FAQ: Code Challenge: Loops - Same Values

3 posts were merged into an existing topic: What is an index?

I had trouble with this initially because I indented the return with the first for loop.
it returned [0, 2, 3, 0, 2, 3, 0, 2, 3, 0, 2, 3, 0, 2, 3] because it repeated the command from the second for loop for the 5 index positions in lst1, correct?

3 posts were split to a new topic: The expected value is beyond the index?

2 posts were split to a new topic: Why do we only need the range for one list?

5 posts were split to a new topic: Code adds multiple copies of each number

4 posts were split to a new topic: Remember to check only index to index

is this the best way of doing it?
if not how can i improve it?

this returns the indices which match but before all that it checks which list is smaller and uses that as the range

def same_values(lst1,lst2):
  index = []
  length = range(len(lst1))
  length2 = range(len(lst2))
  if len(lst1) <= len(lst2):
    for i in length:
      if lst1[i] == lst2[i]:
    return index
    for i in length2:
      if lst1[i] == lst2[i]:
    return index

zipping the two lists resolves the uneven length concern by truncating the longer of the two lists. Values are paired up by their indices. 0:0, 1:1, 2:2, &c. Mind, the instructions do state two lists of numbers of equal size.

With a zipped list we only iterate a single object. Using enumerate we can track the index.

for i, x in enumerate(zip(lst1, lst2)):

x is a tuple, so we can unpack it…

    a, b = x
def same_values(lst1, lst2):
  y = []
  for i, x in enumerate(zip(lst1, lst2)):
    a, b = x
    if a == b:
  return y

The more naive approach is to keep track of the index with a count variable.

    count = 0
    y = []
    for x in zip(lst1, lst2):
        a, b = x
        if a == b:
        count += 1
    return y

We will most certainly come up with other naive approaches if one sticks with the problem awhile. Play it out and see how many you can come up with.

1 Like

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]:
      i = i + 1
    return same

@bitplayer30925 Man. Please stop posting the answers

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.

Hello i believe i have solved the code chalanges (loops) same values. But i can’t pass the subject
here is my code

#Write your function here
def same_val(lst1, lst2, res=[]) :
  fa = lst1[0]
  fb = lst2[0]
  ra = lst1[1:]
  rb = lst2[1:]
  r = []
  if fa == fb :
  if ra == [] :
    return res
  return same_val(ra,rb, res+r)

def same_values(lst1, lst2) : 
  res = same_val(lst1,lst2)
  result = []
  for n in res :
  return result
#Uncomment the line below when your function is done
print(same_values([5, 1, -10, 3, 3], [5, 10, -10, 3, 5]))

I’m using helper function same_val to recursively get the same values and then get the indicies in the same_values function. is this a bug or what? thanks

Hi all,

I figured out the for-loop solution but was trying to find and alternative using list comprehension… and failed. What’s wrong with this / what is the way? (error was list out of range). My sense is that the index object is looking for “i” rather than what I intend which is the first element of lst1 if it is the same as the first element of lst2.

svi = [lst1.index(i) for i in lst1 if lst1[i] == list2[i]]

Thanks for any help.

your variable i contains value from the list, so trying to access by index (lst1[i]) will throw an index error when there is a value in the list which is higher then the highest index (which is quite common)

i wouldn’t use the .index() method, the index method always gives the first match, which is problematic if one value occurs multiple times in a list

Thanks. Is there a way to use list comprehension to solve?

yes, there is. You could use range() to get the indexes/indices. That should help

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

def is_odd(x):
    return x % 2 and True
1 Like

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]