Python类合并已排序的文件,如何改进？

Question

背景:

我正在清理大型(不能保留在内存中)制表符分隔的文件.当我清理输入文件时,我在内存中建立了一个列表; 当它获得1,000,000个条目(内存大约1GB)时,我将其排序(使用下面的默认密钥)并将列表写入文件.此类用于将已排序的文件重新组合在一起.它适用于我迄今遇到的文件.到目前为止,我最大的案例是合并66个已排序的文件.

问题:

1. 我的逻辑中有漏洞(它在哪里脆弱)？
2. 我是否正确实施了合并排序算法？
3. 有没有明显的改进？

示例数据:

这是其中一个文件中的一行的抽象:

'hash_of_SomeStringId\tSome String Id\t\t\twww.somelink.com\t\tOtherData\t\n'

外卖是我'SomeStringId'.lower().replace(' ', '')用作我的排序键.

原始代码:

class SortedFileMerger():
    """ A one-time use object that merges any number of smaller sorted 
        files into one large sorted file.

        ARGS:
            paths - list of paths to sorted files
            output_path - string path to desired output file
            dedup - (boolean) remove lines with duplicate keys, default = True
            key - use to override sort key, default = "line.split('\t')[1].lower().replace(' ', '')"
                  will be prepended by "lambda line: ".  This should be the same 
                  key that was used to sort the files being merged!
    """
    def __init__(self, paths, output_path, dedup=True, key="line.split('\t')[1].lower().replace(' ', '')"):
        self.key = eval("lambda line: %s" % key)
        self.dedup = dedup
        self.handles = [open(path, 'r') for path in paths]
        # holds one line from each file
        self.lines = [file_handle.readline() for file_handle in self.handles]
        self.output_file = open(output_path, 'w')
        self.lines_written = 0
        self._mergeSortedFiles() #call the main method

    def __del__(self):
        """ Clean-up file handles.
        """
        for handle in self.handles:
            if not handle.closed:
                handle.close()
        if self.output_file and (not self.output_file.closed):
            self.output_file.close()

    def _mergeSortedFiles(self):
        """ Merge the small sorted files to 'self.output_file'. This can 
            and should only be called once.
            Called from __init__().
        """
        previous_comparable = ''
        min_line = self._getNextMin()
        while min_line:
            index = self.lines.index(min_line)
            comparable = self.key(min_line)
            if not self.dedup:                      
                #not removing duplicates
                self._writeLine(index)
            elif comparable != previous_comparable: 
                #removing duplicates and this isn't one
                self._writeLine(index)
            else:                                   
                #removing duplicates and this is one
                self._readNextLine(index)
            previous_comparable = comparable
            min_line = self._getNextMin()
        #finished merging
        self.output_file.close()

    def _getNextMin(self):
        """ Returns the next "smallest" line in sorted order.
            Returns None when there are no more values to get.
        """
        while '' in self.lines:
            index = self.lines.index('')
            if self._isLastLine(index):
                # file.readline() is returning '' because 
                # it has reached the end of a file.
                self._closeFile(index)
            else:
                # an empty line got mixed in
                self._readNextLine(index)
        if len(self.lines) == 0:
            return None
        return min(self.lines, key=self.key)

    def _writeLine(self, index):
        """ Write line to output file and update self.lines
        """
        self.output_file.write(self.lines[index])
        self.lines_written += 1
        self._readNextLine(index)

    def _readNextLine(self, index):
        """ Read the next line from handles[index] into lines[index]
        """
        self.lines[index] = self.handles[index].readline()

    def _closeFile(self, index):
        """ If there are no more lines to get in a file, it 
            needs to be closed and removed from 'self.handles'.
            It's entry in 'self.lines' also need to be removed.
        """
        handle = self.handles.pop(index)
        if not handle.closed:
            handle.close()
        # remove entry from self.lines to preserve order
        _ = self.lines.pop(index)

    def _isLastLine(self, index):
        """ Check that handles[index] is at the eof.
        """
        handle = self.handles[index]            
        if handle.tell() == os.path.getsize(handle.name):
            return True
        return False

编辑:实现Brian的建议我提出了以下解决方案:

第二次编辑:根据John Machin的建议更新了代码:

def decorated_file(f, key):
    """ Yields an easily sortable tuple. 
    """
    for line in f:
        yield (key(line), line)

def standard_keyfunc(line):
    """ The standard key function in my application.
    """
    return line.split('\t', 2)[1].replace(' ', '').lower()

def mergeSortedFiles(paths, output_path, dedup=True, keyfunc=standard_keyfunc):
    """ Does the same thing SortedFileMerger class does. 
    """
    files = map(open, paths) #open defaults to mode='r'
    output_file = open(output_path, 'w')
    lines_written = 0
    previous_comparable = ''
    for line in heapq26.merge(*[decorated_file(f, keyfunc) for f in files]):
        comparable = line[0]
        if previous_comparable != comparable:
            output_file.write(line[1])
            lines_written += 1
        previous_comparable = comparable
    return lines_written

粗糙测试

使用相同的输入文件(2.2 GB数据):

• SortedFileMerger课程耗时51分钟(3068.4秒)
• Brian的解决方案耗时40分钟(2408.5秒)
• 添加John Machin的建议后,解决方案代码耗时36分钟(2214.0秒)

Answer 1

请注意,在python2.6中,heapq有一个新的合并函数,它将为您执行此操作.

要处理自定义键函数,您可以使用装饰它的东西包装文件迭代器,以便根据键进行比较,然后将其删除:

def decorated_file(f, key):
    for line in f: 
        yield (key(line), line)

filenames = ['file1.txt','file2.txt','file3.txt']
files = map(open, filenames)
outfile = open('merged.txt')

for line in heapq.merge(*[decorated_file(f, keyfunc) for f in files]):
    outfile.write(line[1])

[编辑] 即使在早期版本的蟒蛇,它可能是值得只是采取合并的从后来的heapq模块的实现.它是纯python,并且在python2.5中运行未修改,并且因为它在使用大量文件时使用堆来获得下一个最小值应该非常有效.

您应该能够简单地从python2.6安装中复制heapq.py,将其作为"heapq26.py"复制到您的源并使用" from heapq26 import merge" - 其中没有使用2.6特定功能.或者,你可以只复制合并功能(重写heappop等要求,以引用的python2.5 heapq模块).