Suppose that I have a string of values corresponding to the height of a group of people
height_str ={'1.76000000000000';
'1.55000000000000';
'1.61000000000000';
'1.71000000000000';
'1.74000000000000';
'1.79000000000000';
'1.74000000000000';
'1.86000000000000';
'1.72000000000000';
'1.82000000000000';
'1.72000000000000';
'1.63000000000000'}
and a single height value.
height_val = 177;
I would like to find the indices of the people that are in the range height_val +- 3cm.
To find the exact match I would do like this
[idx_height,~]=find(ismember(cell2mat(height_str),height_val/100));
How can I include the matches in the previous range (174-180)?
idx_height should be = [1 5 6 7]
You can convert you strings into an numeric array (as #Divakar mentioned) by
height = str2num(char(height_str))*100; % in cm
Then just
idx_height = find(height>=height_val-3 & height<=height_val+3);
Assuming that the precision of heights stays at 0.01cm, you can use a combination of str2double and ismember for a one-liner -
idx_height = find(ismember(str2double(height_str)*100,[height_val-3:height_val+3]))
The magic with str2double is that it works directly with cell arrays to get us a numeric array without resorting to a combined effort of converting that cell array to a char array and then to a numeric array.
After the use of str2double, we can use ismember as you tried in your problem to get us the matches as a logical array, whose indices are picked up with find. That's the whole story really.
Late addition, but for binning my first choice would be to go with bsxfun and logical operations:
idx_height = find(bsxfun(#le,str2double(height_str)*100,height_val+3) & ...
bsxfun(#ge,str2double(height_str)*100,height_val-3))
Related
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
In my cell array test = cell(1,2,20,14); I want to find numeric values in the subset test(:,1,1,1).
For example test(:,:,1,1) looks like this:
>> test(:,:,1,1)
ans =
[ 0] [0.1000] [57]
[0.9000] [0.9500] [73]
I want to find the index of the cell containing 0.9 in the first column, so I can access the third column (in this case value 73). I tried:
find(test{:,:,1,1} == 0.9) which gives:
Error using == Too many input arguments..
How can I find the respective index?
Thanks.
Try this to access that third column value directly -
cell2mat(test(vertcat(test{:,1,1,1})==0.9,3,1,1))
Edit 1: If you would like to test out for match w.r.t. the first two columns of test's subset, use this -
v1 = reshape(vertcat(test{:,[1 2],1,1}),[],2)
cell2mat(test(ismember(v1,[0.9 0.95],'rows'),3,1,1))
Just add brackets [] around test{:,:,1,1}. This wraps the different cell values together to one vector/matrix.
Like this:
[index1, index2] = find([test{:,:,1,1}] == 0.9)
I have CellArray1 with 50 unique strings and CellArray2 with 2000 unique strings (50 of which are the same as the ones in CellArray1). Is there a way to find the positions of all 50 unique strings from the first cell array in the second cell array without using loops?
Yes - the following code demonstrates this:
cellArray1 = {'hello', 'world'};
cellArray2 = {'good', 'morning', 'world'};
overlap = find(ismember(cellArray2, cellArray1)};
This will return the value 3 in overlap since cellArray2{3} appears in cellArray1.
UPDATE
The above code returns the indices, but not in the order of the original. If you need the original order, you can do the following
overlap = cellfun(#(x)find(ismember(cellArray2, x)), cellArray1, 'uniformOutput', false);
overlapSorted = cell2mat(overlap);
It could be argued that cellfun actually has an implicit loop in it (but then all vector operations have implicit loops, really); but one of these constructions will do what you asked for. If you don't need it sorted, the first will be significantly faster, I imagine.
We've got an array of values, and we would like to create another array whose values are not in the first one.
Example:
load('internet.mat')
The first column contains the values in MBs, we have thought in something like:
MB_no = setdiff(v, internet(:,1))
where v is a 0 vector whose length equals to the number of rows in internet.mat. But it just doesn't work.
So, how do we do this?
You need to specify the range of possible values to define what values are not in internet . Say the range is v = 1:10 then setdiff(v,internet(:,1)) will give you the values in 1:10 that are not in the first column of internet.
It seems as if you don't want the first column.
You can simply do:
MB_no=internet(:,2:end);
assuming internet(:,1) has only positive integers and you wish to find which are the integers in [1,...,max( internet(:,1) )] that do not appear in that range you can simply do
app = [];
app( internet(:,1) ) = 1;
MB_no = find( app == 0 );
This is somewhat like bucket sort.
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')