A=[1 3 4 5; 2 3 4 6;1 3 1 5]對矩陣A求平均，mean(A):如果是求平均的話，答案應該是： 1.3333 3.0000 3.0000 5.3333為什么我這顯示的答案是： 4 9 9 16非常費解！請問有人碰到這種情況沒有？ 咋解決？>> A=[1 3 4 5; 2 3 4 6;1 3 1 5];>> B=mean(A)B =4 9 9 16>> C=mean(A,1)??? Error using ==> meanToo many input arguments我剛把這位仁兄的代碼替換了我機子上的mean.m.結果還是如上，瘋了！
3 回答
精慕HU
TA貢獻1845條經驗 獲得超8個贊
>> A=[1 3 4 5; 2 3 4 6;1 3 1 5];
>> mean(A,1)
ans =
1.3333 3.0000 3.0000 5.3333
>> mean(A,2)
ans =
3.2500
3.7500
2.5000
>>
mean(A,1)表示對列取平均，mean(A,2）表示對行取平均，mean(A）則默認為mean(A,1）
慕森王
TA貢獻1777條經驗 獲得超3個贊
在磁盤:\matlb\toolbox\matlab\datafun下可以找到mean，把以下原代碼貼到里面去：
function y = mean(x,dim)
%MEAN Average or mean value.
% For vectors, MEAN(X) is the mean value of the elements in X. For
% matrices, MEAN(X) is a row vector containing the mean value of
% each column. For N-D arrays, MEAN(X) is the mean value of the
% elements along the first non-singleton dimension of X.
%
% MEAN(X,DIM) takes the mean along the dimension DIM of X.
%
% Example: If X = [0 1 2
% 3 4 5]
%
% then mean(X,1) is [1.5 2.5 3.5] and mean(X,2) is [1
% 4]
%
% Class support for input X:
% float: double, single
%
% See also MEDIAN, STD, MIN, MAX, VAR, COV, MODE.
% Copyright 1984-2005 The MathWorks, Inc.
% $Revision: 5.17.4.3 $ $Date: 2005/05/31 16:30:46 $
if nargin==1,
% Determine which dimension SUM will use
dim = min(find(size(x)~=1));
if isempty(dim), dim = 1; end
y = sum(x)/size(x,dim);
else
y = sum(x,dim)/size(x,dim);
end
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