tpm six sigma

category labelThe individual's understanding of the algorithm is that mapping raw data points to K-dimensional data spaces facilitates better classification of data.The following is the algorithm implementation of the MATLAB code, Python code, etc. have time to write and then paste it up.function [AFTER_CLASS_DATA,CLASS_LABEL,ACC] = spectral_clustering (dataset,class_num,sigma,k_near) Affinity_mat = Creat_d (DATASET,K_NEAR,

, mu, sigma] = featureNormalize(X)%FEATURENORMALIZE Normalizes the features in X % FEATURENORMALIZE(X) returns a normalized version of X where% the mean value of each feature is 0 and the standard deviation% is 1. This is often a good preprocessing step to do when% working with learning algorithms.% You need to set these values correctlyX_norm = X;mu = zeros(1, size(X, 2));sigma = zeros(1, size(X, 2

1.1 Population and Samples Overall: the entire study object. One-dimensional or multi-dimensional quantity indicator. Random Variable. Individual: each study object. Sample: part of the population. 1.1.1Simple Random Sample , I. I. d, independent of the same distribution. Unlimited population sampling. Various random numbers in Matlab can be considered to be independent and distributed, that is, simple random samples. The implementation methods in Matlab are listed below. , Evenly distributed Sa

I held a game on HUST and added some simple mathematical questions in Liu lujia's book. Today, I am not going to go to the server. You still don't know what it looks like. Hust will return if the question is handed in.Judging Error 1. Let's just explain this question first. Question This gives a formula A I =( I-1 + I + 1)/2-C I (1 A is A 0 ,A 1 ,..., A N + 1(N I 1000) Yes Sequence of N + 2 Elements . Here you are. A 0 ,A N + 1, C1,..., CNYou need to findA1 Derivation: First,

Function 1: sum (x) returns the Range Sum of the original array [1, x]. Update (x, W) adds W to the number of the original array subscript X.These two functions use the O (logn) time and O (n) Space to complete the single point addition and subtraction, interval summation function. Function 2: Upgrade the tree array.Interval addition and subtraction, single point Query.Consider dividing the original array so that D [I] = A [I]-A [I-1], especially d [1] = A [1].Then the operation of adding K to

.8, Six Sigma means six times times the standard deviation, in the quality of each million bad product rate of less than 3.4.9. What is the core of the Six Sigma management law? And what does DMAIC mean? What are the four elements?The core of the Six Sigma approach is to use all the work as a process, using a quantitative approach to analyze the factors that affe

possible to find the distribution function of Y, calculate the probability density by derivation, and then calculate the mathematical expectation of y according to the mathematical expectation definition.The variance calculation formula is as follows, you can see the requirement variance needs to calculate the expectation first.Sometimes the abnormal integrals are more difficult to calculate or not convergent, then it becomes difficult to solve the problem. Consider the first-order Taylor expan

)1.4.1 Establish quality standard system1.4.2 to monitor the project test1.4.3 will actually compare with the standard1.4.4 Corrective Error CorrectionWhat are the 8 principles of 1.5iso9000 Quality management?1.5.1 Customer-centric1.5.2 Leadership role1.5.3 Full participation1.5.4 Process MethodA systematic approach to 1.5.5 management1.5.6 Continuous Improvement1.5.7 the decision-making method based on factsThe relationship between 1.5.8 and supply-side mutual benefit1.6 Total Quality Manageme

Eight domain search, level set segmentation, watershed segmentation,The first is the level set method, the effect is not goodClear All;close all;img = Imread (' index_1.bmp '); % the same cell image in the paper is used Hereimg=double (IMG (:,:, 1)); sigma=1.5; % parameter in Gaussian kernel for smoothing. G=fspecial (' Gaussian ', 15,sigma); Img_smooth=conv2 (img,g, ' same '); % smooth image by Gaussi

and can quickly converge to the near truth. While the off-line algorithm is accurate to solve the linear equations, no data preprocessing is needed, only the feature vector x is needed to expand a intercept term.But-------off-line algorithm needs to solve the inverse of the matrix, when the amount of data is large, this method is not suitable.CLC Clear All;close all;x= Load ('Ex3x.dat');%Load Data y= Load ('Ex3y.dat');%%%%--------------------Data preprocessing----------------------%%%%%%m=lengt

SVD (Singular value decomposition) singular value decomposition, can be used to simplify the data, remove noise, improve the results of the algorithm.I. SVD and recommendation systemThe dishes are made by the restaurant's food and vegetable master, who can use any integer from 1 to 5 to rate the dish, and if the food master has not tasted a dish, it is rated at 0.Create a new file svdrec.py and add the following code:Def loadexdata (): return[[0, 0, 0, 2, 2], [0, 0, 0, 3, 3],

circle formula, the center is obtained. Then use the nth + 1 equation (set subscript N) Sigma ((xi-oi) ^2) = r^2 n equations to be easily obtained: Sigma ((xi-oi) ^2) = Sigma ((yi-oi) ^2)Both sides expand and then eliminate the oi^2 to get Sigma (xi-yi) *oi) = Sigma (xi^2-y

When calculating the sub-pixel scale of a feature point, there are two rows belowCodeIt takes a long time to understandFeat-> SCL = Sigma * POW (2.0, dData-> octv + intvl/intvls ); DData-> scl_octv = Sigma * POW (2.0, intvl/intvls); Because K = POW (2.0, 1.0/intvls ), the Sigma between the two Ave ave is exactly one of the two sides, so there is a POW (2.0, dDat

Topic Links:http://acm.hdu.edu.cn/showproblem.php?pid=5293Test instructionsTo give you some chain, each chain has its own value, to find the non-overlapping chain can be composed of the maximum value.ExercisesTree-shaped DP,For each chain u,v,w, we only process it at the apex of the LCA (U,V)Let Dp[i] indicate the maximum value of the exponent with the root of I, Sum[i] the and (vi is the Sons of I) that represent DP[VI]There are two decisions in point I, one is not to choose the chain with I as

to discuss a and B. The sum is the sum of the disconnections between a and B and that between a and B.F (n) indicates n knots in the outer ring, and a and B indicate the number of disconnections.G (n) indicates that there are n knots in the outer ring, and a and B are connected together.If a and B are disconnected, if k (plus a itself) is directly connected to a, it is clear that these k must be connected to other, there must be an edge with the center. If there are multiple sides, the ring wil

", 4th ed., 2001, pp. Wuyi, Wuyi. Examples Draw samples from the distribution: >>> >>>mu, Sigma = 0, 0.1 # mean and standard deviation>>>s = NP.Random.Normal(mu, Sigma, +) Verify the mean and the variance: >>> >>> abs ( mu - np mean ( s 0.01 true >>> >>> abs ( sigma - np std ( s ddof = 1

xlrdImport NumPyAs NPImport Matplotlib.pyplotAs Pltdata = Xlrd.open_workbook (' D:\\python Workspace\\data\\cet4.xls ')Table = Data.sheets () [0]#sheet 0COL5 = Table.col_values (5) [1:]#取第5列的成绩, and remove the column property nameCount = [0For IIn range (0,650)]#初始化countx = [IFor IIn range (0,650)]For IIn COL5:num = Int (i)Count[num] + =1 #统计每个人数的人数 plt.xlabel ( ' Score ') plt.ylabel ( ' number of people ') Plt.title ( ' distribution of CET-4 Scores ') plt.ylim ( 0,8) plt.plot ([i for i

:0.75:15); for i=1:1:21 %{%BP bb=a ' *b; C=lamda (i) *ones (2*m,1) +[-bb;bb]; B=[a ' *a,-a ' *a;-a ' *a, a ' *a]; X0=quadprog (b,c,[],[],[],[],zeros (2*m,1)); x0=x0 (1:m)-x0 (m+1:2*m); %}%BPDN bb=a ' *b; C=lamda (i) *ones (2*m,1) +[-bb;bb];% Note the value of lamda, which is taken as 0.05 b=[a ' *a,-a ' *a;-a ' *a, a ' *a]; X0=quadprog (b,c,[],[],[],[],zeros (2*m,1)); X0=x0 (1:m)-x0 (m+1:2*m); 　　3.IRLS Iterative Weighted least squaresL0 relaxation to l1, ge

Workspace \ Data \ cet4.xls ')Table = data. sheets () [0] # sheet 0Col5 = table. col_values (5) [1:] # obtain the score of column 5th and remove the column attribute name.Count = [0 for I in range (0,650)] # initialize countX = [I for I in range (0,650)]For I in col5:Num = int (I)Count [num] + = 1 # count the number of peoplePlt. xlabel ('score ')Plt. ylabel ('number of people ')Plt. title ('stribution of CET-4 scores ')Plt. ylim (0, 8)Plt. Plut ([I for I in range (250,650) if count [I]! = 0],

= 0.3 $. This means that when p=0.3, this result (30,70) is most likely to occur. This is consistent with our common sense. So 0.3 is the maximum likelihood value of the parameter P we have obtained.Example 2 Normal distributionIf there is a set of sample values $ (x_1,x_2,..., x_n) $, we know that it obeys a normal distribution and the standard deviation is known. When this normal distribution is expected, the probability of generating this sampled data is the largest?In this example, the norm