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.6 Answers
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)
TrueSame error can raise by mixing: tabs + spaces.
with open('/foo', 'w') as f: (spaces OR tab) print f <-- success (spaces AND tab) print f <-- failfile = open("filename.txt", newline='')
for row in self.data: print(row)Save data to a variable(file), so you need a with.
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.
I was getting this exception when debugging in PyCharm, given that no breakpoint was being hit. To prevent it, I added a breakpoint just after the with block, and then it stopped happening.
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.
