打开图像

1. 通过路径

# PIL
from PIL import Image

# Image path
im_path = "./a.jpg"

# 1. path
im1 = Image.open(im_path)
print (' From image path {}'.format(im1))

2. 通过打开图像对象

   # PIL
   from PIL import Image
   

Image path

im_path = “./a.jpg”

2. file

with open(im_path, ‘rb’) as f:
im2 = Image.open(f)
print (‘ From image file {}’.format(im2))

3. 打开图像字符串流
```python
# Platform.
import platform
if ('2.' in platform.python_version()):
    from StringIO import StringIO as Bytes2Data
else:
    from io import BytesIO as Bytes2Data

# PIL
from PIL import Image

# Image path
im_path = "./a.jpg"

# 3. Bytes.
with open(im_path, 'rb') as f:
    im_bytes = f.read()
    im3 = Image.open(Bytes2Data(im_bytes))
    print (' From image bytes {}'.format(im3))

4. 打开包含图像的压缩包

   # Platform
   import platform
   if ('2.' in platform.python_version()):
    from StringIO import StringIO as Bytes2Data
   elif ('3.' in platform.python_version()):
    from io import BytesIO as Bytes2Data
   from PIL import Image
   # Zip
   import zipfile
   

Zip path

zip_path = “./z.zip”

4. Zip.

z_file = zipfile.ZipFile(zip_path, “r”)
for filename in z_file.namelist():
# Bytes.
bytes_img = z_file.read(filename)
if (0 != len(bytes_img)):
im4 = Image.open(Bytes2Data(bytes_img))
print(‘ From zip file {}’.format(im4))
else: # directory.
pass