有没有我忽略的明显的方法?我只是想做个缩略图。


当前回答

# Importing Image class from PIL module
from PIL import Image

# Opens a image in RGB mode
im = Image.open(r"C:\Users\System-Pc\Desktop\ybear.jpg")

# Size of the image in pixels (size of original image)
# (This is not mandatory)
width, height = im.size

# Setting the points for cropped image
left = 4
top = height / 5
right = 154
bottom = 3 * height / 5

# Cropped image of above dimension
# (It will not change original image)
im1 = im.crop((left, top, right, bottom))
newsize = (300, 300)
im1 = im1.resize(newsize)
# Shows the image in image viewer
im1.show()

其他回答

基于@tomvon,我完成了以下使用(选择你的案例):

a)调整高度(我知道新的宽度,所以我需要新的高度)

new_width  = 680
new_height = new_width * height / width 

b)调整宽度(我知道新的高度,所以我需要新的宽度)

new_height = 680
new_width  = new_height * width / height

然后:

img = img.resize((new_width, new_height), Image.ANTIALIAS)

一个简单的方法,保持约束比率和通过最大宽度/高度。不是最漂亮的,但可以完成工作,而且很容易理解:

def resize(img_path, max_px_size, output_folder):
    with Image.open(img_path) as img:
        width_0, height_0 = img.size
        out_f_name = os.path.split(img_path)[-1]
        out_f_path = os.path.join(output_folder, out_f_name)

        if max((width_0, height_0)) <= max_px_size:
            print('writing {} to disk (no change from original)'.format(out_f_path))
            img.save(out_f_path)
            return

        if width_0 > height_0:
            wpercent = max_px_size / float(width_0)
            hsize = int(float(height_0) * float(wpercent))
            img = img.resize((max_px_size, hsize), Image.ANTIALIAS)
            print('writing {} to disk'.format(out_f_path))
            img.save(out_f_path)
            return

        if width_0 < height_0:
            hpercent = max_px_size / float(height_0)
            wsize = int(float(width_0) * float(hpercent))
            img = img.resize((max_px_size, wsize), Image.ANTIALIAS)
            print('writing {} to disk'.format(out_f_path))
            img.save(out_f_path)
            return

下面是一个python脚本,使用这个函数来运行批量图像调整。

举个丑陋的例子。

函数获取像pic[0-9a-z]这样的文件。[扩展]",将它们调整为120x120,移动部分到中心,并保存到"ico[0-9a-z]。[扩展]”,肖像和风景作品:

def imageResize(filepath):
    from PIL import Image
    file_dir=os.path.split(filepath)
    img = Image.open(filepath)

    if img.size[0] > img.size[1]:
        aspect = img.size[1]/120
        new_size = (img.size[0]/aspect, 120)
    else:
        aspect = img.size[0]/120
        new_size = (120, img.size[1]/aspect)
    img.resize(new_size).save(file_dir[0]+'/ico'+file_dir[1][3:])
    img = Image.open(file_dir[0]+'/ico'+file_dir[1][3:])

    if img.size[0] > img.size[1]:
        new_img = img.crop( (
            (((img.size[0])-120)/2),
            0,
            120+(((img.size[0])-120)/2),
            120
        ) )
    else:
        new_img = img.crop( (
            0,
            (((img.size[1])-120)/2),
            120,
            120+(((img.size[1])-120)/2)
        ) )

    new_img.save(file_dir[0]+'/ico'+file_dir[1][3:])
######get resize coordinate after resize the image using this function#####
def scale_img_pixel(points,original_dim,resize_dim):
        multi_list = [points]
        new_point_list = []
        multi_list_point = []
        for point in multi_list:
            multi_list_point.append([point[0],point[1]])
            multi_list_point.append([point[2],point[3]])
        for lsingle_point in multi_list_point:
            x1 = int((lsingle_point[0] * (resize_dim[0] / original_dim[0])))
            y1 = int((lsingle_point[1] * (resize_dim[1] / original_dim[1])))
            new_point_list.append(x1)
            new_point_list.append(y1)
            
        return new_point_list
    
    
    points = [774,265,909,409]
    original_dim = (1237,1036)
    resize_dim = (640,480)
    result = scale_img_pixel(points,original_dim,resize_dim)
    print("result: ", result)  

我还推荐使用PIL的缩略图方法,因为它消除了所有的比例麻烦。

不过,有一个重要的提示:替换

im.thumbnail(size)

with

im.thumbnail(size,Image.ANTIALIAS)

默认情况下,PIL使用映像。最近的过滤器调整大小,结果良好的性能,但质量差。