knn算法的训练数据集需要多大

knn算法的训练数据集需要多大

这个不一定。之所以要分训练集和测试集是因为怕过度拟合（overfitting），所以需要一个测试集来检验确定 你建立的模型并不只是适合于这一组数据。我一般都是70%训练集30%测试集。当然，得看数据量有多大，以及复杂程度。只要训练集>=测试集，就不会错，但好不好得具体分析。如果数据量在1000以下的话，最好是k折交叉验证（基本上只要不是特别复杂的数据，都推荐k折交叉验证）。如果要是数据量大于10万的话，最好考虑80：20甚至90:10。

function [ccr,pgroupt]=knnt(x,group,k,dist,xt,groupt) %# %# aim: to classify test set objects or unknown objects with the %# k nearest neighbour method %# %# principle: knn is a supervised, deterministic, non-parametric %# classification method. it uses the majority rule to %# assign new objects to a class. %# it is assumed that the number of objects in each class %# is similar. %# there are no assumptions about the data distribution and %# the variance-covariance matrices of each class. %# there is no limitation of the number of variables when %# the euclidean distance is used. %# however, when the correlation coefficient is used, the %# number of variables must be larger than 1. %# ref: massart d. l., vandeginste b. g. m., deming s. n., %# michotte y. and kaufman l., chemometrics: a textbook, %# chapter 23, 395-397, elsevier science publishers b. v., %# amsterdam 1988. %# %# input: x: (mxn) data matrix with m objects and n variables, %# containing samples of several classes (training set) %# group: (mx1) column vector labelling the m objects from the %# training set %# k: integer, number of nearest neighbours %# dist: integer, %# = 1, euclidean distance %# = 2, correlation coefficient, (no. of variables >1) %# xt: (mtxn) data matrix with mt objects and n variables %# (test set or unknowns) %# groupt: (mtx1) column vector labelling the mt objects from %# the test set %# --> if the new objects are unknown, input []. %# %# output: ccr: scalar, correct classification rate %# pgroupt:row vector, predicted class label for the test set %# 0 means that the object is not classified to any %# class %# %# subroutines: sortlab.m: sorts the group label vector into classes %# %# author: wen wu %# copyright(c) 1997 for chemoac %# fabi, vrije universiteit brussel %# laarbeeklaan 103 1090 jette %# %# version: 1.1 (28/02/1998) %# %# test: andrea candolfi %# function [ccr,pgroupt]=knnt(x,group,k,dist,xt,groupt); if nargin==5, groupt=[]; end % for unknown objects distance=dist; clear dist % change variable if size(group,1)>1, group=group'; % change column vector into row vector groupt=groupt'; % change column vector into row vector end; [m,n]=size(x); % size of the training set if distance==2 & n<2, error('number of variables must > 1'),end % to check the number of variables when using correlation coefficient [mt,n]=size(xt); % size of the test set dis=zeros(mt,m); % initial values for the distance (matrix of zeros) % calculation of the distance for each test set object for i=1:mt for j=1:m % between each training set object and each test set object if distance==1 dis(i,j)=(xt(i,:)-x(j,:))*(xt(i,:)-x(j,:))'; % euclidian distance else r=corrcoef(xt(i,:)',x(j,:)'); % correlation coefficient matrix r=r(1,2); % correlation coefficient dis(i,j)=1-r*r; % 1 - the power of correlation coefficient end end end % finding of the nearest neighbours lab=zeros(1,mt); % initial values of lab for i=1:mt % for each test object [a,b]=sort(dis(i,:)); % sort distances b=b(find(a<=a(k))); % to find the nearest neighbours indices b=group(b); % the nearest neighbours objects [ng,lgroup]=sortlab(b); % calculate the number of objects from each class in the nearest neighbours a=find(ng==max(ng)); % find the class with the maximum number of objects if length(a)==1 % only one class lab(i)=lgroup(a); % class label else lab(i)=0; % more than one class end end % calculation of the success rate if ~isempty(groupt) dif=groupt-lab; % difference between predicted class label and known class label ccr=sum(dif==0)/mt; % success rate end pgroupt=lab; % the output vector

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