I have a .mat file that contains data from the years 2006-2100. Each year, there is a different number of lines. I need to count how many lines are 2006, how many are 2007, etc.
The set up, by column, is: Year, Month, Day, Lat, Long
I just want to count the number of rows containing the same Year entry and get an array back with an array containing that info.
I'm thinking a for or while loop should work, but I don't know how to right it.
If we assume your data are in a numeric matrix, you can just do:
num_lines2006 = sum(data(:,1)==2006);
data2006 = data(data(:,1)==2006),:);
If you want to add a column with number of rows for corresponding year, here is a solution with a loop:
for k=size(data,1):-1:1
num_year(k,1) = sum(data(:,1)==data(k,1));
end
data = [data num_year];
Here is a solution without loop:
[unq_year,~,idx] = unique(data(:,1),'stable');
num_year = grpstats(data(:,1),unq_year,#numel);
data = [data num_year(idx)];
To count numeric entries, you may want to use histc
years = unique(data(:,1);
counts = histc(data(:,1),years);
Since you just want to count the number of rows you could just write something simple like:
years = unique(data(:, 1));
counts = arrayfun(#(year) nnz(data(:, 1) == year), years);
years contains the unique years, and numRows the number of times they are found.
You could also use a one-liner inspired by Jonas' answer:
[counts, years] = hist(data(:,1), unique(data(:,1))');
Related
I have two timetables, each of them have 4 columns, where the first 2 columns are of my particular interest. The first column is a date and the second is an hour.
How can I know which observations (by date an hour) are in the timetable 1 but not in the timetable 2 and, therefore, drop those observations from my timetable 1?
So for example, just by looking I realized that timetable1 included the day 25/05/2015 with hours 1 and 2, but the timetable 2 did not include them, therefore I would like to drop those observations from timetable 1.
I tried using the command groups_timetable1 = findgroups(timetable1.Date,timetable1.Hour);but unfortunately this command does not tell you a lot how to distinguish between observations.
Thank you!
call ismember to find one set of data in another.
to find multiple records as a group in another composite records, you call ismember(..., 'rows').
for example
baseline=[
100, 2.1
200, 7.5
120, 11.0
];
isin=ismember(baseline,[200, 7.5],'rows');
pos=find(isin)
if you have time date strings or datetime objects, please convert those to numerical values, such as by calling datenum or posixtime first.
You can use the timetable method innerjoin to do this. Like so:
% Fabricate some data
dates1 = datetime(2015, 5, ones(10,1));
hours1 = (1:10)';
timetable1 = timetable(dates1(:), hours1, rand(10,1), rand(10,1), ...
'VariableNames', {'Hour', 'Price', 'Volume'});
% Subselect a few rows for timetable2
timetable2 = timetable1([1:3, 6:10],:);
% Use innerjoin to pick rows where Time & Hour intersect:
innerjoin(timetable1, timetable2, 'Keys', {'Time', 'Hour'})
By default, the result of innerjoin contains the table variables from both input tables - that may or may not be what you want.
In MatLab, I have several data vectors that are in text. For example:
speciesname = [species1 species2 species3];
genomelength = [8 10 5];
gonometype = [RNA DNA RNA];
I realise that to make a plot, arrays must be numerical. Is there a quick and easy way to assign unique entries in an array a number, for example so that RNA = 1 and DNA = 2? Note that some arrays might not be binary (i.e. have more than two options).
Thanks!
So there is a quick way to do it, but im not sure that your plots will be very intelligible if you use numbers instead of words.
You can make a unique array like this:
u = unique(gonometype);
and make a corresponding number array is just 1:length(u)
then when you go through your data the number of the current word will be:
find(u == current_name);
For your particular case you will need to utilize cells:
gonometype = {'RNA', 'DNA', 'RNA'};
u = unique(gonometype)
u =
'DNA' 'RNA'
current = 'RNA';
find(strcmp(u, current))
ans =
2
I'm trying to loop through an array of dates/times in matlab, split each column using regexp with the following delimiters ('/' or ':' or '.'), and store each column separately as year, day, hour, min, sec, ss, respectively. Ultimately I'm trying to turn this array of Julian dates and times into a plot-able format in matlab. So far I've been able to loop through my array called 'time' and created a new 1x6 cell called 'clean2_time' which splits each row into 6 columns (year, day, hour, min, sec, ss) based on the delimiters '/' ':' and '.'. My issue is that the loop overwrites 'clean2_time' every iteration and I am left with only the final 1x6 time stamp for the last row. I have tried creating a new variable of all zeros 'z' and setting 'clean2_time' equal to z but have no luck.
Sample of 'time':
'2013/231/21:38:09.856619'
'2013/231/21:38:09.955640'
'2013/231/21:38:10.156685'
'2013/231/21:38:10.356550'
'2013/231/21:38:10.556770'
'2013/231/21:38:10.756565'
'2013/231/21:38:10.955627'
'2013/231/21:38:11.256588'
'2013/231/21:38:11.556649'
'2013/231/21:38:11.955597'
'2013/231/21:38:12.356627'
'2013/231/21:38:12.856557'
'2013/231/21:38:13.356558'
'2013/231/21:38:14.156530'
'2013/231/21:38:14.970500'
'2013/231/21:38:16.256545'
'2013/231/21:38:16.266736'
'2013/231/21:38:18.156398'
Code I've tried so far:
z=zeros(size(time,1),6);
for i = 1:size(time,1) % for i = 1 to 5922
clean2_time = regexp(time{i,1}, '[/:.]', 'split');
z{i,1} = clean2_time(i,1)
z{i,2} = clean2_time(i,2)
z{i,3} = clean2_time(i,3)
z{i,4} = clean2_time(i,4)
z{i,5} = clean2_time(i,5)
z{i,6} = clean2_time(i,6)
end
You are on the right track, however, you don't need the for loop.
Simply doing this would suffice:
clean2_time=regexp(time, '[/:.]', 'split');
Then clean2_time is a cell structure in which every row contains another 1x6 cell array. You can then access the different values with: clean2_time{row}{column}. If you really want clean2_time to be a nx6 numerical matrix instead of this cell array of strings, simply use this to reshape:
clean2_time=cellfun(#str2num,vertcat(clean2_time{:}))
clean2_time=zeros(size(time,1),6);
for i = 1:size(time,1) % for i = 1 to 5922
clean2_time(i,:)=regexp(time{i,1}, '[/:.]', 'split')
end
clean2_time(i,:) indexes the i-th row of the cell.
I simply want to generate a series of dates 1 year apart from today.
I tried this
CurveLength=30;
t=zeros(CurveLength);
t(1)=datestr(today);
x=2:CurveLength-1;
t=addtodate(t(1),x,'year');
I am getting two errors so far?
??? In an assignment A(I) = B, the number of elements in B and
Which I am guessing is related to the fact that the date is a string, but when I modified the string to be the same length as the date dd-mmm-yyyy i.e. 11 letters I still get the same error.
Lsstly I get the error
??? Error using ==> addtodate at 45
Quantity must be a numeric scalar.
Which seems to suggest that the function can't be vectorised? If this is true is there anyway to tell in advance which functions can be vectorised and which can not?
To add n years to a date x, you do this:
y = addtodate(x, n, 'year');
However, addtodate requires the following:
x must be a scalar number, not a string.
n must be a scalar number, not a vector.
Hence the errors you get.
I suggest you use a loop to do this:
CurveLength = 30;
t = zeros(CurveLength, 1);
t(1) = today; % # Whatever today equals to...
for ii = 2:CurveLength
t(ii) = addtodate(t(1), ii - 1, 'year');
end
Now that you have all your date values, you can convert it to strings with:
datestr(t);
And here's a neat one-liner using arrayfun;
datestr(arrayfun(#(n)addtodate(today, n, 'year'), 0:CurveLength))
If you're sequence has a constant known start, you can use datenum in the following way:
t = datenum( startYear:endYear, 1, 1)
This works fine also with months, days, hours etc. as long as the sequence doesn't run into negative numbers (like 1:-1:-10). Then months and days behave in a non-standard way.
Here a solution without a loop (possibly faster):
CurveLength=30;
t=datevec(repmat(now(),CurveLength,1));
x=[0:CurveLength-1]';
t(:,1)=t(:,1)+x;
t=datestr(t)
datevec splits the date into six columns [year, month, day, hour, min, sec]. So if you want to change e.g. the year you can just add or subtract from it.
If you want to change the month just add to t(:,2). You can even add numbers > 12 to the month and it will increase the year and month correctly if you transfer it back to a datenum or datestr.
i have a cell array as below, which are dates. I am wondering how can i extract the year at the last 4 digits? Could anyone teach me how to locate the year in the string? Thank you!
'31.12.2001'
'31.12.2000'
'31.12.2004'
'31.12.2003'
'31.12.2002'
'31.12.2000'
'31.12.1999'
'31.12.1998'
'31.12.1997'
'31.12.2005'
'31.12.2004'
'31.12.2003'
'31.12.2002'
'31.12.2001'
'31.12.2000'
'31.12.1999'
'31.12.1998'
'31.12.2005'
'31.12.2004'
'31.12.2003'
'31.12.2002'
'31.12.2005'
Example cell array:
A = {'31.12.2001'; '31.12.2002'; '31.12.2003'};
Apply some regular expressions:
B = regexp(A, '\d\d\d\d', 'match')
B = [B{:}];
EDIT: I never realized that matlab will "nest" an extra layer of cells until I tested this. I don't like this solution as much now that I know the second line is necessary. Here is an alternative approach that gets you the years in numeric form:
C = datevec(A, 'dd.mm.yyyy');
C = C(:, 1);
SECOND EDIT: Suprisingly, if your cell array has less than 10000 elements, the regexp approach is faster on my machine. But the output of it is another cell array (which takes up much more memory than a numeric matrix). You can use B = cell2mat(B) to get a character array instead, but this brings the two approaches to approximately equal efficiency.
Just to add a fun answer, designed to take the OP to the stranger regions of Matlab:
C = char(C);
y = (D(:,7:end)-'0') * 10.^(3:-1:0).'
which is an order of magnitude faster than anything posted in the other answers :)
Or, to stay a bit closer to home,
y = cellfun(#(x)str2double(x(7:end)),C);
or, yet another regexp variation:
y = str2num(char(regexprep(C, '\d+\.\d+\.','')));
Assuming your matrix with dates is M or a cell array C:
In case your data is in a cell array start with
M = cell2mat(C)
Then get the relevant part
Y=M(:,end-4:end)
If required you can even make the year a number
Year = str2num(Y)
Using regexp this will works also with dates with slightly different formats, like 1.1.2000, which can mess with you offsets
res = regexp(dates, '(?<=\d+\.\d+\.)\d+', 'match')