Welcome to the Treehouse Community
Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.
Looking to learn something new?
Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.
Start your free trial
Chris Komaroff
Courses Plus Student 14,198 PointsIs teacher's answer for Tuples with Functions challenge 1 of 1 correct? "Bummer expected (10,T)" is wrong?
Challenge 1 of 1 code comment:
combo([1, 2, 3], 'abc')
Output:
[(1, 'a'), (2, 'b'), (3, 'c')]
If you use .append(), you'll want to pass it a tuple of new values
That's what my function combo() gives. I tested in workspace. Not following this.
# combo([1, 2, 3], 'abc')
# Output:
# [(1, 'a'), (2, 'b'), (3, 'c')]
# If you use .append(), you'll want to pass it a tuple of new values.
def combo( iter1, iter2 ):
return_list = []
temp_dict = {}
list_len = len( iter1 )
i = 0
while i < list_len:
temp_dict[ iter1[i] ] = iter2[i]
i += 1
for key, value in temp_dict.items():
return_list.append( (key,value) )
return return_list
1 Answer
Chris Freeman
Treehouse Moderator 68,441 PointsThe ordering of items found when iterating on a dictionary can not be guaranteed. By constructing a temp_dict then reading key/value pairs out, you are scrambling the tuple order. Instead you can build add append the tuples directly
def combo( iter1, iter2 ):
return_list = []
list_len = len( iter1 )
i = 0
while i < list_len:
return_list.append((iter1[i], iter2[i]))
i += 1
return return_list
Chris Freeman
Treehouse Moderator 68,441 PointsChris Freeman
Treehouse Moderator 68,441 PointsThere are many ways to improve on this code:
Or....