Setting an image at center of figure using matplotlib Figure.figimage

Question

I'm trying to set an image, a watermark essentially, at the center of a matplotlib Figure. I would like the center of the image to be at the center of the figure. The problem appears to be a difference between the figure size and dpi of 10" x 10" and 960 x 960 pixels and the saved image which has pixel dimensions of 610 x 623. When I add the image via Figure.figimage at 960/2, 960/2, that is offcenter of the finished image. My code and example image are below.

Creating the plot:

f, ax = plt.subplots(1, figsize=(10,10))
f.set_facecolor((1,1,1))
ax.set_title('{} Reflection Integrated Intensity vs Max Intensity of {}'.format(refl, rcmap.sample_number), size=14)
f = add_logo(f, '/home/zeppelin_d/Desktop/company.com_logo.png',x_frac=0.5, y_frac=0.5, scale=0.5, alpha=0.15)
ax = sns.regplot(x='z', y='int_int', data=df, scatter_kws={'alpha':0.4})
ax.set_xlabel('{} Max Intensity'.format(refl), size=14)
ax.set_ylabel('{} Integrated Intensity'.format(refl, size=14))
ax.set_xlim([0,max(df['z']) * 1.05])
ax.set_ylim([0,max(df['int_int']) * 1.05])

and adding the image:

def add_logo(f, path, x_frac=0.5, y_frac=0.5, scale=1, alpha=1):
    """
    Add an image to the figure (not the axes)
    f: a matplotlib figure instance.
    path: the string path to the image to add to the figure.
    x_frac: the fraction of the x dimension of the figure to set the offset to.
        Must be a float.
    y_frac: the fraction of the y dimension of the figure to set the offset to.
        Must be a float.
    scale: the float scale by which to multiply to the image pixel dimensions.
    alpha: the alpha to set the inserted image to

    Set the figure dpi to the same as screen dpi.

    Use this like:
    f = add_logo(f, 'mah_business_logo.png',x_frac=0.5, y_frac=0.5, scale=0.5, alpha=0.15)
    for setting a watermark. This should put the center of the image in the center of the
    figure at half it's original size.
    """
    assert type(x_frac) == float and type(y_frac) == float,  "x_frac and y_frac must be floats."
    im = Image.open(path)
    f.set_dpi(96)
    im.thumbnail((int(im.size[0] * scale), int(im.size[1] * scale)), Image.ANTIALIAS)
    img_x, img_y = im.size[0], im.size[1]
    x_offset = int((f.bbox.xmax * x_frac - img_x/2))
    y_offset = int((f.bbox.ymax * y_frac - img_y/2))
    f.figimage(im, xo=x_offset, yo=y_offset, origin='upper', zorder=10, alpha=alpha)
    return f

How can I know the dimensions of the saved image at any scale? Or why is the finished image not actually 10x10" and 960x960pixels? Thanks.

Answer 1

In order to get a 960x960 pixel figure out of a 10"x10" plot you also have to set the dpi keyword in plt.savefig() to 96 dpi. With eg:

plt.savefig('im.png', dpi=96)

The output is correct:

While with eg 72 dpi the result is: