Parse folder with images into (m, n) shaped numpy array

Question

I am trying to parse a folder with images into a numpy array. I want to obtain an array which looks like this:

import numpy as np
from sklearn.datasets import fetch_mldata

#load 70,000 MNIST images (28 X 28 pixels)
mnist = fetch_mldata("MNIST original")
X = mnist.data

print X.shape

Desired output:

(70000L, 784L)

This is what I tried:

from PIL import Image                                                            
import numpy                                                                     
import matplotlib.pyplot as plt                                                  
import glob

#I have two colour images, each 64 X 64 pixels, in the folder
imageFolderPath = 'C:\\Users\\apples\\Desktop\\t-sne\\pics'
imagePath = glob.glob(imageFolderPath+'/*.JPEG') 

im_array = numpy.array( [numpy.array(Image.open(imagePath[i])) for i in range(len(imagePath))] )

print im_array.shape

But it produces the following output:

(2L, 64L, 64L, 3L)

How can I obtain an array with the following dimensions:

(m, n)

user2379410 · Accepted Answer · 2016-04-27 14:33:22Z

2

PIL loads color images in RGB format (maybe not always), hence the last dimension is size 3 (one for each color channel). So you probably want to convert the image to a different pixel format. Also you need to flatten the image arrays to get to the layout that you want. You could do something like:

def img2array(path):
    img = Image.open(path).convert('F')
    return numpy.array(img).ravel()

im_array = numpy.array([img2array(path) for path in imagePath])

answered Apr 27, 2016 at 14:33

user2379410

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2 Comments

apples-oranges Over a year ago

Thank you, this worked like a charm! Can I ask in what order does this process the images in the folder. I have to make another array with labels corresponding to each image e.g. (0,1...n), therefore I need to make sure the labels match the correct image.

user2379410 Over a year ago

@apples-oranges - glob.glob uses os.listdir which gives paths in arbitrary order. Of course you can wrap it in a call to sorted to fix the order. Or you could parse the filenames.

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