具有多个元素的数组的真值是模棱两可的Python错误

Question

I need to detect if image is pizelized or not. So i use a python code, that was taken from other stackoverflow post:

import numpy as np
from PIL import Image, ImageChops

im = Image.open('img/low2.jpg')    
im2 = im.transform(im.size, Image.AFFINE, (1,0,1,0,1,1))
im3 = ImageChops.subtract(im, im2)
im3 = np.asarray(im3)
im3 = np.sum(im3,axis=0)[:-1]
mean = np.mean(im3)

peak_spacing = np.diff([i for i,v in enumerate(im3) if v > mean*2])

mean_spacing = np.mean(peak_spacing)
std_spacing = np.std(peak_spacing)

I'm getting this error:

File "pixelated.py", line 11, in peak_spacing = np.diff([i for i,v in enumerate(im3) if v > mean*2]) ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

How can i fix this? I'm newbie in Python, please give me any idea or help.

Answer 1

The problem is v > mean*2 which results in an array of boolean values.

The boolean value of such array is ambiguous for if . As the error text advises, you need to tell Python, whether all of values are expected to be True :

(v > mean * 2).all()

or if any of them is enough:

(v > mean * 2).any()

Answer 2

It looks like v is a numpy array. When you're comparing a numpy array with something, a new array of booleans is generated. This means that v > m*2 generates an array (for example [True, False, False, ... True] ). It is impossible to get single boolean value from such a list and use it in the if expression. So, try to use np.any(v > m*2) or np.all(v > m*2) depending on your code logic.

Answer 3

Also it seems that this code works well with grayscale images. But fails in the following way with RGB. So, try to convert the image to grayscale

im = im.convert("L")  

just after the image initialization