python—使用matplotlib和boto将绘图从内存上载到s3

这是我的工作脚本，用于生成绘图，将其本地保存到磁盘，上载到S3并删除文件：

plt.figure(figsize=(6,6))
plt.plot(x, y, 'bo')
plt.savefig('file_location')

conn = boto.s3.connect_to_region(
    region_name=AWS_REGION,
    aws_access_key_id=AWS_ACCESS_KEY_ID,
    aws_secret_access_key=AWS_SECRET_ACCESS_KEY,
    calling_format=boto.s3.connection.OrdinaryCallingFormat()
    )
bucket = conn.get_bucket('bucket_name')
k = Key(bucket)
k.key = 'file_name'
k.set_contents_from_filename('file_location')

os.remove(file_location)

我想要的是跳过磁盘写入，直接从内存上传绘图

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如何实现这一目标，有什么建议吗？

综合起来：

img_data = io.BytesIO()
plt.savefig(img_data, format='png')
img_data.seek(0)

s3 = boto3.resource('s3')
bucket = s3.Bucket(BUCKET_NAME)
bucket.put_object(Body=img_data, ContentType='image/png', Key=KEY)

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谢谢@padraic cunningham和@guyb7的提示

保存到

io.BytesIO（）

，

f=io.BytesIO（），plt.savefig（f）

太好了，谢谢！为了以后上传，我只需调用

f.seek（0）

，然后从文件（f）中设置内容。是的，指针将在写入后结束。上载的图像对我来说是空的：（但我正在尝试上载实际图像，而不是plt图形…这会有区别吗？我如何读取图像？我尝试了img=skio.imread（io.BytesIO（obj.get（）['Body'].read（）），但我得到一个错误：“找不到在“i”模式下读取指定文件的格式。@Raksha请尝试以下操作：将plt.savefig（img_数据，format='png'）替换为plt.imsave（img_数据，my_img）其中my_img是您的图像数据。如果上传的图像如@Raksha所述为空。请检查此答案。是否有人知道此解决方案在我使用AWS signedURL打开它时为什么会裁剪绘图？

import matplotlib.pyplot as plt
import numpy as np
from matplotlib.backends.backend_agg import FigureCanvasAgg
import boto3
import io

# some random plotting. We need the figure object later
fig, ax = plt.subplots(1,1,figsize=(6,6))
ax.plot(np.linspace(0,1,50),
        np.random.normal(0.5,0.5,50))


canvas = FigureCanvas(fig) # renders figure onto canvas
imdata = io.BytesIO() # prepares in-memory binary stream buffer (think of this as a txt file but purely in memory)
canvas.print_png(imdata) # writes canvas object as a png file to the buffer. You can also use print_jpg, alternatively

s3 = boto3.resource('s3',
                    aws_access_key_id='your access key id',
                    aws_secret_access_key='your secret access key',
                    region_name='us-east-1') # or whatever region your s3 is in

s3.Object('yourbucket','picture.png').put(Body=imdata.getvalue(),
                                          ContentType='image/png') 
# this makes a new object in the bucket and puts the file in the bucket
# ContentType parameter makes sure resulting object is of a 'image/png' type and not a downloadable 'binary/octet-stream'

s3.ObjectAcl('yourbucket','picture.png').put(ACL='public-read')
# include this last line if you find the url for the image to be inaccessible