I'm new to python and really struggling with writing a program to download a CSV from this page
So far I have:
import csv
import requests
import os
output_dir = 'C:/Users/Moshe/Downloads'
output_file = 'Covid_19_uk_timeseries.csv'
CSV_URL = 'https://coronavirus.data.gov.uk/downloads/csv/coronavirus-deaths_latest.csv'
assert os.path.exists(output_dir)# test that we can write to output_dir
with requests.Session() as s:
download = s.get(CSV_URL)
decoded_content = download.content.decode('utf-8')
cr = csv.reader(decoded_content.splitlines(), delimiter=',')
my_list = list(cr)
for row in my_list:
print(row)
Its output show that it gets the CSV fine, and has access to the output directory. But for now I just want to save it as it is, and can't seem to work out how to do that.
def main():
import requests
url = "https://coronavirus.data.gov.uk/downloads/csv/coronavirus-deaths_latest.csv"
with requests.get(url, stream=True) as response:
response.raise_for_status()
with open("deaths_latest.csv", "wb") as file:
for chunk in response.iter_content(chunk_size=8192):
file.write(chunk)
file.flush()
print("Done")
return 0
if __name__ == "__main__":
import sys
sys.exit(main())
You can use
csv.writerows()
to write the csv file. Check this site for more. https://www.programiz.com/python-programming/writing-csv-files
I am trying to download data from the iex api using python and currently I am to the point where i get the data, but now i want to format it.
basically i get a lot of data which i do not care about, I just want to have the "float" section.
The data should look like this:
Ticker, Float,
AAPL, 4700000000, (something like that)
The code I am using:
import requests
url = "https://api.iextrading.com/1.0/stock/aapl/stats"
response = requests.get(url).json()
print (response)
I would be verry happy if someone could explain me how to do this.
Kind regards
Right now I have the code:
import requests
url = "https://api.iextrading.com/1.0/stock/aapl/stats"
response = requests.get(url).json()
data = (response['symbol'], response['float'])
import json filename='resp.json'
with open(filename, 'a+') as outfile:
json.dump(data, outfile, indent=4)
import requests
url = "https://api.iextrading.com/1.0/stock/tsla/stats"
response = requests.get(url).json()
data = (response['symbol'], response['float'])
import json filename='resp.json'
with open(filename, 'a+') as outfile:
json.dump(data, outfile, indent=4)
I would like the data to show as:
Ticker, Float,
AAPL, 4700000000,
TSLA, 1700000000,
(Ticker and float do not neceserally have to be placed above, i could do that myself in excel power query anyway).
You can just treat it like a dictionary. response['float'] would give you the float. Similarly for any key.
import requests
url = "https://api.iextrading.com/1.0/stock/aapl/stats"
response = requests.get(url).json()
print (response['float'])
print(response['symbol'])
Output
4705473314
AAPL
Your code is doing exactly what it should do, if you want a certain part of the json, just access it.
import requests
url = "https://api.iextrading.com/1.0/stock/aapl/stats"
response = requests.get(url).json()
print(response['float'])
>4705473314
print(response['symbol'])
>'AAPL'
print(response['symbol'], response['float'])
to store response in a json file, we can do something like this
import json
filename='resp.json'
with open(filename, 'w') as outfile:
json.dump(response, outfile, indent=4)
I'm trying to download and save an image from the web using python's requests module.
Here is the (working) code I used:
img = urllib2.urlopen(settings.STATICMAP_URL.format(**data))
with open(path, 'w') as f:
f.write(img.read())
Here is the new (non-working) code using requests:
r = requests.get(settings.STATICMAP_URL.format(**data))
if r.status_code == 200:
img = r.raw.read()
with open(path, 'w') as f:
f.write(img)
Can you help me on what attribute from the response to use from requests?
You can either use the response.raw file object, or iterate over the response.
To use the response.raw file-like object will not, by default, decode compressed responses (with GZIP or deflate). You can force it to decompress for you anyway by setting the decode_content attribute to True (requests sets it to False to control decoding itself). You can then use shutil.copyfileobj() to have Python stream the data to a file object:
import requests
import shutil
r = requests.get(settings.STATICMAP_URL.format(**data), stream=True)
if r.status_code == 200:
with open(path, 'wb') as f:
r.raw.decode_content = True
shutil.copyfileobj(r.raw, f)
To iterate over the response use a loop; iterating like this ensures that data is decompressed by this stage:
r = requests.get(settings.STATICMAP_URL.format(**data), stream=True)
if r.status_code == 200:
with open(path, 'wb') as f:
for chunk in r:
f.write(chunk)
This'll read the data in 128 byte chunks; if you feel another chunk size works better, use the Response.iter_content() method with a custom chunk size:
r = requests.get(settings.STATICMAP_URL.format(**data), stream=True)
if r.status_code == 200:
with open(path, 'wb') as f:
for chunk in r.iter_content(1024):
f.write(chunk)
Note that you need to open the destination file in binary mode to ensure python doesn't try and translate newlines for you. We also set stream=True so that requests doesn't download the whole image into memory first.
Get a file-like object from the request and copy it to a file. This will also avoid reading the whole thing into memory at once.
import shutil
import requests
url = 'http://example.com/img.png'
response = requests.get(url, stream=True)
with open('img.png', 'wb') as out_file:
shutil.copyfileobj(response.raw, out_file)
del response
How about this, a quick solution.
import requests
url = "http://craphound.com/images/1006884_2adf8fc7.jpg"
response = requests.get(url)
if response.status_code == 200:
with open("/Users/apple/Desktop/sample.jpg", 'wb') as f:
f.write(response.content)
I have the same need for downloading images using requests. I first tried the answer of Martijn Pieters, and it works well. But when I did a profile on this simple function, I found that it uses so many function calls compared to urllib and urllib2.
I then tried the way recommended by the author of requests module:
import requests
from PIL import Image
# python2.x, use this instead
# from StringIO import StringIO
# for python3.x,
from io import StringIO
r = requests.get('https://example.com/image.jpg')
i = Image.open(StringIO(r.content))
This much more reduced the number of function calls, thus speeded up my application.
Here is the code of my profiler and the result.
#!/usr/bin/python
import requests
from StringIO import StringIO
from PIL import Image
import profile
def testRequest():
image_name = 'test1.jpg'
url = 'http://example.com/image.jpg'
r = requests.get(url, stream=True)
with open(image_name, 'wb') as f:
for chunk in r.iter_content():
f.write(chunk)
def testRequest2():
image_name = 'test2.jpg'
url = 'http://example.com/image.jpg'
r = requests.get(url)
i = Image.open(StringIO(r.content))
i.save(image_name)
if __name__ == '__main__':
profile.run('testUrllib()')
profile.run('testUrllib2()')
profile.run('testRequest()')
The result for testRequest:
343080 function calls (343068 primitive calls) in 2.580 seconds
And the result for testRequest2:
3129 function calls (3105 primitive calls) in 0.024 seconds
This might be easier than using requests. This is the only time I'll ever suggest not using requests to do HTTP stuff.
Two liner using urllib:
>>> import urllib
>>> urllib.request.urlretrieve("http://www.example.com/songs/mp3.mp3", "mp3.mp3")
There is also a nice Python module named wget that is pretty easy to use. Found here.
This demonstrates the simplicity of the design:
>>> import wget
>>> url = 'http://www.futurecrew.com/skaven/song_files/mp3/razorback.mp3'
>>> filename = wget.download(url)
100% [................................................] 3841532 / 3841532>
>> filename
'razorback.mp3'
Enjoy.
Edit: You can also add an out parameter to specify a path.
>>> out_filepath = <output_filepath>
>>> filename = wget.download(url, out=out_filepath)
Following code snippet downloads a file.
The file is saved with its filename as in specified url.
import requests
url = "http://example.com/image.jpg"
filename = url.split("/")[-1]
r = requests.get(url, timeout=0.5)
if r.status_code == 200:
with open(filename, 'wb') as f:
f.write(r.content)
There are 2 main ways:
Using .content (simplest/official) (see Zhenyi Zhang's answer):
import io # Note: io.BytesIO is StringIO.StringIO on Python2.
import requests
r = requests.get('http://lorempixel.com/400/200')
r.raise_for_status()
with io.BytesIO(r.content) as f:
with Image.open(f) as img:
img.show()
Using .raw (see Martijn Pieters's answer):
import requests
r = requests.get('http://lorempixel.com/400/200', stream=True)
r.raise_for_status()
r.raw.decode_content = True # Required to decompress gzip/deflate compressed responses.
with PIL.Image.open(r.raw) as img:
img.show()
r.close() # Safety when stream=True ensure the connection is released.
Timing both shows no noticeable difference.
As easy as to import Image and requests
from PIL import Image
import requests
img = Image.open(requests.get(url, stream = True).raw)
img.save('img1.jpg')
This is how I did it
import requests
from PIL import Image
from io import BytesIO
url = 'your_url'
files = {'file': ("C:/Users/shadow/Downloads/black.jpeg", open('C:/Users/shadow/Downloads/black.jpeg', 'rb'),'image/jpg')}
response = requests.post(url, files=files)
img = Image.open(BytesIO(response.content))
img.show()
Here is a more user-friendly answer that still uses streaming.
Just define these functions and call getImage(). It will use the same file name as the url and write to the current directory by default, but both can be changed.
import requests
from StringIO import StringIO
from PIL import Image
def createFilename(url, name, folder):
dotSplit = url.split('.')
if name == None:
# use the same as the url
slashSplit = dotSplit[-2].split('/')
name = slashSplit[-1]
ext = dotSplit[-1]
file = '{}{}.{}'.format(folder, name, ext)
return file
def getImage(url, name=None, folder='./'):
file = createFilename(url, name, folder)
with open(file, 'wb') as f:
r = requests.get(url, stream=True)
for block in r.iter_content(1024):
if not block:
break
f.write(block)
def getImageFast(url, name=None, folder='./'):
file = createFilename(url, name, folder)
r = requests.get(url)
i = Image.open(StringIO(r.content))
i.save(file)
if __name__ == '__main__':
# Uses Less Memory
getImage('http://www.example.com/image.jpg')
# Faster
getImageFast('http://www.example.com/image.jpg')
The request guts of getImage() are based on the answer here and the guts of getImageFast() are based on the answer above.
I'm going to post an answer as I don't have enough rep to make a comment, but with wget as posted by Blairg23, you can also provide an out parameter for the path.
wget.download(url, out=path)
This is the first response that comes up for google searches on how to download a binary file with requests. In case you need to download an arbitrary file with requests, you can use:
import requests
url = 'https://s3.amazonaws.com/lab-data-collections/GoogleNews-vectors-negative300.bin.gz'
open('GoogleNews-vectors-negative300.bin.gz', 'wb').write(requests.get(url, allow_redirects=True).content)
my approach was to use response.content (blob) and save to the file in binary mode
img_blob = requests.get(url, timeout=5).content
with open(destination + '/' + title, 'wb') as img_file:
img_file.write(img_blob)
Check out my python project that downloads images from unsplash.com based on keywords.
You can do something like this:
import requests
import random
url = "https://images.pexels.com/photos/1308881/pexels-photo-1308881.jpeg? auto=compress&cs=tinysrgb&dpr=1&w=500"
name=random.randrange(1,1000)
filename=str(name)+".jpg"
response = requests.get(url)
if response.status_code.ok:
with open(filename,'w') as f:
f.write(response.content)
Agree with Blairg23 that using urllib.request.urlretrieve is one of the easiest solutions.
One note I want to point out here. Sometimes it won't download anything because the request was sent via script (bot), and if you want to parse images from Google images or other search engines, you need to pass user-agent to request headers first, and then download the image, otherwise, the request will be blocked and it will throw an error.
Pass user-agent and download image:
opener=urllib.request.build_opener()
opener.addheaders=[('User-Agent','Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/70.0.3538.102 Safari/537.36 Edge/18.19582')]
urllib.request.install_opener(opener)
urllib.request.urlretrieve(URL, 'image_name.jpg')
Code in the online IDE that scrapes and downloads images from Google images using requests, bs4, urllib.requests.
Alternatively, if your goal is to scrape images from search engines like Google, Bing, Yahoo!, DuckDuckGo (and other search engines), then you can use SerpApi. It's a paid API with a free plan.
The biggest difference is that there's no need to figure out how to bypass blocks from search engines or how to extract certain parts from the HTML or JavaScript since it's already done for the end-user.
Example code to integrate:
import os, urllib.request
from serpapi import GoogleSearch
params = {
"api_key": os.getenv("API_KEY"),
"engine": "google",
"q": "pexels cat",
"tbm": "isch"
}
search = GoogleSearch(params)
results = search.get_dict()
print(json.dumps(results['images_results'], indent=2, ensure_ascii=False))
# download images
for index, image in enumerate(results['images_results']):
# print(f'Downloading {index} image...')
opener=urllib.request.build_opener()
opener.addheaders=[('User-Agent','Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/70.0.3538.102 Safari/537.36 Edge/18.19582')]
urllib.request.install_opener(opener)
# saves original res image to the SerpApi_Images folder and add index to the end of file name
urllib.request.urlretrieve(image['original'], f'SerpApi_Images/original_size_img_{index}.jpg')
-----------
'''
]
# other images
{
"position": 100, # 100 image
"thumbnail": "https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcQK62dIkDjNCvEgmGU6GGFZcpVWwX-p3FsYSg&usqp=CAU",
"source": "homewardboundnj.org",
"title": "pexels-helena-lopes-1931367 - Homeward Bound Pet Adoption Center",
"link": "https://homewardboundnj.org/upcoming-event/black-cat-appreciation-day/pexels-helena-lopes-1931367/",
"original": "https://homewardboundnj.org/wp-content/uploads/2020/07/pexels-helena-lopes-1931367.jpg",
"is_product": false
}
]
'''
Disclaimer, I work for SerpApi.
Here is a very simple code
import requests
response = requests.get("https://i.imgur.com/ExdKOOz.png") ## Making a variable to get image.
file = open("sample_image.png", "wb") ## Creates the file for image
file.write(response.content) ## Saves file content
file.close()
for download Image
import requests
Picture_request = requests.get(url)
I'm using Requests to upload a PDF to an API. It is stored as "response" below. I'm trying to write that out to Excel.
import requests
files = {'f': ('1.pdf', open('1.pdf', 'rb'))}
response = requests.post("https://pdftables.com/api?&format=xlsx-single",files=files)
response.raise_for_status() # ensure we notice bad responses
file = open("out.xls", "w")
file.write(response)
file.close()
I'm getting the error:
file.write(response)
TypeError: expected a character buffer object
I believe all the existing answers contain the relevant information, but I would like to summarize.
The response object that is returned by requests get and post operations contains two useful attributes:
Response attributes
response.text - Contains str with the response text.
response.content - Contains bytes with the raw response content.
You should choose one or other of these attributes depending on the type of response you expect.
For text-based responses (html, json, yaml, etc) you would use response.text
For binary-based responses (jpg, png, zip, xls, etc) you would use response.content.
Writing response to file
When writing responses to file you need to use the open function with the appropriate file write mode.
For text responses you need to use "w" - plain write mode.
For binary responses you need to use "wb" - binary write mode.
Examples
Text request and save
# Request the HTML for this web page:
response = requests.get("https://stackoverflow.com/questions/31126596/saving-response-from-requests-to-file")
with open("response.txt", "w") as f:
f.write(response.text)
Binary request and save
# Request the profile picture of the OP:
response = requests.get("https://i.stack.imgur.com/iysmF.jpg?s=32&g=1")
with open("response.jpg", "wb") as f:
f.write(response.content)
Answering the original question
The original code should work by using wb and response.content:
import requests
files = {'f': ('1.pdf', open('1.pdf', 'rb'))}
response = requests.post("https://pdftables.com/api?&format=xlsx-single",files=files)
response.raise_for_status() # ensure we notice bad responses
file = open("out.xls", "wb")
file.write(response.content)
file.close()
But I would go further and use the with context manager for open.
import requests
with open('1.pdf', 'rb') as file:
files = {'f': ('1.pdf', file)}
response = requests.post("https://pdftables.com/api?&format=xlsx-single",files=files)
response.raise_for_status() # ensure we notice bad responses
with open("out.xls", "wb") as file:
file.write(response.content)
You can use the response.text to write to a file:
import requests
files = {'f': ('1.pdf', open('1.pdf', 'rb'))}
response = requests.post("https://pdftables.com/api?&format=xlsx-single",files=files)
response.raise_for_status() # ensure we notice bad responses
with open("resp_text.txt", "w") as file:
file.write(response.text)
As Peter already pointed out:
In [1]: import requests
In [2]: r = requests.get('https://api.github.com/events')
In [3]: type(r)
Out[3]: requests.models.Response
In [4]: type(r.content)
Out[4]: str
You may also want to check r.text.
Also: https://2.python-requests.org/en/latest/user/quickstart/