I would like to get the Image size in python,as I do it with c++.
int w = src->width;
printf("%d", 'w');
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What's that `->` syntax for? Never seen it before... – Yatharth Agarwal Oct 23 '12 at 14:56
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For accessing an object's properties like in PHP? If so, why don't they just use the dot operator (`.`)? – Yatharth Agarwal Oct 23 '12 at 14:57
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@YatharthROCK In C++ `a->b` is the equivalent of `(*a).b` – Gabi Purcaru Oct 23 '12 at 14:58
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@GabiPurcaru OK, so what does the `*` operator do then? in Python it unpacked a tuple or list... – Yatharth Agarwal Oct 23 '12 at 14:59
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I don't know the technical name for it, but it turns a pointer (don't even ask what that is!) into the thing it points to – Gabi Purcaru Oct 23 '12 at 15:04
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2it's called dereferencing, and allows us to access to the value of the variable pointed by the * operator – Samuele Mattiuzzo Oct 23 '12 at 15:06
8 Answers
151
Using openCV and numpy it is as easy as this:
import cv2
img = cv2.imread('path/to/img',0)
height, width = img.shape[:2]
Arkistarvh Kltzuonstev
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cowhi
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For me the easiest way is to take all the values returned by image.shape:
height, width, channels = img.shape
if you don't want the number of channels (useful to determine if the image is bgr or grayscale) just drop the value:
height, width, _ = img.shape
Stein
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Use the function GetSize from the module cv with your image as parameter. It returns width, height as a tuple with 2 elements:
width, height = cv.GetSize(src)
halex
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13This function doesn't seem to exist in OpenCV 3. Maybe it was removed a while ago? – polm23 Mar 02 '18 at 11:15
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I use numpy.size() to do the same:
import numpy as np
import cv2
image = cv2.imread('image.jpg')
height = np.size(image, 0)
width = np.size(image, 1)
Rodrigo Lara
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from this tutorial: https://www.tutorialkart.com/opencv/python/opencv-python-get-image-size/
import cv2
# read image
img = cv2.imread('/home/ubuntu/Walnut.jpg', cv2.IMREAD_UNCHANGED)
# get dimensions of image
dimensions = img.shape
# height, width, number of channels in image
height = img.shape[0]
width = img.shape[1]
channels = img.shape[2]
from this other tutorial: https://www.pyimagesearch.com/2018/07/19/opencv-tutorial-a-guide-to-learn-opencv/
image = cv2.imread("jp.png")
(h, w, d) = image.shape
Please double check things before posting answers.
Tomer Shetah
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Javi
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Here is a method that returns the image dimensions:
from PIL import Image
import os
def get_image_dimensions(imagefile):
"""
Helper function that returns the image dimentions
:param: imagefile str (path to image)
:return dict (of the form: {width:<int>, height=<int>, size_bytes=<size_bytes>)
"""
# Inline import for PIL because it is not a common library
with Image.open(imagefile) as img:
# Calculate the width and hight of an image
width, height = img.size
# calculat ethe size in bytes
size_bytes = os.path.getsize(imagefile)
return dict(width=width, height=height, size_bytes=size_bytes)
Roei Bahumi
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I believe simply img.shape[-1::-1] would be nicer.
Tomer Shetah
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Harrison Wang
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You can use image.shape to get the dimensions of the image. It returns 3 values. The first value is height of an image, the second is width, and the last one is number of channels. You don't need the last value here so you can use below code to get height and width of image:
height, width = src.shape[:2]
print(width, height)
