169
import csv    

with open('v.csv', 'w') as csvfile:
    cwriter = csv.writer(csvfile, delimiter=' ', quotechar='|', quoting=csv.QUOTE_MINIMAL)

for w, c in p.items():
    cwriter.writerow(w + c)

Here, p is a dictionary, w and c both are strings.

When I try to write to the file it reports the error:

ValueError: I/O operation on closed file.

7 Answers 7

232

Indent correctly; your for statement should be inside the with block:

import csv    

with open('v.csv', 'w') as csvfile:
    cwriter = csv.writer(csvfile, delimiter=' ', quotechar='|', quoting=csv.QUOTE_MINIMAL)

    for w, c in p.items():
        cwriter.writerow(w + c)

Outside the with block, the file is closed.

>>> with open('/tmp/1', 'w') as f:
...     print(f.closed)
... 
False
>>> print(f.closed)
True
1
  • Literally, my issue. Commented Sep 12, 2024 at 14:04
11

Same error can raise by mixing: tabs + spaces.

with open('/foo', 'w') as f:
 (spaces OR  tab) print f       <-- success
 (spaces AND tab) print f       <-- fail
1
  • 1
    True, but this is always the case in python when mixing them up right?
    – Nebulosar
    Commented Apr 20, 2018 at 9:24
1

I also have the same problem. Here is my previous code

csvUsers = open('/content/gdrive/MyDrive/Ada/Users.csv', 'a', newline='', encoding='utf8')
usersWriter = csv.writer(csvUsers)
for t in users:
    NodeWriter=.writerow(users)
csvUsers.close() 

Apparently, I shoud write the usersWriter instead of NodeWriter.

NodeWriter=.writerow(users)
usersWriter=.writerow(users)

Below is my current code and it is working

csvUsers = open('/content/gdrive/MyDrive/Ada/Users.csv', 'a', newline='', encoding='utf8')
usersWriter = csv.writer(csvUsers)
for t in users:
    usersWriter=.writerow(users)
csvUsers.close() 
1
  • Isn't NodeWriter=.writerow(users) a syntax error? =. is not in any Python grammar, is it? Commented Jan 19, 2023 at 11:33
0
file = open("filename.txt", newline='')
for row in self.data:
    print(row)

Save data to a variable(file), so you need a with.

0

I had this problem when I was using an undefined variable inside the with open(...) as f:. I removed (or I defined outside) the undefined variable and the problem disappeared.

0

Another possible cause is the case when, after a round of copypasta, you end up reading two files and assign the same name to the two file handles, like the below. Note the nested with open statement.

with open(file1, "a+") as f:
    # something...
    with open(file2, "a+", f):
        # now file2's handle is called f!

    # attempting to write to file1
    f.write("blah") # error!!

The fix would then be to assign different variable names to the two file handles, e.g. f1 and f2 instead of both f.

0

I ran into a similar problem after using auto-py-to-exe to compile my program.

I used with open(link, "rb") as f: but switching to PdfReader method fixed it.

Just wanted to share in case someone else faces this.

# it diesn't work --> ValueError: I/O operation on closed file.
     for link in links:
                if os.path.isfile(link):
                    with open(link, "rb") as f:
                    pdf_reader = PyPDF2.PdfReader(f)
            pdf_exporter.write("C:\\Temp\\test.pdf")
            os.startfile("C:\\Temp\\test.pdf")

# it does
     for link in links:
                if os.path.isfile(link):
                    # with open(link, "rb") as f:
                    # pdf_reader = PyPDF2.PdfReader(f)
                    pdf_reader = PyPDF2.PdfReader(link)
            pdf_exporter.write("C:\\Temp\\test.pdf")
            os.startfile("C:\\Temp\\test.pdf")

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