Read .csv file from URL into Python 3.x – _csv.Error: iterator should return strings, not bytes (did you open the file in text mode?)

Question

I’ve been struggling with this simple problem for too long, so I thought I’d ask for help. I am trying to read a list of journal articles from National Library of Medicine ftp site into Python 3.3.2 (on Windows 7). The journal articles are in a .csv file.

I have tried the following code:

import csv
import urllib.request

url = "ftp://ftp.ncbi.nlm.nih.gov/pub/pmc/file_list.csv"
ftpstream = urllib.request.urlopen(url)
csvfile = csv.reader(ftpstream)
data = [row for row in csvfile]

It results in the following error:

Traceback (most recent call last):
File "<pyshell#4>", line 1, in <module>
data = [row for row in csvfile]
File "<pyshell#4>", line 1, in <listcomp>
data = [row for row in csvfile]
_csv.Error: iterator should return strings, not bytes (did you open the file in text mode?)

I presume I should be working with strings not bytes? Any help with the simple problem, and an explanation as to what is going wrong would be greatly appreciated.

Asked By: Chris

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Source

Answer 1

urlopen will return a urllib.response.addinfourl instance for an ftp request.

For ftp, file, and data urls and requests explicity handled by legacy
URLopener and FancyURLopener classes, this function returns a
urllib.response.addinfourl object which can work as context manager…

>>> urllib2.urlopen(url)
<addinfourl at 48868168L whose fp = <addclosehook at 48777416L whose fp = <socket._fileobject object at 0x0000000002E52B88>>>

At this point ftpstream is a file like object, using .read() would return the contents however csv.reader requires an iterable in this case:

Defining a generator like so:

def to_lines(f):
    line = f.readline()
    while line:
        yield line
        line = f.readline()

We can create our csv reader like so:

reader = csv.reader(to_lines(ftps))

And with a url

url = "http://pic.dhe.ibm.com/infocenter/tivihelp/v41r1/topic/com.ibm.ismsaas.doc/reference/CIsImportMinimumSample.csv"

The code:

for row in reader: print row

Prints

>>> 
['simpleci']
['SCI.APPSERVER']
['SRM_SaaS_ES', 'MXCIImport', 'AddChange', 'EN']
['CI_CINUM']
['unique_identifier1']
['unique_identifier2']

Answered By: HennyH

Answer 2

The problem relies on urllib returning bytes. As a proof, you can try to download the csv file with your browser and opening it as a regular file and the problem is gone.

A similar problem was addressed here.

It can be solved decoding bytes to strings with the appropriate encoding. For example:

import csv
import urllib.request

url = "ftp://ftp.ncbi.nlm.nih.gov/pub/pmc/file_list.csv"
ftpstream = urllib.request.urlopen(url)
csvfile = csv.reader(ftpstream.read().decode('utf-8'))  # with the appropriate encoding 
data = [row for row in csvfile]

The last line could also be: data = list(csvfile) which can be easier to read.

By the way, since the csv file is very big, it can slow and memory-consuming. Maybe it would be preferable to use a generator.

EDIT:
Using codecs as proposed by Steven Rumbalski so it’s not necessary to read the whole file to decode. Memory consumption reduced and speed increased.

import csv
import urllib.request
import codecs

url = "ftp://ftp.ncbi.nlm.nih.gov/pub/pmc/file_list.csv"
ftpstream = urllib.request.urlopen(url)
csvfile = csv.reader(codecs.iterdecode(ftpstream, 'utf-8'))
for line in csvfile:
    print(line)  # do something with line

Note that the list is not created either for the same reason.

Answered By: Diego Herranz

Answer 3

Even though there is already an accepted answer, I thought I’d add to the body of knowledge by showing how I achieved something similar using the requests package (which is sometimes seen as an alternative to urlib.request).

The basis of using codecs.itercode() to solve the original problem is still the same as in the accepted answer.

import codecs
from contextlib import closing
import csv
import requests

url = "ftp://ftp.ncbi.nlm.nih.gov/pub/pmc/file_list.csv"

with closing(requests.get(url, stream=True)) as r:
    reader = csv.reader(codecs.iterdecode(r.iter_lines(), 'utf-8'))
    for row in reader:
        print row

Here we also see the use of streaming provided through the requests package in order to avoid having to load the entire file over the network into memory first (which could take long if the file is large).

I thought it might be useful since it helped me, as I was using requests rather than urllib.request in Python 3.6.

Some of the ideas (e.g using closing()) are picked from this similar post

Answered By: Irvin H.

Answer 4

I had a similar problem using requests package and csv.
The response from post request was type bytes.
In order to user csv library, first I a stored them as a string file in memory (in my case the size was small), decoded utf-8.

import io
import csv
import requests

response = requests.post(url, data)

# response.content is something like: 
# b'"City","Awb","Total"rn"Bucuresti","6733338850003","32.57"rn'    
csv_bytes = response.content

# write in-memory string file from bytes, decoded (utf-8)
str_file = io.StringIO(csv_bytes.decode('utf-8'), newline='n')
    
reader = csv.reader(str_file)
for row_list in reader:
    print(row_list)

# Once the file is closed,
# any operation on the file (e.g. reading or writing) will raise a ValueError
str_file.close()

Printed something like:

['City', 'Awb', 'Total']
['Bucuresti', '6733338850003', '32.57']

Answered By: Cosmin Dinu

Read .csv file from URL into Python 3.x – _csv.Error: iterator should return strings, not bytes (did you open the file in text mode?)

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