8

背景:

我正在清理大型(不能保存在内存中)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;当它达到 1,000,000 个条目(大约 1GB 内存)时,我对其进行排序(使用下面的默认键)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。它适用于我迄今为止遇到的文件。到目前为止,我最大的案例是合并 66 个排序文件。

问题:

  1. 我的逻辑是否存在漏洞(它在哪里脆弱)?
  2. 我是否正确实现了合并排序算法?
  3. 有什么明显的改进可以做的吗?

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:实施布赖恩的建议我想出了以下解决方案:

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

粗略测试

使用相同的输入文件(2.2 GB 数据):

  • SortedFileMerger 类耗时 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)
4

2 回答 2

16

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。

要处理自定义键功能,您可以用装饰它的东西包装文件迭代器,以便它根据键进行比较,然后将其剥离:

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 python 中,从后面的 heapq 模块中获取合并的实现可能也是值得的。它是纯python,并且在python2.5中未经修改即可运行,并且由于它使用堆来获得下一个最小值,因此在合并大量文件时应该非常有效。

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,您可以只复制合并函数(重写 heappop 等调用以引用 python2.5 heapq 模块)。

于 2009-06-16T13:50:56.457 回答
2

<<这个“答案”是对原始提问者结果代码的评论>>

建议:使用 eval() 是 ummmm 并且您正在做的事情限制调用者使用 lambda - 密钥提取可能需要不止一个单行,并且在任何情况下您不需要相同的函数来进行初步排序步骤吗?

所以替换这个:

def mergeSortedFiles(paths, output_path, dedup=True, key="line.split('\t')[1].lower().replace(' ', '')"):
    keyfunc = eval("lambda line: %s" % key)

有了这个:

def my_keyfunc(line):
    return line.split('\t', 2)[1].replace(' ', '').lower()
    # minor tweaks may speed it up a little

def mergeSortedFiles(paths, output_path, keyfunc, dedup=True):    
于 2009-06-17T00:33:01.747 回答