How to read a CSV file from a URL with Python?

Question:

when I do curl to a API call link http://example.com/passkey=wedsmdjsjmdd

curl 'http://example.com/passkey=wedsmdjsjmdd'

I get the employee output data on a csv file format, like:

"Steve","421","0","421","2","","","","","","","","","421","0","421","2"

how can parse through this using python.

I tried:

import csv 
cr = csv.reader(open('http://example.com/passkey=wedsmdjsjmdd',"rb"))
for row in cr:
    print row

but it didn’t work and I got an error

http://example.com/passkey=wedsmdjsjmdd No such file or directory:

Thanks!

Asked By: mongotop

||

Answers:

You need to replace open with urllib.urlopen or urllib2.urlopen.

e.g.

import csv
import urllib2

url = 'http://winterolympicsmedals.com/medals.csv'
response = urllib2.urlopen(url)
cr = csv.reader(response)

for row in cr:
    print row

This would output the following

Year,City,Sport,Discipline,NOC,Event,Event gender,Medal
1924,Chamonix,Skating,Figure skating,AUT,individual,M,Silver
1924,Chamonix,Skating,Figure skating,AUT,individual,W,Gold
...

The original question is tagged "python-2.x", but for a Python 3 implementation (which requires only minor changes) see below.

Answered By: eandersson

Using pandas it is very simple to read a csv file directly from a url

import pandas as pd
data = pd.read_csv('https://example.com/passkey=wedsmdjsjmdd')

This will read your data in tabular format, which will be very easy to process

Answered By: Kathirmani Sukumar

You could do it with the requests module as well:

url = 'http://winterolympicsmedals.com/medals.csv'
r = requests.get(url)
text = r.iter_lines()
reader = csv.reader(text, delimiter=',')
Answered By: Rodo

To increase performance when downloading a large file, the below may work a bit more efficiently:

import requests
from contextlib import closing
import csv

url = "http://download-and-process-csv-efficiently/python.csv"

with closing(requests.get(url, stream=True)) as r:
    reader = csv.reader(r.iter_lines(), delimiter=',', quotechar='"')
    for row in reader:
        # Handle each row here...
        print row   

By setting stream=True in the GET request, when we pass r.iter_lines() to csv.reader(), we are passing a generator to csv.reader(). By doing so, we enable csv.reader() to lazily iterate over each line in the response with for row in reader.

This avoids loading the entire file into memory before we start processing it, drastically reducing memory overhead for large files.

Answered By: The Aelfinn
import pandas as pd
url='https://raw.githubusercontent.com/juliencohensolal/BankMarketing/master/rawData/bank-additional-full.csv'
data = pd.read_csv(url,sep=";") # use sep="," for coma separation. 
data.describe()

enter image description here

Answered By: user2458922

what you were trying to do with the curl command was to download the file to your local hard drive(HD). You however need to specify a path on HD

curl http://example.com/passkey=wedsmdjsjmdd -o ./example.csv
cr = csv.reader(open('./example.csv',"r"))
for row in cr:
    print row



Answered By: Adeshina Otayo

This question is tagged python-2.x so it didn’t seem right to tamper with the original question, or the accepted answer. However, Python 2 is now unsupported, and this question still has good google juice for "python csv urllib", so here’s an updated Python 3 solution.

It’s now necessary to decode urlopen‘s response (in bytes) into a valid local encoding, so the accepted answer has to be modified slightly:

import csv, urllib.request

url = 'http://winterolympicsmedals.com/medals.csv'
response = urllib.request.urlopen(url)
lines = [l.decode('utf-8') for l in response.readlines()]
cr = csv.reader(lines)

for row in cr:
    print(row)

Note the extra line beginning with lines =, the fact that urlopen is now in the urllib.request module, and print of course requires parentheses.

It’s hardly advertised, but yes, csv.reader can read from a list of strings.

And since someone else mentioned pandas, here’s a pandas rendition that displays the CSV in a console-friendly output:

python3 -c 'import pandas
df = pandas.read_csv("http://winterolympicsmedals.com/medals.csv")
print(df.to_string())'

Pandas is not a lightweight library, though. If you don’t need the things that pandas provides, or if startup time is important (e.g. you’re writing a command line utility or any other program that needs to load quickly), I’d advise that you stick with the standard library functions.

Answered By: TheDudeAbides

I am also using this approach for csv files (Python 3.6.9):

import csv
import io
import requests

r = requests.get(url)
buff = io.StringIO(r.text)
dr = csv.DictReader(buff)
for row in dr:
    print(row)
Answered By: Michal Skop

All the above solutions didn’t work with Python3, I got all the "famous" error messages, like _csv.Error: iterator should return strings, not bytes (did you open the file in text mode?) and _csv.Error: new-line character seen in unquoted field - do you need to open the file in universal-newline mode?. So I was a bit stuck here.

My mistake here, was that I used response.text while response is a requests.models.Response class, while I should have used response.content instead (as the first error suggested), so I was able to decode its UTF-8 correctly and split lines afterwards. So here is my solutions:

response = reqto.get("https://example.org/utf8-data.csv")
# Do some error checks to avoid bad results
if response.ok and len(response.content) > 0:
    reader = csv.DictReader(response.content.decode('utf-8').splitlines(), dialect='unix')
    for row in reader:
        print(f"DEBUG: row={row}")

The above example gives me already a dict back with each row. But with leading # for each dict key, which I may have to live with.

Answered By: Roland
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