蟒蛇类的合并排序的文件，这怎么可能改善？
https://stackoverflow.com/questions/1001569
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05-07-2019 - |
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背景：
我很清洁大(无法将保持在存储器)符分隔的文件。正如我清理输入文件，我建立了一个列表中的存储器；当它获得1,000,000项(约1GB在存储器)I类(使用默认的关键的下文)，并编写的清单的文件。此类是把排序的文件一起回来。它的工作文件，我也遇到过这样远。我最大的情况下，迄今为止，合并66排的文件。
问题：
- 有孔在我的逻辑(其中它是脆弱的)?
- 我实施的合并排序 算法是否正确？
- 是否有任何明显的改进 可以吗？
例的数据：
这是一个抽象的行在这些文件：
'hash_of_SomeStringId\tSome String Id\t\t\twww.somelink.com\t\tOtherData\t\n'
外卖的是，我使用 'SomeStringId'.lower().replace(' ', '') 作为我的排序关键。
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： 执行的建议 布赖恩 我想出了以下方案：
第二编辑： 新代码每 约翰*麦金's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
粗糙 测试
使用相同的输入文件(2.2GB的数据)：
解决方案
请注意，在python2.6,heapq有一个新的 合并 功能，这将为你做这个。
处理该定义关键功能，则可以仅包裹文件迭代的东西装饰它，使它比较的基础上的关键，并带它出去之后:
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，这可能是值得的，只需采取的执行情况的合并从以后heapq模块。这是纯粹的蟒蛇，并运行未经修改的在python2.5，因为它使用一堆获取下一个最低应该是非常有效时，合并大量的文件。
你应该能够简单复制的heapq.py 从python2.6安装、复制到你源作为"heapq26.py"并使用"from heapq26 import merge"-有没有2.6具体特征使用它。或者，你可以复制的合并的功能(重写heappop等呼吁参考的python2.5heapq模块)。
其他提示
<< 这个"答案"是一个评论的原始提问的结果代码>>
建议：eval()是咽和你在做什么限制的呼叫者使用lambda-关键取可能需要超过一个衬垫，并且在任何情况下，不需要同样的功能的初步排序步骤？
所以替代这样的：
def mergeSortedFiles(paths, output_path, dedup=True, key="line.split('\t')[1].lower().replace(' ', '')"):
keyfunc = eval("lambda line: %s" % key)
与此：
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):
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