Question

我有260+文本档案目录,其中载有评分信息。我想就所有这些档案中包含档案名称和每个档案头两条内容的简要文件编制。我的想法是分别编制两份清单,并编制清单。然而,我可以拿到档案名称清单,但我无法把档案的头两条线列入所附清单。我的法典是:

# creating a list of filename
for f in os.listdir("../scores"):
    (pdb, extension) = os.path.splitext(f)
    name.append(pdb[1:5])

# creating a list of the first two lines of each file
for f in os.listdir("../scores"):
    for line in open(f):
        score.append(line)
        b = f.nextline()
        score.append(b)

我发现了一个错误,即<代码>str没有属性nextline。请事先提供帮助。

Answer 1

你重新获得的问题,是试图从记分文档使用文档编号<>(<>为的线>)中抽取一个以上线的结果。这里的快速解决方案(几种方式之一,我确信):

# creating a list of the first two lines of each file for f in os.listdir("../scores"): with open(f) as fh: score.append(fh.readline()) score.append(fh.readline())

<with statement take care of concluding the file for You after Youed, and it given You a filehandle Object (fh), which can grab Line from Manually.

Answer 2

文档物体有<代码>next(>> 方法不nextline()。

Answer 3

发表戴维的评论和回答:

from __future__ import with_statement
from itertools import islice
import os
NUM_LINES = 2
with open( ../dir_summary.txt , w ) as dir_summary:
  for f in os.listdir( . ):
    with open(f) as tf:
      dir_summary.write( %s: %s
  % (f, repr( ,  .join(islice(tf,NUM_LINES)))))

Answer 4

这里,我可能更老的时装版,改用印刷,以方便新线。

## written for Python 2.7, summarize filename and two first lines of files of given filetype
import os
extension =  .txt  ## replace with extension of desired files
os.chdir( . ) ##  ../scores ) ## location of files

summary = open( summary.txt , w )
# creating a list of filenames with right extension
for fn in [fn for fn in os.listdir(os.curdir) if os.path.isfile(fn) and fn.endswith(extension)]:
    with open(fn) as the_file:
        print >>summary,  ** +fn+ ** 
        print >>summary, the_file.readline(), the_file.readline(),
        print >>summary,  - *60
summary.close()
## show resulta
print(open( summary.txt ).read())

友情链接