Pandas read_csv from url

Question

I am using Python 3.4 with IPython and have the following code. I'm unable to read a csv-file from the given URL:

import pandas as pd
import requests

url="https://github.com/cs109/2014_data/blob/master/countries.csv"
s=requests.get(url).content
c=pd.read_csv(s)

I have the following error

"Expected file path name or file-like object, got type"

How can I fix this?

Answer 1

The problem you're having is that the output you get into the variable 's' is not a csv, but a html file. In order to get the raw csv, you have to modify the url to:

'https://raw.githubusercontent.com/cs109/2014_data/master/countries.csv'

Your second problem is that read_csv expects a file name, we can solve this by using StringIO from io module. Third problem is that request.get(url).content delivers a byte stream, we can solve this using the request.get(url).text instead.

End result is this code:

from io import StringIO

import pandas as pd
import requests
url='https://raw.githubusercontent.com/cs109/2014_data/master/countries.csv'
s=requests.get(url).text

c=pd.read_csv(StringIO(s))

output:

>>> c.head()
    Country  Region
0   Algeria  AFRICA
1    Angola  AFRICA
2     Benin  AFRICA
3  Botswana  AFRICA
4   Burkina  AFRICA

Answer 2

As I commented you need to use a StringIO object and decode i.e c=pd.read_csv(io.StringIO(s.decode("utf-8"))) if using requests, you need to decode as .content returns bytes if you used .text you would just need to pass s as is s = requests.get(url).text c = pd.read_csv(StringIO(s)).

A simpler approach is to pass the correct url of the raw data directly to read_csv, you don't have to pass a file like object, you can pass a url so you don't need requests at all:

c = pd.read_csv("https://raw.githubusercontent.com/cs109/2014_data/master/countries.csv")

print(c)

Output:

                              Country         Region
0                             Algeria         AFRICA
1                              Angola         AFRICA
2                               Benin         AFRICA
3                            Botswana         AFRICA
4                             Burkina         AFRICA
5                             Burundi         AFRICA
6                            Cameroon         AFRICA
..................................

From the docs:

filepath_or_buffer :

string or file handle / StringIO The string could be a URL. Valid URL schemes include http, ftp, s3, and file. For file URLs, a host is expected. For instance, a local file could be file ://localhost/path/to/table.csv

Answer 3

Just as the error suggests , pandas.read_csv needs a file-like object as the first argument.

If you want to read the csv from a string, you can use io.StringIO (Python 3.x) or StringIO.StringIO (Python 2.x) .

Also, for the URL - https://github.com/cs109/2014_data/blob/master/countries.csv - you are getting back html response , not raw csv, you should use the url given by the Raw link in the github page for getting raw csv response , which is - https://raw.githubusercontent.com/cs109/2014_data/master/countries.csv

Example -

import pandas as pd
import io
import requests
url="https://raw.githubusercontent.com/cs109/2014_data/master/countries.csv"
s=requests.get(url).content
c=pd.read_csv(io.StringIO(s.decode('utf-8')))

Update

From pandas 0.19.2 you can now just pass the url directly.

Answer 4

In the latest version of pandas (0.19.2) you can directly pass the url

import pandas as pd

url="https://raw.githubusercontent.com/cs109/2014_data/master/countries.csv"
c=pd.read_csv(url)