I recently started using imports to better organize my code in python. My original code in file1.py used the line:
def foo():
files = [f for f in os.listdir('.') if os.path.isfile(f)]
print files
#do stuff here....
Which referenced all the files in the same folder as the code, print files showing the correct output as an array of filenames.
However, I recently changed the directory structure to something like this:
./main.py
./folder1/file1.py
./folder1/data_file1.csv
./folder1/data_file2.csv
./folder1/......
And in main.py, I use:
import imp
file1 = imp.load_source('file1', "./folder1/file1.py")
.
.
.
file1.foo()
Now, files is an empty array. What happened? I have tried absolute filepaths in addition to relative. Directly declaring an array with data_file1.csv works, but I can't get anything else to work with this import.
What's going on here?
When you do os.listdir('.') , you are trying to list the contents of '.' (which is the current directory) , but this does not need to be the directory in which the script resides, it would be the directory in which you were in when you ran the script (unless you used os.chdir() inside the python script).
You should not depend on the current working directory , instead you can use __file__ to access the path of the current script, and then you can use os.path.dirname() to get the directory in which the script resides.
Then you can use os.path.join() to join the paths of the files you get from os.listdir() with the directory of the script and check if those paths are files to create your list of files.
Example -
def foo():
filedir = os.path.dirname(__file__)
files = [f for f in (os.path.join(filedir, fil) for fil in os.listdir(filedir)) if os.path.isfile(f)]
print files
#do stuff here....
Related
Here is my code:
files = [f for f in os.listdir(os.getcwd() + "\\folder") if os.path.isfile(f)]
for file in files:
print("hello")
I am running this from the directory which contains a folder called "folder". This folder has 4 files in it. This should print "hello" four times in my head - but it doesn't.
What have I misunderstood?
PS Do I need to use os.getcwd() here? I figure it would be cleaner to just use a relative path, but that also doesn't work.
With os.path.isfile(f) you're asking if f is a file inside your current directory, not inside folder. Replace your code with:
[f for f in os.listdir(os.path.join(os.getcwd(), "folder")) if os.path.isfile(os.path.join("folder", f))]
I've also taken the liberty of using os.path.join to avoid direct concatenation of file and folder names as strings, since slashes can be a bit iffy.
And for the record, no you don't need to use os.getcwd() here (but I left it there anyways).
I'm trying to print the names of all the files from a folder directory. I have a folder called "a", and in that folder there are 3 NC files, lets call them "b","c","d", whose directory I want to print. How would I do this?
For example, given my path to the folder is
path=r"C:\\Users\\chz08006\\Documents\\Testing\\a"
I want to print the directories to all the files in the folder "a", so the result should print:
C:\\Users\\chz08006\\Documents\\Testing\\a\\b.nc
C:\\Users\\chz08006\\Documents\\Testing\\a\\c.nc
C:\\Users\\chz08006\\Documents\\Testing\\a\\d.nc
So far, I've tried
for a in path:
print(os.path.basename(path))
But that doesn't seem to be right.
I think you're looking for this:
import os
path = r"C:\\Users\\chz08006\\Documents\\Testing\\a"
for root, dirs, files in os.walk(path):
for file in files:
print("{root}\\{file}".format(root=root, file=file))
You can have a list of file names in a folder using listdir().
import os
path = "C:\\Users\\chz08006\\Documents\\Testing\\a"
l = os.listdir(path)
for a in l:
print(path + a)
You made a couple mistakes. You were using os.path.basename, that only returns the name of the file or folder represented at the end of a path after the last file separator.
Instead, use os.path.abspath to get the full path of any file.
The other mistake was one of using the wrong variable inside the loop (print(os.path.basename(path) instead of using the variable a)
Also, dont forget to use os.listdir to list the files inside the folder before looping.
import os
path = r"C:\Users\chz08006\Documents\Testing\a"
for file in os.listdir(path): #using a better name compared to a
print(os.path.abspath(file)) #you wrote path here, instead of a.
#variable names that do not have a meaning
#make these kinds of errors easier to make,
#and harder to spot
I have written the Python script below to find the last modified date of a file:
for root, dirs, files in os.walk(folder_path):
for f in files:
last_modified = datetime.fromtimestamp(os.path.getmtime(f)).day
#do something
This code works fine for normal files like .txt files. However, if it encounters files with extensions like .tar.gz or .run, it throws an error saying "No such file or directory" when finding last_modified.
* EDIT *
I realized that, if I change the code as below, it works fine.
for f in files:
name = os.path.join(root, f)
last_modified = datetime.fromtimestamp(os.path.getmtime(name)).day
#do something
But why does this happen?
os.walk return filenames relative to the folder it is walking. E.g. if it is walking a directory called foo with a file called bar, it will return bar, not foo/bar. However, since your script (presumably) is not in the folder os is walking, it needs an absolute path, or a path relative to itself. Therefore, you must join the folder path to the filename, so that python can find the file.
As the docs say:
Note that the names in the lists contain no path components. To get a full path (which begins with top) to a file or directory in dirpath, do os.path.join(dirpath, name).
Why it happens to work with the text files I'm not sure. Do you have text files with the same name in the folder containing your script?
I'm uploading a zipped folder that contains a folder of text files, but it's not detecting that the folder that is zipped up is a directory. I think it might have something to do with requiring an absolute path in the os.path.isdir call, but can't seem to figure out how to implement that.
zipped = zipfile.ZipFile(request.FILES['content'])
for libitem in zipped.namelist():
if libitem.startswith('__MACOSX/'):
continue
# If it's a directory, open it
if os.path.isdir(libitem):
print "You have hit a directory in the zip folder -- we must open it before continuing"
for item in os.listdir(libitem):
The file you've uploaded is a single zip file which is simply a container for other files and directories. All of the Python os.path functions operate on files on your local file system which means you must first extract the contents of your zip before you can use os.path or os.listdir.
Unfortunately it's not possible to determine from the ZipFile object whether an entry is for a file or directory.
A rewrite or your code which does an extract first may look something like this:
import tempfile
# Create a temporary directory into which we can extract zip contents.
tmpdir = tempfile.mkdtemp()
try:
zipped = zipfile.ZipFile(request.FILES['content'])
zipped.extractall(tmpdir)
# Walk through the extracted directory structure doing what you
# want with each file.
for (dirpath, dirnames, filenames) in os.walk(tmpdir):
# Look into subdirectories?
for dirname in dirnames:
full_dir_path = os.path.join(dirpath, dirname)
# Do stuff in this directory
for filename in filenames:
full_file_path = os.path.join(dirpath, filename)
# Do stuff with this file.
finally:
# ... Clean up temporary diretory recursively here.
Usually to make things handle relative paths etc when running scripts you'd want to use os.path.
It seems to me that you're reading from a Zipfile the items you've not actually unzipped it so why would you expect the file/dirs to exist?
Usually I'd print os.getcwd() to find out where I am and also use os.path.join to join with the root of the data directory, whether that is the same as the directory containing the script I can't tell. Using something like scriptdir = os.path.dirname(os.path.abspath(__file__)).
I'd expect you would have to do something like
libitempath = os.path.join(scriptdir, libitem)
if os.path.isdir(libitempath):
....
But I'm guessing at what you're doing as it's a little unclear for me.
I have a Python script that reads through a text csv file and creates a playlist file. However I can only do one at a time, like:
python playlist.py foo.csv foolist.txt
However, I have a directory of files that need to be made into a playlist, with different names, and sometimes a different number of files.
So far I have looked at creating a txt file with a list of all the names of the file in the directory, then loop through each line of that, however I know there must be an easier way to do it.
for f in *.csv; do
python playlist.py "$f" "${f%.csv}list.txt"
done
Will that do the trick? This will put foo.csv in foolist.txt and abc.csv in abclist.txt.
Or do you want them all in the same file?
Just use a for loop with the asterisk glob, making sure you quote things appropriately for spaces in filenames
for file in *.csv; do
python playlist.py "$file" >> outputfile.txt;
done
Is it a single directory, or nested?
Ex.
topfile.csv
topdir
--dir1
--file1.csv
--file2.txt
--dir2
--file3.csv
--file4.csv
For nested, you can use os.walk(topdir) to get all the files and dirs recursively within a directory.
You could set up your script to accept dirs or files:
python playlist.py topfile.csv topdir
import sys
import os
def main():
files_toprocess = set()
paths = sys.argv[1:]
for p in paths:
if os.path.isfile(p) and p.endswith('.csv'):
files_toprocess.add(p)
elif os.path.isdir(p):
for root, dirs, files in os.walk(p):
files_toprocess.update([os.path.join(root, f)
for f in files if f.endswith('.csv')])
if you have directory name you can use os.listdir
os.listdir(dirname)
if you want to select only a certain type of file, e.g., only csv file you could use glob module.