Tag: str-replace

Is there a method similar to str.replace which can do the following:

>> replace(sequence=[0,1,3], old=[0,1], new=[1,2]) 
[1,2,3]

It should really act like str.replace : replacing a “piece” of a sequence by another sequence, not map elements of “old” with “new” ‘s ones.
Thanks 🙂

Solution:

No, I’m afraid there is no built-in function that does this, however you can create your own!

The steps are really easy, we just need to slide a window over the list where the width of the window is the len(old). At each position, we check if the window == to old and if it is, we slice before the window, insert new and concatenate the rest of the list on after – this can be done simply be assigning directly to the old slice as pointed out by @OmarEinea.

def replace(seq, old, new):
    seq = seq[:]
    w = len(old)
    i = 0
    while i < len(seq) - w + 1:
        if seq[i:i+w] == old:
            seq[i:i+w] = new
            i += len(new)
        else:
            i += 1
    return seq

and some tests show it works:

>>> replace([0, 1, 3], [0, 1], [1, 2])
[1, 2, 3]
>>> replace([0, 1, 3, 0], [0, 1], [1, 2])
[1, 2, 3, 0]
>>> replace([0, 1, 3, 0, 1], [0, 1], [7, 8])
[7, 8, 3, 7, 8]
>>> replace([1, 2, 3, 4, 5], [1, 2, 3], [1, 1, 2, 3])
[1, 1, 2, 3, 4, 5]
>>> replace([1, 2, 1, 2], [1, 2], [3])
[3, 3]

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As pointed out by @user2357112, using a for-loop leads to re-evaluating replaced sections of the list, so I updated the answer to use a while instead.