Let's say that I have 2 3D-Matrix:
A = rand(10,4,100);
B = rand(10,4,100);
L = gt(A,B);
Now I want substitute all elements of B with elements of A only where L==1but this doesn't work:
B(L==1,:,:) = A(L==1,:,:);
Any suggestion?
Our even shorter whithout find
B(L) = A(L);
Sounds like a job for the find() function.
p = find(L);
B(p) = A(p);
EDIT: Just realized you don't need the find() function. Just use logical indexing like this:
B(L==1) = A(L==1);
Related
I am a newbie. I have problem.
I have 20 (1x100) different named vectors. I want to combine these vectors to create a 20x100 matrix with a for loop.
There are the examples of vectors.
namelist=["First","B","New"]
First = [1:100]
B = [1:2:200]
New = [4:4:400]
for i = 1: length(namelist)
new_database(i,1:end) = namelist{i}
end
But, when I want to try this I saw "The end operator must be used within an array index expression." error.
I know I can do same thing with this:
"new_database= [First;B;New]"
but i want to do this with a for loop.
Would you help me how can fix this error? or Would you explain me how can do this?
Your problem is with this line:
new_database(i,1:end) = namelist{i}
the curly braces are used with cells, exclusively and there is no need to use range indexing as you do (i, 1:end)
Generally, it is better practice to assign character arrays or strings to cells.
One question, what are you doing with the 'First', 'New' and 'B' ranges arrays?
Something like:
namelist=["First","B","New"]
First = [1:100];
B = [1:2:200];
New = [4:4:400];
new_database = cell(1, length(namelist));
for i = 1: length(namelist) % or length(new_database)
new_database{i} = namelist(i)
end
which generates this output:
EDIT: My apologies, now I see what you are trying to accomplish. You are building a database from a series of arrays, correct?
Following my previous response, you must consider some points:
1 Your new_database should be square. Regardless of the dimensions of the arrays you are passing to it, if you form a cell from them, you will invariably have empty cells if no data is passed to those rows or columns
2 In some cases, you don't need to use for-loops, where simple indexing might suffice your case problem. Consider the following example using cellstr:
titles = ["Position", "Fruits", "Mythical creatures"]
A = ["One", "Two", "Three"];
B = ["Apple", "Banana", "Durian"];
C = ["Dragon", "Cat", "Hamster"];
db = cell(4, 3);
db(1,:) = cellstr(titles)
db(2:end,1) = cellstr(A)
db(2:end,2) = cellstr(B)
db(2:end,3) = cellstr(C)
which generates this output:
Here's my code:
N = 1:999;
for i = N
if rem(i,3) == 0 || rem(i,5) == 0
v(i,1) = i
end
end
Te problem is that I get an Array with some zeros in, but I just want an an arraywith the values comforming to my conditions.
How can I fix it?
Thank you!
I think the OP is looking for a result like:
v= N( (rem(N,3)==0) | (rem(N,5)==0) );
though without looping... :-)
I'm assuming that you're using a loop for a reason, and am not removing it from my solution. However, loops should be avoided where possible.
If I understand your question, you're trying to store only those values of i which correspond to a true conditional evaluation. You're problem is that you're using i as your index value inside the assignment statement. Use the end index keyword. Like so:
N = 1:999;
v = [];
for i = N
if rem(i,3) == 0 || rem(i,5) == 0
v(end+1) = i
end
end
I have a struct with several fields, some that should be numeric, and some that should be char. However, after my use of regexp I have chars in the fields that I want to use as numbers.
For example:
foo.str = 'one';
foo.data = '1';
foo(2).str = 'two';
foo(2).data = '2';
In my dreams I could do: foo.data = str2double(foo.data), but this does not work.
I could iterate through the struct, but that's only an ok option.
It's a long struct (100,000) with about 20 files.
for i = 1:length(foo)
foo(i).data = str2double(foo(i).data);
end
Any ideas?
Collect all elements from the subfield and call str2double once:
str2double({foo.data})
Here is a fairly compact solution that deals numerical values back into each foo.data:
fdnc = num2cell(str2double({foo.data}));
[foo.data] = deal(fdnc{:});
Remember to clear and redefine foo when testing.
EDIT: Fixed vertcat problem with multiple digits. Thanks, nispio.
A little cumbersome, but this works; no loops:
N = length(foo);
[aux_str{1:N}] = deal(foo.str);
[aux_data{1:N}] = deal(foo.data);
aux_data = mat2cell(str2double(aux_data),1,ones(1,N));
foo = cell2struct([aux_str; aux_data],{'str','data'},1);
Suppose I have a series of strings such as:
a = '101010101010'
b = '010101'
c = '000101010'
is there a way in Matlab to concatenate them and produce the binary number 101010101010010101000101010?
Use the concatenation operator [ ], with horizontal concatenation , (vertical concatenation ; will fail here unless you reshape() into column vectors):
[a,b,c]
However, I suggest storing your variables in a cell array:
s = {'101010101010','010101', '000101010'};
[s{:}]
or
cat(2,s{:})
To concatenate strings, you could say:
out = [a b c];
Alternatively:
out = strcat(a,b,c);
Yet another way:
out = sprintf('%s', a,b,c);
I think that this should work:
res = [a,b,c]
or alternatively call
res = strcat(a,b,c)
or, yet
res = cat(2,a,b,c)
Is there a way that I get all the structure subsubfield values of a subfield in one line ? Something like this :
struct.field(1:end).field
If I understand your question aright, you want to collect all the fields of the second-level structure, with the name 'field', into a single output array. It doesn't quite meet your request for a one-liner, but you can do it like this:
a.field1.a = 1;
a.field1.b = 2;
a.field2.a = 3;
a.field2.b = 4;
result = [];
for x = fieldnames(a)'
result = horzcat(result, a.(x{:}).a);
end
The ending value of result is [1 3]
Simple Structure Example
aStruct.subField = struct('subSubField', {1;2;3;4})
So that
aStruct.subField(1).subSubField == 1
aStruct.subField(1).subSubField == 2
Etc. Then the values of the leaf nodes can be obtained via a one-liner as
valueLeafs = [aStruct.subField.subSubField];
Which can be checked via assert(all(valueLeafs == [1,2,3,4])).
Non-Scalar Structure Example
The above one-liner also works when the leaf node values are non-scalar such that they can be horizontally concatenated. For example
bStruct.subField = struct('subSubField', {[1,2];[3,4]})
valueLeafs_b = [bStruct.subField.subSubField]; % works okay
cStruct.subField = struct('subSubField', {[1,2];[3;4]})
valueLeafs_c = [cStruct.subField.subSubField]; % error: bad arg dims
Distinct Class Structure Example
The one-line solution given previously does not work whenever the leaf node values are different class since they cannot - in general, be concatenated. However, use of arrayfun and a tricky anonymous function typically provide the required indexing technique:
dStruct.subField = struct('subSubField', {[1;2];'myString'});
valueLeafs_d = arrayfun(#(x) x.subSubField, dStruct.subField, 'UniformOutput', false)