Replacing a sublist with another sublist in python

In [39]: a=[1,3,5,10,13]

In [40]: sub_list_start = 1

In [41]: sub_list_end = 3

In [42]: a[sub_list_start : sub_list_end+1] = [9,7]

In [43]: a
Out[43]: [1, 9, 7, 13]

Hope that helps

You can do this nicely with list slicing:

>>> a=[1, 3, 5, 10, 13]
>>> a[1:4] = [9, 7]
>>> a
[1, 9, 7, 13]

So how do we get the indices here? Well, let's start by finding the first one. We scan item by item until we find a matching sublist, and return the start and end of that sublist.

def find_first_sublist(seq, sublist, start=0):
    length = len(sublist)
    for index in range(start, len(seq)):
        if seq[index:index+length] == sublist:
            return index, index+length

We can now do our replacement - we start at the beginning, replace the first one we find, and then try to find another after our newly finished replacement. We repeat this until we can no longer find sublists to replace.

def replace_sublist(seq, sublist, replacement):
    length = len(replacement)
    index = 0
    for start, end in iter(lambda: find_first_sublist(seq, sublist, index), None):
        seq[start:end] = replacement
        index = start + length

Which we can use nicely:

>>> a=[1, 3, 5, 10, 13]
>>> replace_sublist(a, [3, 5, 10], [9, 7])
>>> a
[1, 9, 7, 13]

You need to take a slice from start_index to end_index + 1, and assign your sublist to it.

Just like you can do: - a[0] = 5, you can similarly assign a sublist to your slice: - a[0:5] -> Creates a slice from index 0 to index 4

All you need is to find out the position of the sublist you want to substitute.

>>> a=[1,3,5,10,13]

>>> b_sub = [9, 7]

>>> a[1:4] = [9,7]  # Substitute `slice` from 1 to 3 with the given list

>>> a
[1, 9, 7, 13]
>>> 

As you can see that, substituted sublist don't have to be of same length of the substituting sublist.

In fact you can replace, 4 length list with 2 length list and vice-versa.