I have seen an image of a girl which is made up of multiple images of her.So I want to achieve the same thing using a python script.(I am completely new to image processing)
I am using pil library for this script.
import sys,os
from PIL import Image
img = Image.open("DSC_0149.jpg")
pixels = img.load()
for i in range(img.size[0]):
for j in range(img.size[1]):
pixels[i,j] = (i, j, 100) # I will change this to some pic image.
img.show()
I am trying first just to change the colour of pixel retaining the pic,But this code dint work.
Can anyone guide me how to achieve it.
Edit : I want to fill the picture with multiple pictures and yet RETAIN the original picture.
Something like this : http://www.photoshopessentials.com/photo-effects/photo-fill/ but in a much better way.
So first you need to edit each pixel with this to change the color:
If it is rgb:
img.putpixel((10,15),(r,g,b))
or
faster: pixels[1, 1] = (r, g, b)
otherwise:
Is it possible to change the color of one individual pixel in Python?
After knowing how to edit each pixel you have to create a small copy of your image with a resize like this:
Copy Image:
// Not tested : Make sure it's rgb
img = Image.new( 'RGB', (img.size[0],(img.size[1]), "black") # create a new black image
pixels = img.load() # create the pixel map
for i in range(img.size[0]): # for every pixel:
for j in range(img.size[1]):
pixels[i,j] = other_image_pixel[i,j] # set the colour accordingly
https://opensource.com/life/15/2/resize-images-python
Apply a Color filter to each small image to match the area color you will replace with this image.
The best way to understand the whole process is to take time to read this code in the same language, it's around 200 lines:
https://github.com/codebox/mosaic
Hope it solve your problems
Related
I want to change the pixel value of a grayscale image using OpenCV.
Assume that I have a grayscale image and I want to convert all its pixel to 0 value one at a time. So that the resultant image is completely black. I tried this but there is no change in the image:
image = cv2.imread('test_image.png',0)
for i in range(image.shape[0]):
for j in range(image.shape[1]):
image[i, j] = 0
Result:
display the updated image
In most cases, you want to avoid using double for loops to modify pixel values since it is very slow. A better approach is to use Numpy for pixel modification since OpenCV uses Numpy arrays to display images. To achieve your desired result, you can use np.zeros to create a completely black image with the same shape as the original image.
import cv2
import numpy as np
image = cv2.imread("test_image.png", 0)
black = np.zeros(image.shape, np.uint8)
cv2.imshow('image', image)
cv2.imshow('black', black)
cv2.waitKey(0)
For example with a test image. Original (left), result (right)
I would suggest you to always try manipulating the copy of an image so that the image doesn't get affected in the wrong way. Coming to your question, you can do the following:
import cv2
image = cv2.imread('test_image.png',0)
#Creating a copy of the image to confirm right operation is performed on the image.
image_copy = image.copy()
image_copy[:,:] = [0] #Setting all values to 0.
I want to analyse a specific part of an image, as an example I'd like to focus on the bottom right 200x200 section and count all the black pixels, so far I have:
im1 = Image.open(path)
rgb_im1 = im1.convert('RGB')
for pixel in rgb_im1.getdata():
Whilst you could do this with cropping and a pair of for loops, that is really slow and not ideal.
I would suggest you use Numpy as it is very commonly available, very powerful and very fast.
Here's a 400x300 black rectangle with a 1-pixel red border:
#!/usr/bin/env python3
import numpy as np
from PIL import Image
# Open the image and make into Numpy array
im = Image.open('image.png')
ni = np.array(im)
# Declare an ROI - Region of Interest as the bottom-right 200x200 pixels
# This is called "Numpy slicing" and is near-instantaneous https://www.tutorialspoint.com/numpy/numpy_indexing_and_slicing.htm
ROI = ni[-200:,-200:]
# Calculate total area of ROI and subtract non-zero pixels to get number of zero pixels
# Numpy.count_nonzero() is highly optimised and extremely fast
black = 200*200 - np.count_nonzero(ROI)
print(f'Black pixel total: {black}')
Sample Output
Black pixel total: 39601
Yes, you can make it shorter, for example:
h, w = 200,200
im = np.array(Image.open('image.png'))
black = h*w - np.count_nonzero(ni[-h:,-w:])
If you want to debug it, you can take the ROI and make it into a PIL Image which you can then display. So just use this line anywhere after you make the ROI:
# Display image to check
Image.fromarray(ROI).show()
You can try cropping the Image to the specific part that you want:-
img = Image.open(r"Image_location")
x,y = img.size
img = img.crop((x-200, y-200, x, y))
The above code takes an input image, and crops it to its bottom right 200x200 pixels. (make sure the image dimensions are more then 200x200, otherwise an error will occur)
Original Image:-
Image after Cropping:-
You can then use this cropped image, to count the number of black pixels, where it depends on your use case what you consider as a BLACK pixel (a discrete value like (0, 0, 0) or a range/threshold (0-15, 0-15, 0-15)).
P.S.:- The final Image will always have a dimension of 200x200 pixels.
from PIL import Image
img = Image.open("ImageName.jpg")
crop_area = (a,b,c,d)
cropped_img = img.crop(crop_area)
I have a problem Please help me if I have an image multi 16 in 1 image I wanna find black contour in an image just one show YES if find black contour What should I do? in Image Processing use Python!
Python Image Library can give RGB data of an image. You can get RGB data simply by using
import Image
pic = Image.open('/path/to/file')
rgbdata = pic.load()
width, height = pic.size
def identify_black():
for i in range(width):
for j in range(height):
if rgbdata[i,j] == (0,0,0):
#print rgbdata[i,j]
print "Yes"
return True
break
identify_black()
You can view the data purely in terms of RGB values of ijth pixel in rgbdata[i,j]. width and height help you define the blocks.
RGB value for black is (0,0,0). So if in your image list, a block of pixels give you (0,0,0) you can use that i, j th number to find in which block you have black! In fact, you identify any color!
Hope this helps.
A very simple solution is thresholding, just set any non black pixel(0,0,0) to (255,255,255) and you will have all the contours in an image.
Is there any way to make an image half transparent?
the pseudo code is something like this:
from PIL import Image
image = Image.open('image.png')
image = alpha(image, 0.5)
I googled it for a couple of hours but I can't find anything useful.
I realize this question is really old, but with the current version of Pillow (v4.2.1), there is a function called putalpha. It seems to work fine for me. I don't know if will work for every situation where you need to change the alpha, but it does work. It sets the alpha value for every pixel in the image. It seems, though that you can use a mask: http://www.leancrew.com/all-this/2013/11/transparency-with-pil/.
Use putalpha like this:
from PIL import Image
img = Image.open(image)
img.putalpha(127) # Half alpha; alpha argument must be an int
img.save(dest)
Could you do something like this?
from PIL import Image
image = Image.open('image.png') #open image
image = image.convert("RGBA") #convert to RGBA
rgb = image.getpixel(x,y) #Get the rgba value at coordinates x,y
rgb[3] = int(rgb[3] / 2) or you could do rgb[3] = 50 maybe? #set alpha to half somehow
image.putpixel((x,y), rgb) #put back the modified reba values at same pixel coordinates
Definitely not the most efficient way of doing things but it might work. I wrote the code in browser so it might not be error free but hopefully it can give you an idea.
EDIT: Just noticed how old this question was. Leaving answer anyways for future help. :)
I put together Pecan's answer and cr333's question from this question:
Using PIL to make all white pixels transparent?
... and came up with this:
from PIL import Image
opacity_level = 170 # Opaque is 255, input between 0-255
img = Image.open('img1.png')
img = img.convert("RGBA")
datas = img.getdata()
newData = []
for item in datas:
newData.append((0, 0, 0, opacity_level))
else:
newData.append(item)
img.putdata(newData)
img.save("img2.png", "PNG")
In my case, I have text with black background and wanted only the background semi-transparent, in which case:
from PIL import Image
opacity_level = 170 # Opaque is 255, input between 0-255
img = Image.open('img1.png')
img = img.convert("RGBA")
datas = img.getdata()
newData = []
for item in datas:
if item[0] == 0 and item[1] == 0 and item[2] == 0:
newData.append((0, 0, 0, opacity_level))
else:
newData.append(item)
img.putdata(newData)
img.save("img2.png", "PNG")
I had an issue, where black boxes were appearing around my image when applying putalpha().
This workaround (applying alpha in a copied layer) solved it for me.
from PIL import Image
with Image.open("file.png") as im:
im2 = im.copy()
im2.putalpha(180)
im.paste(im2, im)
im.save("file2.png")
Explanation:
Like I said, putalpha modifies all pixels by setting their alpha value, so fully transparent pixels become only partially transparent. The code I posted above first sets (putalpha) all pixels to semi-transparent in a copy, then copies (paste) all pixels to the original image using the original alpha values as a mask. This means that fully transparent pixels in the original image are skipped during the paste.
Credit: https://github.com/nulano # https://github.com/python-pillow/Pillow/issues/4687#issuecomment-643567573
I just did this by myself...even though my code maybe a little bit weird...But it works fine. So I share it here. Hopes it could help anybody. =)
The idea: To transparent a pic means lower alpha which is the 4th element in the tuple.
my frame code:
from PIL import Image
img=open(image)
img=img.convert('RGBA') #you can make sure your pic is in the right mode by check img.mode
data=img.getdata() #you'll get a list of tuples
newData=[]
for a in data:
a=a[:3] #you'll get your tuple shorten to RGB
a=a+(100,) #change the 100 to any transparency number you like between (0,255)
newData.append(a)
img.putdata(newData) #you'll get your new img ready
img.save(filename.filetype)
I didn't find the right command to fulfil this job automatically, so I write this by myself. Hopes it'll help again. XD
This method helps to reduce opacity of logo with transparency before pasting it over image
# pip install Pillow
# PIL.__version__ is 9.3.0
from PIL import Image, ImageEnhance
im = Image.open('logo.png').convert('RGBA')
alpha = im.split()[3]
alpha = ImageEnhance.Brightness(alpha).enhance(.5)
im.putalpha(alpha)
I am trying to find the exact pixel height and length of a piece of red text on an blank black image. To do this I am taking the pixel values of the red colours and copying and pasting it into a x,y matrix. I then create a new image and paste the results. However, I cannot get the size of the output image to be EXACTLY the height and length of the text that's extracted.
img = Image.open("word.png")
img = img.convert("P")
his = img.histogram()
values = {}
for i in range(256):
values[i] = his[i]
for j,k in sorted(values.items(), key=itemgetter(1), reverse=True)[:10]:
print j,k
img2 = Image.new("P",img.size,255) #THIS SIZE MODE IS WHAT I CAN'T FIX
img = img.convert("P")
temp = {}
for x in range(img.size[1]):
for y in range(img.size[0]):
pix = img.getpixel((y,x))
temp[pix] = pix
if pix == 15 or pix == 11 or pix == 12 or pix == 14:
img2.putpixel((y,x),0)
img2.save("output.png")
In the bit that I have commented:
If I put (0,0) as a blank canvas, it says "image index out of range".
If I put nothing it doesn't work (it needs a size there).
img.size just uses the size from the first img.
If anyone has a better method of doing this whole thing feel free to tell me. All I am trying to do is get the pixel height/length of a word in a specific font+size so I can store it in a database.
Thanks in advance if anyone can help (its indented on this i don't know why it isn't in the code)
The problem you are trying to solve sounds like you want to size your image to a certain size but don't know what that size is in advance.
PIL does not support dynamically sized images, so you would need to create the larger image as you did and then crop it using the img.crop method.