求助
这张图，是我在MATLAB中调试这个程序得到的x=[2.48 2.90 2.5 2.5 2.3 2.39 3.30 2.7 2.8 2.5
2.62 2.90 2.6 2.9 2.4
2.51 2.95 2.6 2.2 2.4]
p=anova1(x)
但是，不知道怎样解读，麻烦各位大神，指点迷津！
function = anova1(x,group,displayopt,extra)
%ANOVA1 One-way analysis of variance (ANOVA).
% ANOVA1 performs a one-way ANOVA for comparing the means of two or more
% groups of data. It returns the p-value for the null hypothesis that the
% means of the groups are equal.
%
% P = ANOVA1(M) for a matrix M treats each column as a separate group,
% and determines whether the population means of the columns are equal.
% This form of ANOVA1 is appropriate when each group has the same number
% of elements (balanced ANOVA).
%
% P = ANOVA1(V,GROUP) groups elements in the vector V according to values
% in the grouping variable GROUP. GROUP must be a categorical variable,
% numeric vector, logical vector, string array, or cell array of strings
% with one group name for each element of X. X values corresponding to
% the same value of GROUP are placed in the same group.
%
% P = ANOVA1(M,GROUP) accepts a character array or cell array of strings,
% with one group name for each column of M. Columns with the same group
% name are treated as part of the same group.
%
% P = ANOVA1(X,GROUP,DISPLAYOPT) controls the display. DISPLAYOPT can be
% 'on' (the default) to display figures containing a standard one-way
% anova table and a boxplot, or 'off' to omit these displays. Note that
% the notches in the boxplot provide a test of group medians (see HELP
% BOXPLOT), and this is not the same as the F test for different means in
% the anova table. X can be either a vector or matrix. If X is a matrix
% and there are no group names, specify GROUP as [].
%
% = ANOVA1(...) returns the ANOVA table values as the
% cell array ANOVATAB.
%
% = ANOVA1(...) returns an additional structure
% of statistics useful for performing a multiple comparison of means
% with the MULTCOMPARE function.
%
% See also ANOVA2, ANOVAN, BOXPLOT, MANOVA1, MULTCOMPARE.
% Reference: Robert V. Hogg, and Johannes Ledolter, Engineering Statistics
% Macmillan 1987 pp. 205-206.
% Copyright 1993-2013 The MathWorks, Inc.
narginchk(1,4);
classical = 1;
nargs = nargin;
if (nargin>0 && strcmp(x,'kruskalwallis'))
% Called via kruskalwallis function, adjust inputs
classical = 0;
if (nargin >= 2), x = group; group = []; end
if (nargin >= 3), group = displayopt; displayopt = []; end
if (nargin >= 4), displayopt = extra; end
nargs = nargs-1;
end
if (nargs < 2), group = []; end
if (nargs < 3), displayopt = 'on'; end
% Note: for backwards compatibility, accept 'nodisplay' for 'off'
willdisplay = ~(strcmp(displayopt,'nodisplay') | strcmp(displayopt,'n') ...
| strcmp(displayopt,'off'));
% Convert group to cell array from character array, make it a column
if (ischar(group) && ~isempty(group))
group = cellstr(group);
end
if (size(group, 1) == 1)
group = group';
end
% If the input is a matrix, it may not be balanced if it contains NaNs or
% if there are repeated grouping values, so turn it into a vector.
needvector = false;
if ~isvector(x)
if (any(isnan(x(:))))
needvector = true;
elseif ~isempty(group) && length(group)~=length(unique(group))
needvector = true;
end
end
% If X is a matrix with NaNs, convert to vector form.
if ~isvector(x)
if needvector
= size(x);
x = x(:);
gi = reshape(repmat((1:m), n, 1), n*m, 1);
if isempty(group) % no group names
group = gi;
elseif (size(group,1) == m)
group = group(gi,:);
else
error(message('stats:anova1:InputSizeMismatch'));
end
end
elseif ~isempty(group) && (size(group,1) ~= length(x))
error(message('stats:anova1:InputSizeMismatch'));
end
% If X is a matrix GROUP is provided with the correct size, use GROUP
% to define groups and to label boxes
if (~isempty(group) && (length(x) < numel(x)) ...
&& (size(x,2) == size(group,1)))
named = 1;
= grp2idx(group);
gnames = gnames(gid);
grouped = 0;
else
named = 0;
gnames = [];
grouped = ~isempty(group);
end
if (grouped)
% Single data vector and a separate grouping variable
x = x(:);
lx = length(x);
if (lx ~= numel(x))
error(message('stats:anova1:VectorRequired'))
end
nonan = ~isnan(x);
x = x(nonan);
% Convert group to indices 1,...,g and separate names
group = group(nonan,:);
= grp2idx(group);
named = 1;
% Remove NaN values
nonan = ~isnan(groupnum);
if (~all(nonan))
groupnum = groupnum(nonan);
x = x(nonan);
end
lx = length(x);
xorig = x; % use uncentered version to make M
groupnum = groupnum(:);
maxi = size(gnames, 1);
if isa(x,'single')
xm = zeros(1,maxi,'single');
else
xm = zeros(1,maxi);
end
countx = xm;
if (classical)
mu = mean(x);
x = x - mu; % center to improve accuracy
xr = x;
else
= tiedrank(x);
end
for j = 1:maxi
% Get group sizes and means
k = find(groupnum == j);
lk = length(k);
countx(j) = lk;
xm(j) = mean(xr(k)); % column means
end
gm = mean(xr); % grand mean
df1 = sum(countx>0) - 1; % Column degrees of freedom
df2 = lx - df1 - 1; % Error degrees of freedom
xc = xm - gm; % centered
xc(countx==0) = 0;
RSS = dot(countx, xc.^2); % Regression Sum of Squares
else
% Data in matrix form, no separate grouping variable
= size(x);
lx = r * c;
if (classical)
xr = x;
mu = mean(xr(:));
xr = xr - mu; % center to improve accuracy
else
= tiedrank(x(:));
xr = reshape(xr, size(x));
end
countx = repmat(r, 1, c);
xorig = x; % save uncentered version for boxplot
xm = mean(xr); % column means
gm = mean(xm); % grand mean
df1 = c-1; % Column degrees of freedom
df2 = c*(r-1); % Error degrees of freedom
RSS = r*(xm - gm)*(xm-gm)'; % Regression Sum of Squares
end
TSS = (xr(:) - gm)'*(xr(:) - gm); % Total Sum of Squares
SSE = TSS - RSS; % Error Sum of Squares
if (df2 > 0)
mse = SSE/df2;
else
mse = NaN;
end
if (classical)
if (SSE~=0)
F = (RSS/df1) / mse;
p = fpval(F,df1,df2); % Probability of F given equal means.
elseif (RSS==0) % Constant Matrix case.
F = NaN;
p = NaN;
else % Perfect fit case.
F = Inf;
p = 0;
end
else
F = (12 * RSS) / (lx * (lx+1));
if (tieadj > 0)
F = F / (1 - 2 * tieadj/(lx^3-lx));
end
p = chi2pval(F,df1);
end
Table=zeros(3,5); %Formatting for ANOVA Table printout
Table(:,1)=[ RSS SSE TSS]';
Table(:,2)=';
Table(:,3)=[ RSS/df1 mse Inf ]';
Table(:,4)=[ F Inf Inf ]';
Table(:,5)=[ p Inf Inf ]';
colheads = {getString(message('stats:anova1:ColHeadSource')), getString(message('stats:anova1:ColHeadSS')), getString(message('stats:anova1:ColHeadDf')), getString(message('stats:anova1:ColHeadMS')), getString(message('stats:anova1:ColHeadF')), getString(message('stats:anova1:ColHeadProbGtF'))};
if (~classical)
colheads{5} = getString(message('stats:anova1:ColHeadChisq'));
colheads{6} = getString(message('stats:anova1:ColHeadProbGtChisq'));
end
rowheads = {getString(message('stats:anova1:RowHeadColumns')), getString(message('stats:anova1:RowHeadError')), getString(message('stats:anova1:RowHeadTotal'))};
if (grouped)
rowheads{1} = getString(message('stats:anova1:RowHeadGroups'));
end
% Create cell array version of table
atab = num2cell(Table);
for i=1:size(atab,1)
for j=1:size(atab,2)
if (isinf(atab{i,j}))
atab{i,j} = [];
end
end
end
atab = ;
atab = ;
if (nargout > 1)
anovatab = atab;
end
% Create output stats structure if requested, used by MULTCOMPARE
if (nargout > 2)
if ~isempty(gnames)
stats.gnames = gnames;
else
stats.gnames = strjust(num2str((1:length(xm))'),'left');
end
stats.n = countx;
if (classical)
stats.source = 'anova1';
stats.means = xm + mu;
stats.df = df2;
stats.s = sqrt(mse);
else
stats.source = 'kruskalwallis';
stats.meanranks = xm;
stats.sumt = 2 * tieadj;
end
end
if (~willdisplay), return; end
digits = [-1 -1 0 -1 2 4];
if (classical)
wtitle = getString(message('stats:anova1:OnewayANOVA'));
ttitle = getString(message('stats:anova1:ANOVATable'));
else
wtitle = getString(message('stats:anova1:KruskalWallisOnewayANOVA'));
ttitle = getString(message('stats:anova1:KruskalWallisANOVATable'));
end
tblfig = statdisptable(atab, wtitle, ttitle, '', digits);
set(tblfig,'tag','table');
f1 = figure('pos',get(gcf,'pos') + ,'tag','boxplot');
ax = axes('Parent',f1);
if (~grouped)
boxplot(ax,xorig,'notch','on');
else
boxplot(ax,xorig,groupnum,'notch','on');
h = get(ax,'XLabel');
set(h,'String',getString(message('stats:anova1:GroupNumber')));
end
% If there are group names, use them
if ~isempty(gnames)
h = get(ax,'XLabel');
if (named)
set(h,'String','');
end
set(ax, 'xtick', (1:size(gnames,1)), 'xticklabel', gnames);
end
在matlab中的ANOVA分为anova1，anova2，anovan共三个函数。
详细说明见官网解释：http://www.mathworks.cn/help/toolbox/stats/anova1.html
anova1： one-way analysis of variance 单因素方差分析
anova2： two-way analysis of variance 两因素方差分析
anovan: n-way analysis of variance 多因素方差分析
一般使用anova1足矣。
特别注意，给出的矩阵X要每个列含有相同个数的元素，做anova时结果会给出p-value，p-value越小，说明各组样品的均值中不全相同，至少有一组的均值有显著性差异。一般的显著性等级（Common significance level）为0.05或者0.01。
若p-value大于Common significance level，则说明各组均值之间没有统计学差异。
急需大神支招！
有点难度
森之张卫东 发表于 2015-10-29 22:45 static/image/common/back.gif
在matlab中的ANOVA分为anova1，anova2，anovan共三个函数。
详细说明见官网解释：http://www.mathworks.cn ...
The standard ANOVA table has six columns:
1. The source of the variability.
2. The sum of squares (SS) due to each source.
3. The degrees of freedom (df) associated with each source.
4. The mean squares (MS) for each source, which is the ratio SS/df.
5. The F-statistic, which is the ratio of the mean squares.
6. The p-value, which is derived from the cdf of F.
The box plot of the columns of X suggests the size of the F-statistic and the p-value. Large differences in the center lines of the boxes correspond to large values of F and correspondingly small values of p.
标准方差分析表有六列:
1。变化的来源。
2。平方和(SS)由于每个源。
3所示。自由度(df)与每个源有关。
4所示。平均为每个源广场(MS),SS / df的比率。
5。的f统计量的比例意味着广场。
6。假定值,这是来自F的运作。
X的列的箱线图表明f统计量和假定值的大小。大箱子对应的中心线的差异大的F值和相应的小p值。
The anova1 function displays two figures, the standard ANOVA table and a box plot of the columns of X.
The standard ANOVA table divides the variability of the data into two parts:
• Variability due to the differences among the column means (variability between groups)
• Variability due to the differences between the data in each column and the column mean (variability within groups)
anova1函数显示两个数据,标准方差分析表和列的箱线图的X。
标准方差分析表中数据的变化分为两个部分:
•可变性由于差异列意味着(组)之间的差异
•可变性将在每一列数据之间的差异和列意味着(组内差异
The small p -value indicates that differences between column means are significant. The probability of this outcome under the null hypothesis (that samples drawn from the same population would have means differing by the amounts seen in X ) is equal to the p -value.
同学，解决没，我想学习学习！
路过的大神，如果可以给出解答，偶将不甚感激！
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