How to return Flask response with multiple dictionary and file

Question

With server.py running:

from flask import Flask, request, Response, json

app = Flask(__name__)

@app.route('/test', methods=['GET','POST'])
def route():  
    open_read =  open(r'C:\log.log').read()
    dict_data = {'str': str(), 'list': list(), 'dict': dict()}
    return Response(json.dumps(dict_data), mimetype='text/plain')

if __name__ == '__main__':
    app.run('0.0.0.0', 5000)

I would like to send a client request and receive back the open_read and dict_data variables data.

Is there a way to return both variables to the client using flask.Response object?

client:

import json, requests  
response = requests.post('http://127.0.0.1:5000/test')
print '...response: %s' % response.json(), type(response.json())

FULLY WORKING SOLUTION IS POSTED BELOW:

Server:

from flask import Flask, request, Response, json
import base64

app = Flask(__name__)

@app.route('/test', methods=['GET','POST'])
def route():  
    open_read =  open('my_file.zip').read()
    encoded = base64.b64encode(open_read)
    dict_data = {'str': str(), 'list': list(), 'dict': dict(), 'encoded': encoded}
    return Response(json.dumps(dict_data), mimetype='text/plain')

if __name__ == '__main__':
    app.run('0.0.0.0', 5000)

Client:

response = requests.post('http://127.0.0.1:5000/test')
response_json = response.json()
encoded = response_json['encoded']

decoded = base64.b64decode(encoded)

dst_filepath = "my_zip.zip"
with open(dst_filepath, 'w') as _file:
    _file.write(decoded)