MATLAB的冒号运算符如何工作？

Question

正如指出的这个答案由山姆·罗伯茨和由gnovice这对方的回答,MATLAB的冒号运算符(start:step:stop)创造价值的在用不同的方式矢量linspace一样.Sam Roberts特别指出:

冒号运算符将增量添加到起始点,并从结束点减去减量以达到中间点.这样,它确保输出向量尽可能对称.

但是,有关The MathWorks的官方文档已从其网站中删除.

如果Sam的描述是正确的,步长中的错误不会是对称的吗？

>> step = 1/3;
>> C = 0:step:5;
>> diff(C) - step
ans =
   1.0e-15 *
  Columns 1 through 10
         0         0    0.0555   -0.0555   -0.0555    0.1665   -0.2776    0.6106   -0.2776    0.1665
  Columns 11 through 15
    0.1665   -0.2776   -0.2776    0.6106   -0.2776

关于冒号运算符的有趣事项:

• 它的值取决于它的长度:

  >> step = 1/3;
  >> C = 0:step:5;
  >> X = 0:step:3;
  >> C(1:10) - X
  ans =
     1.0e-15 *
           0         0         0         0         0   -0.2220         0   -0.4441    0.4441         0
  

• 如果它们被舍入,它可以生成重复值:

  >> E = 1-eps : eps/4 : 1+eps;
  >> E-1
  ans =
     1.0e-15 *
     -0.2220   -0.2220   -0.1110         0         0         0         0    0.2220    0.2220
  

• 最后一个值有一个容差,如果步长在末尾创建一个值,则仍然使用此结束值:

  >> A = 0 : step : 5-2*eps(5)
  A =
    Columns 1 through 10
           0    0.3333    0.6667    1.0000    1.3333    1.6667    2.0000    2.3333    2.6667    3.0000
    Columns 11 through 16
      3.3333    3.6667    4.0000    4.3333    4.6667    5.0000
  >> A(end) == 5 - 2*eps(5)
  ans =
    logical
     1
  >> step*15 - 5
  ans =
       0
  

Answer 1

Sam回答中提到的删除页面仍由Way Back Machine归档.幸运的是,即使附加的M文件colonop也存在.似乎这个函数仍然匹配MATLAB的功能(我在R2017a上):

>> all(0:step:5 == colonop(0,step,5))
ans =
  logical
   1
>> all(-pi:pi/21:pi == colonop(-pi,pi/21,pi))
ans =
  logical
   1

我将在这里复制函数对一般情况的作用(有一些生成整数向量和处理特殊情况的快捷方式).我正在用更有意义的变量替换函数的变量名.输入是start,step和stop.

首先,它计算多少步在两者之间有start和stop.如果最后一步超过stop公差,则不采取以下措施:

n = round((stop-start)/step);
tol = 2.0*eps*max(abs(start),abs(stop));
sig = sign(step);
if sig*(start+n*step - stop) > tol
  n = n - 1;
end

这解释了问题中提到的最后一个观察结果.

接下来,它计算最后一个元素的值,并确保它不超过该stop值,即使它允许在先前的计算中超过它.

last = start + n*step;
if sig*(last-stop) > -tol
   last = stop;
end

这就是为什么A问题中向量中的lasat值实际上具有stop作为最后一个值的值的原因.

接下来,它按两个部分计算输出数组,如公布的那样:数组的左半部分和右半部分是独立填充的:

out = zeros(1,n+1);
k = 0:floor(n/2);
out(1+k) = start + k*step;
out(n+1-k) = last - k*step;

请注意,它们不是通过递增来填充,而是通过计算整数数组并将其乘以步长来linspace完成.这就是E对问题中阵列的观察.不同之处在于通过从值中减去这些值来填充数组的右半部分last.

作为最后一步,对于奇数大小的数组,中间值是单独计算的,以确保它恰好位于两个端点的中间位置:

if mod(n,2) == 0
   out(n/2+1) = (start+last)/2;
end

完整功能colonop复制在底部.

请注意,分别填充数组的左侧和右侧并不意味着步长中的错误应该是完全对称的.这些错误由舍入误差给出.但是,如果stop不能通过步长确切地达到这一点,那就确实存在差异,就像问题中的数组一样A.在这种情况下,稍微缩短的步长是在数组的中间,而不是在结尾:

>> step=1/3;
>> A = 0 : step : 5-2*eps(5);
>> A/step-(0:15)
ans =
   1.0e-14 *
  Columns 1 through 10
         0         0         0         0         0         0         0   -0.0888   -0.4441   -0.5329
  Columns 11 through 16
   -0.3553   -0.3553   -0.5329   -0.5329   -0.3553   -0.5329

但即使在stop精确到达该点的情况下,也会在中间积累一些额外的误差.以C问题中的数组为例.此错误累积不会发生linspace:

C = 0:1/3:5;
lims = eps(C);
subplot(2,1,1)
plot(diff(C)-1/3,'o-')
hold on
plot(lims,'k:')
plot(-lims,'k:')
plot([1,15],[0,0],'k:')
ylabel('error')
title('0:1/3:5')
L = linspace(0,5,16);
subplot(2,1,2)
plot(diff(L)-1/3,'x-')
hold on
plot(lims,'k:')
plot(-lims,'k:')
plot([1,15],[0,0],'k:')
title('linspace(0,5,16)')
ylabel('error')

colonop:

function out = colonop(start,step,stop)
% COLONOP  Demonstrate how the built-in a:d:b is constructed.
%
%   v = colonop(a,b) constructs v = a:1:b.
%   v = colonop(a,d,b) constructs v = a:d:b.
%
%   v = a:d:b is not constructed using repeated addition.  If the
%   textual representation of d in the source code cannot be
%   exactly represented in binary floating point, then repeated
%   addition will appear to have accumlated roundoff error.  In
%   some cases, d may be so small that the floating point number
%   nearest a+d is actually a.  Here are two imporant examples.
%
%   v = 1-eps : eps/4 : 1+eps is the nine floating point numbers
%   closest to v = 1 + (-4:1:4)*eps/4.  Since the spacing of the
%   floating point numbers between 1-eps and 1 is eps/2 and the
%   spacing between 1 and 1+eps is eps,
%   v = [1-eps 1-eps 1-eps/2 1 1 1 1 1+eps 1+eps].
%
%   Even though 0.01 is not exactly represented in binary,
%   v = -1 : 0.01 : 1 consists of 201 floating points numbers
%   centered symmetrically about zero.
%
%   Ideally, in exact arithmetic, for b > a and d > 0,
%   v = a:d:b should be the vector of length n+1 generated by
%   v = a + (0:n)*d where n = floor((b-a)/d).
%   In floating point arithmetic, the delicate computatations
%   are the value of n, the value of the right hand end point,
%   c = a+n*d, and symmetry about the mid-point.

if nargin < 3
    stop = step;
    step = 1;
end

tol = 2.0*eps*max(abs(start),abs(stop));
sig = sign(step);

% Exceptional cases.

if ~isfinite(start) || ~isfinite(step) || ~isfinite(stop)
   out = NaN;
   return
elseif step == 0 || start < stop && step < 0 || stop < start && step > 0
   % Result is empty.
   out = zeros(1,0);
   return
end

% n = number of intervals = length(v) - 1.

if start == floor(start) && step == 1
   % Consecutive integers.
   n = floor(stop) - start;
elseif start == floor(start) && step == floor(step)
   % Integers with spacing > 1.
   q = floor(start/step);
   r = start - q*step;
   n = floor((stop-r)/step) - q;
else
   % General case.
   n = round((stop-start)/step);
   if sig*(start+n*step - stop) > tol
      n = n - 1;
   end
end

% last = right hand end point.

last = start + n*step;
if sig*(last-stop) > -tol
   last = stop;
end

% out should be symmetric about the mid-point.

out = zeros(1,n+1);
k = 0:floor(n/2);
out(1+k) = start + k*step;
out(n+1-k) = last - k*step;
if mod(n,2) == 0
   out(n/2+1) = (start+last)/2;
end