How to store an image in a variable

Question

I would like to store the image generated by matplotlib in a variable raw_data to use it as inline image.

import os
import sys
os.environ['MPLCONFIGDIR'] = '/tmp/'
import matplotlib
matplotlib.use("Agg")
import matplotlib.pyplot as plt

print "Content-type: image/png\n"
plt.plot(range(10, 20))

raw_data = plt.show()

if raw_data:
    uri = 'data:image/png;base64,' + urllib.quote(base64.b64encode(raw_data))
    print '<img src = "%s"/>' % uri
else:
    print "No data"

#plt.savefig(sys.stdout, format='png')

None of the functions suit my use case:

plt.savefig(sys.stdout, format='png') - Writes it to stdout. This does help.. as I have to embed the image in a html file.
plt.show() / plt.draw() does nothing when executed from command line

Paul · Accepted Answer · 2011-03-15 19:40:16Z

33

Have you tried cStringIO or an equivalent?

import os
import sys
import matplotlib
import matplotlib.pyplot as plt
import StringIO
import urllib, base64

plt.plot(range(10, 20))
fig = plt.gcf()

imgdata = StringIO.StringIO()
fig.savefig(imgdata, format='png')
imgdata.seek(0)  # rewind the data

print "Content-type: image/png\n"
uri = 'data:image/png;base64,' + urllib.quote(base64.b64encode(imgdata.buf))
print '<img src = "%s"/>' % uri

edited Mar 15, 2011 at 19:40

answered Mar 15, 2011 at 16:31

Paul

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5 Comments

Ramya Over a year ago

Thanks.. That helped... I was hoping there would be a way to directly get the image instead of using a file handle..

Paul Over a year ago

@Ramya: StringIO does not use filehandles. It is all in-memory storage and there is no OS limit to the number of StringIO instances: stackoverflow.com/questions/1177230/…

arno_v Over a year ago

Don't forget to unquote the string before decoding, because I got some 'Incorrect padding' errors because of the url encoding.

metaperture Over a year ago

FYI, in Python 3 you'll need to use io.BytesIO, eg: buf = io.BytesIO(); plt.gcf().savefig(buf, format='png'); buf.seek(0); return base64.b64encode(buf.read())

CMCDragonkai Over a year ago

Instead of a plot, I have a 2d matrix. How do I return it as JPG image data as a variable? I was hoping imsave had a "return" kind of option.

dgmp88 · Accepted Answer · 2017-07-14 09:48:25Z

23

Complete python 3 version, putting together Paul's answer and metaperture's comment.

import matplotlib.pyplot as plt
import io
import urllib, base64

plt.plot(range(10))
fig = plt.gcf()

buf = io.BytesIO()
fig.savefig(buf, format='png')
buf.seek(0)
string = base64.b64encode(buf.read())

uri = 'data:image/png;base64,' + urllib.parse.quote(string)
html = '<img src = "%s"/>' % uri

answered Jul 14, 2017 at 9:48

dgmp88

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1 Comment

syntheso Over a year ago

Hi, is there any way around the very long urls?

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