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Start your free trialNelson Chuang
1,221 PointsDidn't get the right output. For example, expected (10, 'T') as the first item, got (10, 'T') instead.
looks right but error message seems strange
# 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(x,y):
mylist=[]
for a in x:
for b in y:
mylist.append((a,b))
return mylist
2 Answers
jcorum
71,830 PointsJennifer points out the issue. For each a in x you will get b tuples, so it looks like your total number of tuples will be a * b, a few more than expected.
Here's a second way to do it. One that uses enumerate():
def combo(iter1, iter2):
combo_list = []
for index, value in enumerate(iter1):
tuple = value, iter2[index]
combo_list.append(tuple)
return combo_list
Jennifer Nordell
Treehouse TeacherHi, no those should not be nested. Odd though that you get that error from the challenge about the first element given that the first element is the only piece that will be correct after running your code. I'll show you how I did it:
def combo(x, y):
myList = list()
for z in range(len(x)):
myList.append((x[z], y[z]))
return myList
I took the liberty of editing your code to get to the final solution. In the instructions it explicitly states that you can assume that the length of the two inputs will be the same. So I looked over the range for the length of x. I could have just as easily looked at the length of y given that we know they're the same. Then I go through that list and append the value of x at the z index and y at the value of z index. Then we return the resulting list. Hope that helps!
Nelson Chuang
1,221 PointsNelson Chuang
1,221 Pointsactually realized my two for loops should not be nested. I don't get how to do this one with what was taught.