Generate image on-the-fly using FLask and matplotlib
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0
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I would like to plot an image without having to save it in a static folder.
I've got an example posted here by user Messa working just fine:
import io
import random
from flask import Response
from matplotlib.backends.backend_agg import FigureCanvasAgg as FigureCanvas
from matplotlib.figure import Figure
@app.route('/plot.png')
def plot_png():
fig = create_figure()
output = io.BytesIO()
FigureCanvas(fig).print_png(output)
return Response(output.getvalue(), mimetype='image/png')
def create_figure():
fig = Figure()
axis = fig.add_subplot(1, 1, 1)
xs = range(100)
ys = [random.randint(1, 50) for x in xs]
axis.plot(xs, ys)
return fig
Then in the HTML file:
<img src="/plot.png" alt="my plot">
The result is perfect. I get the image on the fly.
Now I would like to use some parameters like this:
@app.route('/<x_values_source>/<y_values_source>/plot.png')
def plot_png(x_values_source, y_values_source):
fig = create_figure(x_values_source, y_values_source)
output = io.BytesIO()
FigureCanvas(fig).print_png(output)
return Response(output.getvalue(), mimetype='image/png')
def create_figure(x_values_source, y_values_source):
x_values = function_that_returns_x_values_set(x_values_source)
y_values = function_that_returns_y_values_set(y_values_source)
fig = Figure()
axis = fig.add_subplot(1, 1, 1)
axis.plot(x_values, y_values)
return fig
But of course It doesn't work. I'm quite new to Python and I'm still learing a lot of things.
Thanks!
python matplotlib flask
add a comment |
up vote
0
down vote
favorite
I would like to plot an image without having to save it in a static folder.
I've got an example posted here by user Messa working just fine:
import io
import random
from flask import Response
from matplotlib.backends.backend_agg import FigureCanvasAgg as FigureCanvas
from matplotlib.figure import Figure
@app.route('/plot.png')
def plot_png():
fig = create_figure()
output = io.BytesIO()
FigureCanvas(fig).print_png(output)
return Response(output.getvalue(), mimetype='image/png')
def create_figure():
fig = Figure()
axis = fig.add_subplot(1, 1, 1)
xs = range(100)
ys = [random.randint(1, 50) for x in xs]
axis.plot(xs, ys)
return fig
Then in the HTML file:
<img src="/plot.png" alt="my plot">
The result is perfect. I get the image on the fly.
Now I would like to use some parameters like this:
@app.route('/<x_values_source>/<y_values_source>/plot.png')
def plot_png(x_values_source, y_values_source):
fig = create_figure(x_values_source, y_values_source)
output = io.BytesIO()
FigureCanvas(fig).print_png(output)
return Response(output.getvalue(), mimetype='image/png')
def create_figure(x_values_source, y_values_source):
x_values = function_that_returns_x_values_set(x_values_source)
y_values = function_that_returns_y_values_set(y_values_source)
fig = Figure()
axis = fig.add_subplot(1, 1, 1)
axis.plot(x_values, y_values)
return fig
But of course It doesn't work. I'm quite new to Python and I'm still learing a lot of things.
Thanks!
python matplotlib flask
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I would like to plot an image without having to save it in a static folder.
I've got an example posted here by user Messa working just fine:
import io
import random
from flask import Response
from matplotlib.backends.backend_agg import FigureCanvasAgg as FigureCanvas
from matplotlib.figure import Figure
@app.route('/plot.png')
def plot_png():
fig = create_figure()
output = io.BytesIO()
FigureCanvas(fig).print_png(output)
return Response(output.getvalue(), mimetype='image/png')
def create_figure():
fig = Figure()
axis = fig.add_subplot(1, 1, 1)
xs = range(100)
ys = [random.randint(1, 50) for x in xs]
axis.plot(xs, ys)
return fig
Then in the HTML file:
<img src="/plot.png" alt="my plot">
The result is perfect. I get the image on the fly.
Now I would like to use some parameters like this:
@app.route('/<x_values_source>/<y_values_source>/plot.png')
def plot_png(x_values_source, y_values_source):
fig = create_figure(x_values_source, y_values_source)
output = io.BytesIO()
FigureCanvas(fig).print_png(output)
return Response(output.getvalue(), mimetype='image/png')
def create_figure(x_values_source, y_values_source):
x_values = function_that_returns_x_values_set(x_values_source)
y_values = function_that_returns_y_values_set(y_values_source)
fig = Figure()
axis = fig.add_subplot(1, 1, 1)
axis.plot(x_values, y_values)
return fig
But of course It doesn't work. I'm quite new to Python and I'm still learing a lot of things.
Thanks!
python matplotlib flask
I would like to plot an image without having to save it in a static folder.
I've got an example posted here by user Messa working just fine:
import io
import random
from flask import Response
from matplotlib.backends.backend_agg import FigureCanvasAgg as FigureCanvas
from matplotlib.figure import Figure
@app.route('/plot.png')
def plot_png():
fig = create_figure()
output = io.BytesIO()
FigureCanvas(fig).print_png(output)
return Response(output.getvalue(), mimetype='image/png')
def create_figure():
fig = Figure()
axis = fig.add_subplot(1, 1, 1)
xs = range(100)
ys = [random.randint(1, 50) for x in xs]
axis.plot(xs, ys)
return fig
Then in the HTML file:
<img src="/plot.png" alt="my plot">
The result is perfect. I get the image on the fly.
Now I would like to use some parameters like this:
@app.route('/<x_values_source>/<y_values_source>/plot.png')
def plot_png(x_values_source, y_values_source):
fig = create_figure(x_values_source, y_values_source)
output = io.BytesIO()
FigureCanvas(fig).print_png(output)
return Response(output.getvalue(), mimetype='image/png')
def create_figure(x_values_source, y_values_source):
x_values = function_that_returns_x_values_set(x_values_source)
y_values = function_that_returns_y_values_set(y_values_source)
fig = Figure()
axis = fig.add_subplot(1, 1, 1)
axis.plot(x_values, y_values)
return fig
But of course It doesn't work. I'm quite new to Python and I'm still learing a lot of things.
Thanks!
python matplotlib flask
python matplotlib flask
edited Nov 12 at 20:42
asked Nov 12 at 18:35
Luciano
11
11
add a comment |
add a comment |
1 Answer
1
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up vote
-1
down vote
To pass parameters in, it's best to not prefix them to the URI, since that might confuse Flask's route decoding. Instead, if you form the URL like
/plot.png?x=1,2,3,4&y=100,200,300,400
you can pick the args up via request.args
. They'll come in as strings, so you'll need to seperate and convert to ints (or floats, if you're using real numbers)
from flask import request
@app.route('/plot.png')
def plot():
try:
x_values = [int(x) for x in request.args.get('x','').split(',')]
y_values = [int(y) for y in request.args.get('y','').split(',')]
except:
# malformed
I have a working example here of a robust way to handle matplotlib dynamically. The general idea is
from flask import send_file
@app.route('/plot.png')
def plot_png():
fig, ax = plt.subplots()
draw(ax)
return fig_response(fig)
def fig_response(fig):
"""Turn a matplotlib Figure into Flask response"""
img_bytes = BytesIO()
fig.savefig(img_bytes)
img_bytes.seek(0)
return send_file(img_bytes, mimetype='image/png')
where you supply draw()
. Take a look at my example, though, since there are some cache headers you'll want to add.
Hi! Thanks for the reply. I still don't see how could I send the parameters through the route.
– Luciano
Nov 12 at 20:23
I added example code for that above.
– Dave W. Smith
Nov 12 at 20:45
Thanks! I'll give it a try. I've also modified my original question in order to make it more clear.
– Luciano
Nov 12 at 20:51
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
-1
down vote
To pass parameters in, it's best to not prefix them to the URI, since that might confuse Flask's route decoding. Instead, if you form the URL like
/plot.png?x=1,2,3,4&y=100,200,300,400
you can pick the args up via request.args
. They'll come in as strings, so you'll need to seperate and convert to ints (or floats, if you're using real numbers)
from flask import request
@app.route('/plot.png')
def plot():
try:
x_values = [int(x) for x in request.args.get('x','').split(',')]
y_values = [int(y) for y in request.args.get('y','').split(',')]
except:
# malformed
I have a working example here of a robust way to handle matplotlib dynamically. The general idea is
from flask import send_file
@app.route('/plot.png')
def plot_png():
fig, ax = plt.subplots()
draw(ax)
return fig_response(fig)
def fig_response(fig):
"""Turn a matplotlib Figure into Flask response"""
img_bytes = BytesIO()
fig.savefig(img_bytes)
img_bytes.seek(0)
return send_file(img_bytes, mimetype='image/png')
where you supply draw()
. Take a look at my example, though, since there are some cache headers you'll want to add.
Hi! Thanks for the reply. I still don't see how could I send the parameters through the route.
– Luciano
Nov 12 at 20:23
I added example code for that above.
– Dave W. Smith
Nov 12 at 20:45
Thanks! I'll give it a try. I've also modified my original question in order to make it more clear.
– Luciano
Nov 12 at 20:51
add a comment |
up vote
-1
down vote
To pass parameters in, it's best to not prefix them to the URI, since that might confuse Flask's route decoding. Instead, if you form the URL like
/plot.png?x=1,2,3,4&y=100,200,300,400
you can pick the args up via request.args
. They'll come in as strings, so you'll need to seperate and convert to ints (or floats, if you're using real numbers)
from flask import request
@app.route('/plot.png')
def plot():
try:
x_values = [int(x) for x in request.args.get('x','').split(',')]
y_values = [int(y) for y in request.args.get('y','').split(',')]
except:
# malformed
I have a working example here of a robust way to handle matplotlib dynamically. The general idea is
from flask import send_file
@app.route('/plot.png')
def plot_png():
fig, ax = plt.subplots()
draw(ax)
return fig_response(fig)
def fig_response(fig):
"""Turn a matplotlib Figure into Flask response"""
img_bytes = BytesIO()
fig.savefig(img_bytes)
img_bytes.seek(0)
return send_file(img_bytes, mimetype='image/png')
where you supply draw()
. Take a look at my example, though, since there are some cache headers you'll want to add.
Hi! Thanks for the reply. I still don't see how could I send the parameters through the route.
– Luciano
Nov 12 at 20:23
I added example code for that above.
– Dave W. Smith
Nov 12 at 20:45
Thanks! I'll give it a try. I've also modified my original question in order to make it more clear.
– Luciano
Nov 12 at 20:51
add a comment |
up vote
-1
down vote
up vote
-1
down vote
To pass parameters in, it's best to not prefix them to the URI, since that might confuse Flask's route decoding. Instead, if you form the URL like
/plot.png?x=1,2,3,4&y=100,200,300,400
you can pick the args up via request.args
. They'll come in as strings, so you'll need to seperate and convert to ints (or floats, if you're using real numbers)
from flask import request
@app.route('/plot.png')
def plot():
try:
x_values = [int(x) for x in request.args.get('x','').split(',')]
y_values = [int(y) for y in request.args.get('y','').split(',')]
except:
# malformed
I have a working example here of a robust way to handle matplotlib dynamically. The general idea is
from flask import send_file
@app.route('/plot.png')
def plot_png():
fig, ax = plt.subplots()
draw(ax)
return fig_response(fig)
def fig_response(fig):
"""Turn a matplotlib Figure into Flask response"""
img_bytes = BytesIO()
fig.savefig(img_bytes)
img_bytes.seek(0)
return send_file(img_bytes, mimetype='image/png')
where you supply draw()
. Take a look at my example, though, since there are some cache headers you'll want to add.
To pass parameters in, it's best to not prefix them to the URI, since that might confuse Flask's route decoding. Instead, if you form the URL like
/plot.png?x=1,2,3,4&y=100,200,300,400
you can pick the args up via request.args
. They'll come in as strings, so you'll need to seperate and convert to ints (or floats, if you're using real numbers)
from flask import request
@app.route('/plot.png')
def plot():
try:
x_values = [int(x) for x in request.args.get('x','').split(',')]
y_values = [int(y) for y in request.args.get('y','').split(',')]
except:
# malformed
I have a working example here of a robust way to handle matplotlib dynamically. The general idea is
from flask import send_file
@app.route('/plot.png')
def plot_png():
fig, ax = plt.subplots()
draw(ax)
return fig_response(fig)
def fig_response(fig):
"""Turn a matplotlib Figure into Flask response"""
img_bytes = BytesIO()
fig.savefig(img_bytes)
img_bytes.seek(0)
return send_file(img_bytes, mimetype='image/png')
where you supply draw()
. Take a look at my example, though, since there are some cache headers you'll want to add.
edited Nov 12 at 20:43
answered Nov 12 at 19:06
Dave W. Smith
16k22430
16k22430
Hi! Thanks for the reply. I still don't see how could I send the parameters through the route.
– Luciano
Nov 12 at 20:23
I added example code for that above.
– Dave W. Smith
Nov 12 at 20:45
Thanks! I'll give it a try. I've also modified my original question in order to make it more clear.
– Luciano
Nov 12 at 20:51
add a comment |
Hi! Thanks for the reply. I still don't see how could I send the parameters through the route.
– Luciano
Nov 12 at 20:23
I added example code for that above.
– Dave W. Smith
Nov 12 at 20:45
Thanks! I'll give it a try. I've also modified my original question in order to make it more clear.
– Luciano
Nov 12 at 20:51
Hi! Thanks for the reply. I still don't see how could I send the parameters through the route.
– Luciano
Nov 12 at 20:23
Hi! Thanks for the reply. I still don't see how could I send the parameters through the route.
– Luciano
Nov 12 at 20:23
I added example code for that above.
– Dave W. Smith
Nov 12 at 20:45
I added example code for that above.
– Dave W. Smith
Nov 12 at 20:45
Thanks! I'll give it a try. I've also modified my original question in order to make it more clear.
– Luciano
Nov 12 at 20:51
Thanks! I'll give it a try. I've also modified my original question in order to make it more clear.
– Luciano
Nov 12 at 20:51
add a comment |
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