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python-3.x - 如何将图像保存在由 python matplotlib 生成的 AWS Lambda 函数中？

问题描述

我有下面的python代码。它.wav通过邮递员将文件作为输入。它在此处作为 base64 字符串接收，然后从 base64 解码回来。代码进一步处理.wav文件并生成.png图像。我必须将其保存在 AWS S3 中。我在将其保存到 AWS S3 时遇到问题，因为保存在那里的文件没有打开。它说photo viewer doesn't support this file format。知道怎么做吗？

import json
import base64
import boto3
#import scipy.io.wavfile as wav
#import scipy.signal as signal
import numpy as np
from matplotlib import pyplot as plt
from scipy import signal
import shutil
import wavio
import wave
import matplotlib.pylab as plt
from scipy.signal import butter, lfilter
from scipy.io import wavfile
import scipy.signal as sps
from io import BytesIO    

def lambda_handler(event, context):
   s3 = boto3.client("s3")
   
   # retrieving data from event. Which is the wave audio file
   get_file_content_from_postman = event["content"]
   
   # decoding data. Here the wava file is converted back to binary form
   decoded_file_name = base64.b64decode(get_file_content_from_postman)
   
   new_rate = 2000
   
   # Read file
   sample_rate, clip = wavfile.read(BytesIO(decoded_file_name))
   
   # Resample data
   number_of_samples = round(len(clip) * float(new_rate) / sample_rate)
   clip = sps.resample(clip, number_of_samples)
   
   #butter_bandpass_filter is another fuction
   a = butter_bandpass_filter(clip, 20, 400, 2000, order=4)
   
   filtered = 2*((a-min(a))/(max(a)-min(a)))-1
   
   fig = plt.figure(figsize=[1,1])
   ax = fig.add_subplot(212)
   ax.axes.get_xaxis().set_visible(False)
   ax.axes.get_yaxis().set_visible(False)
   ax.set_frame_on(False)
   powerSpectrum, freqenciesFound, time, imageAxis = plt.specgram(filtered, Fs=2000)
   
   #filename is referring to the AWS Lambda /tmp directory
   filename  = '/tmp/' + 'image.png'
   
   plt.savefig(filename, dpi=400, bbox_inches='tight',pad_inches=0)
   
   s3_upload = s3.put_object( Bucket="aaa", Key="filename.png", Body=filename)
   return {
   'statusCode': 200,
   'body': json.dumps("Executed successfully")
   }

标签： python-3.xamazon-web-servicesmatplotlibamazon-s3aws-lambda

解决方案

您正在使用put_object这意味着 Body不是文件名：

  • 正文（字节或可查找的类似文件的对象）——对象数据。

如果你想继续使用put_object，那么它应该是：

with open(filename, 'rb') as file_obj:
   s3_upload = s3.put_object( Bucket="aaa", Key="filename.png", Body=file_obj)

或者使用更直观的upload_file 。

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