Does any one know why i got an error of "FileNotFoundError: [Errno 2] No such file or directory: 'bcs.xlsx'" when i'm loading this file of size 2mb it has around 60,000 rows and 4 columns.
i tried using csv instead of xlsx but i get the same error and i've checked hundreds times that the script and the file are at he same directory.
This is because Python does not find your file, errors are not lying.
But there's a misunderstanding in your question, you checked that the file is in the same directory as your script, but that's not the check you have to do. You have to check the file is in the current working directory of your python script.
To see your current working directory, use:
import os
print(os.getcwd())
And as we're at it you can list this directory:
print(os.listdir())
I don't know how you execute your script, but if you're using a terminal emulator, a typical way to give a file name to a program is by argument, not hardcoding its name, like by using argparse. And if you do this way, your shell completion may help you naming your file properly, like:
import argparse
parser = argparse.ArgumentParser()
parser.add_argument('file', type=argparse.FileType('r'))
args = parser.parse_args()
print(args.file.read())
Now on a shell if you type:
python3 ./thescript.py ./th[TAB]
your shell will autocomplete "./th" to "./thescript.py" (if and only if it exists), highly reducing the probablity of having a typo. Typically if there's a space in the filename like "the script.py", your shell should properly autocomplete the\ script.py.
Also if you use argparse with the argparse.FileType as I did, you'll have a verbose error in case the file does not exist:
thescript.py: error: argument file: can't open 'foo': [Errno 2] No such file or directory: 'foo'
But… you already have a verbose error.
Related
I making a script in python3. this script takes an input file. depends on who is running the script every time the location of this input file is different but always would be in the same directory as the script is. so I want to give the location of the input file to the script but basically the script should find it. my input file always would have the same name (infile.txt). to do so, I am using this way in python3:
path = os.path.join(os.getcwd())
input = path/infile.txt
but it does not return anything. do you know how to fix it?
os.getcwd() return the working directory which can be different to the directory where the script is. The working directory corresponds from where python is executed.
In order to know where is the scipt you should use
`
import os
input = os.path.join(os.path.dirname(os.path.realpath(__file__)), infile.txt)
# and you should use os.path.join
`
If i understand your question properly;
You have python script (sample.py) and a input file (sample_input_file.txt) in a directory say; D:\stackoverflow\sample.y and D:\stackoverflow\sample_input_file.txt respectively.
import os
stackoverflow_dir = os.getcwd()
sample_txt_file_path = os.path.join(stackoverflow_dir, 'sample_input_file.txt')
print(sample_txt_file_path)
os.path.join() takes *args as second argument which must have been your file path to join.
Upon looping a directory to delete txt files ONLY - a message is returned indicating The System cannot find the file specified: 'File.txt'.
I've made sure the txt files that I'm attempting to delete exist in the directory I'm looping. I've also checked my code and to make sure it can see my files by printing them in a list with the print command.
import os
fileLoc = 'c:\\temp\\files'
for files in os.listdir(fileLoc):
if files.endswith('.txt'):
os.unlink(files)
Upon initial execution, I expected to see all txt files deleted except for other non-txt files. The actual result was an error message "FileNotFoundError: [WinError 2] The system cannot find the file specified: 'File.txt'.
Not sure what I'm doing wrong, any help would be appreciated.
It isn't found because the the path you intended to unlink is relative to fileLoc. In fact with your code, the effect is to unlink the file relative to the current working directory. If there were *.txt files
in the cwd then the code would have unfortunate side-effects.
Another way to look at it:
Essentially, by analogy, in the shell what you're trying to do is equivalent to this:
# first the setup
$ mkdir foo
$ touch foo/a.txt
# now your code is equvalent to:
$ rm *.txt
# won't work as intended because it removes the *.txt files in the
# current directory. In fact the bug is also that your code would unlink
# any *.txt files in the current working directory unintentionally.
# what you intended was:
$ rm foo/*.txt
The missing piece was the path to the file in question.
I'll add some editorial: The Old Bard taught us to "when in doubt, print variables". In other words, debug it. I don't see from the OP an attempt to do that. Just a thing to keep in mind.
Anyway the new code:
Revised:
import os
fileLoc = 'c:\\temp\\files'
for file in os.listdir(fileLoc):
if file.endswith('.txt'):
os.unlink(os.path.join(fileLoc,file))
The fix: os.path.join() builds a path for you from parts. One part is the directory (path) where the file exists, aka: fileLoc. The other part is the filename, aka file.
os.path.join() makes a whole valid path from them using whatever OS directory separator is appropriate for your platform.
Also, might want to glance through:
https://docs.python.org/2/library/os.path.html
I'm working on a Python3 script where the code walks through directories and sub-directories to pull out all gzipped warc files.
I'd like to also add that the files are not in my home directory
file_path = os.path.join('/nappa7/pip73/Service')
walk_file(parallel_bulk, file_path)
Perhaps python is not looking where i think it's looking, nevertheless, here is my walk_file functions:
def walk_file(bulk, file_path):
warc = warcat.model.WARC()
try:
for (file_path,dirs,files) in os.walk(file_path):
for filenames in files:
if filenames.endswith('.warc.gz'):
warc.load(filenames)
except ValueError:
pass
When I replace the warc.load(filenames) with a print statement like so:
if filenames.endswith('.warc.gz'):
print(filenames)
The filenames are printed out onto the console as expected. Therefore, It leads me to believe that python was able to succesfully locate all warc.gz files. However, when i try the warc.load(filenames), i get:
FileNotFoundError: [Errno 2] No such file or directory: 'Sample.warc.gz'
I can certainly use some guidance.
Thank you.
So for anyone else who has a similar issue:
changing the code to this worked:
warc.load(os.path.join(file_path, filenames))
You need to use os.path.join(file_path, filenames) instead of just filenames.
Otherwise the operating system will look for the file in the current directory instead of file_path.
(And why is filenames plural when it refers to a single filename?)
I'm trying to read in a text file to work with Word Clouds. Here is the syntax I'm trying:
# Read the whole text.
text = open(r'C:\Users\mswitajski\Desktop\alice.txt').read()
But I keep getting the following error:
FileNotFoundError: [Errno 2] No such file or directory: 'C:\\Users\\mswitajski\\Desktop\\alice.txt'
I've triple checked the file name, tried reading it as a raw file, changed the slashes and everything but I continue to get the same error.
Well, if someone reaches up to here and still could not find the solution then here is the more pythonic way of doing the absolute path in windows.
Instead of using:
text = open(r'C:\Users\mswitajski\Desktop\alice.txt').read()
use os.sep, in conjunction of os.path.join like the following:
import os
text = open(os.path.join('C:', os.sep, 'Users', 'mswitajski', 'Desktop', 'alice.txt')).read()
Try changin the path to this"
'C:\Users\mswitajski\Desktop/alice.txt'
Sometimes windows won't find/recognize the file path when the file is specified like this
'C:\Users\mswitajski\Desktop\alice.txt'
In the answer it shows up as only one \ but you still need 2 like your previous path. The only difference is the last slash /. Hope that works.
At your text raw file (alice.txt) try delete the .txt.
The file probably is named alice.txt.txt
I face the same issue and solve it by deleted the .txt.
I had to use double slashes instead of one, because python interpreted it as a escape sequence. My final string was:
C:\\Users\\ArpitChinmay's\\AppData\\Roaming\\Code\\User\\globalStorage\\moocfi.test-my-
code\\tmcdata\\TMC workspace\\Exercises\\hy\\hy-data-analysis-with-python-
2020\\part02-e04_word_frequencies\\src\\alice.txt
However, it worked this way too,
C:\\Users\\Arpit Chinmay's\\AppData\\Roaming\\Code\\User\\globalStorage\\moocfi.test-
my-code\\tmcdata\\TMC workspace\\Exercises\\hy\\hy-data-analysis-with-python-
2020\\part02-e04_word_frequencies\\src/alice.txt
I have a file in my site-packages directory called wordlist.py that consists of just one line:
f = open("words.txt")
There is a file called words.txt in the same directory. When I run wordlist.py it works fine. However, whenever I use import wordlist, I get an error:
FileNotFoundError: [Errno 2] No such file or directory: 'words.txt'
I am using IDLE for Python 3.4
if you use relative paths for file or directory names python will look for them (or create them) in your current working directory (the $PWD variable in bash).
if you want to have them relative to the current python file, you can use (python 3.4)
import pathlib
HERE = Path(__file__).parent.absolute()
WORDS_PATH = HERE / '../path/to/words.txt'
with WORDS_PATH.open() as file_pointer:
'do something with file_pointer...'