Python PIL：查找图像的大小而不将其写为文件 [英] Python PIL: Find the size of image without writing it as a file

问题描述

编辑：此问题已被标记为重复？我的问题显然是关于优化这个过程，而不是如何去做。我甚至提供了代码来证明我已经找到了后者。在你标记它们之前，你的互联网大厅监视器是否甚至通过标题阅读了这些问题？

我有以下代码块来使用PIL压缩图像，直到所述图像小于一定大小。

I have the following block of code to compress an image using PIL, until said image is under a certain size.

from PIL import Image
import os

def compress(image_file, max_size, scale):
    while os.path.getsize(image_file) > max_size:
        pic = Image.open(image_file)
        original_size = pic.size
        pic = pic.resize((int(original_size[0] * scale),
            int(original_size[1] * scale)),
            Image.ANTIALIAS)
        pic.save(image_file, optimize=True, quality=95)

在这段代码中，我使用 os.path.getsize（image_file）来获取图像的大小。但是，这意味着每次循环运行时，文件必须保存在 pic.save（image_file，optimize = True，quality = 95 中。

In this code, I use os.path.getsize(image_file) to get the size of the image. However, this means that the file must be saved in pic.save(image_file, optimize=True, quality=95 every time the loop runs.

这个过程需要很长时间。

That process takes a long time.

有没有办法通过某种方式获得大小的优化 PIL中的图像 图像对象图片？

Is there a way to optimise this by somehow getting the size of the image in the PIL Image object pic?

推荐答案

使用 io.BytesIO（）将图像保存到内存中。也许最好每次调整原始文件的大小，如下所示：

Use io.BytesIO() to save the image into memory. It is also probably better to resize from your original file each time as follows:

from PIL import Image
import os
import io

def compress(original_file, max_size, scale):
    assert(0.0 < scale < 1.0)
    orig_image = Image.open(original_file)
    cur_size = orig_image.size

    while True:
        cur_size = (int(cur_size[0] * scale), int(cur_size[1] * scale))
        resized_file = orig_image.resize(cur_size, Image.ANTIALIAS)

        with io.BytesIO() as file_bytes:
            resized_file.save(file_bytes, optimize=True, quality=95, format='jpeg')

            if file_bytes.tell() <= max_size:
                file_bytes.seek(0, 0)
                with open(original_file, 'wb') as f_output:
                    f_output.write(file_bytes.read())
                break

compress(r"c:\mytest.jpg", 10240, 0.9) 

因此，这将获取文件并将其缩小 0.9 每次尝试，直到达到合适的大小。然后它会覆盖原始文件。

So this will take the file and scale it down 0.9 each attempt until a suitable size is reached. It then overwrites the original file.

作为替代方法，您可以创建要尝试的比例列表，例如： [0.01,0.02 .... 0.99,1] 然后使用二进制印章确定最接近 max_size的文件大小的比例结果如下：

As an alternative approach, you could create a list of scales to try, e.g. [0.01, 0.02 .... 0.99, 1] and then use a binary chop to determine which scale results in a filesize closest to max_size as follows:

def compress(original_file, max_size):
    save_opts={'optimize':True, 'quality':95, 'format':'jpeg'}
    orig_image = Image.open(original_file)
    width, height = orig_image.size
    scales = [scale / 1000 for scale in range(1, 1001)]  # e.g. [0.001, 0.002 ... 1.0]

    lo = 0
    hi = len(scales)

    while lo < hi:
        mid = (lo + hi) // 2

        scaled_size = (int(width * scales[mid]), int(height * scales[mid]))
        resized_file = orig_image.resize(scaled_size, Image.ANTIALIAS)

        file_bytes = io.BytesIO()
        resized_file.save(file_bytes, **save_opts)
        size = file_bytes.tell()
        print(size, scales[mid])

        if size < max_size: 
            lo = mid + 1
        else: 
            hi = mid

    scale = scales[max(0, lo-1)]
    print("Using scale:", scale)
    orig_image.resize((int(width * scale), int(height * scale)), Image.ANTIALIAS).save(original_file, **save_opts)

因此对于 max_size 10000 ，如果尝试过大 0.251 ，则循环首先尝试 0.501 的比例等等。当 max_size = 1024 时，将尝试以下比例：

So for a max_size of 10000, the loop first tries a scale of 0.501, if too big 0.251 is tried and so on. When max_size=1024 the following scales would be tried:

180287 0.501
56945 0.251
17751 0.126
5371 0.063
10584 0.095
7690 0.079
9018 0.087
10140 0.091
9336 0.089
9948 0.09
Using scale: 0.09

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