[英]Parse varying sized arrays in data structure - Java
As a novice Java programmer, I've run into a rather high hurdle while trying to analyze data for a personal project of mine. 作为Java的新手程序员,在尝试分析我的个人项目的数据时遇到了很大的障碍。 I have a text file with ~33.5k data points in the following format: 我有一个文本文件,带有约33.5k数据点,格式如下:
PointNumber: 33530
Lat: 8.99167773897820e-001
Lon: 6.20173660875318e+000
Alt: 0.00000000000000e+000
NumberOfAccesses: 4
0 4.80784667215499e+003 4.80872732950073e+003
0 1.05264215520092e+004 1.05273043378212e+004
1 1.65167780853593e+004 1.65185840063538e+004
1 6.52228387069902e+004 6.52246514228552e+004
The final rows, ie ones beginning with 0 or 1, correspond to the integer in the Number of Accesses row. 最后的行(即以0或1开头的行)对应于“访问次数”行中的整数。 I would like to parse through the file and print the Lat, Lon, and two values following each access instance of only PointNumbers having more than 1 number of accesses. 我想解析该文件,并在每个PointNumbers具有超过1次访问次数的每个访问实例之后打印Lat,Lon和两个值。
I'm not sure whether to start with a scanner or tokenizer or to compile a pattern. 我不确定是从扫描仪还是标记器开始还是要编译模式。 Which technique makes storing only valid PointNumbers easier? 哪种技术使仅存储有效的PointNumber变得容易? Intuition tells me to parse through the file, placing relevant values into an object: 直觉告诉我解析文件,将相关值放入对象中:
PointNumber num1 = new PointNumber(lat, lon, accesses[]);
Then loop through the objects and print the object's values if the accesses array length is > 1. On the other hand, it would be better to disregard the information that does not meet the requirements. 然后,如果访问数组的长度> 1,则遍历对象并打印对象的值。另一方面,最好忽略不符合要求的信息。 Could this be done by checking the NumberOfAccesses value while parsing then jump ahead to the next string.startsWith("PointNumber:") if the value is <= 1? 是否可以通过在解析时检查NumberOfAccesses值来完成,然后跳转到下一个string.startsWith(“ PointNumber:”)如果值<= 1?
I have a feeling the overwhelming majority of this community will try steering me towards XML or YAML, but I really prefer trying to tackle this in Java. 我有一种感觉,这个社区的绝大多数人会尝试将我引导到XML或YAML,但是我真的更喜欢尝试用Java解决这个问题。 Any advice, direction, or applicable examples is always greatly appreciated. 任何建议,指导或适用示例始终受到人们的赞赏。
You can loop through this while you have input to parse 您可以在输入解析后循环浏览
// Assuming you have a scanner object named s and pointed to your file:
// PointNumber: 33530
int pointNumber = 0;
while(!s.next().equals("PointNumber:")) {
pointNumber = s.nextInt();
}
// Lat: 8.99167773897820e-001
double lat = 0.0;
while(!s.next().equals("Lat:")) {
lat = s.nextDouble();
}
// Lon: 6.20173660875318e+000
double lon = 0.0;
while(!s.next().equals("Lon:")) {
lon = s.nextDouble();
}
// Alt: 0.00000000000000e+000
double alt = 0.0;
while(!s.next().equals("Alt:")) {
alt = s.nextDouble();
}
// NumberOfAccesses: 4
int numberOfAccesses = 0;
while(!s.next().equals("NumberOfAccesses:")) {
numberOfAccesses = s.nextInt();
}
// 0 4.80784667215499e+003 4.80872732950073e+003
// 0 1.05264215520092e+004 1.05273043378212e+004
// 1 1.65167780853593e+004 1.65185840063538e+004
// 1 6.52228387069902e+004 6.52246514228552e+004
// Assuming you have defined an Access class
LinkedList<Access> accesses = new LinkedList<Access>();
for(int i = 0; i < numberOfAccesses; i++) {
// Assuming this is the constructor for the Access class
accesses.add(new Access(s.nextInt(), s.nextDouble(), s.nextDouble()));
}
Then you can add all the data you actually want into a list of "PointNumber" and then print or do whatever from it. 然后,您可以将所需的所有数据添加到“ PointNumber”列表中,然后打印或执行任何操作。
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