我想知道,如果我有一个单行字符串,那么如何使用简单的java代码计算该字符串中特定单词的频率?!
提前致谢..
我正在寻找的是java中的示例逻辑代码,用于搜索句子中的特定单词.我正在构建一个垃圾邮件过滤器,需要读取该行然后对其进行分类.
这是代码片段.
read = new FileReader("trainfiles/"+filenames[i]);
br = new BufferedReader(read);
while((lines = br.readLine())!=null){
st = new StringTokenizer(lines);
while(st.hasMoreTokens()){
bw = new BufferedWriter(new FileWriter("files/file.txt"));
bw.write(st.nextToken());
bw.newLine();
}
}
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编辑:我正在从目录中读取文件.所以,我需要在每个循环中打开阅读器.我做了一些修改,但后来也没有写入该文件.这是代码:
for(i=0;i==0;i++){
if(filenames[i].matches(".*ham.*")){
System.out.println("ham:"+filenames[i]);
read = new FileReader("trainfiles/"+filenames[i]);
br = new BufferedReader(read);
while((lines = br.readLine())!=null){
st = new StringTokenizer(lines);
while(st.hasMoreTokens()){
System.out.println(st.nextToken());
bw.write(st.nextToken());
}
}
bw.close();
br.close();
}else{
System.out.println("spam:"+filenames[i]);
}
}
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编辑:我修改了代码,但没有成功,
while((lines = br.readLine())!=null){
st = new StringTokenizer(lines);
bw = new BufferedWriter(new FileWriter("files/file.txt"));
while(st.hasMoreTokens()){
System.out.println(st.nextToken());
bw.write(st.nextToken());
}
bw.close();
}
br.close();
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我收到这个错误: Exception in thread …
我有一个项目,我需要为用户生成一个唯一的5位数字母数字ID.如何使用codeigniter实现这一目标!?
谢谢

jdom库在那里,我已经删除并一次又一次地添加到构建路径.而且,我已多次清理该项目.但我仍然得到这个错误:
04-27 10:51:27.867: W/ActivityThread(530): Application com.pkg.src.nhs.glos is waiting for the debugger on port 8100...
04-27 10:51:27.936: I/System.out(530): Sending WAIT chunk
04-27 10:51:29.396: I/dalvikvm(530): Debugger is active
04-27 10:51:29.736: I/System.out(530): Debugger has connected
04-27 10:51:29.746: I/System.out(530): waiting for debugger to settle...
04-27 10:51:29.996: I/System.out(530): waiting for debugger to settle...
04-27 10:51:30.262: I/System.out(530): waiting for debugger to settle...
04-27 10:51:30.479: I/System.out(530): waiting for debugger to settle...
04-27 10:51:30.689: I/System.out(530): waiting for debugger to settle...
04-27 10:51:30.896: I/System.out(530): waiting for debugger to settle... …Run Code Online (Sandbox Code Playgroud) 我有一个模式来验证正常的IP地址,即:
private static final String PATTERN =
"^([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\." +
"([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\." +
"([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\." +
"([01]?\\d\\d?|2[0-4]\\d|25[0-5])$";
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以及验证输入是否是有效IP地址的验证方法:
private static boolean validate(final String ip){
Pattern pattern = Pattern.compile(PATTERN);
Matcher matcher = pattern.matcher(ip);
return matcher.matches();
}
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但是,现在我需要为CIDR添加验证(例如84.240.40.0/24)符号和没有主机的IP范围(例如172.24.105),我尝试了许多不同的模式但没有得到具体的东西.有什么建议?
我想要实现的是制作一个弧形的搜索栏.我知道有很多库可以用来实现这一点,但我只是尝试自定义视图.我遇到过几个问题:
我有一个扩展SeekBar的类,我也实现了onDraw和onMeasure方法,但我无法在eclipse中的布局编辑器中查看,这里是自定义视图类的代码:
package com.custom.android.views;
import android.content.Context;
import android.graphics.Canvas;
import android.graphics.Color;
import android.graphics.Paint;
import android.graphics.Path;
import android.graphics.Path.Direction;
import android.graphics.PathMeasure;
import android.util.AttributeSet;
import android.view.MotionEvent;
import android.view.View;
import android.widget.SeekBar;
import android.widget.Toast;
public class CustomSeekBar extends SeekBar {
public CustomSeekBar(Context context) {
super(context);
// TODO Auto-generated constructor stub
}
public CustomSeekBar(Context context, AttributeSet attrs) {
this(context, attrs,0);
}
public CustomSeekBar(Context context, AttributeSet attrs, int defStyle) {
super(context, attrs, defStyle);
}
@Override
public void draw(Canvas canvas) {
// TODO Auto-generated method stub
super.draw(canvas);
}
@Override …Run Code Online (Sandbox Code Playgroud)我正在接受编程挑战并且遇到了这个问题,我不知所措.我得到了一个列表:
[10, 13, 15, 18, 20, 15]
我必须找到这个中的任何元素的总和是否会提供结果30并且输出应该是计数我们可以看到我们有两种可能性,10 + 20 = 30和15 + 15,所以计数是2.但是如何使用for循环或itertools或任何切片或功能解决方案?
python algorithm python-itertools python-2.7 data-structures
我有这个问题,我需要解决,请帮助.给出了由N个整数组成的非空零索引数组A. 该阵列的幅度被定义为其两个元素之间的最大可能差异,即:幅度(A)= max {A [P] - A [Q]:0≤P,Q <N}
写一个函数
功能幅度($ A);
在给定由N个整数组成的非空零索引数组A的情况下,返回其幅度.
假使,假设:
N是[1..1,000,000]范围内的整数; 数组A的每个元素都是[0..5,000,000]范围内的整数.例如,给定数组A就是这样
A [0] = 10 A [1] = 2 A [2] = 44 A [3] = 15 A [4] = 39 A [5] = 20该函数应返回42.
复杂:
预期的最坏情况时间复杂度是O(N); 预期的最坏情况空间复杂度是O(1),超出输入存储(不计入输入参数所需的存储).可以修改输入数组的元素.
我处于这样一种情况,我需要制作自定义身份验证和自定义中间件来验证和授权用户.我必须在POST请求中设置用户名密码params,或者是否设置了cookie用于基于令牌的身份验证.现在,我知道python中不允许函数重载,我怎么能实现它.我将下面的代码放在自定义身份验证和自定义中间件上.
自定义中间件:
from django.contrib.auth import authenticate
class AuthMiddleWare(object):
def process_request(self, request):
if request.path != '/favicon.ico':
print "inside process_request " + request.path
if request.method == 'POST' and request.POST.has_key('username' ) and request.POST.has_key('password'):
authenticate(username = request.POST.get('username'),password = request.POST.get('password'))
if 'SPRING_SECURITY_REMEMBER_ME_COOKIE' in request.COOKIES:
authenticate(token = request.COOKIES.get('SPRING_SECURITY_REMEMBER_ME_COOKIE'))
return None
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和自定义身份验证后端:
from core.api import NcpAPI
class CustomNCPAuthBackend(object):
"""
This is custom authentication backend.
Authenticate against the webservices call.
The method below would override authenticate() of django.contrib.auth
"""
def authenticate(self, username = None, password = None):
print …Run Code Online (Sandbox Code Playgroud) python django django-middleware django-authentication python-2.7
我正在尝试使用 Pulumi 在 aws fargate 上部署简单的 Flask python 应用程序。python应用程序的dockerfile从容器公开端口8000。我如何使用 pulumi 设置负载均衡器?到目前为止,我已经使用index.ts(pulumi)尝试了以下操作:
import * as awsx from "@pulumi/awsx";
// Step 1: Create an ECS Fargate cluster.
const cluster = new awsx.ecs.Cluster("first_cluster");
// Step 2: Define the Networking for our service.
const alb = new awsx.elasticloadbalancingv2.ApplicationLoadBalancer(
"net-lb", { external: true, securityGroups: cluster.securityGroups });
const web = alb.createListener("web", { port: 80, external: true });
// Step 3: Build and publish a Docker image to a private ECR registry.
const img = awsx.ecs.Image.fromPath("app-img", "./app"); …Run Code Online (Sandbox Code Playgroud)