Converting a list into comma-separated string with “and” before the last item - Python 2.7

Question

I have created this function to parse the list:

listy = [\'item1\', \'item2\',\'item3\',\'item4\',\'item5\', \'item6\']


def coma(abc):
    for i in abc[0:-         


        

Answer 1

Might as well round out the solutions with a recursive example.

>>> listy = ['item1', 'item2','item3','item4','item5', 'item6']
>>> def foo(a):
    if len(a) == 1:
        return ', and ' + a[0]
    return a[0] + ', ' + foo(a[1:])

>>> foo(listy)
'item1, item2, item3, item4, item5, , and item6'
>>> 

Answer 2

One more different way to do:

listy = ['item1', 'item2','item3','item4','item5', 'item6']

first way:

print(', '.join('and, ' + listy[item] if item == len(listy)-1 else listy[item]
for item in xrange(len(listy))))

output >>> item1, item2, item3, item4, item5, and, item6

second way:

print(', '.join(item for item in listy[:-1]), 'and', listy[-1])

output >>> (item1, item2, item3, item4, item5, 'and', 'item6')

Answer 3

Correction for Craig’s answer above for a 2-element list (I’m not allowed to comment):

def oxford_comma_join(l):
    if not l:
        return ""
    elif len(l) == 1:
        return l[0]
    elif len(l) == 2:
        return l[0] + " and " + l[1]
    else:
        return ', '.join(l[:-1]) + ", and " + l[-1]

print(oxford_comma_join(['item1', 'item2', 'item3', 'item4', 'item5', 'item6']))

print(oxford_comma_join(['i1', 'i2']))

Results:

item1, item2, item3, item4, item5, and item6
i1 and i2

Answer 4

I cannot take full credit but if you want succinct -- I modified RoadieRich's answer to use f-strings and also made it more concise. It uses the solution by RootTwo given in a comment on that answer:

def join(items):
    *start, last = items
    return f"{','.join(start)}, and {last}" if start else last