 for 365 rows
SO: to recap, I need to get the mean of [12; -15; 13; 2] then [14; 10; 3; 37] and so on.
I have been trying to use the unique() function to loop through which works for getting the number rows to average but incorrect means. Now I need it to do each month(28-31 rows) and column individually. Result should be a 12x30 matrix. I feel like I am missing something SIMPLE. Code:
u = unique(m); %get unique values of m (months) ie) 1-12
for i=1:length(u)
month(i) = mean(obatm(u(i), (2:31)); % the average for each month of each year
end
Appreciate any ideas! Thanks!
You can simply filter the rows for each month and then apply mean, like so:
month = zeros(12, size(obatm, 2));
for k = 1:12
month(k, :) = mean(obatm(obatm(:, 1) == k, :));
end
EDIT:
If you want something fancy, you can also do this:
cols = size(obatm, 2) - 1;
subs = bsxfun(#plus, obatm(:, 1), (0:12:12 * (cols - 1)));
vals = obatm(:, 2:end);
month = reshape(accumarray(subs(:), vals(:), [12 * cols, 1], #mean), 12, cols)
Look, Ma, no loops!
This is a simplistic example of a problem I am facing:
depth = [0:1:20]';
data = rand(1,length(depth))';
d = [depth,data];
d = [d;d;d];
Consider the matrix 'd'. Here we have depth in the first column followed by temperature measurements recorded at that depth in column 2 (in this example we have 3 days of data). How could I alter this matrix so that each column represents a specific depth and each row represents time. So, finally I should have 3 rows with 21 columns.
If I understand correctly your array d has the data for day 1 in rows 1:21, for day 2 in rows 22:42, and so on. Column 1 of d holds the depths (3 times), and column 2 holds the measurements.
One way to get the results in the form you want is to execute:
d2 = reshape(d(:,2),21,3)'; % note the ' for transposition here
This leaves you with an array with 3 rows and 21 columns. Each column represents the measurements for one depth, each row the measurements for one day.
I have 19 cells (19x1) with temperature data for an entire year where the first 18 cells represent 20 days (each) and the last cell represents 5 days, hence (18*20)+5 = 365days.
In each cell there should be 7200 measurements (apart from cell 19) where each measurement is taken every 4 minutes thus 360 measurements per day (360*20 = 7200).
The time vector for the measurements is only expressed as day number i.e. 1,2,3...and so on (thus no decimal day),
which is therefore displayed as 360 x 1's... and so on.
As the sensor failed during some days, some of the cells contain less than 7200 measurements, where one in
particular only contains 858 rows, which looks similar to the following example:
a=rand(858,3);
a(1:281,1)=1;
a(281:327,1)=2;
a(327:328,1)=5;
a(329:330,1)=9;
a(331:498,1)=19;
a(499:858,1)=20;
Where column 1 = day, column 2 and 3 are the data.
By knowing that each day number should be repeated 360 times is there a method for including an additional
amount of every value from 1:20 in order to make up the 360. For example, the first column requires
79 x 1's, 46 x 2's, 360 x 3's... and so on; where the final array should therefore have 7200 values in
order from 1 to 20.
If this is possible, in the rows where these values have been added, the second and third column should
changed to nan.
I realise that this is an unusual question, and that it is difficult to understand what is asked, but I hope I have been clear in expressing what i'm attempting to
acheive. Any advice would be much appreciated.
Here's one way to do it for a given element of the cell matrix:
full=zeros(7200,3)+NaN;
for i = 1:20 % for each day
starti = (i-1)*360; % find corresponding 360 indices into full array
full( starti + (1:360), 1 ) = i; % assign the day
idx = find(a(:,1)==i); % find any matching data in a for that day
full( starti + (1:length(idx)), 2:3 ) = a(idx,2:3); % copy matching data over
end
You could probably use arrayfun to make this slicker, and maybe (??) faster.
You could make this into a function and use cellfun to apply it to your cell.
PS - if you ask your question at the Matlab help forums you'll most definitely get a slicker & more efficient answer than this. Probably involving bsxfun or arrayfun or accumarray or something like that.
Update - to do this for each element in the cell array the only change is that instead of searching for i as the day number you calculate it based on how far allong the cell array you are. You'd do something like (untested):
for k = 1:length(cellarray)
for i = 1:length(cellarray{k})
starti = (i-1)*360; % ... as before
day = (k-1)*20 + i; % first cell is days 1-20, second is 21-40,...
full( starti + (1:360),1 ) = day; % <-- replace i with day
idx = find(a(:,1)==day); % <-- replace i with day
full( starti + (1:length(idx)), 2:3 ) = a(idx,2:3); % same as before
end
end
I am not sure I understood correctly what you want to do but this below works out how many measurements you are missing for each day and add at the bottom of your 'a' matrix additional lines so you do get the full 7200x3 matrix.
nbMissing = 7200-size(a,1);
a1 = nan(nbmissing,3)
l=0
for i = 1:20
nbMissing_i = 360-sum(a(:,1)=i);
a1(l+1:l+nbMissing_i,1)=i;
l = l+nb_Missing_i;
end
a_filled = [a;a1];