我的意思是；是否可以在不指定压缩多少的情况下智能地压缩图像大小示例：-https://compressjpeg.com/
就像这个网站如何自动将图像压缩到特定的图像一样，如果是的话，我如何使用python来做到这一点？
我读了数百个关于这方面的博客，但没有找到一个完美的。。。真的需要你的帮助：(
您的问题是：
"是否可以在不指定压缩多少的情况下智能地压缩图像大小？
如果最终用户没有选择压缩图像的程度，那么其他人会选择。
一个">智能"；自动选择输出图像大小的函数被保证是必须手动指定质量的函数的包装器。
def automagically_intelligently_compress_picture(in_pic):
# output will be 500 pixels by 680 pixels
out_pic = manually_compress(in_pic, (500, 680))
return out_pic
要压缩图片，可以使用PIL(Python图像库(。
以下代码输出的照片几乎总是小于24000字节(24KB或0.024MB(：
import math
from PIL import Image
# Note that when installing the `PIL` library,
# you must write "Pillow", not "PIL"
# EXAMPLE: C:Usersuser>pip install Pillow
big_photo_path_string = "C:\Users\user\Desktop\rrefwf.png"
# Note that when writing paths when using
# the Microsoft Windows Operatings System:
#
# blackslash () is an escape sequence for strings
# For example "n" means:
# 1. leave the program
# 2. insert a new-line character
# 3. return to program"]
#
# str.__repr__() will...
# ... convert new-lines into slash-n ("n")
# ... convert carriage returns into slash-r ("r")
# ... convert tabs into slash-t ("t")
# ... et cetra
big_photo_path_string = repr(big_photo_path_string)[1:-1]
bigphoto = Image.open(big_photo_path_string)
# Establish Target Size
target_byte_count = 24000
target_pixel_count = 2.8114*target_byte_count
# Note that an image which measures
# 1920 by 1080
# pixels is approximately
# 737,581 bytes
# So we have
# 2.8114 pixels per byte
scale_factor = target_pixel_count/math.prod(bigphoto.size)
# if
# the image is already smaller than the target size
# then
# Do not enlarge the image.
# keep the image the same size.
scale_factor = 1 if scale_factor > 1 else scale_factor
# set new photographs size
# `bigphoto.size` is something like (20000, 37400)
sml_size = list(scale_factor*dim for dim in bigphoto.size)
# We cannot have two-and-a-half pixels
# convert decimal numbers into whole numbers
sml_size = [int(number) for number in sml_size]
# cast `sml_size` to the same data-type as `bigphoto.size`
copy_constructor = type(bigphoto.size)
sml_size = copy_constructor(sml_size)
# create new photograph
sml_photo = bigphoto.resize(sml_size, resample=Image.LANCZOS)
# The new (small/compressed) photo will have different
# name than old (large/un-compressed) photo
#
# EXAMPLE:
# OLD NAME: my_photo.jpg
# NEW NAME: my_photo_out.jpg
p = big_photo_path_string
parts = p.split(".")
out_path = "".join(parts[0:1]) + "_out." + parts[-1]
# save new (small) photograph
sml_photo.save(out_path, optimize=True, quality=95)