近端梯度算法

人脸识别2DLDA代码，MATLAB写的。(code written in MATLAB.)

该工具箱主要功能：显示心电信号，分析心电信号中的outlier(ECG signal display，ECG signal analysis)

用语言编写的解线性方程组的直接方法包含Gauss顺序消去法,Gauss列主元素消去法,Doolittle三角分解法(The direct methods for solving linear equations written in language include Gauss sequential elimination method, Gauss column principal element elimination method and Doolittle trigonometric decomposition method.)

脉搏波的多处采集点的波形，可用来进行对比，或是进行血管建模，很适合初学者(The various collection points of pulse wave waveform, can be used to compare, or modeling of blood vessels, it is suitable for beginners)

基于差分法的QRS波检测算法程序（PAN.J, TOMPKINS. W.J： A Real-Time QRS Detection Algorithm 对应的matlab程序）。(QRS detection algorithm program based difference method (PAN.J, TOMPKINS WJ:. " A Real-Time QRS Detection Algorithm" corresponding matlab program).)

ECG的应用，ECG在医疗检测方面发挥着中重要作用，所以，心电信号至关重要(With the application of ECG, ECG plays an important role in medical detection, so ECG signal is very important.)

该工具箱主要功能：显示心电信号，分析心电信号中的outlier(ECG signal display，ECG signal analysis)

说明：  匹配追踪分解为联合压缩和心电图QRS检测(Matching pursuit decomposition on electrocardiograms for joint compression and QRS detection)

可以用FFT频谱对脑电信号进行提取。我们可以利用提取出的各个波段脑电信号，来诊断一些脑部疾病或者对大脑组织的电活动及大脑的功能状态进行分析。 1.将实验测得的脑电数据文件转换为文本文件(已经过50Hz陷波)， 获得在Matlab 平台上可直接使用的脑电信号数据，即0661.txt。 2.在Matlab中导入数据，并提取Fp1通道的脑电信号，通过FFT变换对α，β，θ，δ波段进行提取，并做FFT逆变换变到时域。 3.计算各个波段的功率谱。(FFT spectrum can be extracted on EEG. We can use the extracted EEG each band to diagnose some brain disorders or brain tissue on the brain' s electrical activity and functional status were analyzed. One would experimentally measured EEG data file into a text file (has been 50Hz notch), obtained in Matlab platform can be used directly EEG data, that 0661.txt. (2) import data in Matlab and extract Fp1 channel EEG by FFT transform α, β, θ, δ bands were extracted, and do FFT inverse transform variable to the time domain. 3 calculate the power spectrum of each band.)

ACM资料大集合,超多关于ACM的资料,全是很经典的,要是把这个全搞定,肯定是金奖！(ACM data collection, a lot of information about ACM, all is very classic, if this is all done, is definitely a gold medal!)