How to get an array from RGB values of a bitmap image?

python, python-imaging-library

Solution

You mentioned `numpy`. If you want a numpy array of the image, don't iterate through it, just do `data = np.array(im)`.

E.g.

from PIL import Image
import numpy as np
im = Image.open("/Users/Hugo/green_leaves.jpg")
p = np.array(im)

Building up a numpy array by repeatedly appending to it is very inefficient. Numpy arrays aren't like python lists (python lists serve that purpose very well!!). They're fixed-size, homogenous, memory-efficient arrays.

If you did want to build up a numpy array through appending, use a list (which can be efficiently appended to) and then convert that list to a numpy array.

However, in this case, PIL images support being converted to numpy arrays directly.

On one more note, the example I gave above isn't 100% equivalent to your code. `p` will be a height by width by numbands (3 or 4) array, instead of a numpixels by numbands array as it was in your original example.

If you want to reshape the array into numpixels by numbands, just do:

p = p.reshape(-1, p.shape[2])

(Or equivalently, `p.shape = -1, p.shape[2]`)

This will reshape the array into `width*height` by numbands (either 3 or 4, depending on whether or not there's an alpha channel) array. In other words a sequence of the red,green,blue,alpha pixel values in the image. The `-1` is a placeholder that tells numpy to calculate the appropriate shape for the first axes based on the other sizes that are specified.

Problem

I am running this code ``` from PIL import Image import numpy as np im = Image.open("/Users/Hugo/green_leaves.jpg") im.load() height, widht = im.size p = np.array([0,0,0]) for row in range(height): for col in range(widht): a = im.getpixel((row,col)) p = np.append(a.asarray()) ``` But I am getting the following error ``` Traceback (most recent call last): File "/Users/hugo/PycharmProjects/Meteo API/image.py", line 17, in <module> p = np.append(a.asarray()) AttributeError: 'tuple' object has no attribute 'asarray' ``` Could you help me?

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