Python 404 from valid url URLLIB2

Question

I am getting a 404 from urllib2. Despite the fact that the link works fine in my browser...

import urllib2
import numpy
import scipy.stats
import csv
import httplib

f1 = urllib2.urlopen("http://ichart.finance.yahoo.com/table.csv?s=GOOG&a=01&b=1&c=2010&d=01&e=01&f=2011&g=d&ignore=.csv")
f2 = urllib2.urlopen("http://ichart.finance.yahoo.com/table.csv?s=APPL&a=01&b=1&c=2010&d=01&e=01&f=2011&g=d&ignore=.csv")

a = numpy.genfromtxt(f1, delimiter=',', skip_header=0, names=True);
b = numpy.genfromtxt(f2, delimiter=',', skip_header=0, names=True);

prs = scipy.stats.pearsonr(a['Open'], b['Open'])

print prs

unutbu · Accepted Answer · 2013-06-25 16:50:44Z

2

The stock symbol for Apple is AAPL, not APPL.

Therefore, use:

http://ichart.finance.yahoo.com/table.csv?s=AAPL&a=01&b=1&c=2010&d=01&e=01&f=2011&g=d&ignore=.csv"

answered Jun 25, 2013 at 16:50

unutbu

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lborgav · Accepted Answer · 2013-06-25 16:49:53Z

0

There's something wrong with URL in f2. When I check in my browser: http://ichart.finance.yahoo.com/table.csv?s=APPL&a=01&b=1&c=2010&d=01&e=01&f=2011&g=d&ignore=.csv

I get: Sorry, the page you requested was not found. So you'll get a 404 error in this line

answered Jun 25, 2013 at 16:49

lborgav

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