我写了一个类似于外部排序的程序。我从 this blog 得到了一个好主意。在这里,他们试图仅对数字进行外部排序。我的要求略有不同。 我的输入文件可能有超过一百万条记录,并且很难在内存中对它们进行排序,所以我必须使用我的磁盘。我将输入分成不同的部分,对其进行排序,然后将其存储在临时文件中。然后将排序后的输出合并到一个文件中。下面我可以将其拆分为临时文件,然后仅合并 key 。
我有一个输入文件如下:
key1 abc
key2 world
key1 hello
key3 tom
key7 yankie
key3 apple
key5 action
key7 jack
key4 apple
key2 xon
key1 lemon
假设磁盘上文件的大小为 10,内存缓冲区可以容纳的最大项目为 4,所以我所做的是一次获取 4 条记录并将其存储在 HashMap 中,对我的值以及更新的计数进行排序。此输入将分为 3 个排序文件,如下所示。您可以看到,对于每个键,我都有一个计数以及按字典顺序排列的最高值。
临时文件-0.txt
key1: 2, hello
key2: 1, world
key3: 1, tom
临时文件-1.txt
key5: 1, action
key3: 1, apple
key7: 2, yankie
临时文件-2.txt
key1: 1, lemon
key2: 1, xon
key4: 1, apple
合并所有这 3 个文件后,输出应如下所示:
key1: 3 lemon
key2: 2 xon
key3: 2 world
key5: 1 action
key7: 2 yankie
我不确定将整行与计数以及该键的字典最高值合并的逻辑,我的下面的代码能够给我所有键,如下所示:
key1
key1
key2
key2
key3
key4
key5
key3
key7
在下面的代码中,我打开每个文件并合并它们,然后写回磁盘到一个名为external-sorted.txt
的新单个文件
static int N = 10; // size of the file in disk
static int M = 4; // max items the memory buffer can hold
int slices = (int) Math.ceil((double) N/M);
String tfile = "temp-file-";
//Reading all the 3 temp files
BufferedReader[] brs = new BufferedReader[slices];
String[] topNums = new String[slices];
for(i = 0; i<slices; i++){
brs[i] = new BufferedReader(new FileReader(tfile + Integer.toString(i) + ".txt"));
String t = brs[i].readLine();
String[] kv = t.split(":");
if(t!=null){
topNums[i] = kv[0];
}
//topNums [key1, key5, key1]
}
FileWriter fw = new FileWriter("external-sorted.txt");
PrintWriter pw = new PrintWriter(fw);
for(i=0; i<N; i++){
String min = topNums[0];
System.out.println("min:"+min);
int minFile = 0;
for(j=0; j<slices; j++){
if(min.compareTo(topNums[j])>0)
{
min = topNums[j];
minFile = j;
}
}
pw.println(min);
String t = brs[minFile].readLine();
String[] kv = new String[2];
if (t != null)
kv = t.split(":");
topNums[minFile] = kv[0];
}
for (i = 0; i < slices; i++)
brs[i].close();
pw.close();
fw.close();
}
任何想法都值得赞赏。如果您有任何疑问,请询问。 TIA。
最佳答案
嗯,像这样的东西是有效的,我确信有更好的方法,但目前我还没有能力真正思考:
// Declare Scanner Object to read our file
Scanner in = new Scanner(new File(stringRepresentingLocationOfYourFileHere));
// create Map that will contain keys in sorted order (TreeMap)
// along with last value assigned to the key
Map<String, String> mapa = new TreeMap<>();
// another map to hold keys from first map and number of
// occurrences of those keys (repetitions), this could have been
// done using single Map as well, but whatever
Map<String, Integer> mapaDva = new HashMap<>();
// String array that will hold words of each line of our .txt file
String[] line;
// we loop until we reach end of our .txt file
while(in.hasNextLine()){
// check if map already contains given key, if it does
// increment value by 1 otherwise initialize the value with 1
if (mapa.put((line = in.nextLine().split(" "))[0], line[1]) != null)
mapaDva.put(line[0], mapaDva.get(line[0])+1);
else
mapaDva.put(line[0], 1);
}
// loop through our maps and print out keys, number of
//repetitions, last assigned value
for (Map.Entry<String, String> m : mapa.entrySet()){
System.out.println(m.getKey() + " " + mapaDva.get(m.getKey()) + " " + m.getValue());
}
如果此代码有任何不清楚的具体内容,请询问。
示例输入文件:
key1 abcd
key2 zzz
key1 tommy
key3 world
完成后输出:
key1 2 tommy
key2 1 zzz
key3 1 world
编辑2(处理多个文件时的解决方案):
// array of File objects that hold path to all your files to iterate through
File[] files = {new File("file1.txt"),
new File("file2.txt"),
new File("file3.txt")};
Scanner in;
Map<String, String> mapa = new TreeMap<>();
Map<String, Integer> mapaDva = new HashMap<>();
String[] line;
for (int i = 0; i < files.length; i++) {
// assign new File to Scanner on each iteration (go through our File array)
in = new Scanner(files[i]);
while(in.hasNextLine()){
if (mapa.put((line = in.nextLine().split(" "))[0], line[1]) != null)
mapaDva.put(line[0], mapaDva.get(line[0])+1);
else
mapaDva.put(line[0], 1);
}
}
for (Map.Entry<String, String> m : mapa.entrySet()){
System.out.println(m.getKey() + " " + mapaDva.get(m.getKey()) + " " + m.getValue());
}
因此,我们将所有 File 对象存储在 File 数组中,然后遍历每个对象,组合所有内容并打印出最终结果:
3 个示例输入文件:
文件1.txt
key1 abcd
key2 zzz
key1 tommy
key3 world
file2.txt
key1 abc
key3 xxx
key1 tommy
key6 denver
file3.txt
key5 lol
key8 head
key6 tommy
key6 denver
输出:
key1 4 tommy
key2 1 zzz
key3 2 xxx
key5 1 lol
key6 3 denver
key8 1 head
关于java - 如何使用Java根据键值对合并文件并对其进行排序?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48038178/