Matlab问题,高手请教!!
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发布时间：2022-04-23 02:08
共4个回答
热心网友
时间：2023-08-25 08:22
function [R, ind] = randinterval(N, Interval, Weight)
% RANDINTERVAL - random numbers within multiple intervals
%
% R = RANDINTERVAL(N, INTERVALS) returns a N-by-N matrix R of random numbers
% taken from a uniform distribution over multiple intervals.
%
% Similar to RAND, RANDINTERVAL([N M],..) returns a N-by-M matrix, and
% RANDINTERVAL([N M P ...],..) returns a N-by-M-by-P-by-.. array. Note
% that a notation similar to rand(N,M,P,..) is not available.
%
% INTERVALS is a N-by-2 matrix in which the rows specify the intervals from
% which the numbers are drawn. Each interval is given by a lower bound
% (in the first column of INTERVALS) and an upper bound (2nd column of
% INTERVALS). If the lower bound equals the upper bound, no numbers will
% be drawn from that interval. The upper bound cannot be smaller than the
% lower bound.
% If INTERVALS is a vector (K-by-1 or 1-by-K), the intervals are
% formed as (INTERVALS(1),INTERVALS(2)), (INTERVALS(2),INTERVALS(3)),
% ... , (INTERVALS(K-1),INTERVALS(K)). In this case, the number of
% intervals N is one less than the number of elements K, i.e., N=K-1.
% The relative length of an interval determines the likelihood that a
% number is drawn from that interval. For instance, when INTERVALS is [1
% 2 ; 10 12], the probablity that a number is drawn from the first
% interval (1,2) is 1 out of 3, vs 2 out of 3 for the second interval (10,12).
%
% R = RANDINTERVAL(N, INTERVALS, WEIGHTS) allow for weighting each
% interval specifically. WEIGHTS is vector with N numbers. These numbers
% specify, in combination with the length of the corresponding interval,
% the relative likelihood for each interval (i.e., the weight). Larger
% values of W(k) increases the likelihood that a number is drawn from the
% k-th interval.
% For instance, when INTERVALS is [11 12 ; 20 21] and WEIGHTS is [3 1], the
% probability that a numbers is drawn from the first interval is 3 out of
% 4, vs 1 out of 4 for the second interval. Note that both intervals have
% the same length. When the intervals do not have the same lengths, the
% likelihood that a number is drawn from an interval k, with length L(k)
% and weight W(k) is given by the formula:
% p(k) = (W(k) x L(k)) / sum(W(i) ./ L(i)),i=1:N
%
% [R, IND] = RANDINTERVAL(..) also returns a index array IND, which has the
% same size as R. IND contains numbers between 1 and the number of
% intervals. Intervals with zero length and/or a weight of zero will not
% be present in IND.
%
% Notes:
% - overlapping intervals are allowed, but may proce hard-to-debug
% results. In this case a warning is issued. On the other, such
% overlapping intervals could be (ab-)used to draw from specific
% distributions.
% - for a single interval (A B) this function is equivalent to
% A + (B-A) * rand(N)
%
% Examples:
% % Basic use, no weights
% R = randinterval([1,10000], [1 3 ; 5 7 ; 10 15 ; 20 26]) ;
% % returns a row vector with 10000 numbers between 1 and 3, between
% % 5 and 7, between 10 and 15, or between 20 and 26. Approximately
% % 2/5 of the numbers is between 20 and 26:
% N = histc(R,[-Inf 1 3 5 7 10 15 20 26 Inf]), N = N ./ sum(N), bar(N)
%
% % Weights
% [R,n] = randinterval([10,100], [1 2 ; 100 200 ; -2 -1 ; 12 12], [1 0 2 3]) ;
% % returns a 10-by-100 matrix with values in the intervals (-2,-1)
% % or (1,2). Roughly 666 numbers are negative:
% sum(R(:)<0), sum(n(:)==3)
% % The interval (100,200) has a weight of zero, and the interval
% % (12,12) has zero length; these are "ignored".
% histc(n(:),1:4)
%
% % Adjacent intervals with weights:
% [R,n] = randinterval([10000,1], [-2:2],[1 4 2 8]) ;
% bar(-2:2,histc(R,-2:2)/100,'histc') ; ylabel('Percentage')
%
% See also RAND, RANDN, RANDPERM
% RANDSAMPLE (Stats Toolbox)
% RANDP (File Exchange)
% for Matlab R13 and up
% version 1.0 (oct 2008)
% (c) Jos van der Geest
% email: jos@jasen.nl
% History
% Created: oct 2008
% Revisions: -
% Argument checks
error(nargchk(2,3,nargin)) ;
if ndims(Interval) ~= 2 || ~isnumeric(Interval) || numel(Interval)<2,
error([mfilename ':IntervalWrongSize'],...
'INTERVALS should be a N-by-2 (or N-by-1, with N>1) numeric matrix.')
end
if min(size(Interval)) == 1,
% Interval is a vector. Make the bounds column vectors
LowerBound = reshape(Interval(1:end-1),[],1) ;
UpperBound = reshape(Interval(2:end),[],1) ;
elseif size(Interval,2) ~= 2
error([mfilename ':IntervalWrongSize'],...
'INTERVALS should be a N-by-2 (or N-by-1) numeric matrix.')
else
% Get the lower and upper bounds of the intervals
LowerBound = Interval(:,1) ;
UpperBound = Interval(:,2) ;
end
if any(UpperBound < LowerBound),
error([mfilename ':NegativeInterval'],...
'Intervals cannot be smaller than 0') ;
end
if nargin==3 && (~isempty(Weight) || ~isnumeric(Weight)),
if numel(Weight) ~= numel(LowerBound),
error([mfilename ':WeightsWrongSize'],...
'The number of weights should match the number of intervals.') ;
end
if any(Weight < 0)
error([mfilename ':WeightsNegative'],...
'Weights cannot be negative.')
end
else
% default, all intervals are equally likely
Weight = 1 ;
end
tmp = [LowerBound UpperBound] ;
if any(isnan(tmp(:)) | isinf(tmp(:))) || any(isnan(Weight(:)) | isinf(Weight(:))),
error('Intervals or weights cannot contain NaNs or Infs.') ;
end
% It is nice to give a warning when intervals overlap. If they do, the
% lowerbound of an interval is smaller than the upperbound of a previous
% interval, when the intervals are in sorted order
tmp = sortrows(tmp) ;
if any(tmp(2:end,1) < tmp(1:end-1,2)),
warning([mfilename ':OverlappingIntervals'], ...
'Intervals overlap.') ;
end
% The trick is to put all the intervals next to each other. Intervals with
% larger weights should become larger, so they are more likely. The whole
% range spans numbers between 0 and IntervalEdges(end):
IntervalSize = (UpperBound - LowerBound) ;
IntervalEdges = [0 ; cumsum(Weight(:) .* IntervalSize(:))].' ;
if IntervalEdges(end) == 0,
% all intervals had a zero length and/or a zero weight, so there is
% nothing to do!
error([mfilename ':AllZero'],...
'All intervals have a length and/or weight of zero.') ;
end
try
% let RAND catch the errors in the size argument
R = rand(N) ; % random numbers between 0 and 1
catch
% rephrase the error
ERR = lasterror ;
ERR.message = strrep(ERR.message,'rand',[mfilename ' (in a call to RAND) ']) ;
rethrow(ERR) ;
end
if isempty(R),
% nothing to do, e.g. when N contains zero.
ind = [] ;
return
end
% To which interval do these random random numbers belong
[ind,ind] = histc(IntervalEdges(end) * R(:),IntervalEdges) ; %#ok, first output is not used
% now map a new series of random numbers between 0 and 1 to their
% respective interval. We have to create a new series, otherwise not the
% all values in an interval are possible.
R(:) = LowerBound(ind) + IntervalSize(ind) .* rand(numel(R),1) ;
% also return interval numbers in same shape as R
ind = reshape(ind,size(R)) ;
% %#debug plots
% E = linspace(min(Interval(:)), max(Interval(:)),100) ;
% N2 = histc(R(:),E) ;
% figure ;
% subplot(2,1,1) ; bar(E,N2,'histc')
% subplot(2,1,2) ; plot(sort(R(:)),'k.')
%%%%%%%%%%%%%%%%%%%%%%%%%这是该函数，保存后直接调用！
>> [k1,k2]=randinterval([1 10],[1 2;5 6],[3 1])
k1 =
5.6154 1.7919 1.9218 1.7382 5.1763 5.4057 1.9355 1.9169 5.4103 1.36
k2 =
2 1 1 1 2 2 1 1 2 1追问额，看不太懂呢。。。
能发一个.m文件么？？
liuxuejian190221@163.com 谢谢。。。
追答你把上面的函数贴到新建的m文件里，保存后，就是m文件！
热心网友
时间：2023-08-25 08:22
1.心形线是圆的外摆线的一种，它是这样形成的：一个动圆沿一个定圆的圆周外部无滑动的滚动（两个圆的半径相等），动圆的圆周上的一固定点运动的轨迹就是心形线。程序如下：
function [xcar,ycar]=drawcardioid(varargin)
% A demonstration to show the form of the cardioid
error(nargchk(0,3,nargin));
nin = nargin;
if nin==0
cx1=0; cy1=0; % The coordinate of the center point.
r=1; % The radius of the circle.
elseif nin==1
cx1=0; cy1=0;
r=varargin;
elseif nin==2
cx1=0;
cy1= vararin;
r = varargin;
else
cx1=vararin;
cy1=vararin;
r=vararin;
end
if nargout==0
flag=1;
else
flag=0;
end
theta = linspace(0,2*pi,120);
x0 = r*cos(theta);
y0 = r*sin(theta);
x1 = x0+cx1;
y1 = y0+cy1;
cx2 = 2*x0+cx1;
cy2 = 2*y0+cy1;
x3=x0.*cos(theta)-y0.*sin(theta);
y3=x0.*sin(theta)+y0.*cos(theta);
for k=1:120
x2(k,:) = cx2(k)+x0;
y2(k,:) = cy2(k)+y0;
xx(k)=cx2(k)+x3(k);
yy(k)=cy2(k)+y3(k);
end
xcar = xx;
ycar = yy;
if flag
plot(x1,y1);
axis([cx1-3.2*r cx1+3.2*r cy1-3.2*r cy1+3.2*r]);
axis manual;
hold on
daspect([1 1 1]);
set(gcf,'doublebuffer','on');
for k = 1:120
plot(x2(k,:),y2(k,:),'g',xx(k),yy(k),'r*');
pause(0.05);
%plot(x2,y2,'color','w');
end
end
2.求a的平方根：
function y=sq(a)
ep=1.0E-5;
x1=1;
x2=(x1+a/x1)/2;
while(abs(x1-x2)>ep)
x1=x2;
x2=(x1+a/x1)/2;
end
y=x2;
3.多种方法迭代：该方程有一实根：-0.7684531615
方法一：
function y=st1()
ep=1.0E-5;
x1=-2;
x2=(-x1^5+2*x1-1)/(5*x1^2);
while(abs(x1-x2)>ep)
x1=x2;
x2=(-x1^5+2*x1-1)/(5*x1^2);
end
y=x2;
该方法发散
方法二：
function y=st2()
ep=1.0E-5;
x1=-2;
x2=(-5*x1^3+2*x1-1)/x1^4;
while(abs(x1-x2)>ep)
x1=x2;
x2=(-5*x1^3+2*x1-1)/x1^4;
end
y=x2;
该方法不稳定。
方法三：
function y=sq()
ep=1.0E-5;
x1=-1;
tim=-x1^5+2*x1-1;
x2=sign(tim/5)*(abs(tim/5))^(1/3);
while(abs(x1-x2)>ep)
x1=x2;
tim=-x1^5+2*x1-1;
x2=sign(tim/5)*(abs(tim/5))^(1/3);
end
y=x2;
该算法稳定收敛。
由于时间有限，最后一题留给你自己做吧。
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热心网友
时间：2023-08-25 08:23
帮助中没有呀 搜索rand其中的相关也没有这个函数
热心网友
时间：2023-08-25 08:23
人家那个是自定义的函数，当然你没法运行了
randinterval .m 加载在默认路径下，试一试
