I am trying to download a vcard to a specific location on my desktop, with a specific file name (which I define).
I have the code the can download the file to my desktop.
url = "http://www.kirkland.com/vcard.cfm?itemid=10485&editstatus=0"
fp = webdriver.FirefoxProfile()
fp.set_preference("browser.download.folderList",2)
fp.set_preference("browser.download.manager.showWhenStarting",False)
fp.set_preference("browser.download.dir", os.getcwd())
fp.set_preference("browser.helperApps.neverAsk.saveToDisk", "text/x-vcard")
browser = webdriver.Firefox(firefox_profile=fp)
browser.get(url)
Note, the URL above is a link to a vcard.
This is saving to the same directory where the code itself exists, and using a file name that was generated by the site I am downloading from.
I want to specify the directory where the file goes, and the name of the file.
Specifically, I would like to call the file something.txt
Also Note, I realize there are much easier ways to do this (using urllib, or urllib2). I need to do it this specific way (if possible) b/c some links are javascript, which require me to use Selenium. I used the above URL as an example to simplify the situation. I can provide other examples/code to show more complex situations if necessary.
Finally, thank you very much for the help I am sure I will get for this post, and for all the help you have provided me for the last year. I dont know how I would have learned all I have learned in this last year had it not been for this community.
I have code that works. Its more of a hack than a solution, but here it is:
# SET FIREFOX PROFILE
fp = webdriver.FirefoxProfile()
fp.set_preference("browser.download.folderList",2)
fp.set_preference("browser.download.manager.showWhenStarting",False)
fp.set_preference("browser.download.dir", os.getcwd())
fp.set_preference("browser.helperApps.neverAsk.saveToDisk", "text/x-vcard")
#OPEN URL
browser = webdriver.Firefox(firefox_profile=fp)
browser.get(url)
#FIND MOST RECENT FILE IN (YOUR) DIR AND RENAME IT
os.chdir("DIR-STRING")
files = filter(os.path.isfile, os.listdir("DIR-STRING"))
files = [os.path.join("DIR-STRING", f) for f in files]
files.sort(key=lambda x: os.path.getmtime(x))
newest_file = files[-1]
os.rename(newest_file, "NEW-FILE-NAME"+"EXTENSION")
#GET THE STRING, AND DELETE THE FILE
f = open("DIR-STRING"+"NEW-FILE-NAME"+"EXTENSION", "r")
string = f.read()
#DO WHATEVER YOU WANT WITH THE STRING/TEXT FROM THE DOWNLOAD
f.close()
os.remove("DIR-STRING"+"NEW-FILE-NAME"+"EXTENSION")
DIR-STRING is the path to the directory where the file is saved
NEW-FILE-NAME is the name of the file you want
EXTENSION is the .txt, etc.
Related
I am working on an idea. I need a bit help here, as I don't have in depth knowledge of python modules(I mean I don't know all the python modules). Take a look at the code
file = open('Data.txt', 'w')
a = input('Enter your name in format F|M|L: ')
file.write(a)
file.close()
The above code opens a file which writes my data to it. However I want to edit the document only through python and not through opening it from saved location. Shortly I want to disable editions done by opening the file in text-editors.
If your OS is Windows, the most straightforward option is to make the file read-only when your script is done. And set the read-only flag to false while your script is running. There are some ways to modify file permissions using pywin32 library, but it's complicated and hard to find good examples.
import os
from stat import S_IWRITE, S_IREAD
fname = 'test.txt'
# if file exists, reset read only to false (allow write)
if os.path.isfile(fname):
os.chmod(fname, S_IWRITE)
fid = open(fname, 'w')
fid.write('shoobie doobie')
fid.close()
It's already been pointed out in the comments, this technique won't stop a determined person from changing the read only attribute.
If text-editor is all your concern, use 'wb' instead of 'w'.
Use 'rb' to open those files too.
I have the following structure. I want to iterate through sub folders (machine, gunshot) and process .wav files and build mfccresult folder in each category and the .csv file in it. I have the following code and the MFCC folder is keep forming in already formed MFCC folder.
parent_dir = 'sound'
for subdirs, dirs, files in os.walk(parent_dir):
resultsDirectory = subdirs + '/MFCC/'
if not os.path.isdir("resultsDirectory"):
os.makedirs(resultsDirectory)
for filename in os.listdir(subdirs):
if filename.endswith('.wav'):
(rate,sig) = wav.read(subdirs + "/" +filename)
mfcc_feat = mfcc(sig,rate)
fbank_feat = logfbank(sig,rate)
outputFile = resultsDirectory + "/" + os.path.splitext(filename)[0] + ".csv"
file = open(outputFile, 'w+')
numpy.savetxt(file, fbank_feat, delimiter=",")
file.close()
What version of python are you using? Not sure if this has changed in the past, but os.walk does not return "subdirs" as the first of the tuple, but the dirpath. See here for python 3.6.
I don't know your absolute path, but seeing as you are passing in the path sound as a relative reference, I assume it is a folder inside the directory where you run your python code. So for example, lets say you are running this file (lets call it mycode.py) from
/home/username/myproject/mycode.py
and you have some subdirectory:
/home/username/myproject/sound/
So:
resultsDirectory = subdirs + '/MFCC/'
as written in your code above would resolve to:
/home/username/myproject/sound/MFCC/
So your first if statement will be entered since this is not an existing directory. Thereby you create a new directory:
/home/username/myproject/sound/MFCC/
From there, you take
filename in os.listdir(subdirs)
This is also appears to be a misunderstanding of the output of this function. os.listdir() will return directories not files. See here for the man on that.
So now you are looping through the directories in:
/home/username/myproject/sound/
Here, I assume you have some of the directories from your diagram already made. So I assume you have:
/home/username/myproject/sound/machine_sound
/home/username/myproject/sound/gun_shot_sound
or something along those lines.
So the if statement will never be entered, since your directory names to not end with '.wav'.
Even if it did enter, you'd still have issues asfilename will actually be equal to machine_sound on the first loop, and gun_shot_sound in the second time through.
Maybe you are using some other wav library, but the python built-in is called wave and you need to call the wave.open() on the file not wav.read(). See here for the docs.
I'm not sure what you were trying to achieve with the call to os.path.splitext(filename)[0], but you can read about it here You will end up with the same thing that went in in this case though, so machine_sound and gun_shot_sound.
Your output file will thus result in:
/home/username/myproject/sound/MFCC/machine_sound.csv
on the first loop, and
/home/username/myproject/sound/MFCC/gun_shot_sound.csv
the second time through.
So in conclusion, I'm not sure what is happening when you say "MFCC folder is keep forming in already formed MFCC folder" but you definitely have a lot of reading ahead of you before you can understand your own code, and have any hope of fixing it to do what you want. Assuming you read through the links I provided, you should be able to do that though. Good luck!
Additionally, you had quite few typos in your code that I edited, include the immensely important whitespace characters. You should clean that up and ensure your code runs before posting it here, then double check that your copy/paste action did not result in any errors. People will be much more willing to help if you clean up your presentation a bit.
for subdir,dirs,files in os.walk(parent_dir):
for folder in next(os.walk(parent_dir))[1]:
resultsDirectory= folder + '/MFCC'
absPath = os.path.join(parent_dir, resultsDirectory)
if not os.path.isdir(absPath):
os.makedirs(absPath)
for filename in os.listdir(subdir):
print('listdir')
if filename.endswith('.wav'):
print("csv file writing")
(rate,sig) = wav.read(subdir + "/" +filename)
mfcc_feat = mfcc(sig,rate)
fbank_feat = logfbank(sig,rate)
print("fbank_feat")
outputFile =subdir + "/MFCC"+"/" + os.path.splitext(filename)[0] + ".csv"
file = open(outputFile, "w+")
numpy.savetxt(file, fbank_feat, delimiter=",")
file.close()
Here the csv file is stored in the subdirectory not in mfcc folder for each category.
I have issue with output path file.
I'm putting together a Python script that uses pandas to read data from a CSV file, sort and filter that data, then save the file to another location.
This is something I have to run regularly - at least weekly if not daily. The original file is updated every day and is placed in a folder but each day the file name changes and the old file is removed so there is only one file in the directory.
I am able to make all this work by specifying the file location and name in the script, but since the name of the file changes each day, I'd rather not have to edit the script every every time I want to run it.
Is there a way to read that file based solely on the location? As I mentioned, it's the only file in the directory. Or is there a way to use a wildcard in the name? The name of the file is always something like: ABC_DEF_XXX_YYY.csv where XXX and YYY change daily.
I appreciate any help. Thanks!
from os import listdir
CSV_Files = [file for file in listdir('<path to folder>') if file.endswith('.csv')
If there is only 1 CSV file in the folder, you can do
CSV_File = CSV_Files[0]
afterwards.
To get the file names solely based on the location:
import os, glob
os.chdir("/ParentDirectory")
for file in glob.glob("*.csv"):
print(file)
Assume that dirName holds the directory holding your file.
A call to os.listdir(dirName) gives you files or child directories in this directory (of course, you must earlier import os).
To limit the list to just files, we must write a little more, e.g.
[f for f in os.listdir(dirName) if os.path.isfile(os.path.join(dirName, f))]
So we have a full list of files. To get the first file, add [0] to the
above expression, so
fn = [f for f in os.listdir(dirName) if os.path.isfile(os.path.join(dirName, f))][0]
gives you the name of the first file, but without the directory.
To have the full path, use os.path.join(dirname, fn)
So the whole script, adding check for proper extension, can be:
import os
dirName = r"C:\Users\YourName\whatever_path_you_wish"
fn = [f for f in os.listdir(dirName)\
if f.endswith('.csv') and os.path.isfile(os.path.join(dirName, f))][0]
path = os.path.join(dirName, fn)
Then you can e.g. open this file or make any use of them, as you need.
Edit
The above program will fail if the directory given does not contain any file
with the required extension. To make the program more robust, change it to
something like below:
fnList = [f for f in os.listdir(dirName)\
if f.endswith('.csv') and os.path.isfile(os.path.join(dirName, f))]
if len(fnList) > 0:
fn = fnList[0]
path = os.path.join(dirName, fn)
print(path)
# Process this file
else:
print('No such file')
I'm attempting to create a script that looks into a specific directory and then lists all the files of my chosen types in addition to all folders within the original location.
I have managed the first part of listing all the files of the chosen types, however am encountering issues listing the folders.
The code I have is:
import datetime, os
now = datetime.datetime.now()
myFolder = 'F:\\'
textFile = 'myTextFile.txt'
outToFile = open(textFile, mode='w', encoding='utf-8')
filmDir = os.listdir(path=myFolder)
for file in filmDir:
if file.endswith(('avi','mp4','mkv','pdf')):
outToFile.write(os.path.splitext(file)[0] + '\n')
if os.path.isdir(file):
outToFile.write(os.path.splitext(file)[0] + '\n')
outToFile.close()
It is successfully listing all avi/mp4/mkv/pdf files, however isn't ever going into the if os.path.isdir(file): even though there are multiple folders in my F: directory.
Any help would be greatly appreciated. Even if it is suggesting a more effective/efficient method entirely that does the job.
Solution found thanks to Son of a Beach
if os.path.isdir(file):
changed to
if os.path.isdir(os.path.join(myFolder, file)):
os.listdir returns the names of the files, not the fully-qualified paths to the files.
You should use a fully qualified path name in os.path.isdir() (unless you've already told Python where to look).
Eg, instead of using if os.path.isdir(file): you could use:
if os.path.isdir(os.path.join(myFolder, file)):
I am a Python newbie and need to create a script that will do parse some files and put them into a SQL db. So I am trying to create smaller scripts that do what I want, then combine them into a larger script.
To that end, I am trying run this code:
import os
fileList = []
testDir = "/home/me/somedir/dir1/test"
for i in os.listdir(testDir):
if os.path.isfile(i):
fileList.append(i)
for fileName in fileList:
print(fileName)
When I look at the output, I do not see any files listed. I tried the path without quotes and got stack errors. So searching showed I need the double quotes.
Where did I go wrong?
I found this code that works fine:
import os
in_path = "/home/me/dir/"
for dir_path, subdir_list, file_list in os.walk(in_path):
for fname in file_list:
full_path = os.path.join(dir_path, fname)
print(full_path)
I can use full_path to do my next step.
If anyone has any performance tips, feel free to share them. Or point me in the right direction.
that is because you're most likely ejecuting your script from a folder outside your testdir, os.path.isfile need the full path name of the file so it can check is that is a lile or not (os.listdir return the names), if the full path is not provide then it will check is there is a file with the given name in the same folder from which the script is executed, to fix this you need to give the full path name of that file, you can do it with os.path.join like this
for name in os.listdir(testDir):
if os.path.isfile( os.path.join(testDir,name) ):
fileList.append(name)
or if you also want the full path
for name in os.listdir(testDir):
path = os.path.join(testDir,name)
if os.path.isfile(path):
fileList.append(path)