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

a = [1,2,3,4,5,1,2,3,4,5,1,12]
for i in range (len(a)):
    if a[i]==2:
        a[i]=123

You can use a for and or while loop; however if u know the builtin Enumerate function, then it is recommended to use Enumerate.1

Answer 2

>>> a=[1,2,3,4,5,1,2,3,4,5,1]
>>> item_to_replace = 1
>>> replacement_value = 6
>>> indices_to_replace = [i for i,x in enumerate(a) if x==item_to_replace]
>>> indices_to_replace
[0, 5, 10]
>>> for i in indices_to_replace:
...     a[i] = replacement_value
... 
>>> a
[6, 2, 3, 4, 5, 6, 2, 3, 4, 5, 6]
>>> 

Answer 3

To replace easily all 1 with 10 in a = [1,2,3,4,5,1,2,3,4,5,1]one could use the following one-line lambda+map combination, and 'Look, Ma, no IFs or FORs!' :

# This substitutes all '1' with '10' in list 'a' and places result in list 'c':

c = list(map(lambda b: b.replace("1","10"), a))

Answer 4

Here's a cool and scalable design pattern that runs in O(n) time ...

a = [1,2,3,4,5,6,7,6,5,4,3,2,1]

replacements = {
    1: 10,
    2: 20,
    3: 30,
}

a = [replacements.get(x, x) for x in a]

print(a)
# Returns [10, 20, 30, 4, 5, 6, 7, 6, 5, 4, 30, 20, 10]

Answer 5

>>> a= [1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1]
>>> for n, i in enumerate(a):
...   if i == 1:
...      a[n] = 10
...
>>> a
[10, 2, 3, 4, 5, 10, 2, 3, 4, 5, 10]

Answer 6

Find & replace just one item

your_list = [1,2,1]     # replace the first 1 with 11

loc = your_list.index(1)
your_list.remove(1)
your_list.insert(loc, 11)