here’s my solution
def same_values(lst1, lst2):
same = []
i = 0
if len(lst1) == len(lst2):
for num in lst1:
if lst1[i] == lst2[i]:
same.append(i)
i = i + 1
return same
here’s my solution
def same_values(lst1, lst2):
same = []
i = 0
if len(lst1) == len(lst2):
for num in lst1:
if lst1[i] == lst2[i]:
same.append(i)
i = i + 1
return same
Sorry - didn’t realize that’s a faux pas. Just wanted to document one method of completion, since I like looking at how others solve.
I can obscure the solutions.
Yes, that would be good idea. Please do so.
How to obscure solutions?
Use [details="name"]
to create a show/hide block. Close it with [/details]
.
Use [spoiler]
to convert code to a PNG that cannot be copied. Close it with [/spoiler]
def is_odd(x):
return x % 2 and True
[details="is_odd"] [spoiler] ``` def is_odd(x): return x % 2 and True ``` [/spoiler] [/details]
Please critique my solution. After a lot of tinkering, I managed to get it to work in a list comprehension:
def same_values(lst1, lst2):
return [lst1.index(x) for x, y in zip(lst1, lst2) if x == y]
list.index()
will always give the index of the first instance in the list. It won’t reach any other duplicates.