背景
我正在清除大型（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：
示例数据：
这是其中一个文件中一行的抽象：
'hash_of_SomeStringId\tSome String Id\t\t\twww.somelink.com\t\tOtherData\t\n'
外卖是我使用'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
修改：实施Brian的建议我提出了以下解决方案：
第二次修改：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）：
答案 0 :(得分：16)
请注意，在python2.6中，heapq有一个新的merge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python中，仅仅从后面的heapq模块中获取合并的实现也是值得的。它是纯python，并且在python2.5中运行时未经修改，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” - 没有使用2.6特定功能它。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）。
答案 1 :(得分：2)
＆LT;＆LT;这个“答案”是对原始提问者的结果代码＆gt;＆gt;
的评论建议：使用eval（）是嗯，你正在做什么限制调用者使用lambda - 密钥提取可能需要多个单行，并且在任何情况下你不需要相同的功能初步排序步骤？
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)
用这个：
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):