resize ipython notebook output window

Define a maximum size. Then, compute a resize ratio by taking min(maxwidth/width, maxheight/height).

The proper size is oldsize*ratio.

There is of course also a library method to do this: the method Image.thumbnail.
Below is an (edited) example from the PIL documentation.

import os, sys
import Image

size = 128, 128

for infile in sys.argv[1:]:
    outfile = os.path.splitext(infile)[0] + ".thumbnail"
    if infile != outfile:
        try:
            im = Image.open(infile)
            im.thumbnail(size, Image.ANTIALIAS)
            im.save(outfile, "JPEG")
        except IOError:
            print "cannot create thumbnail for "%s"" % infile

This script will resize an image (somepic.jpg) using PIL (Python Imaging Library) to a width of 300 pixels and a height proportional to the new width. It does this by determining what percentage 300 pixels is of the original width (img.size[0]) and then multiplying the original height (img.size[1]) by that percentage. Change "basewidth" to any other number to change the default width of your images.

from PIL import Image

basewidth = 300
img = Image.open("somepic.jpg")
wpercent = (basewidth/float(img.size[0]))
hsize = int((float(img.size[1])*float(wpercent)))
img = img.resize((basewidth,hsize), Image.ANTIALIAS)
img.save("somepic.jpg")

If you wish to use CV2, you need to use the resize function.

For example, this will resize both axes by half:

small = cv2.resize(image, (0,0), fx=0.5, fy=0.5) 

and this will resize the image to have 100 cols (width) and 50 rows (height):

resized_image = cv2.resize(image, (100, 50)) 

Another option is to use scipy module, by using:

small = scipy.misc.imresize(image, 0.5)

There are obviously more options you can read in the documentation of those functions (cv2.resize, scipy.misc.imresize).

Update:
According to the SciPy documentation:

imresize is deprecated in SciPy 1.0.0, and will be removed in 1.2.0.
Use skimage.transform.resize instead.

Note that if you"re looking to resize by a factor, you may actually want skimage.transform.rescale.

Locale unaware

"{:,}".format(value)  # For Python ‚â•2.7
f"{value:,}"  # For Python ‚â•3.6

Locale aware

import locale
locale.setlocale(locale.LC_ALL, "")  # Use "" for auto, or force e.g. to "en_US.UTF-8"

"{:n}".format(value)  # For Python ‚â•2.7
f"{value:n}"  # For Python ‚â•3.6

Reference

Per Format Specification Mini-Language,

The "," option signals the use of a comma for a thousands separator. For a locale aware separator, use the "n" integer presentation type instead.

I got this to work:

>>> import locale
>>> locale.setlocale(locale.LC_ALL, "en_US")
"en_US"
>>> locale.format("%d", 1255000, grouping=True)
"1,255,000"

Sure, you don"t need internationalization support, but it"s clear, concise, and uses a built-in library.

P.S. That "%d" is the usual %-style formatter. You can have only one formatter, but it can be whatever you need in terms of field width and precision settings.

P.P.S. If you can"t get locale to work, I"d suggest a modified version of Mark"s answer:

def intWithCommas(x):
    if type(x) not in [type(0), type(0L)]:
        raise TypeError("Parameter must be an integer.")
    if x < 0:
        return "-" + intWithCommas(-x)
    result = ""
    while x >= 1000:
        x, r = divmod(x, 1000)
        result = ",%03d%s" % (r, result)
    return "%d%s" % (x, result)

Recursion is useful for the negative case, but one recursion per comma seems a bit excessive to me.

my_list = ["a", "b", "c", "d"]
my_string = ",".join(my_list)

"a,b,c,d"

This won"t work if the list contains integers

And if the list contains non-string types (such as integers, floats, bools, None) then do:

my_string = ",".join(map(str, my_list))