how to save a pylab figure into in-memory file which can be read into PIL image?

Question:

The following is my first shot which never works:

import cStringIO
import pylab
from PIL import Image
pylab.figure()
pylab.plot([1,2])
pylab.title("test")
buffer = cStringIO.StringIO()
pylab.savefig(buffer, format='png')
im = Image.open(buffer.read())
buffer.close()

the error says,

Traceback (most recent call last):
  File "try.py", line 10, in <module>
    im = Image.open(buffer.read())
  File "/awesomepath/python2.7/site-packages/PIL/Image.py", line 1952, in open
    fp = __builtin__.open(fp, "rb")

any ideas? I don’t want the solution to involve extra packages.

Asked By: nye17

||

Answers:

Remember to call buf.seek(0) so Image.open(buf) starts reading from the
beginning of the buf:

import io
from PIL import Image
import matplotlib.pyplot as plt

plt.figure()
plt.plot([1, 2])
plt.title("test")
buf = io.BytesIO()
plt.savefig(buf, format='png')
buf.seek(0)
im = Image.open(buf)
im.show()
buf.close()
Answered By: unutbu

I like having it encapsulated in a function:

def fig2img(fig):
    """Convert a Matplotlib figure to a PIL Image and return it"""
    import io
    buf = io.BytesIO()
    fig.savefig(buf)
    buf.seek(0)
    img = Image.open(buf)
    return img

Then I can call it easily this way:

import numpy as np
import matplotlib.pyplot as plt
from PIL import Image

x = np.arange(-3,3)
plt.plot(x)
fig = plt.gcf()

img = fig2img(fig)
img.show()
Answered By: kotchwane