Create a list with initial capacity in Python

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Code like this often happens:

l = []
while foo:
    # baz
    l.append(bar)
    # qux

This is really slow if you"re about to append thousands of elements to your list, as the list will have to be constantly resized to fit the new elements.

In Java, you can create an ArrayList with an initial capacity. If you have some idea how big your list will be, this will be a lot more efficient.

I understand that code like this can often be refactored into a list comprehension. If the for/while loop is very complicated, though, this is unfeasible. Is there an equivalent for us Python programmers?

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Create a list with initial capacity in Python resize: Questions

How do I resize an image using PIL and maintain its aspect ratio?

5 answers

saturdayplace By saturdayplace

Is there an obvious way to do this that I"m missing? I"m just trying to make thumbnails.

526

Answer #1

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

526

Answer #2

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")

Create a list with initial capacity in Python resize: Questions

How to resize an image with OpenCV2.0 and Python2.6

5 answers

I want to use OpenCV2.0 and Python2.6 to show resized images. I used and adopted this example but unfortunately, this code is for OpenCV2.1 and does not seem to be working on 2.0. Here my code:

import os, glob
import cv

ulpath = "exampleshq/"

for infile in glob.glob( os.path.join(ulpath, "*.jpg") ):
    im = cv.LoadImage(infile)
    thumbnail = cv.CreateMat(im.rows/10, im.cols/10, cv.CV_8UC3)
    cv.Resize(im, thumbnail)
    cv.NamedWindow(infile)
    cv.ShowImage(infile, thumbnail)
    cv.WaitKey(0)
    cv.DestroyWindow(name)

Since I cannot use

cv.LoadImageM

I used

cv.LoadImage

instead, which was no problem in other applications. Nevertheless, cv.iplimage has no attribute rows, cols or size. Can anyone give me a hint, how to solve this problem?

194

Answer #1

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.

We hope this article has helped you to resolve the problem. Apart from Create a list with initial capacity in Python, check other resize-related topics.

Want to excel in Python? See our review of the best Python online courses 2022. If you are interested in Data Science, check also how to learn programming in R.

By the way, this material is also available in other languages:



Chen Innsbruck

Shanghai | 2022-11-28

StackOverflow is always a bit confusing 😭 Create a list with initial capacity in Python is not the only problem I encountered. I just hope that will not emerge anymore

Davies Schteiner

Massachussetts | 2022-11-28

Simply put and clear. Thank you for sharing. Create a list with initial capacity in Python and other issues with StackOverflow was always my weak point 😁. Will get back tomorrow with feedback

Davies Krasiko

Shanghai | 2022-11-28

I was preparing for my coding interview, thanks for clarifying this - Create a list with initial capacity in Python in Python is not the simplest one. Will use it in my bachelor thesis

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