如何在Python中找到String中的模式/组合？(How do you find patterns/combinations within a String in Python? [closed])

 显然，如果你可以将所有数据缓冲到内存中（如果主机系统可以支持一个工作集页面，这个页面足够大，可以让所有数据实际上保留在RAM中而不需要分页，那么这可能是最快的可能的方法。 
 
 但是，要非常小心地考虑由虚拟内存分页子系统引起的“隐藏磁盘I / O”！ 如果请求的“内存中”数据实际上不是 “在内存中”，则会发生页面错误，并且您的进程将停止在其轨道中，直到检索到页面为止。 （如果发生“页面窃取”，那么，你就遇到了麻烦。你的“内存”策略刚刚变成了一个非常低效的（！）基于磁盘的策略。如果密钥是随机分配的，那么你的进程就有了一个巨大的工作集，它正在随机访问。如果所有的内存实际上都不在内存中， 并且会留在那里，那你就麻烦了。 
 
 如果要对大文件进行更新，请考虑在处理update-delta文件之前对其进行排序 ，以便所有相同的键都相邻。 您现在可以编写COBOL程序以利用此功能（当然，如果检测到abend记录，则会abend ！）。 如果“this”记录中的键与“previous”记录中的键相同，则无需重新读取记录。 （并且，在密钥确实发生变化之前，您实际上不需要编写旧记录。）由于索引文件访问方法带有一系列密钥，因此每个密钥可能“接近”先前的密钥 -请求，以便一些必要的索引树页面已经在内存中。 显然，您需要对此进行基准测试，但是对文件进行排序所花费的时间远远少于在索引查找中花费的时间。 （实际上可能相当大。） 

Obviously, if you can buffer all of the data into memory (and if the host system can support a working-set of pages which is big enough to allow all of it to actually remain in RAM without paging, then this would probably be the fastest possible approach.
 
But, be very careful to consider "hidden disk-I/O" caused by the virtual-memory paging subsystem! If the requested "in-memory" data is, in fact, not "in memory," a page-fault will occur and your process will stop in its tracks until the page has been retrieved. (And if "page stealing" occurs, well, you're in trouble. Your "in-memory" strategy just turned into a possibly very-inefficient(!) disk-based one. If keys are distributed randomly, then your process has a gigantic working-set that it is accessing randomly. If all of that memory is not actually in memory, and will stay there, you're in trouble.
 
If you are making updates to a large file, consider sorting the updates-delta file before processing it, so that all occurrences of the same key will be adjacent. You can now write your COBOL program to take advantage of this (and, of course, to abend if an out-of-sequence record is ever detected!). If the key in "this" record is identical to the key of the "previous" one, then you do not need to re-read the record. (And, you do not actually need to write the old record, until the key does change.) As the indexed-file access method is presented with the succession of keys, each key is likely to be "close to" the one previously-requested, such that some of the necessary index-tree pages will already be in-memory. Obviously, you will need to benchmark this, but the amount of time spent sorting the file can be far less than the amount of time spent in index-lookups. (Which actually can be considerable.)