In-place replacement of all occurrences of an element in a list in python

python list replace in-place
lifebalance picture lifebalance · Jun 13, 2014 · Viewed 25.5k times · Source

Assume I have a list:

myl = [1, 2, 3, 4, 5, 4, 4, 4, 6]

What is the most efficient and simplest pythonic way of in-place (double emphasis) replacement of all occurrences of 4 with 44?

I'm also curious as to why there isn't a standard way of doing this (especially, when strings have a not-in-place replace method)?

Answer

user2357112 supports Monica picture user2357112 supports Monica · Jun 13, 2014 
myl[:] = [x if x != 4 else 44 for x in myl]

Perform the replacement not-in-place with a list comprehension, then slice-assign the new values into the original list if you really want to change the original.

Related questions

finding and replacing elements in a list

I have to search through a list and replace all occurrences of one element with another. So far my attempts in code are getting me nowhere, what is the best way to do this? For example, suppose my list has …

Read More
Replace values in list using Python

I have a list where I want to replace values with None where condition() returns True. [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] For example, if condition checks bool(item%2) should return: [None, 1, None, 3, None, 5, None, 7, None, 9, None] What is the most efficient way to do this?

Read More
Removing a list of characters in string

I want to remove characters in a string in python: string.replace(',', '').replace("!", '').replace(":", '').replace(";", '')... But I have many characters I have to remove. I thought about a list list = [',', '!', …

Read More