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Pig简单的代码实例：报表统计行业中的点击和曝光量
, xxx_ad_id:chararray, guid:chararray, id:chararray, create_time:chararray, action_time :chararray, lat:chararray, lon:chararray, cell_id:chararray, lac:chararray, mcc:chararray :chararray, openudid:chararray, mac_address:chararray, uid:chararray, density:chararray :chararray, city_region_id:chararray, ip_lat:chararray, ip_lon:chararray, quadkey:chararray , sex:chararray, age:chararray, income:chararray, edu:chararray, hobby:chararray); --extract needed column
76310编辑于 2022-07-03 - 来自专栏全栈程序员必看
pig询问top k，每个返回hour和ad_network_id最大的两个记录（SUBSTRING，order，COUNT_STAR，limit）
, wizad_ad_id:chararray, guid:chararray, id:chararray, create_time:chararray :chararray, location_accuracy:chararray, lat:chararray, lon:chararray, cell_id :chararray, lac:chararray, mcc:chararray, mnc:chararray, ip:chararray, :chararray, udid:chararray, openudid:chararray, idfa:chararray, mac_address :chararray, user_agent:chararray, app_id:chararray, app_category_id:chararray,
53510编辑于 2022-07-05 - 来自专栏全栈程序员必看
pig中使用的一些实例语法
)$34==’2′ OR (chararray)$34==’3′ OR (chararray)$34==’1′ OR (chararray)$34==’4′) AND (INDEXOF((chararray )$35,’.’)>0) AND ((chararray)$36==’1′ OR (chararray)$36==”) ), log_43 IF ( ((chararray)$34==’1′ OR (chararray )$34==’2′) AND ((chararray)$35==’1′ OR (chararray)$35==’2′ OR (chararray)$35==’3′ OR (chararray)$35== )$1 AS wizad_ad_id:chararray, (chararray)$2 AS guid:chararray, (chararray)$6 AS log_type:chararray , (chararray)$18 AS imei:chararray, (chararray)$22 AS idfa:chararray, (chararray)$23 AS mac_address:chararray
39130编辑于 2022-07-14 - 来自专栏tea9的博客
android安全题目KGB Messenger 解题
[i]; charArray[i] = (char) (charArray[(charArray.length - i) - 1] ^ '2'); charArray ++) { char c = charArray[i]; charArray[i] = (char) (charArray[(charArray.length (char) ((charArray[i] >> (i % 8)) ^ charArray[i]); } for (int i2 = 0; i2 < charArray.length / 2; i2++) { char c = charArray[i2]; charArray[i2] = charArray[(charArray.length char c = charArray[i2]; charArray[i2] = charArray[(charArray.length - i2) - 1];
94520编辑于 2022-09-08 - 来自专栏Java架构师必看
pig基本语法——join
======= 2、innerjoin grunt> A = load '/root/xytest/pig/data/demodata' using PigStorage(',') as (name:chararray age:int,gpa:float); grunt> B = load '/root/xytest/pig/data/demodata2' using PigStorage(',') as (name:chararray ,score:int,address:chararray); grunt> C = join A by name,B by name; grunt> dump C; 输出结果： (aaa,13,1.1, ,score:int,address:chararray); grunt> D = join A by name left,B by name; grunt> dump D; 输出结果： (aaa,13,1.1 ,score:int,address:chararray); grunt> E = join A by name right,B by name; grunt> dump E; 输出结果： (aaa,13,1.1
53030发布于 2021-04-22 - 来自专栏Java架构师必看
pig基本语法——join
======= 2、innerjoin grunt> A = load '/root/xytest/pig/data/demodata' using PigStorage(',') as (name:chararray age:int,gpa:float); grunt> B = load '/root/xytest/pig/data/demodata2' using PigStorage(',') as (name:chararray ,score:int,address:chararray); grunt> C = join A by name,B by name; grunt> dump C; 输出结果： (aaa,13,1.1, ,score:int,address:chararray); grunt> D = join A by name left,B by name; grunt> dump D; 输出结果： (aaa,13,1.1 ,score:int,address:chararray); grunt> E = join A by name right,B by name; grunt> dump E; 输出结果： (aaa,13,1.1
39420发布于 2021-05-14 - 来自专栏AI那点小事
波形图（人人网2017春招真题）
static String Waveform_2_Record(String str){ StringBuffer s = new StringBuffer(); char[] charArray = str.toCharArray(); for ( int i = 0 ; i < charArray.length ; ){ if ( charArray[ i] == '_' && charArray[i+1] == '_'){ s.append('-'); i += 2; continue; }else if ( charArray[i] == '/' && charArray[i+1] == '\\'){ s.append = str.toCharArray(); for ( int i = 0 ; i < charArray.length ; i++){ if ( charArray
61820发布于 2020-04-20 剑指offer第四天
result; } public void permutation(char[] charArray,int beginIdx){ if(beginIdx >= charArray.length ) return; if(beginIdx == charArray.length-1){ result.add(String.valueOf(charArray charSet.contains(charArray[i])){ char temp = charArray[beginIdx]; charArray [beginIdx] = charArray[i]; charArray[i] = temp; permutation(charArray ,beginIdx+1); temp = charArray[beginIdx]; charArray[beginIdx] = charArray
85760发布于 2018-05-09- 来自专栏开源部署
Pig的cogroup详解
,guid:chararray,sex:chararray,log_type:chararray); sample_data = limit industry_existed_Data 20; --STORE , industry_existed_Data:{industryId: chararray,guid: chararray,sex: chararray,log_type: chararray}, industry_current_data ::log_type: chararray} } look_for_cogroup: { group: chararray, industry_current_data: {joined_ad_campaign_data ::industryId: chararray,joined_Orgin_sex_data::distinct_origin_historical_sex::guid: chararray,joined_Orgin_sex_data ::social_sex::sex: chararray,joined_Orgin_sex_data::distinct_origin_historical_sex::log_type: chararray
68020编辑于 2022-07-03 - 来自专栏开源部署
Pig的limit无效（返回所有记录）sample有效
, guid: chararray, Android_id: chararray, imei: chararray, app_category_id: chararray } g_log: { group : (android_id: chararray,app_category_id: chararray), test_data: { wizad_ad_id: chararray, guid: chararray ,app_category_id: chararray), test_data: {wizad_ad_id: chararray,guid: chararray,android_id: chararray ,imei: chararray,app_category_id: chararray} } myts: { group: (android_id: chararray,app_category_id: chararray), test_data: {wizad_ad_id: chararray,guid: chararray,android_id: chararray,imei: chararray
2K20编辑于 2022-07-03 - 来自专栏Java那些事
每天一道leetcode151-反转字符串里的单词
char temp = charArray[i]; charArray[i] = charArray[j]; charArray[j] = temp; (String s1) { char [] charArray = s1.toCharArray(); int begin = 0;int end = s1.length () - 1; while(begin < charArray.length && charArray[begin] == ' ') begin++; while(end >= 0 && charArray[end] == ' ') end--; String temp = ""; while ( begin <= end) { if(charArray[begin] !
78110发布于 2019-09-17 - 来自专栏人邮电数据采集与预处理
项目四 pig预处理北京公交线路
并指定字段及其类型bus_info = LOAD '/pig/Processed_Beijing_Bus_Info.csv' USING PigStorage(',') AS ( bus_name:chararray , bus_type:chararray, bus_time:chararray, tieck:chararray, licheng:chararray , gongsi:chararray, gengxin:chararray, wang_info:chararray, wang_buff:chararray , fan_info:chararray, fan_buff:chararray );# 输出加载的数据以供检查DUMP bus_info;# 通过 DISTINCT
22621编辑于 2024-09-23 - 来自专栏二进制文集
LeetCode 0131. 分割回文串[动态规划详解]
check(charArray, index, i)) { continue; } // [index, i] 为回文，则加入结果 dfs(charArray, i + 1, length, path, ans); path.removeLast(); } } private boolean check(char[] charArray, int start, int end) { while (start < end) { if ( charArray[start++] ! = charArray[end--]) { return false; } } return true;
49320发布于 2021-03-29 - 来自专栏云同步
387. 字符串中的第一个唯一字符
// 用于存储出现过一次的字符的下标 List<Integer> list = new ArrayList<>(); // 将字符串转换成字符数组，方便遍历 char[] charArray = s.toCharArray(); for (int i = 0; i < charArray.length; i++) { // 获取这个字符在 26个字符中的位置 byte index = (byte)(charArray[i] - 'a'); // 获取这个字符出现的次数 byte num = chars[index]; = s.toCharArray(); for (int i = 0; i < charArray.length; i++) { // 获取这个字符在 26个字符中的位置 byte index = (byte)(charArray[i] - 'a'); // 获取这个字符出现的次数 byte num = chars[index];
12910编辑于 2025-08-01 - 来自专栏Java那些事
每天一道leetcode-125验证回文串
Solution { public boolean isPalindrome(String s) { s = s.toLowerCase(); char [] charArray = s.toCharArray(); String temp = ""; for(int i=;i<charArray.length;i++) { if( ((int)charArray[i] >= && (int)charArray[i] <= ) || ((int)charArray[i] >= && (int)charArray[i] < = )) { temp += charArray[i]; } } char [] resultArray
32610发布于 2019-09-17 - 来自专栏后台技术底层理解
Pig介绍和相对于Hive的优势
chararray常量是以加单引号的一系列字符来表示，例如’fred’。 bytearray:一团或者一组字节。 map:是一种chararray和数据元素之间的键值对映射，其中数据元素可以是任意的pig类型，包括复杂类型。chararray被称为键，它作为查找对应元素的索引，相应的数据元素被称为值。 , symbol:chararray, date:chararray, dividends:float); startswithcm = filter divs by symbol matches ' CM.*'; Distinct（去重）： -- dictinct.pig daily = load 'NYSE_daily' as (exchange:chararray, symbol:chararray , symbol:chararray, date:chararray, open:float, high:float, low:float, close
1.4K10发布于 2020-08-04 - 来自专栏艳艳代码杂货店
[LeetCode]415. Add Strings(计算两个字符串表示的数字的和)
[i]^c) == 0){// ^ 优先级小于 ==， 必须在charArray[i]^c外加括号 return i; } } return -1; } char intToChar(int i){//整数转化为字符 char charArray[10] = {'0', '1', '2', '3' [i]^c外加括号 return charArray[m]; } } return '-1'; } }; return -1; } char intToChar(int i){//整数转化为字符 char charArray[10] = {'0', '1', '2', '3' [i]^c外加括号 return charArray[m]; } } return '-1'; } };
70810发布于 2021-09-19 - 来自专栏后端知识体系
LeetCode-5-最长回文字串
是否是回文串 // i是子串的左边界，j是子串的右边界 boolean[][] dp = new boolean[len][len]; char[] charArray (int j = 1; j < len; j++) { // 列 for (int i = 0; i < j; i++) {// 行 if (charArray 2) { return s; } int maxLen = 1; int begin = 0; char[] charArray ); // 枚举所有回文中心 for(int i=0;i<len-1;i++){ int oddLen = expandAroudCenter(charArray [i]==charArray[j]){ i--; j++; }else{ break
27910编辑于 2022-07-14 - 来自专栏Java那些事
每天一道leetcode-557反转字符串中的单词 III
return result.toString(); } public String reverse(String s) {//反转字符串 char [] charArray = s.toCharArray();//转字符数组 int i =0;int j = charArray.length - 1; while(i < j) { char ch = charArray[i]; charArray[i] = charArray[j]; charArray[ /交换前后两个字符 i++; j--;//移动到下一对需要交换的字符上 } return String.valueOf(charArray
43020发布于 2019-09-17 剑指offer 第十天
for(int i = 0;i < charArray.length;i++){ if(charArray[i] == ' '){ end = i -1; Reverse(charArray,start,end); start = i+1; } } Reverse(charArray,start,charArray.length-1); return String.valueOf(charArray); char temp = charArray[start]; charArray[start] = charArray[end]; charArray[end ] = temp; start++; end--; } return charArray; } }
66680发布于 2018-05-09
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