74

私のアプリケーション: 画像を回転させようとしています (OpenCV と Python を使用)

画像の回転

現時点では、入力画像を回転させて黒い境界線でパディングし、A を取得する以下のコードを開発しました。私が欲しいのは、B - 回転した画像内で可能な最大のエリア ト​​リミング ウィンドウです。これを、軸に沿った境界付きボックスと呼びます。

これは本質的にRotate and crop と同じですが、その質問に対する答えが得られません。さらに、その答えは明らかに正方形の画像に対してのみ有効です。私の画像は長方形です。

A に与えるコード:

import cv2
import numpy as np


def getTranslationMatrix2d(dx, dy):
    """
    Returns a numpy affine transformation matrix for a 2D translation of
    (dx, dy)
    """
    return np.matrix([[1, 0, dx], [0, 1, dy], [0, 0, 1]])


def rotateImage(image, angle):
    """
    Rotates the given image about it's centre
    """

    image_size = (image.shape[1], image.shape[0])
    image_center = tuple(np.array(image_size) / 2)

    rot_mat = np.vstack([cv2.getRotationMatrix2D(image_center, angle, 1.0), [0, 0, 1]])
    trans_mat = np.identity(3)

    w2 = image_size[0] * 0.5
    h2 = image_size[1] * 0.5

    rot_mat_notranslate = np.matrix(rot_mat[0:2, 0:2])

    tl = (np.array([-w2, h2]) * rot_mat_notranslate).A[0]
    tr = (np.array([w2, h2]) * rot_mat_notranslate).A[0]
    bl = (np.array([-w2, -h2]) * rot_mat_notranslate).A[0]
    br = (np.array([w2, -h2]) * rot_mat_notranslate).A[0]

    x_coords = [pt[0] for pt in [tl, tr, bl, br]]
    x_pos = [x for x in x_coords if x > 0]
    x_neg = [x for x in x_coords if x < 0]

    y_coords = [pt[1] for pt in [tl, tr, bl, br]]
    y_pos = [y for y in y_coords if y > 0]
    y_neg = [y for y in y_coords if y < 0]

    right_bound = max(x_pos)
    left_bound = min(x_neg)
    top_bound = max(y_pos)
    bot_bound = min(y_neg)

    new_w = int(abs(right_bound - left_bound))
    new_h = int(abs(top_bound - bot_bound))
    new_image_size = (new_w, new_h)

    new_midx = new_w * 0.5
    new_midy = new_h * 0.5

    dx = int(new_midx - w2)
    dy = int(new_midy - h2)

    trans_mat = getTranslationMatrix2d(dx, dy)
    affine_mat = (np.matrix(trans_mat) * np.matrix(rot_mat))[0:2, :]
    result = cv2.warpAffine(image, affine_mat, new_image_size, flags=cv2.INTER_LINEAR)

    return result
4

12 に答える 12

108

このソリューション/実装の背後にある数学は、類似の質問のこのソリューションと同等ですが、式は単純化されており、特異点が回避されています。これは、他のソリューションと同じインターフェイスを持つ python コードですがlargest_rotated_rect、ほとんどすべての場合でより大きな領域を提供します (常に実証済みの最適です)。

def rotatedRectWithMaxArea(w, h, angle):
  """
  Given a rectangle of size wxh that has been rotated by 'angle' (in
  radians), computes the width and height of the largest possible
  axis-aligned rectangle (maximal area) within the rotated rectangle.
  """
  if w <= 0 or h <= 0:
    return 0,0

  width_is_longer = w >= h
  side_long, side_short = (w,h) if width_is_longer else (h,w)

  # since the solutions for angle, -angle and 180-angle are all the same,
  # if suffices to look at the first quadrant and the absolute values of sin,cos:
  sin_a, cos_a = abs(math.sin(angle)), abs(math.cos(angle))
  if side_short <= 2.*sin_a*cos_a*side_long or abs(sin_a-cos_a) < 1e-10:
    # half constrained case: two crop corners touch the longer side,
    #   the other two corners are on the mid-line parallel to the longer line
    x = 0.5*side_short
    wr,hr = (x/sin_a,x/cos_a) if width_is_longer else (x/cos_a,x/sin_a)
  else:
    # fully constrained case: crop touches all 4 sides
    cos_2a = cos_a*cos_a - sin_a*sin_a
    wr,hr = (w*cos_a - h*sin_a)/cos_2a, (h*cos_a - w*sin_a)/cos_2a

  return wr,hr

関数と他のソリューションの比較を次に示します。

>>> wl,hl = largest_rotated_rect(1500,500,math.radians(20))
>>> print (wl,hl),', area=',wl*hl
(828.2888697391496, 230.61639227890998) , area= 191016.990904
>>> wm,hm = rotatedRectWithMaxArea(1500,500,math.radians(20))
>>> print (wm,hm),', area=',wm*hm
(730.9511000407718, 266.044443118978) , area= 194465.478358

angle回転した画像のバウンディング ボックスの角度[0,pi/2[(幅w、高さh) の寸法は次のとおりです。

  • w_bb = w*cos_a + h*sin_a
  • 身長h_bb = w*sin_a + h*cos_a

がトリミングされた画像の計算された最適な幅と高さである場合w_rh_r境界ボックスからのインセットは次のようになります。

  • 水平方向:(w_bb-w_r)/2
  • 垂直方向:(h_bb-h_r)/2

証拠:

最大面積を持つ 2 つの平行線の間の軸に沿った長方形を探すことは、たとえばx次の図のように、1 つのパラメーターを使用した最適化問題です。 アニメートされたパラメータ

2 本の平行s線の間の距離を示します (回転した長方形の短辺になります)。次に、求められている長方形の辺a、は、 、b、それぞれに対して一定の比率を持ちます。つまり、x = a sin α および (sx) = b cos α です。xs-x

ここに画像の説明を入力

したがって、面積a*bを最大化することは を最大化することを意味しx*(s-x)ます。直角三角形の「高さの定理」により、x*(s-x) = p*q = h*h. したがって、最大面積は で達成されますx = s-x = s/2。つまり、平行線の間の 2 つの角 E、G は中線上にあります。

ここに画像の説明を入力

このソリューションは、この最大の長方形が回転した長方形に収まる場合にのみ有効です。したがって、対角線は、回転した長方形EGの反対側よりも長くなってはなりません。l以来

EG = AF + DH = s/2*(cot α + tan α) = s/(2 sin α cos α) = s/sin 2*α

s ≤ l sin 2 α という条件があります。ここで、s と l は、回転した長方形の短辺と長辺です。

s > l sin 2 α の場合、パラメータxは (s/2 よりも) 小さくする必要があり、必要な四角形のすべての角は、回転した四角形のそれぞれの側にあります。これにより、次の式が得られます。

x*cot α + (sx)*tan α = l

x = sin α*(l cos α - s sin α)/cos 2*α を与えます。a = x/sin α と b = (sx)/cos α から、上記の式が得られます。

于 2013-05-27T18:40:05.273 に答える
32

したがって、主張されている多くの解決策を調査した後、最終的に機能する方法を見つけました。回転した四角形で最大の四角形を計算する に関するAndriMagnus Hoffによる回答。

以下の Python コードには、目的のメソッドlargest_rotated_rectと短いデモが含まれています。

import math
import cv2
import numpy as np


def rotate_image(image, angle):
    """
    Rotates an OpenCV 2 / NumPy image about it's centre by the given angle
    (in degrees). The returned image will be large enough to hold the entire
    new image, with a black background
    """

    # Get the image size
    # No that's not an error - NumPy stores image matricies backwards
    image_size = (image.shape[1], image.shape[0])
    image_center = tuple(np.array(image_size) / 2)

    # Convert the OpenCV 3x2 rotation matrix to 3x3
    rot_mat = np.vstack(
        [cv2.getRotationMatrix2D(image_center, angle, 1.0), [0, 0, 1]]
    )

    rot_mat_notranslate = np.matrix(rot_mat[0:2, 0:2])

    # Shorthand for below calcs
    image_w2 = image_size[0] * 0.5
    image_h2 = image_size[1] * 0.5

    # Obtain the rotated coordinates of the image corners
    rotated_coords = [
        (np.array([-image_w2,  image_h2]) * rot_mat_notranslate).A[0],
        (np.array([ image_w2,  image_h2]) * rot_mat_notranslate).A[0],
        (np.array([-image_w2, -image_h2]) * rot_mat_notranslate).A[0],
        (np.array([ image_w2, -image_h2]) * rot_mat_notranslate).A[0]
    ]

    # Find the size of the new image
    x_coords = [pt[0] for pt in rotated_coords]
    x_pos = [x for x in x_coords if x > 0]
    x_neg = [x for x in x_coords if x < 0]

    y_coords = [pt[1] for pt in rotated_coords]
    y_pos = [y for y in y_coords if y > 0]
    y_neg = [y for y in y_coords if y < 0]

    right_bound = max(x_pos)
    left_bound = min(x_neg)
    top_bound = max(y_pos)
    bot_bound = min(y_neg)

    new_w = int(abs(right_bound - left_bound))
    new_h = int(abs(top_bound - bot_bound))

    # We require a translation matrix to keep the image centred
    trans_mat = np.matrix([
        [1, 0, int(new_w * 0.5 - image_w2)],
        [0, 1, int(new_h * 0.5 - image_h2)],
        [0, 0, 1]
    ])

    # Compute the tranform for the combined rotation and translation
    affine_mat = (np.matrix(trans_mat) * np.matrix(rot_mat))[0:2, :]

    # Apply the transform
    result = cv2.warpAffine(
        image,
        affine_mat,
        (new_w, new_h),
        flags=cv2.INTER_LINEAR
    )

    return result


def largest_rotated_rect(w, h, angle):
    """
    Given a rectangle of size wxh that has been rotated by 'angle' (in
    radians), computes the width and height of the largest possible
    axis-aligned rectangle within the rotated rectangle.

    Original JS code by 'Andri' and Magnus Hoff from Stack Overflow

    Converted to Python by Aaron Snoswell
    """

    quadrant = int(math.floor(angle / (math.pi / 2))) & 3
    sign_alpha = angle if ((quadrant & 1) == 0) else math.pi - angle
    alpha = (sign_alpha % math.pi + math.pi) % math.pi

    bb_w = w * math.cos(alpha) + h * math.sin(alpha)
    bb_h = w * math.sin(alpha) + h * math.cos(alpha)

    gamma = math.atan2(bb_w, bb_w) if (w < h) else math.atan2(bb_w, bb_w)

    delta = math.pi - alpha - gamma

    length = h if (w < h) else w

    d = length * math.cos(alpha)
    a = d * math.sin(alpha) / math.sin(delta)

    y = a * math.cos(gamma)
    x = y * math.tan(gamma)

    return (
        bb_w - 2 * x,
        bb_h - 2 * y
    )


def crop_around_center(image, width, height):
    """
    Given a NumPy / OpenCV 2 image, crops it to the given width and height,
    around it's centre point
    """

    image_size = (image.shape[1], image.shape[0])
    image_center = (int(image_size[0] * 0.5), int(image_size[1] * 0.5))

    if(width > image_size[0]):
        width = image_size[0]

    if(height > image_size[1]):
        height = image_size[1]

    x1 = int(image_center[0] - width * 0.5)
    x2 = int(image_center[0] + width * 0.5)
    y1 = int(image_center[1] - height * 0.5)
    y2 = int(image_center[1] + height * 0.5)

    return image[y1:y2, x1:x2]


def demo():
    """
    Demos the largest_rotated_rect function
    """

    image = cv2.imread("lenna_rectangle.png")
    image_height, image_width = image.shape[0:2]

    cv2.imshow("Original Image", image)

    print "Press [enter] to begin the demo"
    print "Press [q] or Escape to quit"

    key = cv2.waitKey(0)
    if key == ord("q") or key == 27:
        exit()

    for i in np.arange(0, 360, 0.5):
        image_orig = np.copy(image)
        image_rotated = rotate_image(image, i)
        image_rotated_cropped = crop_around_center(
            image_rotated,
            *largest_rotated_rect(
                image_width,
                image_height,
                math.radians(i)
            )
        )

        key = cv2.waitKey(2)
        if(key == ord("q") or key == 27):
            exit()

        cv2.imshow("Original Image", image_orig)
        cv2.imshow("Rotated Image", image_rotated)
        cv2.imshow("Cropped Image", image_rotated_cropped)

    print "Done"


if __name__ == "__main__":
    demo()

画像回転のデモ

この画像(非正方形の画像で動作することを示すためにトリミングされています) を上記のファイルと同じディレクトリに配置して、実行します。

于 2013-05-27T09:39:04.580 に答える
15

大作おめでとうございます!あなたのコードを OpenCV で C++ ライブラリと共に使用したかったので、次の変換を行いました。たぶん、このアプローチは他の人に役立つかもしれません。

#include <iostream>
#include <opencv.hpp>

#define PI 3.14159265359

using namespace std;

double degree_to_radian(double angle)
{
    return angle * PI / 180;
}

cv::Mat rotate_image (cv::Mat image, double angle)
{
    // Rotates an OpenCV 2 image about its centre by the given angle
    // (in radians). The returned image will be large enough to hold the entire
    // new image, with a black background

    cv::Size image_size = cv::Size(image.rows, image.cols);
    cv::Point image_center = cv::Point(image_size.height/2, image_size.width/2);

    // Convert the OpenCV 3x2 matrix to 3x3
    cv::Mat rot_mat = cv::getRotationMatrix2D(image_center, angle, 1.0);
    double row[3] = {0.0, 0.0, 1.0};
    cv::Mat new_row = cv::Mat(1, 3, rot_mat.type(), row);
    rot_mat.push_back(new_row);


    double slice_mat[2][2] = {
        {rot_mat.col(0).at<double>(0), rot_mat.col(1).at<double>(0)},
        {rot_mat.col(0).at<double>(1), rot_mat.col(1).at<double>(1)}
    };

    cv::Mat rot_mat_nontranslate = cv::Mat(2, 2, rot_mat.type(), slice_mat);

    double image_w2 = image_size.width * 0.5;
    double image_h2 = image_size.height * 0.5;

    // Obtain the rotated coordinates of the image corners
    std::vector<cv::Mat> rotated_coords;

    double image_dim_d_1[2] = { -image_h2, image_w2 };
    cv::Mat image_dim = cv::Mat(1, 2, rot_mat.type(), image_dim_d_1);
    rotated_coords.push_back(cv::Mat(image_dim * rot_mat_nontranslate));


    double image_dim_d_2[2] = { image_h2, image_w2 };
    image_dim = cv::Mat(1, 2, rot_mat.type(), image_dim_d_2);
    rotated_coords.push_back(cv::Mat(image_dim * rot_mat_nontranslate));


    double image_dim_d_3[2] = { -image_h2, -image_w2 };
    image_dim = cv::Mat(1, 2, rot_mat.type(), image_dim_d_3);
    rotated_coords.push_back(cv::Mat(image_dim * rot_mat_nontranslate));


    double image_dim_d_4[2] = { image_h2, -image_w2 };
    image_dim = cv::Mat(1, 2, rot_mat.type(), image_dim_d_4);
    rotated_coords.push_back(cv::Mat(image_dim * rot_mat_nontranslate));


    // Find the size of the new image
    vector<double> x_coords, x_pos, x_neg;
    for (int i = 0; i < rotated_coords.size(); i++)
    {
        double pt = rotated_coords[i].col(0).at<double>(0);
        x_coords.push_back(pt);
        if (pt > 0)
            x_pos.push_back(pt);
        else
            x_neg.push_back(pt);
    }

    vector<double> y_coords, y_pos, y_neg;
    for (int i = 0; i < rotated_coords.size(); i++)
    {
        double pt = rotated_coords[i].col(1).at<double>(0);
        y_coords.push_back(pt);
        if (pt > 0)
            y_pos.push_back(pt);
        else
            y_neg.push_back(pt);
    }


    double right_bound = *max_element(x_pos.begin(), x_pos.end());
    double left_bound = *min_element(x_neg.begin(), x_neg.end());
    double top_bound = *max_element(y_pos.begin(), y_pos.end());
    double bottom_bound = *min_element(y_neg.begin(), y_neg.end());

    int new_w = int(abs(right_bound - left_bound));
    int new_h = int(abs(top_bound - bottom_bound));

    // We require a translation matrix to keep the image centred
    double trans_mat[3][3] = {
        {1, 0, int(new_w * 0.5 - image_w2)},
        {0, 1, int(new_h * 0.5 - image_h2)},
        {0, 0, 1},
    };


    // Compute the transform for the combined rotation and translation
    cv::Mat aux_affine_mat = (cv::Mat(3, 3, rot_mat.type(), trans_mat) * rot_mat);
    cv::Mat affine_mat = cv::Mat(2, 3, rot_mat.type(), NULL);
    affine_mat.push_back(aux_affine_mat.row(0));
    affine_mat.push_back(aux_affine_mat.row(1));

    // Apply the transform
    cv::Mat output;
    cv::warpAffine(image, output, affine_mat, cv::Size(new_h, new_w), cv::INTER_LINEAR);

    return output;
}

cv::Size largest_rotated_rect(int h, int w, double angle)
{
    // Given a rectangle of size wxh that has been rotated by 'angle' (in
    // radians), computes the width and height of the largest possible
    // axis-aligned rectangle within the rotated rectangle.

    // Original JS code by 'Andri' and Magnus Hoff from Stack Overflow

    // Converted to Python by Aaron Snoswell (https://stackoverflow.com/questions/16702966/rotate-image-and-crop-out-black-borders)
    // Converted to C++ by Eliezer Bernart

    int quadrant = int(floor(angle/(PI/2))) & 3;
    double sign_alpha = ((quadrant & 1) == 0) ? angle : PI - angle;
    double alpha = fmod((fmod(sign_alpha, PI) + PI), PI);

    double bb_w = w * cos(alpha) + h * sin(alpha);
    double bb_h = w * sin(alpha) + h * cos(alpha);

    double gamma = w < h ? atan2(bb_w, bb_w) : atan2(bb_h, bb_h);

    double delta = PI - alpha - gamma;

    int length = w < h ? h : w;

    double d = length * cos(alpha);
    double a = d * sin(alpha) / sin(delta);
    double y = a * cos(gamma);
    double x = y * tan(gamma);

    return cv::Size(bb_w - 2 * x, bb_h - 2 * y);
}

// for those interested in the actual optimum - contributed by coproc
#include <algorithm>
cv::Size really_largest_rotated_rect(int h, int w, double angle)
{
  // Given a rectangle of size wxh that has been rotated by 'angle' (in
  // radians), computes the width and height of the largest possible
  // axis-aligned rectangle within the rotated rectangle.
  if (w <= 0 || h <= 0)
    return cv::Size(0,0);

  bool width_is_longer = w >= h;
  int side_long = w, side_short = h;
  if (!width_is_longer)
    std::swap(side_long, side_short);

  // since the solutions for angle, -angle and pi-angle are all the same,
  // it suffices to look at the first quadrant and the absolute values of sin,cos:
  double sin_a = fabs(sin(angle)), cos_a = fabs(cos(angle));
  double wr,hr;
  if (side_short <= 2.*sin_a*cos_a*side_long)
  {
    // half constrained case: two crop corners touch the longer side,
    // the other two corners are on the mid-line parallel to the longer line
    double x = 0.5*side_short;
    wr = x/sin_a;
    hr = x/cos_a;
    if (!width_is_longer)
      std::swap(wr,hr);
  }
  else
  { 
    // fully constrained case: crop touches all 4 sides
    double cos_2a = cos_a*cos_a - sin_a*sin_a;
    wr = (w*cos_a - h*sin_a)/cos_2a;
    hr = (h*cos_a - w*sin_a)/cos_2a;
  }

  return cv::Size(wr,hr);
}

cv::Mat crop_around_center(cv::Mat image, int height, int width)
{
    // Given a OpenCV 2 image, crops it to the given width and height,
    // around it's centre point

    cv::Size image_size = cv::Size(image.rows, image.cols);
    cv::Point image_center = cv::Point(int(image_size.height * 0.5), int(image_size.width * 0.5));

    if (width > image_size.width)
        width = image_size.width;

    if (height > image_size.height)
        height = image_size.height;

    int x1 = int(image_center.x - width  * 0.5);
    int x2 = int(image_center.x + width  * 0.5);
    int y1 = int(image_center.y - height * 0.5);
    int y2 = int(image_center.y + height * 0.5);


    return image(cv::Rect(cv::Point(y1, x1), cv::Point(y2,x2)));
}

void demo(cv::Mat image)
{
    // Demos the largest_rotated_rect function
    int image_height = image.rows;
    int image_width = image.cols;

    for (float i = 0.0; i < 360.0; i+=0.5)
    {
        cv::Mat image_orig = image.clone();
        cv::Mat image_rotated = rotate_image(image, i);

        cv::Size largest_rect = largest_rotated_rect(image_height, image_width, degree_to_radian(i));
        // for those who trust math (added by coproc):
        cv::Size largest_rect2 = really_largest_rotated_rect(image_height, image_width, degree_to_radian(i));
        cout << "area1 = " << largest_rect.height * largest_rect.width << endl;
        cout << "area2 = " << largest_rect2.height * largest_rect2.width << endl;

        cv::Mat image_rotated_cropped = crop_around_center(
                    image_rotated,
                    largest_rect.height,
                    largest_rect.width
                    );

        cv::imshow("Original Image", image_orig);
        cv::imshow("Rotated Image", image_rotated);
        cv::imshow("Cropped image", image_rotated_cropped);

        if (char(cv::waitKey(15)) == 'q')
            break;
    }

}

int main (int argc, char* argv[])
{
    cv::Mat image = cv::imread(argv[1]);

    if (image.empty())
    {
        cout << "> The input image was not found." << endl;
        exit(EXIT_FAILURE);
    }

    cout << "Press [s] to begin or restart the demo" << endl;
    cout << "Press [q] to quit" << endl;

    while (true)
    {
        cv::imshow("Original Image", image);
        char opt = char(cv::waitKey(0));
        switch (opt) {
        case 's':
            demo(image);
            break;
        case 'q':
            return EXIT_SUCCESS;
        default:
            break;
        }
    }

    return EXIT_SUCCESS;
}
于 2014-11-25T21:29:10.030 に答える
5

優れたimutilsライブラリを利用する簡潔さのための小さな更新。

def rotated_rect(w, h, angle):
    """
    Given a rectangle of size wxh that has been rotated by 'angle' (in
    radians), computes the width and height of the largest possible
    axis-aligned rectangle within the rotated rectangle.

    Original JS code by 'Andri' and Magnus Hoff from Stack Overflow

    Converted to Python by Aaron Snoswell
    """
    angle = math.radians(angle)
    quadrant = int(math.floor(angle / (math.pi / 2))) & 3
    sign_alpha = angle if ((quadrant & 1) == 0) else math.pi - angle
    alpha = (sign_alpha % math.pi + math.pi) % math.pi

    bb_w = w * math.cos(alpha) + h * math.sin(alpha)
    bb_h = w * math.sin(alpha) + h * math.cos(alpha)

    gamma = math.atan2(bb_w, bb_w) if (w < h) else math.atan2(bb_w, bb_w)

    delta = math.pi - alpha - gamma

    length = h if (w < h) else w

    d = length * math.cos(alpha)
    a = d * math.sin(alpha) / math.sin(delta)

    y = a * math.cos(gamma)
    x = y * math.tan(gamma)

    return (bb_w - 2 * x, bb_h - 2 * y)

def crop(img, w, h):
    x, y = int(img.shape[1] * .5), int(img.shape[0] * .5)

    return img[
        int(np.ceil(y - h * .5)) : int(np.floor(y + h * .5)),
        int(np.ceil(x - w * .5)) : int(np.floor(x + h * .5))
    ]

def rotate(img, angle):
    # rotate, crop and return original size
    (h, w) = img.shape[:2]
    img = imutils.rotate_bound(img, angle)
    img = crop(img, *rotated_rect(w, h, angle))
    img = cv2.resize(img,(w,h),interpolation=cv2.INTER_AREA)
    return img
于 2018-09-25T20:06:53.850 に答える
3

Coprox の素晴らしい業績に触発されて、Coprox のコードと一緒に完全なソリューションを形成する関数を作成しました (そのため、簡単にコピー & ペーストすることで使用できます)。以下のrotate_max_area関数は、黒い境界線のない回転した画像を返します。

def rotate_bound(image, angle):
    # CREDIT: https://www.pyimagesearch.com/2017/01/02/rotate-images-correctly-with-opencv-and-python/
    (h, w) = image.shape[:2]
    (cX, cY) = (w // 2, h // 2)
    M = cv2.getRotationMatrix2D((cX, cY), -angle, 1.0)
    cos = np.abs(M[0, 0])
    sin = np.abs(M[0, 1])
    nW = int((h * sin) + (w * cos))
    nH = int((h * cos) + (w * sin))
    M[0, 2] += (nW / 2) - cX
    M[1, 2] += (nH / 2) - cY
    return cv2.warpAffine(image, M, (nW, nH))


def rotate_max_area(image, angle):
    """ image: cv2 image matrix object
        angle: in degree
    """
    wr, hr = rotatedRectWithMaxArea(image.shape[1], image.shape[0],
                                    math.radians(angle))
    rotated = rotate_bound(image, angle)
    h, w, _ = rotated.shape
    y1 = h//2 - int(hr/2)
    y2 = y1 + int(hr)
    x1 = w//2 - int(wr/2)
    x2 = x1 + int(wr)
    return rotated[y1:y2, x1:x2]
于 2018-01-04T18:37:19.590 に答える