finding and replacing elements in a list

Question

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?<

Answer 1

You can simply use list comprehension in python:

def replace_element(YOUR_LIST, set_to=NEW_VALUE):
    return [i
            if SOME_CONDITION
            else NEW_VALUE
            for i in YOUR_LIST]

for your case, where you want to replace all occurrences of 1 with 10, the code snippet will be like this:

def replace_element(YOUR_LIST, set_to=10):
    return [i
            if i != 1  # keeps all elements not equal to one
            else set_to  # replaces 1 with 10
            for i in YOUR_LIST]

Answer 2

I know this is a very old question and there's a myriad of ways to do it. The simpler one I found is using numpy package.

import numpy

arr = numpy.asarray([1, 6, 1, 9, 8])
arr[ arr == 8 ] = 0 # change all occurrences of 8 by 0
print(arr)

Answer 3

List comprehension works well, and looping through with enumerate can save you some memory (b/c the operation's essentially being done in place).

There's also functional programming. See usage of map:

>>> a = [1,2,3,2,3,4,3,5,6,6,5,4,5,4,3,4,3,2,1]
>>> map(lambda x: x if x != 4 else 'sss', a)
[1, 2, 3, 2, 3, 'sss', 3, 5, 6, 6, 5, 'sss', 5, 'sss', 3, 'sss', 3, 2, 1]

Answer 4

This can be easily done by using enumerate function

code-

lst=[1,2,3,4,1,6,7,9,10,1,2]
for index,item in enumerate(lst):
    if item==1:
        lst[index]=10 #Replaces the item '1' in list with '10'

print(lst)