OSError [Errno 22] invalid argument when use open() in Python

Question

def choose_option(self):
        if self.option_picker.currentRow() == 0:
            description = open(":/description_files/program_description.txt","r")
            self.information_shower.setText(description.read())
        elif self.option_picker.currentRow() == 1:
            requirements = open(":/description_files/requirements_for_client_data.txt", "r")
            self.information_shower.setText(requirements.read())
        elif self.option_picker.currentRow() == 2:
            menus = open(":/description_files/menus.txt", "r")
            self.information_shower.setText(menus.read())

I am using resource files and something is going wrong when i am using it as argument in open function, but when i am using it for loading of pictures and icons everything is fine.

For someone else that gets a similar error, you may have invalid characters (for example `:` or `?`) in the filename. — George Ogden, Sep 10 '20 at 09:26
Another scenario (that I just ran into) is if you are trying to write to a file inside a DropBox folder and you just had that file open very recently, you can run into this same error caused by DropBox has detected the changes and is attempting to process your new file. — royce3, May 29 '21 at 03:37

Cory Kramer · Answer 1 · 2018-06-04T18:41:38.963

52

That is not a valid file path. You must either use a full path

open(r"C:\description_files\program_description.txt","r")

Or a relative path

open("program_description.txt","r")

edited Jun 04 '18 at 18:41

answered Aug 30 '14 at 15:47

Cory Kramer

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What does the initial r in open(r"C:\file-path\file.txt", "r") signify? – Calydon Oct 17 '19 at 19:52
9

@Calydon The leading r denotes a python raw string. It treats \ as a literal character. No need to escape it. – Stacksatty Oct 27 '19 at 17:34

score 14 · Answer 2 · edited Mar 18 '20 at 08:54

14

Add 'r' in starting of path:

path = r"D:\Folder\file.txt"

That works for me.

edited Mar 18 '20 at 08:54

סטנלי גרונן

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answered Mar 18 '20 at 07:47

shailu

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what is the use of "r" in prefix, what does that means? – Roshan Yadav Jul 01 '20 at 07:21
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@RoshanYadav 'r' refers to 'raw string'. Check this out https://stackoverflow.com/questions/38742691/python-path-and-raw-strings – DC007744 Jul 02 '20 at 03:09

score 12 · Answer 3 · edited Dec 10 '18 at 13:51

12

I also ran into this fault when I used open(file_path). My reason for this fault was that my file_path had a special character like "?" or "<".

edited Dec 10 '18 at 13:51

tuomastik

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answered Jul 20 '18 at 16:20

wolfog

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I had the same problem. Thank you – Erfan Salavati Nov 10 '21 at 06:42

score 8 · Answer 4 · answered Jul 13 '18 at 17:57

8

I received the same error when trying to print an absolutely enormous dictionary. When I attempted to print just the keys of the dictionary, all was well!

answered Jul 13 '18 at 17:57

duhaime

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score 5 · Answer 5 · answered Dec 01 '22 at 18:58

For me this issue was caused by trying to write a datetime to file.

Note: this doesn't work: myFile = open(str(datetime.now()),"a")

The datetime.now() object contains the colon ''':''' character

To fix this, use a filename which avoid restricted special characters. Note this resource on detecting and replacing invalid characters: https://stackoverflow.com/a/13593932/9053474

For completeness, replace unwanted characters with the following:

import re

re.sub(r'[^\w_. -]', '_', filename)

Note these are Windows restricted characters and invalid characters differ by platform.

score 4 · Answer 6 · answered Mar 06 '20 at 11:55

4

In my case, I was using an invalid string prefix.

Wrong:

path = f"D:\Folder\file.txt"

Right:

path = r"D:\Folder\file.txt"

answered Mar 06 '20 at 11:55

Rafs

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Nick Green · Answer 7 · 2021-05-07T15:53:37.040

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In my case the error was due to lack of permissions to the folder path. I entered and saved the credentials and the issue was solved.

edited May 07 '21 at 15:53

answered Sep 14 '20 at 14:49

Nick Green

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score 3 · Answer 8 · answered Nov 01 '19 at 15:03

3

I had the same problem It happens because files can't contain special characters like ":", "?", ">" and etc. You should replace these files by using replace() function:

filename = filename.replace("special character to replace", "-")

answered Nov 01 '19 at 15:03

Lio Ak

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score 2 · Answer 9 · edited Mar 15 '18 at 12:17

2

you should add one more "/" in the last "/" of path, that is: open('C:\Python34\book.csv') to open('C:\Python34\\book.csv'). For example:

import csv
with open('C:\Python34\\book.csv', newline='') as csvfile:
    spamreader = csv.reader(csvfile, delimiter='', quotechar='|')
    for row in spamreader:
        print(row)

edited Mar 15 '18 at 12:17

Charlie Joynt

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answered Jan 05 '16 at 15:58

Hiep Tran

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Welcome to Stack Overflow! Whilst this may theoretically answer the question, [it would be preferable](//meta.stackoverflow.com/q/8259) to include the essential parts of the answer here, and provide the link for reference. – Enamul Hassan Jan 05 '16 at 16:14
7

There's a set of problems with this answer: 1) `open('C:\Python34\book.csv')` to `open('C:\Python34\book.csv')` -> There is no difference between the two lines. Did you mean `open('C:\Python34\book.csv')` to `open('C:\\Python34\\book.csv')`? 2) Both the code in the example and in the first line is wrong, as you're not escaping backslashes (or, in the case of your example, all of them.) `'C:\Python34\\book.csv'` should be `'C:\\Python34\\book.csv'` or `r'C:\Python34\book.csv'` – GPhilo May 31 '17 at 09:41
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If you click Edit you'll see that @hiep-tran did type double slashes, but SO rendered these as single slashes. I fixed the formatting with backticks around the example commands. – Charlie Joynt Mar 15 '18 at 10:59

score 2 · Answer 10 · edited Sep 06 '18 at 19:48

2

In Windows-Pycharm: If File Location|Path contains any string like \t then need to escape that with additional \ like \\t

edited Sep 06 '18 at 19:48

stderr

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answered Apr 18 '18 at 04:35

Abhijeet

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score 2 · Answer 11 · answered May 24 '18 at 07:21

2

Just replace with "/" for file path :

   open("description_files/program_description.txt","r")

answered May 24 '18 at 07:21

P113305A009D8M

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I am facing same issues like this. But , i replace with "/", it is work well. – P113305A009D8M May 24 '18 at 07:22

score 2 · Answer 12 · edited Mar 25 '21 at 01:57

2

just use single quotation marks only and use 'r' raw string upfront and a single '/'

for eg

f = open(r'C:/Desktop/file.txt','r')
print(f.read())

edited Mar 25 '21 at 01:57

Yu Hao

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answered Mar 24 '21 at 17:24

Anuragh Ragidimilli

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score 1 · Answer 13 · answered Oct 22 '22 at 01:24

I had special characters like '' in my strings, for example for one location I had a file Varzea*, then when I tried to save ('Varzea.csv') with f-string Windows complained. I just "sanitized" the string and all got back to normal.

The best way in my case was to let the strings with just letters, without special characters!

score 0 · Answer 14 · answered Jan 11 '19 at 23:38

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for folder, subs, files in os.walk(unicode(docs_dir, 'utf-8')):
    for filename in files:
        if not filename.startswith('.'):
            file_path = os.path.join(folder, filename)

answered Jan 11 '19 at 23:38

Prostak

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score 0 · Answer 15 · answered Jan 14 '19 at 12:25

0

In my case,the problem exists beacause I have not set permission for drive "C:\" and when I change my path to other drive like "F:\" my problem resolved.

answered Jan 14 '19 at 12:25

Mohsen Hrt

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score 0 · Answer 16 · answered Mar 03 '20 at 05:20

0

import pandas as pd
df = pd.read_excel ('C:/Users/yourlogin/new folder/file.xlsx')
print (df)

answered Mar 03 '20 at 05:20

I Iv

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It would be better if you could add more instructions to the answer, rather than just pasting the code. – Calos Mar 03 '20 at 05:46
If you didn't get it, mostly it means use / instead of \. – Rafs Mar 06 '20 at 11:56

score 0 · Answer 17 · answered Nov 18 '20 at 03:35

0

I got this error because old server instance was running and using log file, hence new instance was not able to write to log file. Post deleting log file this issue got resolved.

answered Nov 18 '20 at 03:35

Abhishek Gaur

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score 0 · Answer 18 · answered Oct 06 '22 at 00:48

0

When I copy the path by right clicking the file---> properties-->security, it shows the error. The working method for this is to copy path and filename separately.

answered Oct 06 '22 at 00:48

Shui Cai Luo

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score 0 · Answer 19 · edited Dec 30 '22 at 16:13

0

I had faced same issue while working with pandas and trying to open a big csv file:

wrong_df = pd.read_csv("D:\Python Projects\ML\titanic.csv")
right_df = pd.read_csv("D:\Python Projects\ML\\titanic.csv")

edited Dec 30 '22 at 16:13

Martin Faucheux

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answered Dec 30 '22 at 10:08

Sangeeta Kundu

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As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Dec 30 '22 at 19:28

score 0 · Answer 20 · answered Jun 11 '23 at 11:38

0

I had the same problem

My fault was: with open(f'file{a}{b}{c}{d}{e}{f},'w') as f: <code>

i tried to open the file as f containing f in filename

caution using with and as <variable> when opening a file

answered Jun 11 '23 at 11:38

Qwerth

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ArunC · Answer 21 · 2023-07-07T00:15:29.587

I faced the same issue, I have resolved it by creating a function that converts backslashes to forward slashes. Probably, this may not be the best way but it fixed my issue.

bs = r"\""
fs = "\\"    
def bs_to_fs(folder_path):
    folder_path_with_forw_Slash = ""
    for i in folder_path:
        if i == bs:
            folder_path_with_forw_Slash = folder_path_with_forw_Slash+'/'
        else:
            folder_path_with_forw_Slash = folder_path_with_forw_Slash+i
    folder_path_with_forw_Slash = folder_path_with_forw_Slash + fs[:1] 
    return(folder_path_with_forw_Slash)

Folder_path = r"E:\Random_Files\Project1"
open_file_name = "Some_File_Name.txt"

FILE_TO_OPEN = bs_to_fs(Folder_path) + op_file_name

with  open(FILE_TO_OPEN,'w') as file1:

Then Continue with rest of your code

OSError [Errno 22] invalid argument when use open() in Python

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