Mean of every n number of doubles in a cell

2 次查看（过去 30 天）
Adnan Habib
Adnan Habib 2023-2-28
评论： Adnan Habib 2023-2-28
I have a cell with 990 doubles of 300 by 300 matrices. Lets call it T
I want to create a new cell with 26 doubles of 300 by 300 matrices each of which are the mean values of every 39 doubles from T (the 26th double of the new cell will be the mean of the final 15 doubles of T). i.e. (39 X 25) + (15 X 1) = 990.
Kindly help me with the code for this.
  2 个评论
Dyuman Joshi
Dyuman Joshi 2023-2-28
How do you want to group them?
#1 - [1-39], [40-78], [79-117], ...
#2 - [1,39,78,117, ...], [2, 40, 79, 118, ...], [3, 41, 80, 119, ...]
Adnan Habib
Adnan Habib 2023-2-28
Hi Dyuman Joshi, I want the first one 1-39, 40-78 etc.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Jan
Jan 2023-2-28
编辑：Jan 2023-2-28
I've reduced the test data size to 30x30 matrices - use the original sizes for your implementation:
T = squeeze(num2cell(rand(30, 30, 990), 1:2)); % Some test data, {990 x 1} cell
nT = numel(T);
R = cell(26, 1);
iR = 0;
for iT = 1:39:nT % Initial index of this block
fT = min(nT, iT + 38); % Final index of this block
tmp = 0;
for k = iT:fT % Accumulate elements of T
tmp = tmp + T{k};
end
iR = iR + 1; % Next output
R{iR} = tmp / (fT - iT + 1); % Mean value
end
Alternative approach:
nT = numel(T);
% Create [1, 40, 79, ...] with the last element is nT+1:
w = 39;
iT = 1:w:nT;
iT(end + (iT(end) == nT)) = nT + 1; % Consider nT is divisable by w
nR = numel(iT) - 1;
R2 = cell(nR, 1);
for iR = 1:nR
fT = iT(iR + 1) - 1; % Final index of this block
tmp = 0;
for k = iT(iR):fT % Accumulate elements of T
tmp = tmp + T{k};
end
R2{iR} = tmp / (fT - iT(iR) + 1); % Mean value
end
  1 个评论
Adnan Habib
Adnan Habib 2023-2-28
Thanks a lot Jan. The first one works. I didn't even try the alternative approach after that.

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Inputs 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by