Question

我正在尝试将图像旋转一定角度然后在窗口中显示它。我的想法是旋转，然后在一个新窗口中显示它，窗口的新宽度和高度根据旧的宽度和高度计算：

new_width = x * cos angle + y * sin angle
new_height = y * cos angle + x * sin angle

我期待结果如下所示：

enter image description here

但事实证明结果如下：

enter image description here

我的代码在这里：

#!/usr/bin/env python -tt
#coding:utf-8

import sys
import math
import cv2
import numpy as np 

def rotateImage(image, angel):#parameter angel in degrees

    if len(image.shape) > 2:#check colorspace
        shape = image.shape[:2]
    else:
        shape = image.shape
    image_center = tuple(np.array(shape)/2)#rotation center

    radians = math.radians(angel)

    x, y = im.shape
    print 'x =',x
    print 'y =',y
    new_x = math.ceil(math.cos(radians)*x + math.sin(radians)*y)
    new_y = math.ceil(math.sin(radians)*x + math.cos(radians)*y)
    new_x = int(new_x)
    new_y = int(new_y)
    rot_mat = cv2.getRotationMatrix2D(image_center,angel,1.0)
    print 'rot_mat =', rot_mat
    result = cv2.warpAffine(image, rot_mat, shape, flags=cv2.INTER_LINEAR)
    return result, new_x, new_y

def show_rotate(im, width, height):
#    width = width/2
#    height = height/2
#    win = cv2.cv.NamedWindow('ro_win',cv2.cv.CV_WINDOW_NORMAL)
#    cv2.cv.ResizeWindow('ro_win', width, height)
    win = cv2.namedWindow('ro_win')
    cv2.imshow('ro_win', im)
    if cv2.waitKey() == '\x1b':
        cv2.destroyWindow('ro_win')

if __name__ == '__main__':

    try:
        im = cv2.imread(sys.argv[1],0)
    except:
        print '\n', "Can't open image, OpenCV or file missing."
        sys.exit()

    rot, width, height = rotateImage(im, 30.0)
    print width, height
    show_rotate(rot, width, height)

我的代码中肯定会有一些愚蠢的错误导致这个问题，但我无法弄明白...... 而且我知道我的代码不够pythonic :( ..RESry for that ..

任何人都可以帮助我吗？

最佳，

bearzk

Answer 1

正如BloodyD的回答所说，cv2.warpAffine不会自动居中转换的图像。相反，它只是使用变换矩阵变换每个像素。（这可以将像素移动到笛卡尔空间中的任何位置，包括原始图像区域之外的任何位置。）然后，当您指定目标图像大小时，它会抓取该大小的区域，从（0,0）开始，即左上角原始框架。变换后的图像中不包含该区域的任何部分都将被截断。

这是旋转和缩放图像的Python代码，结果居中：

def rotateAndScale(img, scaleFactor = 0.5, degreesCCW = 30):
    (oldY,oldX) = img.shape #note: numpy uses (y,x) convention but most OpenCV functions use (x,y)
    M = cv2.getRotationMatrix2D(center=(oldX/2,oldY/2), angle=degreesCCW, scale=scaleFactor) #rotate about center of image.

    #choose a new image size.
    newX,newY = oldX*scaleFactor,oldY*scaleFactor
    #include this if you want to prevent corners being cut off
    r = np.deg2rad(degreesCCW)
    newX,newY = (abs(np.sin(r)*newY) + abs(np.cos(r)*newX),abs(np.sin(r)*newX) + abs(np.cos(r)*newY))

    #the warpAffine function call, below, basically works like this:
    # 1. apply the M transformation on each pixel of the original image
    # 2. save everything that falls within the upper-left "dsize" portion of the resulting image.

    #So I will find the translation that moves the result to the center of that region.
    (tx,ty) = ((newX-oldX)/2,(newY-oldY)/2)
    M[0,2] += tx #third column of matrix holds translation, which takes effect after rotation.
    M[1,2] += ty

    rotatedImg = cv2.warpAffine(img, M, dsize=(int(newX),int(newY)))
    return rotatedImg

Answer 2

当你得到这样的旋转矩阵时：

rot_mat = cv2.getRotationMatrix2D(image_center,angel,1.0)

您的“缩放”参数设置为1.0，因此如果您使用它将图像矩阵转换为相同大小的结果矩阵，则必须剪切它。

您可以改为获得这样的旋转矩阵：

rot_mat = cv2.getRotationMatrix2D(image_center,angel,0.5)

它会旋转和收缩，在边缘留下空间（你可以先将它放大，这样你最终仍会得到一个大图像）。

此外，看起来您对图像大小的numpy和OpenCV约定感到困惑。 OpenCV使用（x，y）表示图像大小和点坐标，而numpy使用（y，x）。这可能就是你从纵向到横向纵横比的原因。

我倾向于明确这样说：

imageHeight = image.shape[0]
imageWidth = image.shape[1]
pointcenter = (imageHeight/2, imageWidth/2)

等...

最终，这对我来说很好：

def rotateImage(image, angel):#parameter angel in degrees
    height = image.shape[0]
    width = image.shape[1]
    height_big = height * 2
    width_big = width * 2
    image_big = cv2.resize(image, (width_big, height_big))
    image_center = (width_big/2, height_big/2)#rotation center
    rot_mat = cv2.getRotationMatrix2D(image_center,angel, 0.5)
    result = cv2.warpAffine(image_big, rot_mat, (width_big, height_big), flags=cv2.INTER_LINEAR)
    return result

<强>更新

这是我执行的完整脚本。只是cv2.imshow（“winname”，image）和cv2.waitkey（）没有参数来保持它打开：

import cv2

def rotateImage(image, angel):#parameter angel in degrees
    height = image.shape[0]
    width = image.shape[1]
    height_big = height * 2
    width_big = width * 2
    image_big = cv2.resize(image, (width_big, height_big))
    image_center = (width_big/2, height_big/2)#rotation center
    rot_mat = cv2.getRotationMatrix2D(image_center,angel, 0.5)
    result = cv2.warpAffine(image_big, rot_mat, (width_big, height_big), flags=cv2.INTER_LINEAR)
    return result

imageOriginal = cv2.imread("/Path/To/Image.jpg")
# this was an iPhone image that I wanted to resize to something manageable to view
# so I knew beforehand that this is an appropriate size
imageOriginal = cv2.resize(imageOriginal, (600,800))
imageRotated= rotateImage(imageOriginal, 45)

cv2.imshow("Rotated", imageRotated)
cv2.waitKey()

那里真的不是很多......如果你正在使用的是一个真正的模块，你肯定是正确使用if __name__ == '__main__':。

Answer 3

嗯，这个问题似乎不是最新的，但我遇到了同样的问题并花了一段时间来解决它，而没有上下缩放原始图像。我将发布我的解决方案（不幸的是C ++代码，但如果需要它可以很容易地移植到python）：

#include <math.h>
#define PI 3.14159265
#define SIN(angle) sin(angle * PI / 180)
#define COS(angle) cos(angle * PI / 180)

void rotate(const Mat src, Mat &dest, double angle, int borderMode, const Scalar &borderValue){

    int w = src.size().width, h = src.size().height;

    // resize the destination image
    Size2d new_size = Size2d(abs(w * COS((int)angle % 180)) + abs(h * SIN((int)angle % 180)), abs(w * SIN((int)angle % 180)) + abs(h * COS((int)angle % 180)));
    dest = Mat(new_size, src.type());

    // this is our rotation point
    Size2d old_size = src.size();
    Point2d rot_point = Point2d(old_size.width / 2.0, old_size.height / 2.0);

    // and this is the rotation matrix
    // same as in the opencv docs, but in 3x3 form
    double a = COS(angle), b = SIN(angle);
    Mat rot_mat   = (Mat_<double>(3,3) << a, b, (1 - a) * rot_point.x - b * rot_point.y, -1 * b, a, b * rot_point.x + (1 - a) * rot_point.y, 0, 0, 1);

    // next the translation matrix
    double offsetx = (new_size.width - old_size.width) / 2,
           offsety = (new_size.height - old_size.height) / 2;
    Mat trans_mat = (Mat_<double>(3,3) << 1, 0, offsetx , 0, 1, offsety, 0, 0, 1);

    // multiply them: we rotate first, then translate, so the order is important!
    // inverse order, so that the transformations done right 
    Mat affine_mat = Mat(trans_mat * rot_mat).rowRange(0, 2);

    // now just apply the affine transformation matrix
    warpAffine(src, dest, affine_mat, new_size, INTER_LINEAR, borderMode, borderValue);
}

一般的解决方案是将旋转并将旋转的图片转换为正确的位置。因此，我们创建了两个转换矩阵（第一个用于旋转，第二个用于转换）并将它们乘以最终的仿射变换。由于opencv的getRotationMatrix2D返回的矩阵只有2x3，我必须手动创建3x3格式的矩阵，所以它们可以乘以。然后只需前两行并应用仿射变换。

编辑：我创建了一个Gist，因为我在不同的项目中经常需要这个功能。还有一个Python版本：https://gist.github.com/BloodyD/97917b79beb332a65758

旋转窗口后的Python 2.7.3 + OpenCV 2.4不适合Image

3 个答案: