Make image Instagram compatible using Python

Image processing is quite easy and fast in Python with libraries available. I recently was trying to make images Instagram-friendly using Python script.

What do I mean Instagram Friendly Images:

Basically, Instagram supports all kinds of images ( https://help.instagram.com/1631821640426723) but a square-shaped photo fits the best.

So, We have to resize the photo to a 1080 x 1080 pixels size.

There are multiple use cases are possible like

1. The original photo is already square but smaller in size.
   

2. The original photo is a rectangle and smaller in size
   

1. The original photo is already square but larger in size.
   

2. The original photo is a rectangle and larger in size.

So the idea here is to use the maximum space from the original image only but if there is an area remaining fill it with a color.

We will use the Pillow (Python’s image manipulation library) to process the image. If not install use below command

pip install Pillow
pip install Pillow

pip install Pillow

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from PIL import Image
import requests
def make_square(im, min_size=1080, fill_color=(0, 0, 0, 0)):
    x, y = im.size
    print("Original size "+str(x)+" X "+str(y))
    size = max(min_size, x, y)
    new_im = Image.new('RGB', (size, size), fill_color)    
    if(x > y):
        mod = size/x
        x = int(x * mod)
        y = int(y * mod)
    elif(x < y):
        mod = size/y
        x = int(x * mod)
        y = int(y * mod)
    new_im.paste(im.resize((x,y)), (int((size - x) / 2), int((size - y) / 2)))
    x, y = new_im.size
    print("New size "+str(x)+" X "+str(y))
    return new_im
def processImage(url):
    img = Image.open(requests.get(url, stream=True).raw)
    new_img = make_square(img)
    new_img.show()
processImage("https://images.pexels.com/photos/771742/pexels-photo-771742.jpeg?auto=compress&cs=tinysrgb&w=1260&h=750&dpr=1")
from PIL import Image
import requests


def make_square(im, min_size=1080, fill_color=(0, 0, 0, 0)):
    x, y = im.size
    print("Original size "+str(x)+" X "+str(y))
    size = max(min_size, x, y)
    new_im = Image.new('RGB', (size, size), fill_color)    
    if(x > y):
        mod = size/x
        x = int(x * mod)
        y = int(y * mod)
    elif(x < y):
        mod = size/y
        x = int(x * mod)
        y = int(y * mod)

    new_im.paste(im.resize((x,y)), (int((size - x) / 2), int((size - y) / 2)))
    x, y = new_im.size
    print("New size "+str(x)+" X "+str(y))
    return new_im



def processImage(url):
    img = Image.open(requests.get(url, stream=True).raw)
    new_img = make_square(img)
    new_img.show()


processImage("https://images.pexels.com/photos/771742/pexels-photo-771742.jpeg?auto=compress&cs=tinysrgb&w=1260&h=750&dpr=1")
from PIL import Image
import requests


def make_square(im, min_size=1080, fill_color=(0, 0, 0, 0)):
    x, y = im.size
    print("Original size "+str(x)+" X "+str(y))
    size = max(min_size, x, y)
    new_im = Image.new('RGB', (size, size), fill_color)    
    if(x > y):
        mod = size/x
        x = int(x * mod)
        y = int(y * mod)
    elif(x < y):
        mod = size/y
        x = int(x * mod)
        y = int(y * mod)

    new_im.paste(im.resize((x,y)), (int((size - x) / 2), int((size - y) / 2)))
    x, y = new_im.size
    print("New size "+str(x)+" X "+str(y))
    return new_im



def processImage(url):
    img = Image.open(requests.get(url, stream=True).raw)
    new_img = make_square(img)
    new_img.show()


processImage("https://images.pexels.com/photos/771742/pexels-photo-771742.jpeg?auto=compress&cs=tinysrgb&w=1260&h=750&dpr=1")

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Original Image:

Resized Image:

It will return the squared image.

Happy Coding <3

原文链接：Make image Instagram compatible using Python