Get Image Size Without Downloading It In Python

How can I get dimensions of image without actually downloading it? Is it even possible? I have a list of urls of images and I want to assign width and size to it. I know there is a

Solution 1:

I found the solution on this site to work well:

import urllib
import ImageFile

def getsizes(uri):
    # get file size *and* image size (None if not known)
    file = urllib.urlopen(uri)
    size = file.headers.get("content-length")
    if size: size = int(size)
    p = ImageFile.Parser()
    while1:
        data = file.read(1024)
        if not data:
            break
        p.feed(data)
        if p.image:
            return size, p.image.size
            break
    file.close()
    return size, None

print getsizes("http://www.pythonware.com/images/small-yoyo.gif")
# (10965, (179, 188))

Solution 2:

This is based on ed's answer mixed with other things I found on the web. I ran into the same issue as grotos with .read(24). Download getimageinfo.py from here and download ReSeekFile.py from here.

importurllib2imgdata= urllib2.urlopen(href)
image_type,width,height = getimageinfo.getImageInfo(imgdata)

Modify getimageinfo as such...

import ReseekFile

defgetImageInfo(datastream):
    datastream = ReseekFile.ReseekFile(datastream)
    data = str(datastream.read(30))

#Skipping to jpeg# handle JPEGselif (size >= 2) and data.startswith('\377\330'):
    content_type = 'image/jpeg'
    datastream.seek(0)
    datastream.read(2)
    b = datastream.read(1)
    try:
        while (b andord(b) != 0xDA):
            while (ord(b) != 0xFF): b = datastream.read(1)
            while (ord(b) == 0xFF): b = datastream.read(1)
            if (ord(b) >= 0xC0andord(b) <= 0xC3):
                datastream.read(3)
                h, w = struct.unpack(">HH", datastream.read(4))
                breakelse:
                datastream.read(int(struct.unpack(">H", datastream.read(2))[0])-2)
            b = datastream.read(1)
        width = int(w)
        height = int(h)
    except struct.error:
        passexcept ValueError:
        pass

Solution 3:

This is just a Python 3+ adaptation of an earlier answer here.

from urllib import request as ulreq
from PIL import ImageFile
 
def getsizes(uri):
    # get file size *and* image size (None if not known)
    file = ulreq.urlopen(uri)
    size = file.headers.get("content-length")
    if size: 
        size = int(size)
    p = ImageFile.Parser()
    while True:
        data = file.read(1024)
        if not data:
            break
        p.feed(data)
        if p.image:
            return size, p.image.size
            break
    file.close()
    return(size, None)

Solution 4:

If you're willing to download the first 24 bytes of each file, then this function (mentioned in johnteslade's answer to the question you mention) will work out the dimensions.

That's probably the least downloading necessary to do the job you want.

importurllib2start= urllib2.urlopen(image_url).read(24)

Edit (1):

In the case of jpeg files it seems to need more bytes. You could edit the function so that instead of reading a StringIO.StringIO(data) it instead reads the file handle from urlopen. Then it will read exactly as much of the image as it needs to find out the width and height.

Solution 5:

Since getimageinfo.py mentioned above doesn't work in Python3. Pillow is used instead of it.

Pillow can be found in pypi, or installed by using pip: pip install pillow.

from io import BytesIO
from PIL import Image
import requests
hrefs = ['https://farm4.staticflickr.com/3894/15008518202_b016d7d289_m.jpg','https://farm4.staticflickr.com/3920/15008465772_383e697089_m.jpg','https://farm4.staticflickr.com/3902/14985871946_86abb8c56f_m.jpg']
RANGE = 5000
for href in hrefs:
    req  = requests.get(href,headers={'User-Agent':'Mozilla5.0(Google spider)','Range':'bytes=0-{}'.format(RANGE)})
    im = Image.open(BytesIO(req.content))

    print(im.size)