I have a matrix with the size 4*n, lets say for instance (4*3000)
So what is the fastest way to store and read the elements from the matrix
I have tried two solutions that have given me the same time approximately.
one array with a size of 12000 elements (2D --> 1D) read by (i+ width*j)
4 arrays with the size 1*3000 and then by using (IF ELSE or Switch case) statement to decide which array i should read
Thus, is there another solution to be used.
Furthermore, how to use the shift technique >> to solve the problem if its applicable for this case
First technique should be faster.
Also, you can improve performance by accessing elements inside of a loop in a row (...arr[11] = ...; arr[12] = ...; arr[13] = ...;...).
Related
I ran into a big of a problem with a tetris program I'm writing currently in C.
I am trying to use a 4D multi-dimensional array e.g.
uint8_t shape[7][4][4][4]
but I keep getting seg faults when I try that, I've read around and it seems to be that I'm using up all the stack memory with this kind of array (all I'm doing is filling the array with 0s and 1s to depict a shape so I'm not inputting a ridiculously high number or something).
Here is a version of it (on pastebin because as you can imagine its very ugly and long).
If I make the array smaller it seems to work but I'm trying to avoid a way around it as theoretically each "shape" represents a rotation as well.
https://pastebin.com/57JVMN20
I've read that you should use dynamic arrays so they end up on the heap but then I run into the issue how someone would initialize a dynamic array in such a way as linked above. It seems like it would be a headache as I would have to go through loops and specifically handle each shape?
I would also be grateful for anybody to let me pick their brain on dynamic arrays how best to go about them and if it's even worth doing normal arrays at all.
Even though I have not understood why do you use 4D arrays to store shapes for a tetris game, and I agree with bolov's comment that such an array should not overflow the stack (7*4*4*4*1 = 448 bytes), so you should probably check other code you wrote.
Now, to your question on how to manage 4D (N-Dimensional)dynamically sized arrays. You can do this in two ways:
The first way consists in creating an array of (N-1)-Dimensional arrays. If N = 2 (a table) you end up with a "linearized" version of the table (a normal array) which dimension is equal to R * C where R is the number of rows and C the number of columns. Inductively speaking, you can do the very same thing for N-Dimensional arrays without too much effort. This method has some drawbacks though:
You need to know beforehand all the dimensions except one (the "latest") and all the dimensions are fixed. Back to the N = 2 example: if you use this method on a table of C columns and R rows, you can change the number of rows by allocating C * sizeof(<your_array_type>) more bytes at the end of the preallocated space, but not the number of columns (not without rebuilding the entire linearized array). Moreover, different rows must have the same number of columns C (you cannot have a 2D array that looks like a triangle when drawn on paper, just to get things clear).
You need to carefully manage the indicies: you cannot simply write my_array[row][column], instead you must access that array with my_array[row*C + column]. If N is not 2, then this formula gets... interesting
You can use N-1 arrays of pointers. That's my favourite solution because it does not have any of the drawbacks from the previous solution, although you need to manage pointers to pointers to pointers to .... to pointers to a type (but that's what you do when you access my_array[7][4][4][4].
Solution 1
Let's say you want to build an N-Dimensional array in C using the first solution.
You know the length of each dimension of the array up to the (N-1)-th (let's call them d_1, d_2, ..., d_(N-1)). We can build this inductively:
We know how to build a dynamic 1-dimensional array
Supposing we know how to build a (N-1)-dimensional array, we show that we can build a N-Dimensional array by putting each (N-1)-dimensional array we have available in a 1-Dimensional array, thus increasing the available dimensions by 1.
Let's also assume that the data type that the arrays must hold is called T.
Let's suppose we want to create an array with R (N-1)-dimensional arrays inside it. For that we need to know the size of each (N-1)-dimensional array, so we need to calculate it.
For N = 1 the size is just sizeof(T)
For N = 2 the size is d_1 * sizeof(T)
For N = 3 the size is d_2 * d_1 * sizeof(T)
You can easily inductively prove that the number of bytes required to store R (N-1)-dimensional arrays is R*(d_1 * d_2 * ... * d_(n-1) * sizeof(T)). And that's done.
Now, we need to access a random element inside this massive N-dimensional array. Let's say we want to access the item with indicies (i_1, i_2, ..., i_N). For this we are going to repeat the inductive reasoning:
For N = 1, the index of the i_1 element is just my_array[i_1]
For N = 2, the index of the (i_1, i_2) element can be calculated by thinking that each d_1 elements, a new array begins, so the element is my_array[i_1 * d_1 + i_2].
For N = 3, we can repeat the same process and end up having the element my_array[d_2 * ((i_1 * d_1) + i_2) + i_3]
And so on.
Solution 2
The second solution wastes a bit more memory, but it's more straightforward, both to understand and to implement.
Let's just stick to the N = 2 case so that we can think better. Imagine to have a table and to split it row by row and to place each row in its own memory slot. Now, a row is a 1-dimensional array, and to make a 2-dimensional array we only need to be able to have an ordered array with references to each row. Something like the following drawing shows (the last row is the R-th row):
+------+
| R1 -------> [1,2,3,4]
|------|
| R2 -------> [2,4,6,8]
|------|
| R3 -------> [3,6,9,12]
|------|
| .... |
|------|
| RR -------> [R, 2*R, 3*R, 4*R]
+------+
In order to do that, you need to first allocate the references array (R elements long) and then, iterate through this array and assign to each entry the pointer to a newly allocated memory area of size d_1.
We can easily extend this for N dimensions. Simply build a R dimensional array and, for each entry in this array, allocate a new 1-Dimensional array of size d_(N-1) and do the same for the newly created array until you get to the array with size d_1.
Notice how you can easily access each element by simply using the expression my_array[i_1][i_2][i_3]...[i_N].
For example, let's suppose N = 3 and T is uint8_t and that d_1, d_2 and d_3 are known (and not uninitialized) in the following code:
size_t d1 = 5, d2 = 7, d3 = 3;
int ***my_array;
my_array = malloc(d1 * sizeof(int**));
for(size_t x = 0; x<d1; x++){
my_array[x] = malloc(d2 * sizeof(int*));
for (size_t y = 0; y < d2; y++){
my_array[x][y] = malloc(d3 * sizeof(int));
}
}
//Accessing a random element
size_t x1 = 2, y1 = 6, z1 = 1;
my_array[x1][y1][z1] = 32;
I hope this helps. Please feel free to comment if you have questions.
Is it possible to create arrays based of their index as in
int x = 4;
int y = 5;
int someNr = 123;
int foo[x][y] = someNr;
dynamically/on the run, without creating foo[0...3][0...4]?
If not, is there a data structure that allow me to do something similar to this in C?
No.
As written your code make no sense at all. You need foo to be declared somewhere and then you can index into it with foo[x][y] = someNr;. But you cant just make foo spring into existence which is what it looks like you are trying to do.
Either create foo with correct sizes (only you can say what they are) int foo[16][16]; for example or use a different data structure.
In C++ you could do a map<pair<int, int>, int>
Variable Length Arrays
Even if x and y were replaced by constants, you could not initialize the array using the notation shown. You'd need to use:
int fixed[3][4] = { someNr };
or similar (extra braces, perhaps; more values perhaps). You can, however, declare/define variable length arrays (VLA), but you cannot initialize them at all. So, you could write:
int x = 4;
int y = 5;
int someNr = 123;
int foo[x][y];
for (int i = 0; i < x; i++)
{
for (int j = 0; j < y; j++)
foo[i][j] = someNr + i * (x + 1) + j;
}
Obviously, you can't use x and y as indexes without writing (or reading) outside the bounds of the array. The onus is on you to ensure that there is enough space on the stack for the values chosen as the limits on the arrays (it won't be a problem at 3x4; it might be at 300x400 though, and will be at 3000x4000). You can also use dynamic allocation of VLAs to handle bigger matrices.
VLA support is mandatory in C99, optional in C11 and C18, and non-existent in strict C90.
Sparse arrays
If what you want is 'sparse array support', there is no built-in facility in C that will assist you. You have to devise (or find) code that will handle that for you. It can certainly be done; Fortran programmers used to have to do it quite often in the bad old days when megabytes of memory were a luxury and MIPS meant millions of instruction per second and people were happy when their computer could do double-digit MIPS (and the Fortran 90 standard was still years in the future).
You'll need to devise a structure and a set of functions to handle the sparse array. You will probably need to decide whether you have values in every row, or whether you only record the data in some rows. You'll need a function to assign a value to a cell, and another to retrieve the value from a cell. You'll need to think what the value is when there is no explicit entry. (The thinking probably isn't hard. The default value is usually zero, but an infinity or a NaN (not a number) might be appropriate, depending on context.) You'd also need a function to allocate the base structure (would you specify the maximum sizes?) and another to release it.
Most efficient way to create a dynamic index of an array is to create an empty array of the same data type that the array to index is holding.
Let's imagine we are using integers in sake of simplicity. You can then stretch the concept to any other data type.
The ideal index depth will depend on the length of the data to index and will be somewhere close to the length of the data.
Let's say you have 1 million 64 bit integers in the array to index.
First of all you should order the data and eliminate duplicates. That's something easy to achieve by using qsort() (the quick sort C built in function) and some remove duplicate function such as
uint64_t remove_dupes(char *unord_arr, char *ord_arr, uint64_t arr_size)
{
uint64_t i, j=0;
for (i=1;i<arr_size;i++)
{
if ( strcmp(unord_arr[i], unord_arr[i-1]) != 0 ){
strcpy(ord_arr[j],unord_arr[i-1]);
j++;
}
if ( i == arr_size-1 ){
strcpy(ord_arr[j],unord_arr[i]);
j++;
}
}
return j;
}
Adapt the code above to your needs, you should free() the unordered array when the function finishes ordering it to the ordered array. The function above is very fast, it will return zero entries when the array to order contains one element, but that's probably something you can live with.
Once the data is ordered and unique, create an index with a length close to that of the data. It does not need to be of an exact length, although pledging to powers of 10 will make everything easier, in case of integers.
uint64_t* idx = calloc(pow(10, indexdepth), sizeof(uint64_t));
This will create an empty index array.
Then populate the index. Traverse your array to index just once and every time you detect a change in the number of significant figures (same as index depth) to the left add the position where that new number was detected.
If you choose an indexdepth of 2 you will have 10² = 100 possible values in your index, typically going from 0 to 99.
When you detect that some number starts by 10 (103456), you add an entry to the index, let's say that 103456 was detected at position 733, your index entry would be:
index[10] = 733;
Next entry begining by 11 should be added in the next index slot, let's say that first number beginning by 11 is found at position 2023
index[11] = 2023;
And so on.
When you later need to find some number in your original array storing 1 million entries, you don't have to iterate the whole array, you just need to check where in your index the first number starting by the first two significant digits is stored. Entry index[10] tells you where the first number starting by 10 is stored. You can then iterate forward until you find your match.
In my example I employed a small index, thus the average number of iterations that you will need to perform will be 1000000/100 = 10000
If you enlarge your index to somewhere close the length of the data the number of iterations will tend to 1, making any search blazing fast.
What I like to do is to create some simple algorithm that tells me what's the ideal depth of the index after knowing the type and length of the data to index.
Please, note that in the example that I have posed, 64 bit numbers are indexed by their first index depth significant figures, thus 10 and 100001 will be stored in the same index segment. That's not a problem on its own, nonetheless each master has his small book of secrets. Treating numbers as a fixed length hexadecimal string can help keeping a strict numerical order.
You don't have to change the base though, you could consider 10 to be 0000010 to keep it in the 00 index segment and keep base 10 numbers ordered, using different numerical bases is nonetheless trivial in C, which is of great help for this task.
As you make your index depth become larger, the amount of entries per index segment will be reduced
Please, do note that programming, especially lower level like C consists in comprehending the tradeof between CPU cycles and memory use in great part.
Creating the proposed index is a way to reduce the number of CPU cycles required to locate a value at the cost of using more memory as the index becomes larger. This is nonetheless the way to go nowadays, as masive amounts of memory are cheap.
As SSDs' speed become closer to that of RAM, using files to store indexes is to be taken on account. Nevertheless modern OSs tend to load in RAM as much as they can, thus using files would end up in something similar from a performance point of view.
I want to load a csv file to Matlab using testread(), since the data in it has more than 2 million records, so I should preallocate the array for those data.
Suppose I cannot know the exact length of arrays, the docs of MATLAB v6.5 recommend me to use repmat() for my expanding array. The original words in the doc is below:
"In cases where you cannot preallocate, see if you can increase the
size of your array using the repmat function. repmat tries to get you
a contiguous block of memory for your expanding array".
I really don't know how to use the repmat for expanding?
Does it mean by estimating a rough number of the length for repmat() to preallocating, and then remove the empty elements?
If so, how is that different from preallocating using zeros() or cell()?
The documentation also says:
When you preallocate a block of memory to hold a matrix of some type
other than double, it is more memory efficient and sometimes faster to
use the repmat function for this.
The statement below uses zeros to preallocate a 100-by-100 matrix of
uint8. It does this by first creating a full matrix of doubles, and
then converting the matrix to uint8. This costs time and uses memory
unnecessarily.
A = int8(zeros(100));
Using repmat, you create only one double, thus reducing your memory
needs.
A = repmat(int8(0), 100, 100);
Therefore, the advantage is if you want a datatype other than doubles, you can use repmat to replicate a non-double datatype.
Also see: http://undocumentedmatlab.com/blog/preallocation-performance, which suggests:
data1(1000,3000) = 0
instead of:
data1 = zeros(1000,3000)
to avoid initialisation of other elements.
As for dynamic resizing, repmat can be used to concisely double the size of your array (a common method which results in amortized O(1) appends for each element):
data = [0];
i = 1;
while another element
...
if i > numel(data)
data = repmat(data,1,2); % doubles the size of data
end
data(i) = element
i = i + 1;
end
And yes, after you have gathered all your elements, you can resize the array to remove empty elements at the end.
I'm trying to create a cell array of size N,
where every cell is a randomized Matrix of size M,
I've tried using deal or simple assignments, but the end result is always N identical Matrices of size M
for example:
N=20;
M=10;
CellArray=cell(1,N);
CellArray(1:20)={rand(M)};
this yields identical matrices in each cell, iv'e tried writing the assignment like so:
CellArray{1:20}={rand(M)};
but this yields the following error:
The right hand side of this assignment has too few values to satisfy the left hand side.
the ends results should be a set of transition probability matrices to be used for a model i'm constructing,
there's a currently working version of the model, but it uses loops to create the matrices, and works rather slowly,
i'd be thankful for any help
If you don't want to use loops because you are interested in a low execution time, get rid of the cells.
RandomArray=rand(M,M,N)
You can access each slice, which is your intended MxM matrix, using RandomArray(:,:,index)
Use cellfun:
N = 20;
M = 10;
CellArray = cellfun(#(x) rand(M), cell(1,N), 'uni',0)
For every cell it newly calls rand(M) - unlike before, you were assigning the same rand(M) to every cell, which was just computed once.
I have the above loop running on the above variables:
A is a 2d array of size mxn.
mask is a 1d logical array of size 1xn
results is a 1d array of size 1xn
B is a vector of the form mx1
C is a mxm matrix, m is the same as the above.
Edit: expanded foo(x) into the function.
here is the code:
temp = (B.'*C*B);
for k = 1:n
x = A(:,k);
if(mask(k) == 1)
result(k) = (B.'*C*x)^2 / (temp*(x.'*C*x)); %returns scalar
end
end
take note, I am already successfully using the above code as a parfor loop instead of for. I was hoping you would be able to suggest some way to use meshgrid or the sort to yield better performance improvement. I don't think I have RAM problems so a solution can also be expensive memory wise.
Many thanks.
try this:
result=(B.'*C*A).^2./diag(temp*(A.'*C*A))'.*mask;
This vectorization via matrix multiplication will also make sure that result is a 1xn vector. In the code you provided there can be a case where the last elements in mask are zeros, in this case your code will truncate result to a smaller length, whereas, in the answer it'll keep these elements zero.
If your foo admits matrix input, you could do:
result = zeros(1,n); % preallocate result with zeros
mask = logical(mask); % make mask logical type
result(mask) = foo(A(mask),:); % compute foo for all selected columns