如何在python中拆分一个巨大的文本文件

时间:2008-11-14 23:12:14

标签: python text-files

我有一个巨大的文本文件(~1GB),遗憾的是我使用的文本编辑器不会读取这么大的文件。但是,如果我可以将它分成两个或三个部分,我会没事的,所以,作为一个练习,我想在python中编写一个程序来完成它。

我认为我希望程序要做的是找到文件的大小,将该数字分成几部分,对于每个部分,以块的形式读取到该点,写入文件名 .nnn输出文件,然后读取到下一个换行符并写入,然后关闭输出文件等。显然,最后一个输出文件只是复制到输入文件的末尾。

你能帮我解决关键文件系统相关的部分:文件大小,读取和写入块并读取换行符吗?

我将首先编写此代码测试,所以没有必要给我一个完整的答案,除非它是一行代码; - )

15 个答案:

答案 0 :(得分:32)

linux有一个拆分命令

拆分-l 100000 file.txt

将拆分为等于100,000行大小的文件

答案 1 :(得分:15)

查看os.stat()的文件大小和file.readlines([sizehint])。这两个函数应该是阅读部分所需要的,希望你知道如何写作:)

答案 2 :(得分:9)

作为替代方法,使用日志库:

>>> import logging.handlers
>>> log = logging.getLogger()
>>> fh = logging.handlers.RotatingFileHandler("D://filename.txt", 
     maxBytes=2**20*100, backupCount=100) 
# 100 MB each, up to a maximum of 100 files
>>> log.addHandler(fh)
>>> log.setLevel(logging.INFO)
>>> f = open("D://biglog.txt")
>>> while True:
...     log.info(f.readline().strip())

您的文件将显示如下:

  

filename.txt(文件结尾)
  filename.txt.1
  filename.txt.2
  ...
  filename.txt.10(文件开头)

这是一种快速简便的方法,可以使大型日志文件与RotatingFileHandler实施相匹配。

答案 3 :(得分:5)

这种生成器方法是一种(慢速)方式来获取一条线而不会耗尽你的记忆。

import itertools

def slicefile(filename, start, end):
    lines = open(filename)
    return itertools.islice(lines, start, end)

out = open("/blah.txt", "w")
for line in slicefile("/python27/readme.txt", 10, 15):
    out.write(line)

答案 4 :(得分:4)

您可以使用wcsplit(请参阅相应的联机帮助页)以获得所需的效果。在bash

split -dl$((`wc -l 'filename'|sed 's/ .*$//'` / 3 + 1)) filename filename-chunk.

生成相同行数的3个部分(当然,最后一个包含舍入错误),名为filename-chunk.00filename-chunk.02

答案 5 :(得分:4)

不要忘记seek()mmap()随机访问文件。

def getSomeChunk(filename, start, len):
    fobj = open(filename, 'r+b')
    m = mmap.mmap(fobj.fileno(), 0)
    return m[start:start+len]

答案 6 :(得分:4)

虽然Ryan Ginstrom's answer是正确的,但它确实需要更长的时间(正如他已经指出的那样)。这是通过连续迭代打开的文件描述符来绕过对itertools.islice的多次调用的方法:

def splitfile(infilepath, chunksize):
    fname, ext = infilepath.rsplit('.',1)
    i = 0
    written = False
    with open(infilepath) as infile:
        while True:
            outfilepath = "{}{}.{}".format(fname, i, ext)
            with open(outfilepath, 'w') as outfile:
                for line in (infile.readline() for _ in range(chunksize)):
                    outfile.write(line)
                written = bool(line)
            if not written:
                break
            i += 1

答案 7 :(得分:4)

现在,有一个pypi模块可用于将任何大小的文件拆分成块。看看这个

https://pypi.org/project/filesplit/

答案 8 :(得分:2)

我编写了程序,似乎工作正常。感谢Kamil Kisiel让我开始。
(注意FileSizeParts()是这里没有显示的函数)
稍后我可能会做一个执行二进制读取的版本以查看它是否更快。

def Split(inputFile,numParts,outputName):
    fileSize=os.stat(inputFile).st_size
    parts=FileSizeParts(fileSize,numParts)
    openInputFile = open(inputFile, 'r')
    outPart=1
    for part in parts:
        if openInputFile.tell()<fileSize:
            fullOutputName=outputName+os.extsep+str(outPart)
            outPart+=1
            openOutputFile=open(fullOutputName,'w')
            openOutputFile.writelines(openInputFile.readlines(part))
            openOutputFile.close()
    openInputFile.close()
    return outPart-1

答案 9 :(得分:2)

usage - split.py filename splitsizeinkb

import os
import sys

def getfilesize(filename):
   with open(filename,"rb") as fr:
       fr.seek(0,2) # move to end of the file
       size=fr.tell()
       print("getfilesize: size: %s" % size)
       return fr.tell()

def splitfile(filename, splitsize):
   # Open original file in read only mode
   if not os.path.isfile(filename):
       print("No such file as: \"%s\"" % filename)
       return

   filesize=getfilesize(filename)
   with open(filename,"rb") as fr:
    counter=1
    orginalfilename = filename.split(".")
    readlimit = 5000 #read 5kb at a time
    n_splits = filesize//splitsize
    print("splitfile: No of splits required: %s" % str(n_splits))
    for i in range(n_splits+1):
        chunks_count = int(splitsize)//int(readlimit)
        data_5kb = fr.read(readlimit) # read
        # Create split files
        print("chunks_count: %d" % chunks_count)
        with open(orginalfilename[0]+"_{id}.".format(id=str(counter))+orginalfilename[1],"ab") as fw:
            fw.seek(0) 
            fw.truncate()# truncate original if present
            while data_5kb:                
                fw.write(data_5kb)
                if chunks_count:
                    chunks_count-=1
                    data_5kb = fr.read(readlimit)
                else: break            
        counter+=1 

if __name__ == "__main__":
   if len(sys.argv) < 3: print("Filename or splitsize not provided: Usage:     filesplit.py filename splitsizeinkb ")
   else:
       filesize = int(sys.argv[2]) * 1000 #make into kb
       filename = sys.argv[1]
       splitfile(filename, filesize)

答案 10 :(得分:1)

这对我有用

import os

fil = "inputfile"
outfil = "outputfile"

f = open(fil,'r')

numbits = 1000000000

for i in range(0,os.stat(fil).st_size/numbits+1):
    o = open(outfil+str(i),'w')
    segment = f.readlines(numbits)
    for c in range(0,len(segment)):
        o.write(segment[c]+"\n")
    o.close()

答案 11 :(得分:0)

或者,一个python版本的wc和split:

lines = 0
for l in open(filename): lines += 1

然后一些代码将第一行/ 3读入一个文件,下一行/ 3读入另一个,等等。

答案 12 :(得分:0)

我要求将csv文件拆分为导入到Dynamics CRM中,因为导入的文件大小限制为8MB,而我们收到的文件要大得多。此程序允许用户输入FileNames和LinesPerFile,然后将指定的文件拆分为请求的行数。我无法相信它有多快!

# user input FileNames and LinesPerFile
FileCount = 1
FileNames = []
while True:
    FileName = raw_input('File Name ' + str(FileCount) + ' (enter "Done" after last File):')
    FileCount = FileCount + 1
    if FileName == 'Done':
        break
    else:
        FileNames.append(FileName)
LinesPerFile = raw_input('Lines Per File:')
LinesPerFile = int(LinesPerFile)

for FileName in FileNames:
    File = open(FileName)

    # get Header row
    for Line in File:
        Header = Line
        break

    FileCount = 0
    Linecount = 1
    for Line in File:

        #skip Header in File
        if Line == Header:
            continue

        #create NewFile with Header every [LinesPerFile] Lines
        if Linecount % LinesPerFile == 1:
            FileCount = FileCount + 1
            NewFileName = FileName[:FileName.find('.')] + '-Part' + str(FileCount) + FileName[FileName.find('.'):]
            NewFile = open(NewFileName,'w')
            NewFile.write(Header)

        NewFile.write(Line)
        Linecount = Linecount + 1

    NewFile.close()

答案 13 :(得分:0)

这是一个python脚本,可用于使用subprocess分割大文件:

"""
Splits the file into the same directory and
deletes the original file
"""

import subprocess
import sys
import os

SPLIT_FILE_CHUNK_SIZE = '5000'
SPLIT_PREFIX_LENGTH = '2'  # subprocess expects a string, i.e. 2 = aa, ab, ac etc..

if __name__ == "__main__":

    file_path = sys.argv[1]
    # i.e. split -a 2 -l 5000 t/some_file.txt ~/tmp/t/
    subprocess.call(["split", "-a", SPLIT_PREFIX_LENGTH, "-l", SPLIT_FILE_CHUNK_SIZE, file_path,
                     os.path.dirname(file_path) + '/'])

    # Remove the original file once done splitting
    try:
        os.remove(file_path)
    except OSError:
        pass

您可以在外部调用它:

import os
fs_result = os.system("python file_splitter.py {}".format(local_file_path))

您也可以导入subprocess并直接在程序中运行。

这种方法的问题在于高内存使用率:subprocess创建一个内存占用内存大小与您的进程相同的分支,如果您的进程内存已经很大,它会在运行时加倍。与os.system相同的事情。

这是另一种纯粹的python方式,虽然我还没有在大文件上测试它,它会变慢但是更精简内存:

CHUNK_SIZE = 5000

def yield_csv_rows(reader, chunk_size):
    """
    Opens file to ingest, reads each line to return list of rows
    Expects the header is already removed
    Replacement for ingest_csv
    :param reader: dictReader
    :param chunk_size: int, chunk size
    """
    chunk = []
    for i, row in enumerate(reader):
        if i % chunk_size == 0 and i > 0:
            yield chunk
            del chunk[:]
        chunk.append(row)
    yield chunk

with open(local_file_path, 'rb') as f:
    f.readline().strip().replace('"', '')
    reader = unicodecsv.DictReader(f, fieldnames=header.split(','), delimiter=',', quotechar='"')
    chunks = yield_csv_rows(reader, CHUNK_SIZE)
    for chunk in chunks:
        if not chunk:
            break
        # Do something with your chunk here

以下是使用readlines()的另一个示例:

"""
Simple example using readlines()
where the 'file' is generated via:
seq 10000 > file
"""
CHUNK_SIZE = 5


def yield_rows(reader, chunk_size):
    """
    Yield row chunks
    """
    chunk = []
    for i, row in enumerate(reader):
        if i % chunk_size == 0 and i > 0:
            yield chunk
            del chunk[:]
        chunk.append(row)
    yield chunk


def batch_operation(data):
    for item in data:
        print(item)


with open('file', 'r') as f:
    chunks = yield_rows(f.readlines(), CHUNK_SIZE)
    for _chunk in chunks:
        batch_operation(_chunk)

答案 14 :(得分:0)

您可以实现将任何文件拆分为如下所示的块,这里CHUNK_SIZE为500000字节(500kb),内容可以是任何文件:

for idx,val in enumerate(get_chunk(content, CHUNK_SIZE)):
    data=val
    index=idx

def get_chunk(content,size):
        for i in range(0,len(content),size):
            yield content[i:i+size]