基于遗传算法解决柔性车间调度问题代码中的代码解释，对于这个板块
% RANKING.M (RANK-based fitness assignment)
%
% This function performs ranking of individuals.
%函数的功能是将个体进行排列
%
% Syntax: FitnV = ranking(ObjV, RFun, SUBPOP)
%
% This function ranks individuals represented by their associated
% cost, to be *minimized*, and returns a column vector FitnV
% containing the corresponding individual fitnesses. For multiple
% subpopulations the ranking is performed separately for each
% subpopulation.
%函数将个体按照最小输出进行排列，返回一个包含相应个体适合度的列向量FitnV。
%对于多个子种群来说，此排列将每个给提分离开
%
% Input parameters:
% ObjV - Column vector containing the objective values of the
% individuals in the current population (cost values).
%
% RFun - (optional) If RFun is a scalar in [1, 2] linear ranking is
% assumed and the scalar indicates the selective pressure.
% If RFun is a 2 element vector:
% RFun(1): SP - scalar indicating the selective pressure
% RFun(2): RM - ranking method
% RM = 0: linear ranking
% RM = 1: non-linear ranking
% If RFun is a vector with length(Rfun) > 2 it contains
% the fitness to be assigned to each rank.
% the same length as ObjV. Usually RFun is monotonously
% increasing.
% If RFun is omitted or NaN, linear ranking
% and a selective pressure of 2 are assumed.
% SUBPOP - (optional) Number of subpopulations
% if omitted or NaN, 1 subpopulation is assumed
%
% Output parameters:
% FitnV - Column vector containing the fitness values of the
% individuals in the current population.
%
% Author: Hartmut Pohlheim (Carlos Fonseca)
% History: 01.03.94 non-linear ranking
% 10.03.94 multiple populations
% , RFun, SUBPOP
function FitnV = ranking(ObjV)
% Identify the vector size (Nind)
[Nind,~] = size(ObjV);
if nargin < 2
RFun = [];
end
if nargin > 1
if isnan(RFun)
RFun = [];
end
end
if numel(RFun) == 2
if RFun(2) == 1
NonLin = 1;
els eif RFun(2) == 0
NonLin = 0;
else
error('Parameter for ranking method must be 0 or 1');
end
RFun = RFun(1);
if isnan(RFun)
RFun = 2;
end
else if numel(RFun) > 2
if numel(RFun) ~= Nind
error('ObjV and RFun disagree');
end
end
if nargin < 3
SUBPOP = 1;
end
if nargin > 2
if isempty(SUBPOP)
SUBPOP = 1;
elseif isnan(SUBPOP)
SUBPOP = 1;
elseif length(SUBPOP) ~= 1
error('SUBPOP must be a scalar');
end
end
if (Nind/SUBPOP) ~= fix(Nind/SUBPOP)
error('ObjV and SUBPOP disagree');
end
Nind = Nind/SUBPOP; % Compute number of individuals per subpopulation
% Check ranking function and use default values if necessary
if isempty(RFun)
% linear ranking with selective pressure 2
RFun = 2*(0:Nind-1)/(Nind-1);
elseif numel(RFun) ==1
if NonLin ==1
% non-linear ranking
if RFun(1)<1
error('Selective pressure must be greater than 1');
elseif RFun(1) > Nind-2
error('Selective pressure too big');
end
Root1 = roots([RFun(1)-Nind [RFun(1)*ones(1,Nind-1)]]);
RFun = (abs(Root1(1)) * ones(Nind,1)) .^ [(0:Nind-1)'];
RFun = RFun / sum(RFun) * Nind;
else
% linear ranking with SP between 1 and 2
if (RFun(1) < 1 | RFun(1) > 2)
error('Selective pressure for linear ranking must be between 1 and 2');
end
RFun = 2-RFun + 2*(RFun-1)*[0:Nind-1]'/(Nind-1);
end
end
FitnV = [];
% loop over all subpopulations
for irun = 1:SUBPOP
% Copy objective values of actual subpopulation
ObjVSub = ObjV((irun-1)*Nind+1:irun*Nind);
% Sort does not handle NaN values as required. So, find those...
NaNix = isnan(ObjVSub);
Validix = find(~NaNix);
% ... and sort only numeric values (smaller is better).
[ans,ix] = sort(-ObjVSub(Validix));
% Now build indexing vector assuming NaN are worse than numbers,
% (including Inf!)...
ix = [find(NaNix) ; Validix(ix)];
% ... and obtain a sorted version of ObjV
Sorted = ObjVSub(ix);
% Assign fitness according to RFun.
i = 1;
FitnVSub = zeros(Nind,1);
for j = [find(Sorted(1:Nind-1) ~= Sorted(2:Nind)); Nind]'
FitnVSub(i:j) = sum(RFun(i:j)) * ones(j-i+1,1) / (j-i+1);
i =j+1;
end
% Finally, return unsorted vector.
[ans,uix] = sort(ix);
FitnVSub = FitnVSub(uix);
% Add FitnVSub to FitnV
FitnV = [FitnV; FitnVSub];
end
% End of function
注释加在代码片里了：
% RANKING.M (RANK-based fitness assignment) %基于排列的适应度赋值
%函数的功能是将个体进行排列
% Syntax: FitnV = ranking(ObjV, RFun, SUBPOP) %函数名ranking
%输入参数：ObjV, RFun, SUBPOP；返回适应度：FitnV
%函数将个体按照最小输出进行排列，返回一个包含相应个体适合度的列向量FitnV。
%对于多个子种群来说，此排列将每个给提分离开
% Input parameters:
% ObjV - Column vector containing the objective values of the
% individuals in the current population (cost values).
% 目标参数，包含当前种群中个体目标值的列向量
% RFun - (optional) If RFun is a scalar in [1, 2] linear ranking is
% assumed and the scalar indicates the selective pressure.
% 如果RFun的取值在[1,2]内，则假定为线性排序，并且代表选择权重
% If RFun is a 2 element vector: 如果RFun是一个包含两个元素的向量
% RFun(1): SP - scalar indicating the selective pressure 第一个值代表选择权重
% RFun(2): RM - ranking method 第二个值代表排序方法
% RM = 0: linear ranking 第二个值为0时为线性排序
% RM = 1: non-linear ranking 第二个值为1时为非线性排序
% If RFun is a vector with length(Rfun) > 2 it contains 如果RFun是一个长度大于2的向量
% the fitness to be assigned to each rank. 那么它包含要分配给每组排列的适应度
% the same length as ObjV. Usually RFun is monotonously
% increasing. 与ObjV相同长度的RFun通常是单调递增的
% If RFun is omitted or NaN, linear ranking 如果RFun被忽略或者为Nan，则默认为线性排序
% and a selective pressure of 2 are assumed. 且默认权重为2
% SUBPOP - (optional) Number of subpopulations 子种群数量
% if omitted or NaN, 1 subpopulation is assumed 如果被忽略或者为Nan，默认值为1
%
% Output parameters:
% FitnV - Column vector containing the fitness values of the 返回值是个列向量
% individuals in the current population. 表示为当前种群中个体的适应度
%
% Author: Hartmut Pohlheim (Carlos Fonseca)
% History: 01.03.94 non-linear ranking
% 10.03.94 multiple populations
% , RFun, SUBPOP
function FitnV = ranking(ObjV) %函数声明
% Identify the vector size (Nind) 获取向量大小
[Nind,~] = size(ObjV);
if nargin < 2
RFun = []; %RFun忽略
end
if nargin > 1
if isnan(RFun) %如果RFun为Nan
RFun = []; %RFun忽略
end
end
if numel(RFun) == 2 %如果RFun中的元素个数为2
if RFun(2) == 1 %如果RFun中第二个元素为1
NonLin = 1; %非线性排序
els eif RFun(2) == 0
NonLin = 0; %否则线性排序
else %参数异常报错
error('Parameter for ranking method must be 0 or 1');
end
%以下均按照函数摘要声明处的步骤逐一实现，注释工作量过大，免费问答不再逐行赘述
RFun = RFun(1);
if isnan(RFun)
RFun = 2;
end
else if numel(RFun) > 2
if numel(RFun) ~= Nind
error('ObjV and RFun disagree');
end
end
if nargin < 3
SUBPOP = 1;
end
if nargin > 2
if isempty(SUBPOP)
SUBPOP = 1;
elseif isnan(SUBPOP)
SUBPOP = 1;
elseif length(SUBPOP) ~= 1
error('SUBPOP must be a scalar');
end
end
if (Nind/SUBPOP) ~= fix(Nind/SUBPOP)
error('ObjV and SUBPOP disagree');
end
Nind = Nind/SUBPOP; % Compute number of individuals per subpopulation
% Check ranking function and use default values if necessary 检查排序函数，必要时使用默认值
if isempty(RFun)
% linear ranking with selective pressure 2
RFun = 2*(0:Nind-1)/(Nind-1);
elseif numel(RFun) ==1
if NonLin ==1
% non-linear ranking
if RFun(1)<1
error('Selective pressure must be greater than 1');
elseif RFun(1) > Nind-2
error('Selective pressure too big');
end
Root1 = roots([RFun(1)-Nind [RFun(1)*ones(1,Nind-1)]]);
RFun = (abs(Root1(1)) * ones(Nind,1)) .^ [(0:Nind-1)'];
RFun = RFun / sum(RFun) * Nind;
else
% linear ranking with SP between 1 and 2 参数值在[1,2]之间采取线性排序
if (RFun(1) < 1 | RFun(1) > 2)
error('Selective pressure for linear ranking must be between 1 and 2');
end
RFun = 2-RFun + 2*(RFun-1)*[0:Nind-1]'/(Nind-1);
end
end
FitnV = [];
% loop over all subpopulations 遍历所有子种群
for irun = 1:SUBPOP
% Copy objective values of actual subpopulation
ObjVSub = ObjV((irun-1)*Nind+1:irun*Nind);
% Sort does not handle NaN values as required. So, find those...
NaNix = isnan(ObjVSub);
Validix = find(~NaNix);
% ... and sort only numeric values (smaller is better).
[ans,ix] = sort(-ObjVSub(Validix));
% Now build indexing vector assuming NaN are worse than numbers,
% (including Inf!)...
ix = [find(NaNix) ; Validix(ix)];
% ... and obtain a sorted version of ObjV
Sorted = ObjVSub(ix);
% Assign fitness according to RFun. 根据RFun分配适合度
i = 1;
FitnVSub = zeros(Nind,1);
for j = [find(Sorted(1:Nind-1) ~= Sorted(2:Nind)); Nind]'
FitnVSub(i:j) = sum(RFun(i:j)) * ones(j-i+1,1) / (j-i+1);
i =j+1;
end
% Finally, return unsorted vector.
[ans,uix] = sort(ix);
FitnVSub = FitnVSub(uix);
% Add FitnVSub to FitnV
FitnV = [FitnV; FitnVSub]; %返回适应度
end %函数终止标志
%如对您理解程序有帮助，请点个采纳谢谢！
% End of function