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Python error: FileNotFoundError: [Errno 2] No such file or directory

I am trying to open the file from folder and read it but it's not locating it. I am using Python3
Here is my code:
import os
import glob
prefix_path = "C:/Users/mpotd/Documents/GitHub/Python-Sample-
codes/Mayur_Python_code/Question/wx_data/"
target_path = open('MissingPrcpData.txt', 'w')
file_array = [os.path.abspath(f) for f in os.listdir(prefix_path) if
f.endswith('.txt')]
file_array.sort() # file is sorted list
for f_obj in range(len(file_array)):
file = os.path.abspath(file_array[f_obj])
join_file = os.path.join(prefix_path, file) #whole file path
for filename in file_array:
log = open(filename, 'r')#<---- Error is here
Error: FileNotFoundError: [Errno 2] No such file or directory: 'USC00110072.txt'

You are not giving the full path to a file to the open(), just its name - a relative path.
Non-absolute paths specify locations in relation to current working directory (CWD, see os.getcwd).
You would have to either os.path.join() correct directory path to it, or os.chdir() to the directory that the files reside in.
Also, remember that os.path.abspath() can't deduce the full path to a file just by it's name. It will only prefix its input with the path of the current working directory, if the given path is relative.
Looks like you are forgetting to modify the the file_array list. To fix this, change the first loop to this:
file_array = [os.path.join(prefix_path, name) for name in file_array]
Let me reiterate.
This line in your code:
file_array = [os.path.abspath(f) for f in os.listdir(prefix_path) if f.endswith('.txt')]
is wrong. It will not give you a list with correct absolute paths. What you should've done is:
import os
import glob
prefix_path = ("C:/Users/mpotd/Documents/GitHub/Python-Sample-"
"codes/Mayur_Python_code/Question/wx_data/")
target_path = open('MissingPrcpData.txt', 'w')
file_array = [f for f in os.listdir(prefix_path) if f.endswith('.txt')]
file_array.sort() # file is sorted list
file_array = [os.path.join(prefix_path, name) for name in file_array]
for filename in file_array:
log = open(filename, 'r')

You are using relative path where you should be using an absolute one. It's a good idea to use os.path to work with file paths. Easy fix for your code is:
prefix = os.path.abspath(prefix_path)
file_list = [os.path.join(prefix, f) for f in os.listdir(prefix) if f.endswith('.txt')]
Note that there are some other issues with your code:
In python you can do for thing in things. You did for thing in range(len(things)) it's much less readable and unnecessary.
You should use context managers when you open a file. Read more here.

Using terminal command to search through files for a specific string within a python script

I have a parent directory, and I'd like to go through that directory and grab each file with a specific string for editing in python. I have been using grep -r 'string' filepath in terminal, but I want to be able to do everything using python. I'm hoping to get all the files into an array and go through each of them to edit them.
Is there a way to do this by only running a python script?

changing current folder to parent
import os
os.chdir("..")
changing folder
import os
os.chdir(dir_of_your_choice)
finding files with a rule in the current folder
import glob
import os
current_dir = os.getcwd()
for f in glob.glob('*string*'):
do_things(f)

import os
#sourceFolder is the folder you're going to be looking inside for backslashes are a special character in python so they're escaped as double backslashes
sourceFolder = "C:\\FolderBeingSearched\\"
myFiles = []
# Find all files in the directory
for file in os.listdir(sourceFolder):
myFiles.append(file)
#open them for editing
for file in myFiles:
try:
open(sourceFolder + file,'r')
except:
continue
#run whatever code you need to do on each open file here
print("opened %s" % file)
EDIT: If you want to separate all files that contain a string (this just prints the list at the end currently):
import os
#sourceFolder is the folder you're going to be looking inside for backslashes are a special character in python so they're escaped as double backslashes
sourceFolder = "C:\\FolderBeingSearched\\"
myFiles = []
filesContainString = []
stringsImLookingFor = ['first','second']
# Find all files in the directory
for file in os.listdir(sourceFolder):
myFiles.append(file)
#open them for editing
for file in myFiles:
looking = sourceFolder + file
try:
open(looking,'r')
except:
continue
print("opened %s" % file)
found = 0
with open(looking,encoding="latin1") as in_file:
for line in in_file:
for x in stringsImLookingFor:
if line.find(x) != -1:
#do whatever you need to do to the file or add it to a list like I am
filesContainString.append(file)
found = 1
break
if found:
break
print(filesContainString)

Trying to access a file located on a flashdrive

I have made a simple test code in python that reads from a text file, and then preforms an action if the text file contains a line "on".
My code works fine if i run the script on my hardive with the text file in the same folder. Example, (C:\Python27\my_file.txt, and C:\Python27\my_scipt.py).
However, if I try this code while my text file is located on my flashdrive and my script is still on my hardrive it won't work even though I have the correct path specified. Example, (G:\flashdrive_folder\flashdrive_file.txt, and C:\Python27\my_scipt.py).
Here is the code I have written out.
def locatedrive():
file = open("G:\flashdrive_folder\flashdrive_file.txt", "r")
flashdrive_file = file.read()
file.close()
if flashdrive_file == "on":
print "working"
else:
print"fail"
while True:
print "trying"
try:
locatedrive()
break
except:
pass
break

The backslash character does double duty. Windows uses it as a path separator, and Python uses it to introduce escape sequences.
You need to escape the backslash (using a backslash!), or use one of the other techniques below:
file = open("G:\\flashdrive_folder\\flashdrive_file.txt", "r")
or
file = open(r"G:\flashdrive_folder\flashdrive_file.txt", "r")
or
file = open("G:/flashdrive_folder/flashdrive_file.txt", "r")

cd /media/usb0
import os
path = "/media/usb0"
#!/usr/bin/python
import os
path = "/usr/tmp"
# Check current working directory.
retval = os.getcwd()
print "Current working directory %s" % retval
# Now change the directory
os.chdir( path )
# Check current working directory.
retval = os.getcwd()
print "Directory changed successfully %s" % retval

Use:
import os
os.chdir(path_to_flashdrive)

How to extract a file within a folder within a zip?

I need to extract a file called Preview.pdf from a folder called QuickLooks inside of a zip file.
Right now my code looks a little like this:
with ZipFile(newName, 'r') as newName:
newName.extract(\QuickLooks\Preview.pdf)
newName.close()
(In this case, newName has been set equal to the full path to the zip).
It's important to note that the backslash is correct in this case because I'm on Windows.
The code doesn't work; here's the error it gives:
Traceback (most recent call last):
File "C:\Users\Asit\Documents\Evam\Python_Scripts\pageszip.py", line 18, in <module>
ZF.extract("""QuickLooks\Preview.pdf""")
File "C:\Python33\lib\zipfile.py", line 1019, in extract
member = self.getinfo(member)
File "C:\Python33\lib\zipfile.py", line 905, in getinfo
'There is no item named %r in the archive' % name)
KeyError: "There is no item named 'QuickLook/Preview.pdf' in the archive"
I'm running the Python script from inside Notepad++, and taking the output from its console.
How can I accomplish this?
Alternatively, how could I extract the whole QuickLooks folder, move out Preview.pdf, and then delete the folder and the rest of it's contents?
Just for context, here's the rest of the script. It's a script to get a PDF of a .pages file. I know there are bonified converters out there; I'm just doing this as an excercise with some sort of real-world application.
import os.path
import zipfile
from zipfile import *
import sys
file = raw_input('Enter the full path to the .pages file in question. Please note that file and directory names cannot contain any spaces.')
dir = os.path.abspath(os.path.join(file, os.pardir))
fileName, fileExtension = os.path.splitext(file)
if fileExtension == ".pages":
os.chdir(dir)
print (dir)
fileExtension = ".zip"
os.rename (file, fileName + ".zip")
newName = fileName + ".zip" #for debugging purposes
print (newName) #for debugging purposes
with ZipFile(newName, 'w') as ZF:
print("I'm about to list names!")
print(ZF.namelist()) #for debugging purposes
ZF.extract("QuickLook/Preview.pdf")
os.rename('Preview.pdf', fileName + '.pdf')
finalPDF = fileName + ".pdf"
print ("Check out the PDF! It's located at" + dir + finalPDF + ".")
else:
print ("Sorry, this is not a valid .pages file.")
sys.exit
I'm not sure if the import of Zipfile is redundant; I read on another SO post that it was better to use from zipfile import * than import zipfile. I wasn't sure, so I used both. =)
EDIT: I've changed the code to reflect the changes suggested by Blckknght.

Here's something that seems to work. There were several issues with your code. As I mentioned in a comment, the zipfile must be opened with mode 'r' in order to read it. Another is that zip archive member names always use forward slash / characters in their path names as separators (see section 4.4.17.1 of the PKZIP Application Note). It's important to be aware that there's no way to extract a nested archive member to a different subdirectory with Python's currentzipfilemodule. You can control the root directory, but nothing below it (i.e. any subfolders within the zip).
Lastly, since it's not necessary to rename the .pages file to .zip — the filename you passZipFile() can have any extension — I removed all that from the code. However, to overcome the limitation on extracting members to a different subdirectory, I had to add code to first extract the target member to a temporary directory, and then copy that to the final destination. Afterwards, of course, this temporary folder needs to deleted. So I'm not sure the net result is much simpler...
import os.path
import shutil
import sys
import tempfile
from zipfile import ZipFile
PREVIEW_PATH = 'QuickLooks/Preview.pdf' # archive member path
pages_file = input('Enter the path to the .pages file in question: ')
#pages_file = r'C:\Stack Overflow\extract_test.pages' # hardcode for testing
pages_file = os.path.abspath(pages_file)
filename, file_extension = os.path.splitext(pages_file)
if file_extension == ".pages":
tempdir = tempfile.gettempdir()
temp_filename = os.path.join(tempdir, PREVIEW_PATH)
with ZipFile(pages_file, 'r') as zipfile:
zipfile.extract(PREVIEW_PATH, tempdir)
if not os.path.isfile(temp_filename): # extract failure?
sys.exit('unable to extract {} from {}'.format(PREVIEW_PATH, pages_file))
final_PDF = filename + '.pdf'
shutil.copy2(temp_filename, final_PDF) # copy and rename extracted file
# delete the temporary subdirectory created (along with pdf file in it)
shutil.rmtree(os.path.join(tempdir, os.path.split(PREVIEW_PATH)[0]))
print('Check out the PDF! It\'s located at "{}".'.format(final_PDF))
#view_file(final_PDF) # see Bonus below
else:
sys.exit('Sorry, that isn\'t a .pages file.')
Bonus: If you'd like to actually view the final pdf file from the script, you can add the following function and use it on the final pdf created (assuming you have a PDF viewer application installed on your system):
import subprocess
def view_file(filepath):
subprocess.Popen(filepath, shell=True).wait()

errno 2 in Python - No such file or directory

I'm trying to compute the maximum from each file in my data directory, with the following code:
from os import listdir
def max_files(dir):
l = listdir(dir)
for n in l:
list_num(n)
def list_num(file):
f = open(file)
lines = f.readlines()
v=[]
for n in lines:
for s in n.split():
v.append(float(s))
mx = v[0]
maxi=[]
for i in v:
if i > mx:
mx = i
maxi.append(mx)
continue
continue
return maxi
print max_files(path)
I also checked my path, and it's completely correct. The error is:
f = open(file)
IOError: [Errno 2] No such file or directory: 'bvp.txt'
bvp.txt is the first file listed in the data directory.
Why does this problem occur, and how do I fix it?

You've run into a common confusion when using the return value of functions that return lists of files in a directory.
listdir is just returning the list of files in that directory. It's not returning the path to those files, just the file name. So unless the directory on which you're operating is the current directory, this won't work; you're trying to open each file in the current directory.
Whenever using the results of listdir, if you're not going to change working directories into that directory, you need to add the directory name back to the file before you open it. Therefore, pass the full path to the file to list_num instead of just the file name:
list_num(dir + '/' + n)