如何将S3存储桶中的映像文件直接读入内存？

Question

我有以下代码

import matplotlib.pyplot as plt
import matplotlib.image as mpimg
import numpy as np
import boto3
s3 = boto3.resource('s3', region_name='us-east-2')
bucket = s3.Bucket('sentinel-s2-l1c')
object = bucket.Object('tiles/10/S/DG/2015/12/7/0/B01.jp2')
object.download_file('B01.jp2')
img=mpimg.imread('B01.jp2')
imgplot = plt.imshow(img)
plt.show(imgplot)

它的工作原理.但它首先将文件下载到当前目录的问题.是否可以直接在RAM中读取文件并将其解码为图像？

Answer 1

我建议使用io模块将文件直接读入内存,而不必使用临时文件.

例如:

import matplotlib.pyplot as plt
import matplotlib.image as mpimg
import numpy as np
import boto3
import io

s3 = boto3.resource('s3', region_name='us-east-2')
bucket = s3.Bucket('sentinel-s2-l1c')
object = bucket.Object('tiles/10/S/DG/2015/12/7/0/B01.jp2')

file_stream = io.StringIO()
object.download_fileobj(file_stream)
img = mpimg.imread(file_stream)
# whatever you need to do

io.BytesIO如果您的数据是二进制的,您也可以使用.

Answer 2

从 Greg Merritt 的回答中进一步发展以解决评论部分中的所有错误，使用BytesIO代替StringIO，使用 PILImage代替matplotlib.image。

以下函数适用于python3和boto3。同样，write_image_to_s3功能也是一种奖励。

from PIL import Image
from io import BytesIO
import numpy as np

def read_image_from_s3(bucket, key, region_name='ap-southeast-1'):
    """Load image file from s3.

    Parameters
    ----------
    bucket: string
        Bucket name
    key : string
        Path in s3

    Returns
    -------
    np array
        Image array
    """
    s3 = boto3.resource('s3', region_name='ap-southeast-1')
    bucket = s3.Bucket(bucket)
    object = bucket.Object(key)
    response = object.get()
    file_stream = response['Body']
    im = Image.open(file_stream)
    return np.array(im)

def write_image_to_s3(img_array, bucket, key, region_name='ap-southeast-1'):
    """Write an image array into S3 bucket

    Parameters
    ----------
    bucket: string
        Bucket name
    key : string
        Path in s3

    Returns
    -------
    None
    """
    s3 = boto3.resource('s3', region_name)
    bucket = s3.Bucket(bucket)
    object = bucket.Object(key)
    file_stream = BytesIO()
    im = Image.fromarray(img_array)
    im.save(file_stream, format='jpeg')
    object.put(Body=file_stream.getvalue())

Answer 3

格雷格梅里特的答案如下是更好的方法.

我想建议在模块中使用Python的NamedTemporaryFiletempfile.它会创建临时文件,在文件关闭时将被删除(感谢@NoamG)

import matplotlib.pyplot as plt
import matplotlib.image as mpimg
import numpy as np
import boto3
import tempfile

s3 = boto3.resource('s3', region_name='us-east-2')
bucket = s3.Bucket('sentinel-s2-l1c')
object = bucket.Object('tiles/10/S/DG/2015/12/7/0/B01.jp2')
tmp = tempfile.NamedTemporaryFile()

with open(tmp.name, 'wb') as f:
    object.download_fileobj(f)
    img=mpimg.imread(tmp.name)
    # ...Do jobs using img

Answer 4

使用客户端的方法略有不同：

import boto3
import io
from matplotlib import pyplot as plt

client = boto3.client("s3")

bucket='my_bucket'
key= 'my_key'

outfile = io.BytesIO()
client.download_fileobj(bucket, key, outfile)
outfile.seek(0)
img = plt.imread(outfile)

plt.imshow(img)
plt.show()

Answer 5

通过指定文件格式,可以流式传输图像imread().

import boto3
from io import BytesIO
import matplotlib.image as mpimg
import matplotlib.pyplot as plt

resource = boto3.resource('s3', region_name='us-east-2')
bucket = resource.Bucket('sentinel-s2-l1c')

image_object = bucket.Object('tiles/10/S/DG/2015/12/7/0/B01.jp2')
image = mpimg.imread(BytesIO(image_object.get()['Body'].read()), 'jp2')

plt.figure(0)
plt.imshow(image)

Answer 6

object = bucket.Object('tiles/10/S/DG/2015/12/7/0/B01.jp2')
img_data = object.get().get('Body').read()