I just started to study Python and I am stuck at this one.
basically I would like to find out the add numbers in the odd index number.
here is my code.
def odd_ones(lst):
total = []
for i in lst:
if i % 2 == 1:
total.append(i)
return total
print(odd_ones([1,2,3,4,5,6,7,8]))
Output is
[1, 3, 5, 7] instead of [2, 4, 6, 8]
can someone please help me with this?
The output is correct. You iterate over the list of values and not its indices. Condition i % 2 == 1 gives following:
1 % 2 = 1 (true)
2 % 2 = 0 (false)
3 % 2 = 1 (true)
4 % 2 = 0 (false)
5 % 2 = 1 (true)
6 % 2 = 0 (false)
7 % 2 = 1 (true)
8 % 2 = 0 (false)
So the output is (1,3,5,7)
You want to find the odd inedx ,but what you really do is to find the odd element
for i in lst: #(i ---->the element in lst)
if i % 2 == 1:
so you should try this
for i in range(len(lst)): #( i ---> the index of lst)
if i % 2 == 1:
if you wont to get the odd number into your array you need to change your condition, so the code most be like that:
def odd_ones(lst):
total = []
for i in lst:
if i % 2 == 0:
total.append(i)
return total
print(odd_ones([1,2,3,4,5,6,7,8]))
output:[2, 4, 6, 8]
as required odd index number, enumerate provides counter/index
def odd_ones_index(lst):
total = []
for x,i in enumerate(lst):
if i % 2 == 1: ## checking i is odd or not
total.append(x) ## appending index as you want index
return total
print(odd_ones_index([1,2,3,4,5,6,7,8]))
for i in lstis iterating over the elements, not the indices. You needfor i, x in enumerate(lst): if i % 2 == 1: total.append(x)for i in range(len(lst))would be enough.lstandfor i in range(len(lst)): # anything with lst[i]is a huge code smell.for i, _would be bad practice. Wait... I think you edited?for i, x in enumerate(lst)looks good enough. If that was what you wrote in the first place, my apologies.