cumulative average for each element
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hey, i need to build a function who receives a vector and return the cumulative average for each element . for ex: vecMean([1 2 3 4])= [1 1.5 2 2.5].
the problem is that im reciving a row vector but sometimes i want to put column vec and recive a column also.
% the function calculates the cumulative average of elements in a vector.
% each element in the output vector is the average of this element and all
% previous elements in the input.
% the function gets a vector of numbers (=vecOfNum)
% and return output vector (=vecPrevMean),
% each element is the average of all previous elements.
function [vecPrevMean]= vecMean(vecOfNum)
count=1; %count the amount of elements in the vector
sumNum=0; %sum the elements
for i=1:length(vecOfNum);
if iscolumn(vecOfNum)==0
sumNum= sumNum+ vecOfNum(i);
vecPrevMean(i)=sumNum./count;
count=count+1;
elseif iscolumn(vecOfNum)==1
sumNum= sumNum+ vecOfNum(i);
vecPrevMean(i)=(sumNum./count)';
count=count+1;
end
end
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採用された回答
madhan ravi
2018 年 11 月 26 日
編集済み: madhan ravi
2018 年 11 月 26 日
no loops needed:
>> a=1:4
desired_result=cumsum(a)./(1:numel(a))
a =
1 2 3 4
desired_result =
1.0000 1.5000 2.0000 2.5000
>>
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その他の回答 (1 件)
fakhri
2022 年 11 月 30 日
hey,
Write a function called matMean that calculates the cumulative average of the rows or the columns of a matrix. Your function should get two input arguments: • First input argument: a matrix of numbers. • Second input argument: Type of cumulative average to calculate - can get one of two values: the number 1 for calculating the cumulative average of the rows of the matrix, or the number 2 for calculating the cumulative average of the columns of the matrix.
For this question you should write your own algorithm for calculating the cumulative averages of the arrays - do not use built-in statistical MATLAB functions such as: mean, sum, cumsum, etc
1 件のコメント
DGM
2022 年 12 月 1 日
What part of this is an answer to the question?
What have you done other than pasting your assignment?
" For this question you should write your own algorithm"
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