python iterate list and substring

Here is a simplified version of your question, illustrating what I think you're looking for based on the pre-edit question, and using the full alphabet instead of DNA to make positions clearer. (Please see the help files regarding how to write a useful minimal example.)

dContigData = {
    "chromo_1": "abcdefghij",
    "chromo_2": "ABCDEFGHIJ"
}

e = {
    "mgg_1": ["chromo_1", 2, 4, 7, 9],
    "mgg_2": ["chromo_2", 1, 5, 8, 10]
}

Desired output:

mgg_1
bcd...ghi
mgg_2
ABCDE...HIJ

If this is what you mean, this Python 3 code will produce that output. Note that dictionary keys are not in any particular order. You may prefer to use a list of lists for e, rather than a dictionary of lists, since you seem to only be iterating over it anyway.

for mgg in sorted(e):
    lst = e[mgg]
    chrom = lst[0]
    substrings = []
    for i in range(1, len(lst), 2):
        startpos, endpos = lst[i:i+2]
        substrings.append(dContigData[chrom][startpos-1:endpos])
    print("{}\n{}".format(mgg, "".join(substrings)))

If I understand your question correctly, this should do it:

for key, value in e.items():
    print(
        key,
        dContigData[value[0]][value[1]-1:value[-1]]
    )

A more general solution:

import itertools


def group(lst, n):
    """Group an iterable into an n-tuples iterable. Incomplete tuples
    are discarded e.g.

    >>> list(group(range(10), 2))
    [(0, 1), (2, 3), (4, 5), (6, 7), (8, 9)]
    >>> list(group(range(10), 3))
    [(0, 1, 2), (3, 4, 5), (6, 7, 8)]
    """
    return itertools.izip(*[itertools.islice(lst, i, None, n)
                          for i in range(n)])


for key in e:
    sub_str_list = []
    contigID = e[key][0]
    for start, end in group(e[key][1:], 2):
        sub_str_list.append(dContigData[contigID][start-1:end])
    print(contigID, '...'.join(sub_str_list))

Based on your edits, I think this should do what you need. I have used Tom's simplified version of your question which was much more explicit to explain things

dContigData = {
    "chromo_1": "abcdefghij",
    "chromo_2": "ABCDEFGHIJ"
}

e = {
    "mgg_1": ["chromo_1", 2, 4, 7, 9],
    "mgg_2": ["chromo_2", 1, 5, 8, 10]
}

# Iterate over the items (keys/values) of e dictionary
for key, value in e.items():
    # Store in a variable for easier understanding
    string = dContigData[value[0]]
    # Get a list of tuples of (start, end) positions for the substrings
    # Example for mgg_1: zip([2,7], [4,9]) = [(2,4), (7,9)] 
    subPositions = zip(value[1::2], value[2::2])
    # Join the substrings for all these pairs
    # (most efficient string concatenation)
    res = ''.join([string[val[0]-1:val[1]] for val in subPositions])
    print key
    print res

Output:

mgg_2
ABCDEHIJ
mgg_1
bcdghi

This won't guarantee the order of iteration so if this is somethings important to you, you can simply use a sorted iterator iter(sorted(e.items()))

"Any idea to substring a string and then concatenate back to a string while looping?"

don't know if i understood your question, but

# sep = "..."
for key in e:
    for contigID in dContigData:
        if e[key][0] == contigID:
            dnaSeq = ''
            starts = [x-1 for x in e[key][1::2]]
            ends =  e[key][2::2]
            for i in range(len(starts)):
                dnaSeq += dContigData[contigID][starts[i]:ends[i]]
                #if i<len(starts)-1:
                #   dnaSeq += sep 
            print (key, '\n', dnaSeq)

should bring you the supposed result.

Update: Taking into account your last edits, you might skip the 'sep' steps and you'll get dnaSeq consisting of the pieces without any separators between the parts.