以下是MATLAB mean函数的问题，麻烦帮忙看看咋解决！

Question

A=[1 3 4 5; 2 3 4 6;1 3 1 5]对矩阵A求平均，mean(A):如果是求平均的话，答案应该是： 1.3333 3.0000 3.0000 5.3333为什么我这显示的答案是： 4 9 9 16非常费解！请问有人碰到这种情况没有？ 咋解决？>> 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我刚把这位仁兄的代码替换了我机子上的mean.m.结果还是如上，疯了！

Answer 1

在命令行输入which mean
把结果发上来
============
你先把which mean的结果发上来，我怀疑你自己写过一个叫mean的函数

Answer 2

>> 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)表示对列取平均，mean(A,2）表示对行取平均，mean(A）则默认为mean(A,1）

Answer 3

在磁盘:\matlb\toolbox\matlab\datafun下可以找到mean，把以下原代码贴到里面去：
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