Rotate image by 90, 180 or 270 degrees

Question

I need to rotate an image by either 90, 180 or 270 degrees. In OpenCV4Android I can use:

Imgproc.getRotationMatrix2D(new Point(center, center), degrees, 1);         


        

Answer 1

Here's a function to rotate by any angle [-360 ... 360]

def rotate_image(image, angle):
    # Grab the dimensions of the image and then determine the center
    (h, w) = image.shape[:2]
    (cX, cY) = (w / 2, h / 2)

    # Grab the rotation matrix (applying the negative of the
    # angle to rotate clockwise), then grab the sine and cosine
    # (i.e., the rotation components of the matrix)
    M = cv2.getRotationMatrix2D((cX, cY), -angle, 1.0)
    cos = np.abs(M[0, 0])
    sin = np.abs(M[0, 1])

    # Compute the new bounding dimensions of the image
    nW = int((h * sin) + (w * cos))
    nH = int((h * cos) + (w * sin))

    # Adjust the rotation matrix to take into account translation
    M[0, 2] += (nW / 2) - cX
    M[1, 2] += (nH / 2) - cY

    # Perform the actual rotation and return the image
    return cv2.warpAffine(image, M, (nW, nH))

Usage

import cv2
import numpy as np

image = cv2.imread('1.png')
rotate = rotate_image(image, angle=90)

Answer 2

This will rotate an image any number of degrees, using the most efficient means for multiples of 90.

    void
    rotate_cw(const cv::Mat& image, cv::Mat& dest, int degrees)
    {
        switch (degrees % 360) {
            case 0:
                dest = image.clone();
                break;
            case 90:
                cv::flip(image.t(), dest, 1);
                break;
            case 180:
                cv::flip(image, dest, -1);
                break;
            case 270:
                cv::flip(image.t(), dest, 0);
                break;
            default:
                cv::Mat r = cv::getRotationMatrix2D({image.cols/2.0F, image.rows/2.0F}, degrees, 1.0);
                int len = std::max(image.cols, image.rows);
                cv::warpAffine(image, dest, r, cv::Size(len, len));
                break; //image size will change
        }
    }

But with opencv 3.0, this is done by just via the cv::rotate command:

cv::rotate(image, dest, e.g. cv::ROTATE_90_COUNTERCLOCKWISE);

Answer 3

I wrote this Python version using Numpy only, which are much faster than using cv2.transpose() and cv2.flip().

def rotate_image_90(im, angle):
    if angle % 90 == 0:
        angle = angle % 360
        if angle == 0:
            return im
        elif angle == 90:
            return im.transpose((1,0, 2))[:,::-1,:]
        elif angle == 180:
            return im[::-1,::-1,:]
        elif angle == 270:
            return im.transpose((1,0, 2))[::-1,:,:]

    else:
        raise Exception('Error')

Answer 4

You can rotate image using numpy rot90 function

like

def rotate_image(image,deg):
    if deg ==90:
        return np.rot90(image)
    if deg ==180:
        return np.rot90(image,2)
    if deg == 270:
        return np.rot90(image,-1) #Reverse 90 deg rotation

Hope this help ..

Answer 5

In python:

# import the necessary packages
import numpy as np
import cv2

# initialize the camera and grab a reference to the raw camera capture
vs = cv2.VideoCapture(0)
(ret, image_original) = vs.read()
image_rotated_90 = np.rot90(image_original)
image_rotated_180 = np.rot90(image_rotated_90)

# show the frame and press any key to quit the image frame
cv2.imshow("Frame", image_rotated_180)
cv2.waitKey(0)