How to open .csv file from a url with Python?
11,408
Solution 1
Download the stream, then process:
import urllib2
url = "http://httpbin.org/get"
response = urllib2.urlopen(url)
data = response.read()
read = csv.DictReader(data)
Solution 2
I recommend pandas for this:
import pandas as pd
read = pandas.io.parsers.read_csv("http://....", ...)
please see the documentation.
Solution 3
You can do the following :
import csv
import urllib2
url = 'http://winterolympicsmedals.com/medals.csv'
response = urllib2.urlopen(url)
cr = csv.reader(response)
for row in cr:
print row
Author by
Admin
Updated on June 14, 2022Comments
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Admin almost 2 years
I'm trying to open a csv file from a url but for some reason I get an error saying that there is an invalid mode or filename. I'm not sure what the issue is. Help?
url = "http://...." data = open(url, "r") read = csv.DictReader(data)