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时间序列聚类在股票市场分析中的实践与优化 时间序列聚类是一种面向动态行为建模的无监督学习方法,其核心在于通过语义感知的距离度量(如DTW)捕捉价格形态的相位弹性与结构相似性,而非依赖静态数值或标签驱动。相比K-Means等传统算法,它能有效缓解金融时序中普遍存在的噪声敏感、对齐失效与形态失真问题,从而提升聚类结果的市场可解释性与策略鲁棒性。该技术已广泛应用于市场状态识别、行业行为分群、异常交易监测等场景,并成为构建可验证量化分析框架的关键预处理层。本文聚焦A股真实行情数据,详解从原始OHLCV到可聚类时间序列的完整链路,涵盖DTW距离计算、层次聚类 阅读详情
/****************************************************************************************************
    去年天津区预赛水题,刚开始没想法,后来想应该是直接dfs就行了,然后就写了这个屎一样的代码。。。写得真难看啊。。。
****************************************************************************************************/
#include <iostream>
#include <functional>
#include <algorithm>
//#include <hash_map>		//Visual C++ lib
#include <complex>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <sstream>
#include <utility>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <limits>
#include <memory>
#include <string>
#include <vector>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
using namespace std;

#define LOWBIT(x) ( (x) & ( (x) ^ ( (x) - 1 ) ) )
#define CLR(x, k) memset((x), (k), sizeof(x))
#define CPY(t, s) memcpy((t), (s), sizeof(s))
#define LEN(s) static_cast<int>( strlen((s)) )
#define SZ(s) static_cast<int>( (s).sizeof() )

typedef double LF;
typedef long long LL;		//GNU C++
typedef unsigned long long ULL;
//typedef __int64 LL;		//Visual C++ 2010
//typedef unsigned __int64 ULL;

typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef pair<double, double> PDD;

//typedef hash_map<int, int>::iterator HMI;
typedef map<int, int>::iterator MI;
typedef vector<int>::iterator VI;
typedef list<int>::iterator LI;
typedef set<int>::iterator SI;

const int INF_INT = 0x3f3f3f3f;
const LL INF_LL = 0x7fffffffffffffffLL;		//15f
const double oo = 10e9;
const double eps = 10e-9;
const double PI = acos(-1.0);

const int MAXP = 23;

struct Player
{
	string name, pos;
	int key;
}player[MAXP];
map<string, int>NP;
map<string, int>SP;
int pcnt[6], flag[6], maze[MAXP][MAXP];
bool sign[MAXP];

bool inline cmp(const Player &pa, const Player &pb)
{
	return SP[ pa.pos ] < SP[ pb.pos ];
}
void preprocess()
{
	SP.clear();
	SP.insert( pair<string, int>("goalkeeper", 0) );
	SP.insert( pair<string, int>("defender", 1) );
	SP.insert( pair<string, int>("midfielder", 2) );
	SP.insert( pair<string, int>("striker", 3) );
	return ;
}
int maxi;
void dfs(int crt, int state, int tot, int gk_cnt, int df_cnt, int mf_cnt, int sk_cnt)
{
	if (0 == state && gk_cnt >= 1)
	{
		crt = flag[state] - 1;
		++state;
	}
	if (1 == state && df_cnt >= 4)
	{
		crt = flag[state] - 1;
		++state;
	}
	if (2 == state && mf_cnt >= 4)
	{
		crt = flag[state] - 1;
		++state;
	}
	if (3 == state && sk_cnt >= 2)
	{
		crt = flag[state] - 1;
		++state;
	}
	if (4 == state)
	{
		maxi = max(maxi, tot);
		return ;
	}
	for (int nxt = crt + 1; nxt < flag[state]; ++nxt)
	{
		int buf = tot + player[nxt].key;
		for (int ind = 0; ind < nxt; ++ind)
		{
			if (sign[ind])
			{
				buf += maze[ind][nxt];
			}
		}
		sign[nxt] = true;
		dfs(nxt, state, buf, 0 == state ? gk_cnt + 1 : gk_cnt, 1 == state ? df_cnt + 1 : df_cnt, 2 == state ? mf_cnt + 1 : mf_cnt, 3 == state ? sk_cnt + 1 : sk_cnt);
		sign[nxt] = false;
	}
	return ;
}
void ace()
{
	int m, val;
	string na, nb;
	while (cin >> player[0].name >> player[0].key >> player[0].pos)
	{
		for (int i = 1; i < 23; ++i)
		{
			cin >> player[i].name >> player[i].key >> player[i].pos;
		}
		sort(player, player + 23, cmp);
		NP.clear();
		CLR(pcnt, 0);
		for (int i = 0; i < 23; ++i)
		{
			++pcnt[ SP[ player[i].pos ] ];
			NP.insert( pair<string, int>(player[i].name, i) );
		}
		CLR(maze, 0);
		for (cin >> m; m--; )
		{
			cin >> na >> nb >> val;
			maze[ NP[na] ][ NP[nb] ] = maze[ NP[nb] ][ NP[na] ] = val;
		}
		if (pcnt[0] < 1 || pcnt[1] < 4 || pcnt[2] < 4 || pcnt[3] < 2)
		{
			puts("impossible");
			continue ;
		}
		flag[0] = pcnt[0];
		for (int i = 1; i < 4; ++i)
		{
			flag[i] = flag[i - 1] + pcnt[i];
		}
		CLR(sign, false);
		maxi = -INF_INT;
		dfs(-1, 0, 0, 0, 0, 0, 0);
		printf("%d\n", maxi);
	}
	return ;
}
int main()
{
	preprocess();
	ace();
	return 0;
}

数据去重复_京东商品评论情感分析|文本数据预处理 本文爬取了十款热销手机的文本评论数据后,首先对文本评论数据做文本预处理,文本评论数据中存在大量的无价值信息,引入无用的文本数据,对其作情感分析,显然没有价值,得出的文本挖掘结果也不尽人意。所以对于文本评论数据先进行文本预处理是必不可少的一步,把大量的、冗余的、无价值的信息去除,可以极大提高文本数据的质量,精确文本分析结果。1 文本去重1.1文本去重的原因文本数据预处理时,首先应去除文本评论数据中反... 阅读详情

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plot(cnn.rL);assert(er<0.12, 'Too big error');```Example: Neural Networks---------------------```matlabfunction test_example_NNload mnist_uint8;train_x = double(train_x) / 255;test_x = double(test_x) / 255;train_y = double(train_y);test_y = double(test_y);% normalize[train_x, mu, sigma] = zscore(train_x);test_x = normalize(test_x, mu, sigma);%% ex1 vanilla neural netrand('state',0)nn = nnsetup([784 100 10]);opts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samples[nn, L] = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.08, 'Too big error');%% ex2 neural net with L2 weight decayrand('state',0)nn = nnsetup([784 100 10]);nn.weightPenaltyL2 = 1e-4; % L2 weight decayopts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex3 neural net with dropoutrand('state',0)nn = nnsetup([784 100 10]);nn.dropoutFraction = 0.5; % Dropout fraction opts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex4 neural net with sigmoid activation functionrand('state',0)nn = nnsetup([784 100 10]);nn.activation_function = 'sigm'; % Sigmoid activation functionnn.learningRate = 1; % Sigm require a lower learning rateopts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex5 plotting functionalityrand('state',0)nn = nnsetup([784 20 10]);opts.numepochs = 5; % Number of full sweeps through datann.output = 'softmax'; % use softmax outputopts.batchsize = 1000; % Take a mean gradient step over this many samplesopts.plot = 1; % enable plottingnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex6 neural net with sigmoid activation and plotting of validation and training error% split training data into training and validation datavx = train_x(1:10000,:);tx = train_x(10001:end,:);vy = train_y(1:10000,:);ty = train_y(10001:end,:);rand('state',0)nn = nnsetup([784 20 10]); nn.output = 'softmax'; % use softmax outputopts.numepochs = 5; % Number of full sweeps through dataopts.batchsize = 1000; % Take a mean gradient step over this many samplesopts.plot = 1; % enable plottingnn = nntrain(nn, tx, ty, opts, vx, vy); % nntrain takes validation set as last two arguments (optionally)[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');```[![Bitdeli Badge](https://d2weczhvl823v0.cloudfront.net/rasmusbergpalm/deeplearntoolbox/trend.png)](https://bitdeli.com/free "Bitdeli Badge")

hdu 3720——Arranging Your Team

#include #include #include #include #include using namespace std; struct Player { string name; int score; int pos; Player(){} Player(string nn,int ss,int pp) { name=nn;

990

hdu 3720 Arranging Your Team 枚举

不可能解可以直接判断。 搭配产生的附加分可以用一个二维数组保存。 枚举1442,4种类型的人,因为总人数只有23个,所以可以搜索暴力枚举,然后保存最优解。 #include #include #include #include #include #include #include #include #include #include #include

1404

搜索-hdu-3720-Arranging Your Team

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3720 题目大意: 有23个人,告诉每个人的名字,能力值,以及踢球的位置。要求选出1个守门员,2个前锋,4个中场,4个后卫。给m个关系,该关系的两个人如果同时在场,就会得到相应的分数值va(注意va可以为负值)。求怎样安排使得在场的人总价值最大。 解题思路: 先处理出每种位置的人分布,然

cc_again的专栏 1120

HDU 3720 Arranging Your Team

先分组,然后暴力;注意 初始化时不要为0 会有负数;我直接二进制枚举; dfs是正解;呵呵 #include <iostream> #include <cstdio> #include <cstring> #include <string> #include <vector> #include <map>...

weixin_30477293的博客 84

HDU 3720 深搜 枚举

DES:从23个队员中选出4—4—2—1共4种11人来组成比赛队伍。给出每个人对每个职位的能力值。给出m组人在一起时会产生的附加效果。问你整场比赛人员的能力和最高是多少。 用深搜暴力枚举每种类型的人选择情况。感觉是这个深搜写的很机智。 在vector中存结构体就会很慢。TLE。直接存序号就AC了。以后还是尽量少用结构体吧。 #include<stdio.h> ...

weixin_30496431的博客 88

hdu3270Arranging Your Team(dfs)

http://acm.hdu.edu.cn/showproblem.php?pid=3720 1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> ...

weixin_30596165的博客 135
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