Python类合并排序的文件，这可怎么改进？(Python class to merge sorte

背景：

我在大清洗（不能在内存中保存）制表符分隔的文件。当我清理输入文件，我建立在内存中的列表; 当它（在内存约1GB）获得100万个词条我有点它（使用下面的默认密钥）和列表写入文件。这个类是把排序的文件重新走到一起。它适用于我迄今遇到的文件。我最大的情况下，到目前为止，被合并66页排序的文件。

问题：

有没有在我的逻辑孔（这里是脆弱的）？
我有没有正确实现合并排序算法？
是否有可能作出任何明显的改进？

实施例的数据：

这是在这些文件中的一个行的抽象：

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

外卖的是，我用'SomeStringId'.lower().replace(' ', '')作为我的排序键。

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：从实施意见布莱恩我想出了以下解决方案：

第二个编辑：每秒更新的代码约翰·马金的建议：

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

粗略的测试

使用相同的输入文件（2.2 GB的数据）：

SortedFileMerger类了51分钟（3068.4秒）
布赖恩的解决方案用了40分钟（2408.5秒）
加入后约翰·马金的建议，解决方案代码了36分钟（2214.0秒）

Answer 1:

需要注意的是在python2.6的，heapq有一个新的合并功能，这会为你做到这一点。

为了处理自定义按键功能，你可以换东西的文件迭代器，这样它比较基础上，关键是装饰它，之后剥离出来：

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，并在运行的python2.5修改，并且由于它采用的是堆获得下一个最低也要合并大量文件时是非常有效的。

你应该能够简单地heapq.py从python2.6的安装复制，将它复制到你的源为““heapq26.py”，并使用from heapq26 import merge ” -有它没有使用2.6的特定功能。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）。

Answer 2:

<<这个“答案”是在原有的提问者得到的代码中的注释>>

建议：使用eval（）是ummmm和你正在做什么限制呼叫者使用lambda - 密钥提取可能需要比一个班轮多，并且在任何情况下你不需要的初步排序步骤同样的功能？

所以替换此：

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):