Dynamically changing a list while looping

user_17066023 0
Viewed 91

I have a list:

nums = [0, 1, 2, 3]

what I would like to do is loop though this list but then change the list. Basically when I loop through nums = [0, 1, 2, 3] nums would need to change to:

nums = [current_element, 0, 1, 2]
nums = [0, current_element, 1, 2]
nums = [0, 1, current_element, 2]
nums = [0, 1, 2, current_element]

Is there a way of changing nums like this? I feel like there is a simple solution to this, but I've been stumped for hours.

arrays python
4 Answers

You can pass a map() object to the for loop that generates the list you need at each iteration. For each index, we generate a new list using list slicing to replace the element at that index:

current_element = 42
nums = [0, 1, 2, 3]

for lst in map(lambda x: nums[:x] + [current_element] + nums[x+1:], range(len(nums))):
    print(lst)

This outputs:

[42, 1, 2, 3]
[0, 42, 2, 3]
[0, 1, 42, 3]
[0, 1, 2, 42]

This approach has two notable advantages:

  1. This doesn't mutate the original list. If you need to access the original list for any reason, you can.
  2. You don't need to worry about resetting values at the start/end of each iteration. If you need to implement this pattern several times, you don't need to remember to replace the values each time.
user_177698150

Yes, there's a simple solution.

for i in range(len(nums)):
    nums[i] = current_element
    do_processing(nums)
    nums[i] = i

The other way, of course, is just to make a copy of the list.

for i in range(len(nums)):
    n1 = nums[:]
    n1[i] = current_element
    do_processing(n1)
user_18833160

No fancy functions, just a for loop and a temporary variable.

for i in range(len(nums)):
  old_value = nums[i]
  nums[i] = current_element
  ...
  nums[i] = old_value
user_22886590 
nums.pop()
for i in range(len(nums)+1):
  nums.insert(i,current_element)
  print(nums)
  nums.remove(current_element)
user_142402480
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