I currently have a 4x3500 cell array. First row is a single number, 2 row is a single string, 3rd and 4th rows are also single numbers.
Ex:
1 1 2 3 3 4 5 5 5 6
hi no ya he ........ % you get the idea
28 34 18 0 3 ......
55 2 4 42 24 .....
I would like to be able to select all columns that have a certain value in the first row. ie if I wanted '1' as the first row value, it would return
1 1
hi no
28 34
55 2
Then I would like to sort based on the 2nd row's string. ie if I wanted to have'hi', it would return:
1
hi
28
55
I have attempted to do:
variable = cellArray{:,find(cellArray{1,:} == 1)}
However I keep getting:
Error using find
Too many input arguments.
or
Error using ==
Too many input arguments.
Any help would be much appreciated! :)
{} indexing will return a comma separated list which will provide multiple outputs. When you pass this to find, it's the same as passing each element of your cell array as a separate input. This is what leads to the error about to many input arguments.
You will want to surround the comma-separated list with [] to create an array or numbers. Also, you don't need find because you can just use logical indexing to grab the columns you want. Additionally, you will want to index using () to grab the relevant rows, again to avoid the comma-separated list.
variable = cellArray(:, [cellArray{1,:}] == 1)
Related
I want to be able to construct (+; (+; `a; `b); `c) given a list of `a`b`c
Similarly if I have a list of `a`b`c`d, I want to be able to construct another nest and so on and so fourth.
I've been trying to use scan but I cant get it right
q)fsum:(+;;)/
enlist[+;;]/
q)fsum `a`b`c`d
+
(+;(+;`a;`b);`c)
`d
If you only want the raw parse tree output, one way is to form the equivalent string and use parse. This isn't recommended for more complex examples, but in this case it is clear.
{parse "+" sv string x}[`a`b`c`d]
+
`d
(+;`c;(+;`b;`a))
If you are looking to use this in a functional select, we can use +/ instead of adding each column individually, like how you specified in your example
q)parse"+/[(a;b;c;d)]"
(/;+)
(enlist;`a;`b;`c;`d)
q)f:{[t;c] ?[t;();0b;enlist[`res]!enlist (+/;(enlist,c))]};
q)t:([]a:1 2 3;b:4 5 6;c:7 8 9;d:10 11 12)
q)f[t;`a`b`c]
res
---
12
15
18
q)f[t;`a`b]
res
---
5
7
9
q)f[t;`a`b`c]~?[t;();0b;enlist[`res]!enlist (+;(+;`a;`b);`c)]
1b
You can also get the sum by indexing directly to return a list of each column values and sum over these. We use (), to turn any input into a list, otherwise it will sum the values in that single column and return only a single value
q)f:{[t;c] sum t (),c}
q)f[t;`a`b`c]
12 15 18
I have a table 'X' like this:
name value score
joy 3 60
rony 8 50
macheis 20 20
joung 2 80
joy 8 3
joy 90 0
joung 4 78
machies 3 23
joy 7 99
I want to select 2 random rows(with name, value, score) where the name is 'joy'.
I applied something like this:
mnew = datasample(find(X.name=='joy'),2); but it does not work! and gives me the error: Undefined operator '==' for input arguments of type 'cell'.
The rows should be selected randomly (with all columns values) where the name is joy.
Does anyone any other solution of this problem? how can i do it in MATLAB?
You have the right idea, but in order to check for the presence of a string within a cell array of strings, you need to use strcmp, ismember, or another method for comparing a string to a cell array.
You probably also want to specify that you don't want to use replacement when calling datasample so you don't get the same row twice.
subx = X(datasample(find(strcmp(X.name, 'joy')), 2, 'Replace', false),:);
I've data in the format of
a{1}(1,1)=1
a{1}(1,2)=3
a{1}(1,3)=0.5
a{1}(2,1)=1
a{1}(2,2)=5
a{1}(3,1)=2
a{1}(3,2)=7
...
Now I'd like to place item 1 & 2 of all rows in this cell into another cell array from an specific index on. How should I do this? for example to place column 1 & 2 of these unknown number of row in the cell array 'b' from index 5 on without repetition I used:
b{1}(1,(5:end + 1)) = unique(a{1}(:,(1:2)))
I'd like the output of cell array 'b{1}' from 5th elements becomes '1 2 3 5 7'.
However, it is false and gives the "Subscripted assignment dimension mismatch." error. I dont know where is the problem. Is my snippet right?
Any help is appreciated
I have a matrix (4096x4) containing all possible combinations of four values taken from a pool of 8 numbers.
...
3 63 39 3
3 63 39 19
3 63 39 23
3 63 39 39
...
I am only interested in the rows of the matrix that contain four unique values. In the above section, for example, the first and last row should be removed, giving us -
...
3 63 39 19
3 63 39 23
...
My current solution feels inelegant-- basically, I iterate across every row and add it to a result matrix if it contains four unique values:
result = [];
for row = 1:size(matrix,1)
if length(unique(matrix(row,:)))==4
result = cat(1,result,matrix(row,:));
end
end
Is there a better way ?
Approach #1
diff and sort based approach that must be pretty efficient -
sortedmatrix = sort(matrix,2)
result = matrix(all(diff(sortedmatrix,[],2)~=0,2),:)
Breaking it down to few steps for explanation
Sort along the columns, so that the duplicate values in each row end up next to each other. We used sort for this task.
Find the difference between consecutive elements, which will catch those duplicate after sorting. diff was the tool for this purpose.
For any row with at least one zero indicates rows with duplicate rows. To put it other way, any row with no zero would indicate rows with no duplicate rows, which we are looking to have in the output. all got us the job done here to get a logical array of such matches.
Finally, we have used matrix indexing to select those rows from matrix to get the expected output.
Approach #2
This could be an experimental bsxfun based approach as it won't be memory-efficient -
matches = bsxfun(#eq,matrix,permute(matrix,[1 3 2]))
result = matrix(all(all(sum(matches,2)==1,2),3),:)
Breaking it down to few steps for explanation
Find a logical array of matches for every element against all others in the same row with bsxfun.
Look for "non-duplicity" by summing those matches along dim-2 of matches and then finding all ones elements along dim-2 and dim-3 getting us the same indexing array as had with our previous diff + sort based approach.
Use the binary indexing array to select the appropriate rows from matrix for the final output.
Approach #3
Taking help from MATLAB File-exchange's post combinator
and assuming you have the pool of 8 values in an array named pool8, you can directly get result like so -
result = pool8(combinator(8,4,'p'))
combinator(8,4,'p') basically gets us the indices for 8 elements taken 4 at once and without repetitions. We use these indices to index into the pool and get the expected output.
For a pool of a finite number this will work. Create is unique array, go through each number in pool, count the number of times it comes up in the row, and only keep IsUnique to 1 if there are either one or zero numbers found. Next, find positions where the IsUnique is still 1, extract those rows and we finish.
matrix = [3,63,39,3;3,63,39,19;3,63,39,23;3,63,39,39;3,63,39,39;3,63,39,39];
IsUnique = ones(size(matrix,1),1);
pool = [3,63,39,19,23,6,7,8];
for NumberInPool = 1:8
Temp = sum((matrix == pool(NumberInPool))')';
IsUnique = IsUnique .* (Temp<2);
end
UniquePositions = find(IsUnique==1);
result = matrix(UniquePositions,:)
So, presume a matrix like so:
20 2
20 2
30 2
30 1
40 1
40 1
I want to count the number of times 1 occurs for each unique value of column 1. I could do this the long way by [sum(x(1:2,2)==1)] for each value, but I think this would be the perfect use for the UNIQUE function. How could I fix it so that I could get an output like this:
20 0
30 1
40 2
Sorry if the solution seems obvious, my grasp of loops is very poor.
Indeed unique is a good option:
u=unique(x(:,1))
res=arrayfun(#(y)length(x(x(:,1)==y & x(:,2)==1)),u)
Taking apart that last line:
arrayfun(fun,array) applies fun to each element in the array, and puts it in a new array, which it returns.
This function is the function #(y)length(x(x(:,1)==y & x(:,2)==1)) which finds the length of the portion of x where the condition x(:,1)==y & x(:,2)==1) holds (called logical indexing). So for each of the unique elements, it finds the row in X where the first is the unique element, and the second is one.
Try this (as specified in this answer):
>>> [c,~,d] = unique(a(a(:,2)==1))
c =
30
40
d =
1
3
>>> counts = accumarray(d(:),1,[],#sum)
counts =
1
2
>>> res = [c,counts]
Consider you have an array of various integers in 'array'
the tabulate function will sort the unique values and count the occurances.
table = tabulate(array)
look for your unique counts in col 2 of table.