I have a directory containing mutliple files with similar names and subdirectories named after these so that files with like-names are located in that subdirectory. I'm trying to concatenate all the .sdf files in a given subdirectory to a single .sdf file.
import os
from os import system
for ele in os.listdir(Path):
if ele.endswith('.sdf'):
chdir(Path + '/' + ele[0:5])
system('cat' + ' ' + '*.sdf' + '>' + ele[0:5] + '.sdf')
However when I run this, the concatenated file includes every .sdf file from the original directory rather than just the .sdf files from the desired one. How do I alter my script to concatenate the files in the subdirectory only?
this is a very clumsy way of doing it. Using chdir is not recommended, and system either (deprecated, and overkill to call cat)
Let me propose a pure python implementation using glob.glob to filter the .sdf files, and read each file one by one and write to the big file opened before the loop:
import glob,os
big_sdf_file = "all_data.sdf" # I'll let you compute the name/directory you want
with open(big_sdf_file,"wb") as fw:
for sdf_file in glob.glob(os.path.join(Path,"*.sdf")):
with open(sdf_file,"rb") as fr:
fw.write(fr.read())
I left big_sdf_file not computed, I would not recommend to put it in the same directory as the other files, since running the script twice would result in taking the output as input as well.
Note that the drawback of this approach is that if the files are big, they're read fully into memory, which can cause problems. In that case, replace
fw.write(fr.read())
by:
shutil.copyfileobj(fr,fw)
(importing shutil is necessary in that case). That allows packet copy instead of full-file read/write.
I'll add that it's probably not the full solution you're expecting, since there seem to be something about scanning the sub-directories of Path to create 1 big .sdf file per sub-directory, but with the provided code which doesn't use any system command or chdir, it should be easier to adapt to your needs.
Related
Say I have a file that contains the different locations where some '.wav' files are present on a server. For example say the content of the text file location.txt containing the locations of the wav files is this
/home/user/test_audio_folder_1/audio1.wav
/home/user/test_audio_folder_2/audio2.wav
/home/user/test_audio_folder_3/audio3.wav
/home/user/test_audio_folder_4/audio4.wav
/home/user/test_audio_folder_5/audio5.wav
Now what I want to do is that I want to copy these files from different locations within the server to a particular directory within that server, for example say /home/user/final_audio_folder/ and this directory will contain all the audio files from audio1.wav to audio5.wav
I am trying to perform this task by using shutil, but the main problem with shutil that I am facing is that while copying the files, I need to name the file. I have written a demo version of what I am trying to do, but dont know how to scale it when I will be reading the paths of the '.wav' files from the txt file and copy them to my desired location using a loop.
My code for copying a single file goes as follows,
import shutil
original = r'/home/user/test_audio_folder_1/audio1.wav'
target=r'/home/user/final_audio_folder_1/final_audio1.wav'
shutil.copyfile(original,target)
Any suggestions will be really helpful. Thank you.
import shutil
i=0
with open(r'C:/Users/turing/Desktop/location.txt', "r") as infile:
for t in infile:
i+=1
x="audio"+str(i)+".wav"
t=t.rstrip('\n')
original= r'{}'.format(t)
target=r'C:/Users/turing/Desktop/audio_in/' + x
shutil.copyfile(original, target)
Use the built-in string's split() method within a for loop on the location.txt contents & split the name of the directory on the '/' character, then the last element in a new list would be your filename.
I am trying to automatically create a HDF5 structure by using the file paths on my local pc. I want to read through the subdirectories and create a HDF5 structure to match, that I can then save files to. Thank you
You can do this by combining os.walk() and h5py create_group(). The only complications are handling Linux vs Windows (and drive letter on Windows). Another consideration is relative vs absolute path. (I used absolute path, but my example can be modified. (Note: it's a little verbose so you can see what's going on.) Here is the example code (for Windows):
with h5py.File('SO_73879694.h5','w') as h5f:
cwd = os.getcwd()
for root, dirs, _ in os.walk(cwd, topdown=True):
print(f'ROOT: {root}')
# for Windows, modify root: remove drive letter and replace backslashes:
grp_name = root[2:].replace( '\\', '/')
print(f'grp_name: {grp_name}\n')
h5f.create_group(grp_name)
This is actually quite easy to do using HDFql in Python. Here is a complete script that does that:
# import HDFql module (make sure it can be found by the Python interpreter)
import HDFql
# create and use (i.e. open) an HDF5 file named 'my_file.h5'
HDFql.execute("CREATE AND USE FILE my_file.h5")
# get all directories recursively starting from the root of the file system (/)
HDFql.execute("SHOW DIRECTORY / LIKE **")
# iterate result set and create group in each iteration
while HDFql.cursor_next() == HDFql.SUCCESS:
HDFql.execute("CREATE GROUP \"%s\"" % HDFql.cursor_get_char())
I am trying to save all entries of os.listdir("./oldcsv") separately in a list but I don't know how to manipulate the output before it is processed.
What I am trying to do is generate a list containing the absolute pathnames of all *.csv files in a folder, which can later be used to easily manipulate those files' contents. I don't want to put lots of hardcoded pathnames in the script, as it is annoying and hard to read.
import os
for file in os.listdir("./oldcsv"):
if file.endswith(".csv"):
print(os.path.join("/oldcsv", file))
Normally I would use a loop with .append but in this case I cannot do so, since os.listdir just seems to create a "blob" of content. Probably there is an easy solution out there, but my brain won't think of it.
There's a glob module in the standard library that can solve your problem with a single function call:
import glob
csv_files = glob.glob("./*.csv") # get all .csv files from the working dir
assert isinstance(csv_files, list)
I am new to Python and, although having been reading and enjoying it so far, have ∂ experience, where ∂ → 0.
I have a folder tree and each folder at the bottom of the tree's branches contains many files. For me, this whole tree in the input.
I would to perform several steps of analysis (I believe these are irrelavant to this question), the results of which I would like to have returned in an identical tree to that of the input, called output.
I have two ideas:
Read through each folder recursively using os.walk() and for each file to perform the analysis, and
Use a function such as shutil.copytree() and perform the analysis somewhere along the way. So actually, I do not want to COPY the tree at all, rather replicate it's structure but with new files. I thought this might be a kind of 'hack' as I do actually want to use each input file to create the output file, so instead of a copycommand, I need an analyse command. The rest should remain unchanged as far as my imagination allows me to understand.
I have little experience with option 1 and zero experience with option 2.
For smaller trees up until now I have been hard-coding the paths, which has become too time-consuming at this point.
I have also seen more mundane ways, such as using glob to first find all the files I would like and work on them, but I don't know how this might help find a shortcut in recreating the input tree for my output.
My attempt at option 1 looks like this:
import os
for root, dirs, files in os.walk('/Volumes/Mac OS Drive/Data/input/'):
# I have no actual need to print these, it just helps me see what is happening
print root, "\n"
print dirs, "\n"
# This is my actual work going on
[analysis_function(name) for name in files]
however I fear this is going to be very slow, I would also like to do some kind of filtering on files too - for example the .DS_Store files created in mac trees are included in the results of the above. I would attempt to use the fnmatch module to filter only the files I want.
I have seen in the copytree function that it is possible to ignore files according to a pattern, which would be helpful, however I do not understand from the documentation where I could put my analysis function in on each file.
You can use both options: you could provide your custom copy_function that performs analysis instead of the default shutil.copy2 to shutil.copytree() (it is a more of a hack) or you could use os.walk() to have a finer control over the process.
You don't need to create parent directories manually either way. copytree() creates the parent directories for you and os.makedirs(root) can create parent directories if you use os.walk():
#!/usr/bin/env python2
import fnmatch
import itertools
import os
ignore_dir = lambda d: d in ('.git', '.svn', '.hg')
src_dir = '/Volumes/Mac OS Drive/Data/input/' # source directory
dst_dir = '/path/to/destination/' # destination directory
for root, dirs, files in os.walk(src_dir):
for input_file in fnmatch.filter(files, "*.input"): # for each input file
output_file = os.path.splitext(input_file)[0] + '.output'
output_dir = os.path.join(dst_dir, root[len(src_dir):])
if not os.path.isdir(output_dir):
os.makedirs(output_dir) # create destination directories
analyze(os.path.join(root, input_file), # perform analysis
os.path.join(output_dir, output_file))
# don't visit ignored subtrees
dirs[:] = itertools.ifilterfalse(ignore_dir, dirs)
If I am to read a number of files in Python 3.2, say 30-40, and i want to keep the file references in a list
(all the files are in a common folder)
Is there anyway how i can open all the files to their respective file handles in the list, without having to individually open every file via the file.open() function
This is simple, just use a list comprehension based on your list of file paths. Or if you only need to access them one at a time, use a generator expression to avoid keeping all forty files open at once.
list_of_filenames = ['/foo/bar', '/baz', '/tmp/foo']
open_files = [open(f) for f in list_of_filenames]
If you want handles on all the files in a certain directory, use the os.listdir function:
import os
open_files = [open(f) for f in os.listdir(some_path)]
I've assumed a simple, flat directory here, but note that os.listdir returns a list of paths to all file objects in the given directory, whether they are "real" files or directories. So if you have directories within the directory you're opening, you'll want to filter the results using os.path.isfile:
import os
open_files = [open(f) for f in os.listdir(some_path) if os.path.isfile(f)]
Also, os.listdir only returns the bare filename, rather than the whole path, so if the current working directory is not some_path, you'll want to make absolute paths using os.path.join.
import os
open_files = [open(os.path.join(some_path, f)) for f in os.listdir(some_path)
if os.path.isfile(f)]
With a generator expression:
import os
all_files = (open(f) for f in os.listdir(some_path)) # note () instead of []
for f in all_files:
pass # do something with the open file here.
In all cases, make sure you close the files when you're done with them. If you can upgrade to Python 3.3 or higher, I recommend you use an ExitStack for one more level of convenience .
The os library (and listdir in particular) should provide you with the basic tools you need:
import os
print("\n".join(os.listdir())) # returns all of the files (& directories) in the current directory
Obviously you'll want to call open with them, but this gives you the files in an iterable form (which I think is the crux of the issue you're facing). At this point you can just do a for loop and open them all (or some of them).
quick caveat: Jon Clements pointed out in the comments of Henry Keiter's answer that you should watch out for directories, which will show up in os.listdir along with files.
Additionally, this is a good time to write in some filtering statements to make sure you only try to open the right kinds of files. You might be thinking you'll only ever have .txt files in a directory now, but someday your operating system (or users) will have a clever idea to put something else in there, and that could throw a wrench in your code.
Fortunately, a quick filter can do that, and you can do it a couple of ways (I'm just going to show a regex filter):
import os,re
scripts=re.compile(".*\.py$")
files=[open(x,'r') for x in os.listdir() if os.path.isfile(x) and scripts.match(x)]
files=map(lambda x:x.read(),files)
print("\n".join(files))
Note that I'm not checking things like whether I have permission to access the file, so if I have the ability to see the file in the directory but not permission to read it then I'll hit an exception.