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How do I download videos from Pexels API?

I have this code that can pull images off of Pexels, but I don't know how to change it to video. I haven't seen anyone do this before and any help greatly appreciated. I tried switching all the photo tags to videos but that seemed not to work. I've also tried adding more libraries but that doesn't seem to work either.
import argparse
import json
import os
import time
import requests
import tqdm
from pexels_api import API
PEXELS_API_KEY = os.environ['PEXELS_KEY']
MAX_IMAGES_PER_QUERY = 100
RESULTS_PER_PAGE = 10
PAGE_LIMIT = MAX_IMAGES_PER_QUERY / RESULTS_PER_PAGE
def get_sleep(t):
def sleep():
time.sleep(t)
return sleep
def main(args):
sleep = get_sleep(args.sleep)
api = API(PEXELS_API_KEY)
query = args.query
page = 1
counter = 0
photos_dict = {}
# Step 1: Getting urls and meta information
while page <= PAGE_LIMIT:
api.search(query, page=page, results_per_page=RESULTS_PER_PAGE)
photos = api.get_entries()
for photo in tqdm.tqdm(photos):
photos_dict[photo.id] = vars(photo)['_Photo__photo']
counter += 1
if not api.has_next_page:
break
page += 1
sleep()
print(f"Finishing at page: {page}")
print(f"Images were processed: {counter}")
# Step 2: Downloading
if photos_dict:
os.makedirs(args.path, exist_ok=True)
# Saving dict
with open(os.path.join(args.path, f'{query}.json'), 'w') as fout:
json.dump(photos_dict, fout)
for val in tqdm.tqdm(photos_dict.values()):
url = val['src'][args.resolution]
fname = os.path.basename(val['src']['original'])
image_path = os.path.join(args.path, fname)
if not os.path.isfile(image_path): # ignore if already downloaded
response = requests.get(url, stream=True)
with open(image_path, 'wb') as outfile:
outfile.write(response.content)
else:
print(f"File exists: {image_path}")
if __name__ == '__main__':
parser = argparse.ArgumentParser()
parser.add_argument('--query', type=str, required=True)
parser.add_argument('--path', type=str, default='./results_pexels')
parser.add_argument('--resolution', choices=['original', 'large2x', 'large',
'medium', 'small', 'portrait',
'landscape', 'tiny'], default='original')
parser.add_argument('--sleep', type=float, default=0.1)
args = parser.parse_args()
main(args)

sorry for bumping into the question. I just faced a similar situation when downloading the videos from Pexels using the python API, pexelsPy. This may be helpful:
I retrieved the ID of the videos and then created the downloading URL that has the following structure: "https://www.pexels.com/video/"+ ID +"/download".
See the following example:
def download_video(type_of_videos):
video_tag = random.choice(type_of_videos)
PEXELS_API = '-' #please add your API Key here
api = API(PEXELS_API)
retrieved_videos = read_already_download_files('downloaded_files.txt')
video_found_flag = True
num_page = 1
while video_found_flag:
api.search_videos(video_tag, page=num_page, results_per_page=10)
videos = api.get_videos()
for data in videos:
if data.width > data.height: #look for horizontal orientation videos
if data.url not in retrieved_videos:
# write_file('downloaded_files.txt', data.url)
url_video = 'https://www.pexels.com/video/' + str(data.id) + '/download' #create the url with the video id
r = requests.get(url_video)
with open(data.url.split('/')[-2]+'.mp4', 'wb') as outfile:
outfile.write(r.content)
return data.url.split('/')[-2]+'.mp4' #download the video
num_page += 1
download_video function takes an array of strings with several tags, e.g.: ['happy','sad','relax']. Then it randomly chooses one of these tags.
PEXELS_API should contain your API Key.
read_already_download_files('downloaded_files.txt'): Retrieves already downloaded files to check if the current found file is already downloaded.

from pypexels import PyPexels
import requests
api_key = 'api id'
# instantiate PyPexels object
py_pexel = PyPexels(api_key=api_key)
search_videos_page = py_pexel.videos_search(query="love", per_page=40)
# while True:
for video in search_videos_page.entries:
print(video.id, video.user.get('name'), video.url)
data_url = 'https://www.pexels.com/video/' + str(video.id) + '/download'
r = requests.get(data_url)
print(r.headers.get('content-type'))
with open('sample.mp4', 'wb') as outfile:
outfile.write(r.content)
# if not search_videos_page.has_next:
break
# search_videos_page = search_videos_page.get_next_page()

I just tried to do the same. When I was looking for it, I wanted a simple example. All other fancy stuff I was sure I could add myself. So, I built upon inou's answer. The shown example is very basic and requests one page with only 5 results using the 'Tiger' tag in the search query. I download the first video using its id provided by the response and simply write it to the source folder. The api is provided by pexelsPy and the request is executed using the standard requests package. To get access to the API, you need to create a key on pexels website (see here). Once you get your own API key, you should be able to simply substitute the shown example key and run the code as a test.
import pexelsPy
import requests
PEXELS_API = '16gv62567257256iu78krtuzwqsddudrtjberzabzwzjsrtgswnr'
api = pexelsPy.API(PEXELS_API)
api.search_videos('Tiger', page=1, results_per_page=5)
videos = api.get_videos()
url_video = 'https://www.pexels.com/video/' + str(videos[0].id) + '/download'
r = requests.get(url_video)
with open('test.mp4', 'wb') as outfile:
outfile.write(r.content)

You can download multiple videos with this code :
import pexelsPy
import requests
PEXELS_API = '-'
api = pexelsPy.API(PEXELS_API)
api.search_videos('nature', page=2, results_per_page=100, orientation='landscape')
videos = api.get_videos()
for i, video in enumerate(videos):
url_video = 'https://www.pexels.com/video/' + str(video.id) + '/download'
r = requests.get(url_video)
with open(f'test_{i}.mp4', 'wb') as outfile:
outfile.write(r.content)
This will download 100 videos, with each video being written to a separate file named test_0.mp4, test_1.mp4, ..., test_99.mp4.

Uploading local images to microsoft cognitive face

Error Screenshot
import sys
import os, time
import cognitive_face as CF
import global_variables as global_var
import urllib
import sqlite3
import requests
from requests.packages.urllib3.exceptions import InsecureRequestWarning
requests.packages.urllib3.disable_warnings(InsecureRequestWarning)
Key = global_var.key
CF.Key.set(Key)
BASE_URL = global_var.BASE_URL # Replace with your regional Base URL
CF.BaseUrl.set(BASE_URL)
def get_person_id():
person_id = ''
extractId = str(sys.argv[1])[-2:]
connect = sqlite3.connect("Face-DataBase")
c = connect.cursor()
cmd = "SELECT * FROM Students WHERE ID = " + extractId
c.execute(cmd)
row = c.fetchone()
person_id = row[3]
connect.close()
return person_id
if len(sys.argv) is not 1:
currentDir = os.path.dirname(os.path.abspath(__file__))
imageFolder = os.path.join(currentDir, "dataset/" + str(sys.argv[1]))
person_id = get_person_id()
for filename in os.listdir(imageFolder):
if filename.endswith(".jpg"):
print(filename)
imgurl = urllib.request.pathname2url(os.path.join(imageFolder, filename))
imgurl = imgurl[3:]
print("imageurl = {}".format(imgurl))
res = CF.face.detect(imgurl)
if len(res) != 1:
print("No face detected in image")
else:
res = CF.person.add_face(imgurl, global_var.personGroupId, person_id)
print(res)
time.sleep(6)
else:
print("supply attributes please from dataset folder")
I hope images should be converted to byte array but I don't know how to do it. Local images have to be uploaded into cognitive API. Tried many ways but cannot solve the error.
imgurl = urllib.request.pathname2url(os.path.join(imageFolder, filename))
Above line is where error exists

Welcome to Stack Overflow, #arun.
First of all, as per here, the API you're using is deprecated, and you should switch to this one instead.
Second, in this new API, there is a method called detect_with_stream (ref here), that will make a request to the Face Recognition endpoint, using the byte stream instead of an URL (it will use different request headers than the URL-based method). This method accepts a stream of bytes containing your image. I've worked with another cognitive services API that performs text recognition, and so I've faced this problem of sending an image URL or the image byte stream. You can generate a bytestream from the file as follows:
image_data = open(image_path, "rb").read()
The variable image_data can be passed to the method.
Edit: Instructions on how to use the new API with the image bytestream
First, install the following pip package:
pip install azure-cognitiveservices-vision-face
Then, you can try this approach.
import sys
import os, time
import global_variables as global_var
from azure.cognitiveservices.vision.face import FaceClient
from msrest.authentication import CognitiveServicesCredentials
import urllib
import sqlite3
import requests
from requests.packages.urllib3.exceptions import InsecureRequestWarning
requests.packages.urllib3.disable_warnings(InsecureRequestWarning)
KEY = global_var.key
ENDPOINT = global_var.endpoint
face_client = FaceClient(ENDPOINT,CognitiveServicesCredentials(KEY))
def get_person_id():
person_id = ''
extractId = str(sys.argv[1])[-2:]
connect = sqlite3.connect("Face-DataBase")
c = connect.cursor()
cmd = "SELECT * FROM Students WHERE ID = " + extractId
c.execute(cmd)
row = c.fetchone()
person_id = row[3]
connect.close()
return person_id
if len(sys.argv) is not 1:
currentDir = os.path.dirname(os.path.abspath(__file__))
imageFolder = os.path.join(currentDir, "dataset/" + str(sys.argv[1]))
person_id = get_person_id()
for filename in os.listdir(imageFolder):
if filename.endswith(".jpg"):
print(filename)
img_data = open(filename, "rb").read()
res = face_client.face.detect_with_stream(img_data)
if not res:
print('No face detected from image {}'.format(filename))
continue
res = face_client.person_group_person.add_face_from_stream(global_var.personGroupId, person_id, img_data)
print(res)
time.sleep(6)
else:
print("supply attributes please from dataset folder")
Edit 2: Notes on traversing all the files in a directory
Ok #arun, your current problem stems from the fact that you're using os.listdir which only lists the filenames, so you don't have their paths. The quickest solution would be to open each image inside the loop with:
img_data = open(os.path.join(imageFolder, filename), "rb").read()

Recording youtube live stream to file in python

I want to record youtube live stream and save it to file using python.
I tried with pytube library but it probably works for videos, not for live streams.
I want to record stream and save it to file with video format as avi or something like this.

Modification based on #wownis 's answer.
(I tried that answer, however, it doesn't work.)
# pip install urllib
# pip install m3u8
# pip install streamlink
import urllib
import m3u8
import streamlink
def get_stream(url):
"""
Get upload chunk url
"""
streams = streamlink.streams(url)
stream_url = streams["best"]
m3u8_obj = m3u8.load(stream_url.args['url'])
return m3u8_obj.segments[0]
def dl_stream(url, filename, chunks):
"""
Download each chunks
"""
pre_time_stamp = 0
for i in range(chunks+1):
stream_segment = get_stream(url)
cur_time_stamp = \
stream_segment.program_date_time.strftime("%Y%m%d-%H%M%S")
if pre_time_stamp == cur_time_stamp:
pass
else:
print(cur_time_stamp)
file = open(filename + '_' + str(cur_time_stamp) + '.ts', 'ab+')
with urllib.request.urlopen(stream_segment.uri) as response:
html = response.read()
file.write(html)
pre_time_stamp = cur_time_stamp
url = "https://www.youtube.com/watch?v=2U3JnFbD-es"
dl_stream(url, "live", 15)
Output like this:
./
live_20200713-103739.ts
live_20200713-103744.ts
...

I found a solution and i put my code in python:
import urllib
import m3u8
import streamlink
def record_stream(url,filename,iterations):
last_part = 0
for i in range(iterations+1):
streams = streamlink.streams(url)
stream_url = streams["best"]
print(stream_url.args['url'])
m3u8_obj = m3u8.load(stream_url.args['url'])
previous_part_time = last_part
last_part = m3u8_obj.segments[-1].program_date_time
if i >= 1:
for j in range(1, len(m3u8_obj.segments)):
if m3u8_obj.segments[-j].program_date_time == previous_part_time:
break
print(j)
file = open(filename + ".ts", "ab+")
for i in range(j-1,0,-1):
with urllib.request.urlopen(m3u8_obj.segments[-i].uri) as response:
html = response.read()
file.write(html)
url = "https://www.youtube.com/watch?v=BgKGctL0u1U"
record_stream(url,"file",10)
10 means 10 iterations if chunks have 2s it means that records 20s of stream

Pull random image from directory, encode to base64 and then print

I'm having difficult time trying to work these two together. It's frustrating me a little, so I hope to find ideas/solutions.
The complete works (what I'm planning on) should grab a random image from an online directory, encode it to base64 then print the base64. I've had total curl madness going all day and now I'm turning to python. Onwards!
These are kinda just notes at the minute but should explain the process.
import random, os
import base64
def search(): #get file
path = r"/Users/Impshum/Pictures" #should be able to http
random_filename = random.choice([
x for x in os.listdir(path)
if os.path.isfile(os.path.join(path, x))
])
print(random_filename) #not printing full location
def encode(): #encode to base64
image = open('heaven.jpg', 'rb')
image_read = image.read()
image_64_encode = base64.encodestring(image_read)
print image_64_encode
search() #notes
encode() #notes
Many thanks in advance.

You have most of the code you need
import random, os
import base64
def search(path): #get file
random_filename = random.choice([
x for x in os.listdir(path)
if os.path.isfile(os.path.join(path, x))
])
return os.path.join(path, random_filename)
def encode(path):
image = open(path, 'rb')
image_read = image.read()
image.close()
image_64_encode = base64.encodestring(image_read)
return image_64_encode
print(encode(search(r"/Users/Impshum/Pictures")))
There are things you can do to make this "nicer", but this should get you started.
For instance, you might want to use glob instead of os.listdir / os.path.join, etc. And using a context manager
import glob
import base64
import random
def search(path): #get file
random_filename = random.choice(glob.glob(path))
return random_filename
def encode(path):
with open(path, 'rb') as image:
image_read = image.read()
image_64_encode = base64.encodestring(image_read)
return image_64_encode
print(encode(search(r"/Users/Impshum/Pictures/*")))
Error handling is left as an exercise to the OP

Sending a PNG image from a SleekXMPP client to Pidgin

I have googled a lot but unfortunately could not come up with a working solution. I have a Python client using SleekXMPP modules and I would like to send a PNG file to a Pidgin client.
AFAIK, there is no Jingle extension implemented yet but a few methods using IBB, OOB and BOB did not work for me. I have tried XEPs 0047, 0231, 0095, and 0096 for negotiating.
I set my Pidgin to auto-accept files from a certain JID.
Is there any way doing that using SleekXMPP?
p.s. The book XMPP: The definitive guide did not give me any clue too :/
Thanks.
CODE
def upload_files_tgz(self, archivename, files, removearch=True):
# tar file name and mimetype (b for binary)
bfilename = "{0}.tar.gz".format(archivename)
bfilemime = "application/x-gzip";
# create a compressed tgz archive
tar = tarfile.open(bfilename, "w:gz" )
if files:
for f in files:
tar.add(f)
tar.close()
with open(bfilename, "rb") as compressed_file:
bin_output = compressed_file.read()
cid = self.xmpp_bot['xep_0231'].set_bob(bin_output, bfilemime)
msg = self.xmpp_bot.Message()
msg['to'] = self.get_xmpp_server_full_jid()
msg['bob']['cid'] = cid
msg['bob']['type'] = bfilemime
msg['bob']['data'] = bin_output
msg.send()
if removearch == True:
os.remove(bfilename)
def upload_image(self, filename, mimetype='image/png', message=None):
m = self.xmpp_bot.Message()
m['to'] = self.get_xmpp_server_full_jid()
m['type'] = 'chat'
with open(filename, 'rb') as img_file:
img = img_file.read()
if img:
cid = self.xmpp_bot['xep_0231'].set_bob(img, mimetype)
if message:
m['body'] = message
m['html']['body'] = '<img src="cid:%s" />' % cid
m.send()