大数据排序与搜索

发布于 2024-09-27 11:39:05 字数 183 浏览 9 评论 0原文

我有两个数据文件,每个文件有 100 个字符行。文件 A:108 行,文件 B:106 行。我需要找到文件 B 中不在文件 A 中的所有字符串。
起初我想将这两个文件输入 mysql,但看起来它永远无法完成在 108 记录上创建唯一键。

我正在等待您对此的建议。

I have two files of data, 100 char lines each. File A: 108 lines, file B: 106 lines. And I need to find all the strings from file B that are not in file A.
At first I was thinking feeding both files to mysql, but it looks like it won't ever finish creating an unique key on 108 records.

I'm waiting for your suggestions on this.

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终止放荡 2024-10-04 11:39:05

您可以在没有数据库的情况下执行此操作。关键是减小 A 的大小,因为 A 比 B 大得多。具体操作方法如下:

使用适当的哈希函数对 B 文件中的字符串计算 64 位哈希值。将它们存储在内存中(在哈希表中),您可以这样做,因为 B 很小。然后逐行对 A 文件中的所有字符串进行哈希处理,并查看每个字符串是否与 B 文件的哈希值匹配。任何具有匹配哈希值的行(从 B 到其中之一)都应存储在文件 C 中。

此过程完成后,文件 C 将具有 A 的可能匹配字符串(到 B)的小子集。现在您有一个小得多的文件 C,您需要将其与 B 的行进行比较。这将问题简化为实际上可以将 C 中的所有行加载到内存中(作为哈希表)并比较 B 的每一行以查看它是否在 C 中的问题。

You can perform this operation without a database. The key is to reduce the size of A, since A is much larger than B. Here is how to do this:

Calculate 64-bit hashes using a decent hash function for the strings in the B file. Store these in memory (in a hash table), which you can do because B is small. Then hash all of the strings in your A file, line by line, and see if each one matches a hash for your B file. Any lines with matching hashes (to one from B), should be stored in a file C.

When this process is complete file C will have the small subset of A of potentially matching strings (to B). Now you have a much smaller file C that you need to compare lines of B with. This reduces the problem to a problem where you can actually load all of the lines from C into memory (as a hash table) and compare each line of B to see if it is in C.

晨光如昨 2024-10-04 11:39:05

您可以稍微改进@michael-goldshteyn的答案(https://stackoverflow.com/a/3926745/179529)。由于您需要找到 B 中不在 A 中的所有字符串,因此当您将 B 的元素与 A 中的元素进行比较并找到匹配项时,您可以从 B 的元素的哈希表中删除任何项目。留在哈希表中的是文件 A 中未找到的元素。

You can slightly improve on @michael-goldshteyn's answer (https://stackoverflow.com/a/3926745/179529). Since you need to find all the strings in B that are not in A, you can remove any item from the Hash Table of the elements of B, when you compare and find a match for it with the elements in A. The elements that will remain in the Hash Table are the elements that were not found in file A.

反差帅 2024-10-04 11:39:05

对于你提到的大小,你应该能够一次将所有 B 保留在内存中,这样你就可以做 Goldshteyn 答案的简化版本; python 中是这样的:

#!/usr/bin/python3

import sys

if __name__=='__main__':
  b = open(sys.argv[2],'r')
  bs = set()
  for l in b:
    bs.add(l.strip())
  b.close()
  a = open(sys.argv[1],'r')
  for l in a:
    l = l.strip()
    if l in bs:
      bs.remove(l)
  for x in bs:
    print(x)

我已经在原子处理器上对两个大小为 10^5 和 10^7 的文件(每行约 8 个字符)进行了测试。 /usr/bin/time 的输出:

25.15user 0.27system 0:25.80elapsed 98%CPU (0avgtext+0avgdata 56032maxresident)k
0inputs+0outputs (0major+3862minor)pagefaults 0swaps
  60298   60298  509244

For the sizes you mention you should be able to keep all of B in memory at once, so you could do a simplified version of Goldshteyn's answer; something like this in python:

#!/usr/bin/python3

import sys

if __name__=='__main__':
  b = open(sys.argv[2],'r')
  bs = set()
  for l in b:
    bs.add(l.strip())
  b.close()
  a = open(sys.argv[1],'r')
  for l in a:
    l = l.strip()
    if l in bs:
      bs.remove(l)
  for x in bs:
    print(x)

I've tested this on two files of 10^5 and 10^7 in size with ~8 chars per line on an atom processor. Output from /usr/bin/time:

25.15user 0.27system 0:25.80elapsed 98%CPU (0avgtext+0avgdata 56032maxresident)k
0inputs+0outputs (0major+3862minor)pagefaults 0swaps
  60298   60298  509244
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