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How to handle a large binary vector?

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Phan Dang Thoai
Phan Dang Thoai on 26 Jul 2021
Commented: Jan on 8 Aug 2021
I need to handle a frame which incluses 230400 bits. If I use an array to include element of 0 or 1 in unit8 type then I need memory of about 225 Mb. If I use an element of int64 then I need about 3.5 Mb. What is the best practice to handle this large binary vector. In either ways I mentions, it needs almost Mbs to store a frame.
Update: Actually I will process further with 230400 bits. There few blocks where this 230400 bits frame will go through. It is convenient for me to process as a vector of 230400 bits. But I am not sure how normally this problem to be handled in Simulink. Do we need to divide it and process each byte or a portion of 230400 bits each time.
  4 Comments
Phan Dang Thoai
Phan Dang Thoai on 7 Aug 2021
Sorry James for the late reply/ I just want to know how do you process each bit as a single element? My block gets input as a vecor of 230400 bits, then arranged in a matrix of 75*3072 bits, then each couple of two bits are transfromed to a comlex number [(1-2*bit1)+j(1-2*bit2)]/sqrt(2). Bit1 is element (l,n) and bit2 is element (l,n+K), l: from 1 to 75, n: from 1 to 3072/2, K=3072/2.

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Answers (1)

Jan
Jan on 26 Jul 2021
UINT8 oder Logical use 1 byte per value. 225 MB are usually fine. You can store the bits packed in UINT8 also, but this is harder to access. So it depends on what you want to do with the data.
How do you import the data of the "frame"?
Maybe this helps:
Bit = PackBits([1,0,1,0,1,0,1,1, 1,1,1,0,1,1,1,0])
Bit = 1×2
213 119
UnpackBits(Bit)
ans = 1×16
1 0 1 0 1 0 1 1 1 1 1 0 1 1 1 0
function Byte = PackBits(Bit)
% Byte = PackBits(Bit)
% Input: Numerical vector which is 1 for a set bit and 0 otherwise.
% Treated as vector. NUMEL(Bit) must be a multiple of 8.
% Output: UINT8 using all 8 bits per Byte. LSB order.
% Failing in R2021a: bitshift(uint8(1), 0:7) * reshape(uint8(Bit), 8, [])
Byte = sum(bitshift(uint8(1), 0:7).' .* reshape(uint8(Bit), 8, []), 1);
end
function Bit = UnpackBits(Byte)
% Bits = UnpackBits(Byte)
% Input: UINT8 using all 8 bits per Byte. LSB order. Treated as vector.
% Output: UINT8, 1 for set bits, 0 otherwise. NUMEL(Bit) = 8*NUMEL(Byte)
% Failing in R2021a: bitshift(uint8(1), 0:7) * reshape(uint8(Bit), 8, [])
Byte = Byte(:).';
Bit = [bitget(Byte, 1); bitget(Byte, 2); bitget(Byte, 3); bitget(Byte, 4); ...
bitget(Byte, 5); bitget(Byte, 6); bitget(Byte, 7); bitget(Byte, 8)];
Bit = reshape(Bit, 1, []);
end
% License: Jan-DWYW: Do what you want with this code.
% Don't blame me and don't mention my name, if your code contains bugs.
@all readers: Do oyu have a more efficient idea for UnpackBits?
  4 Comments
Jan
Jan on 8 Aug 2021
@Phan Dang Thoai: I cannot follow your descriptions.

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