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Danish Saleem
7,965 Pointscombo challenge
Hello ,
solved this problem using the following code
def combo(iter1,iter2): output = [] for index,item in enumerate(iter2): a = (iter1[index],iter2[index]) output.append(a) return output
however i am wondering if there is a way to use *args as a parameter in combo function than passing separate two iter parameters
# combo([1, 2, 3], 'abc')
# Output:
# [(1, 'a'), (2, 'b'), (3, 'c')]
4 Answers
Ari Misha
19,323 PointsHiya Danish! This might help you to understand the question and the logic:
def combo(x, y):
new_list = []
count = 0
for i in x:
result = x[count], y[count]
new_list.append(result)
count += 1
return new_list
Alfonso Lopez
6,089 PointsI used the following, which will return the desired result, but it does not pass the task for some reason.
def combo(*args):
return zip(*args)
print combo('abc', 'def')
# output: [('a', 'd'), ('b', 'e'), ('c', 'f')]
Herman Brummer
6,414 Pointsdef combo(input1, input2):
counter = 0
new_list = []
for item in input1:
x = item, str(input2[counter])
new_list.append(x)
counter = counter + 1
return new_list
I am trying the following, it has the correct output, but not accepted.
sandeep Maheshwari
443 Pointsdef combo(a, b): final = [] items_in_a = [] items_in_b = [] for item in a: items_in_a.append(item) for item in b: items_in_b.append(item) x = 0 for item in range(len(a)): final.append((a[x], b[x])) x += 1 return final