Question

How do replace a value in a list?

Hello all,

I am currently working on a programming project and am having a mind blank as far as changing the value of a list. I am attempting to implement bubble sort into a program.

Any help would be useful.

Thank you, and have a good day.

Show comments

Submit an answer

This textbox defaults to using Markdown to format your answer.

You can type !ref in this text area to quickly search our full set of tutorials, documentation & marketplace offerings and insert the link!

Sign In or Sign Up to Answer

These answers are provided by our Community. If you find them useful, show some love by clicking the heart. If you run into issues leave a comment, or add your own answer to help others.

Want to learn more? Join the DigitalOcean Community!

Join our DigitalOcean community of over a million developers for free! Get help and share knowledge in Q&A, subscribe to topics of interest, and get courses and tools that will help you grow as a developer and scale your project or business.

@stiltonwalrus432

@MattIPv4 is right, we need a little more detail unless that cut/paste from stack over flow above helps you…

This comment has been deleted

Build a new list with a list comprehension:

new_items = [x if x % 2 else None for x in items]

You can modify the original list in-place if you want, but it doesn’t actually save time:

items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
for index, item in enumerate(items):
    if not (item % 2):
        items[index] = None

Here are (Python 3.6.3) timings demonstrating the non-timesave:

In [1]: %%timeit
   ...: items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
   ...: for index, item in enumerate(items):
   ...:     if not (item % 2):
   ...:         items[index] = None
   ...:
1.06 µs ± 33.7 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [2]: %%timeit
   ...: items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
   ...: new_items = [x if x % 2 else None for x in items]
   ...:
891 ns ± 13.6 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

And Python 2.7.6 timings:

In [1]: %%timeit
   ...: items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
   ...: for index, item in enumerate(items):
   ...:     if not (item % 2):
   ...:         items[index] = None
   ...: 
1000000 loops, best of 3: 1.27 µs per loop
In [2]: %%timeit
   ...: items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
   ...: new_items = [x if x % 2 else None for x in items]
   ...: 
1000000 loops, best of 3: 1.14 µs per loop