csv.Error: iterator should return strings, not bytes

Sample.csv contains the following:

NAME    Id   No  Dept
Tom     1    12   CS
Hendry  2    35   EC
Bahamas 3    21   IT
Frank   4    61   EE

And the Python file contains the following code:

import csv
ifile  = open('sample.csv', "rb")
read = csv.reader(ifile)
for row in read :
    print (row)

When I run the above code in Python, I get the following exception:

File “csvformat.py”, line 4, in
for row in read :
_csv.Error: iterator should return strings, not bytes (did you open the file in text mode?)

How can I fix it?

Answers:

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Method 1

You open the file in text mode.

More specifically:

ifile  = open('sample.csv', "rt", encoding=<theencodingofthefile>)

Good guesses for encoding is “ascii” and “utf8”. You can also leave the encoding off, and it will use the system default encoding, which tends to be UTF8, but may be something else.

Method 2

The reason it is throwing that exception is because you have the argument rb, which opens the file in binary mode. Change that to r, which will by default open the file in text mode.

Your code:

import csv
ifile  = open('sample.csv', "rb")
read = csv.reader(ifile)
for row in read :
    print (row)

New code:

import csv
ifile  = open('sample.csv', "r")
read = csv.reader(ifile)
for row in read :
    print (row)

Method 3

In Python3, csv.reader expects, that passed iterable returns strings, not bytes. Here is one more solution to this problem, that uses codecs module:

import csv
import codecs
ifile  = open('sample.csv', "rb")
read = csv.reader(codecs.iterdecode(ifile, 'utf-8'))
for row in read :
    print (row)

Method 4

Your problem is you have the b in the open flag.
The flag rt (read, text) is the default, so, using the context manager, simply do this:

with open('sample.csv') as ifile:
    read = csv.reader(ifile) 
    for row in read:
        print (row)

The context manager means you don’t need generic error handling (without which you may get stuck with the file open, especially in an interpreter), because it will automatically close the file on an error, or on exiting the context.

The above is the same as:

with open('sample.csv', 'r') as ifile:
    ...

or

with open('sample.csv', 'rt') as ifile:
    ...

Method 5

I had this error when running an old python script developped with Python 2.6.4

When updating to 3.6.2, I had to remove all ‘rb’ parameters from open calls in order to fix this csv reading error.

Method 6

This is a GOTCHA in Django. The open that comes with filefield always opens the file in byte mode as far as I can tell. You need to open with Python’s open instead.

So with avccont_datafile being an instance of a model with a field called datafile = django.db.models.FileField(), The following code….

with avccount_datafile.datafile.file.open('rt') as fp:
    self.avc_data = csv.DictReader(fp)

gives the error

_csv.Error: iterator should return strings, not bytes (did you open the file in text mode?)

but this fixes it

filename = avccount_datafile.datafile.file.name
with open(filename, 'rt') as fp:
        self.avc_data = csv.DictReader(fp)

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0