Merge some list items in a Python List

On what basis should the merging take place? Your question is rather vague. Also, I assume a, b, ..., f are supposed to be strings, that is, 'a', 'b', ..., 'f'.

>>> x = ['a', 'b', 'c', 'd', 'e', 'f', 'g']
>>> x[3:6] = [''.join(x[3:6])]
>>> x
['a', 'b', 'c', 'def', 'g']

Check out the documentation on sequence types, specifically on mutable sequence types. And perhaps also on string methods.

That example is pretty vague, but maybe something like this?

items = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']
items[3:6] = [''.join(items[3:6])]

It basically does a splice (or assignment to a slice) operation. It removes items 3 to 6 and inserts a new list in their place (in this case a list with one item, which is the concatenation of the three items that were removed.)

For any type of list, you could do this (using the + operator on all items no matter what their type is):

items = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']
items[3:6] = [reduce(lambda x, y: x + y, items[3:6])]

This makes use of the reduce function with a lambda function that basically adds the items together using the + operator.

just a variation

alist=["a", "b", "c", "d", "e", 0, "g"]
alist[3:6] = [''.join(map(str,alist[3:6]))]
print alist

Of course @Stephan202 has given a really nice answer. I am providing an alternative.

def compressx(min_index = 3, max_index = 6, x = ['a', 'b', 'c', 'd', 'e', 'f', 'g']):
    x = x[:min_index] + [''.join(x[min_index:max_index])] + x[max_index:]
    return x
compressx()

>>>['a', 'b', 'c', 'def', 'g']

You can also do the following.

x = x[:min_index] + [''.join(x[min_index:max_index])] + x[max_index:]
print(x)

>>>['a', 'b', 'c', 'def', 'g']

my telepathic abilities are not particularly great, but here is what I think you want:

def merge(list_of_strings, indices):
    list_of_strings[indices[0]] = ''.join(list_of_strings[i] for i in indices)
    list_of_strings = [s for i, s in enumerate(list_of_strings) if i not in indices[1:]]
    return list_of_strings

I should note, since it might be not obvious, that it's not the same as what is proposed in other answers.