背景:
我正在清理制表符分隔的大文件(无法保存在内存中)。清理输入文件时，我会在内存中建立一个列表。当它达到1,000,000个条目(大约1GB的内存)时，我对其进行排序(使用下面的默认键)并将列表写入文件。此类用于将排序后的文件放回原处。它可以处理到目前为止我遇到的文件。到目前为止，我最大的案例是合并66个排序的文件。
问题:
算法正确吗？
那可以做到吗？
示例数据:
这是这些文件之一中的一行的抽象:
'hash_of_SomeStringId\tSome String Id\t\t\twww.somelink.com\t\tOtherData\t\n'要点是我使用
'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
编辑:实现Brian的建议后，我想到了以下解决方案:
第二次编辑:更新了根据John Machin的建议的代码:
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
粗测
使用相同的输入文件(2.2 GB数据):
分钟(3068.4秒)
解决方案代码耗时36分钟
(2214.0秒)
最佳答案
请注意，在python2.6中，heapq有一个新的merge函数，它将为您完成此操作。
要处理自定义键功能，您只需将文件迭代器包装上装饰它的东西，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] 即使在较早版本的python中，简单地从更高版本的heapq模块中实现合并也是可能值得的。它是纯python，在python2.5中未经修改即可运行，并且由于它使用堆来获取下一个最小值，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”复制到源代码中，并使用“
from heapq26 import merge”-其中没有使用2.6特定功能。另外，您也可以复制合并功能(重写heappop等调用以引用python2.5 heapq模块)。关于合并分类文件的Python类，该如何改进？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/1001569/