List2

Aladdin

Sep 16, 2019 15:21

How to reverse list elements

John Rennie

@Aladdin In place? Or into a new list?

Aladdin

In place

John Rennie

Suppose you have a list [a,b,c,d,e]

What you do is start at the end and work towards the centre swapping elements as you go.

Aladdin

I did

John Rennie

So we use a for loop for i in range (0,len(mylist)/2)

That goes from the left end to the middle.

I would set a new variable j = len(mylist) - i - 1

Aladdin

Sep 16, 2019 15:29

Lst=[1, 2,3,4]

John Rennie

Then you'll swap the elements [i] and [j].

Aladdin

for I in range (lena(lst)):

Lst[I]==Lst[-i-1]

print (lst)

@JohnRennie but this prints the same list...

John Rennie

Well let's go through it by hand. len(lst) is 4 so your four loop will execute 4 times with i set to 0, 1, 2 and 3. OK so far?

Aladdin

Ok

John Rennie

the second line is lst[i] = lst[-i-1]. Is that correct? On the right hand side you have a negative number as the index i.e. -i-1.

Did you mean lst[i] = lst[len(lst)-i-1] ?

Aladdin

Sep 16, 2019 15:34

Yes

John Rennie

Also, did you really mean == ?

That's the comparison operator not the assignment operator.

Aladdin

I meant =

John Rennie

So let's just check this. Your code is:

lst=[1,2,3,4]
for i in range(0,len(lst)):
    lst[i] = lst[len(lst)-i-1]
print(lst)

Is that definitely correct?

Aladdin

Lst[I]=Lst [-i-1]

My code had this

John Rennie

That can't be correct because -i-1 is a negative number. You can't have a negative number as a list index.

We need to agree on what code you are running, then you need to try running the code we've agreed on to see what happens.

Ah, my bad, Python does allow negative indexes.

Ignore what I said

A negative index means you count from the right end of the list.

@Aladdin OK. So the code is:

lst=[1,2,3,4]
for i in range(0,len(lst)):
    lst[i] = lst[-i-1]
print(lst)

Yes?

Aladdin

Sep 16, 2019 15:49

Yed

But the problem is it's not printing backwards

I thought it could...

John Rennie

OK. Let's start with lst=[1,2,3,4]. On our first pass through the loop i=0 so the command in the loop does lst[0] = lst[-1]. Yes?

Aladdin

Yes

John Rennie

That changes lst to [4,2,3,4]. OK so far?

Aladdin

Yed

Ah I can see my mistake

This won't work

John Rennie

Yes :-)

You need an intermediate variable to do the swap.

Aladdin

Sep 16, 2019 15:53

So we use a for loop for i in range (0,len(mylist)/2)
That goes from the left end to the middle.
I would set a new variable j = len(mylist) - i - 1

Why did u take my list/2

Can u explain ur code

John Rennie

Because you are swapping the left and right halves of the list, so you only need to work your way through half the list.

Let's take your list [1,2,3,4] and see how this works. The first swap is with i=0 so j = 3. That swaps elements 0 and 3 so the list is now: [4,2,3,1]. OK so far?

Aladdin

What if length of list is odd

Ok

John Rennie

If the length is odd we don't process the middle element. But the middle element won't move anyway because it's in the middle.

So len(lst)//2 always works.

(I forgot we need // to do integer division)

Aladdin

Yep

John Rennie

My full code is:

lst = [1,2,3,4]

for i in range(0,len(lst)//2):
    j = len(lst) - i - 1
    x = lst[i]
    lst[i] = lst[j]
    lst[j] = x

print(lst)

Do you want to try running that on your PC?

Aladdin

Sep 16, 2019 16:02

Yes . On it

Yep it's working

John Rennie

:-)

Aladdin

Tbh I think these codes are more maths than coding

John Rennie

They are teaching you how to analyse a problem and construct an algorithm to solve it.

Aladdin

Sep 16, 2019 16:26

@JohnRennie hi

John Rennie

@Aladdin hi

Aladdin

imgur.com/a/AnJcWT6

I need to add elements of l1 and l2 into p

Such that no elements repeat

Replace c with p

@JohnRennie hello?

John Rennie

@Aladdin hi, sorry I was busy for a moment.

Aladdin

Oh okay. Ping me when u are back

John Rennie

I'm free now!

The way I would do this is to first put all elements of l1 into c.

Then I would step through l2, and I would append the elements of l2 to c only if that element isn't in l1.

Aladdin

Sep 16, 2019 16:34

Ah right

No need to make it complicated like mine

John Rennie

So that's my algorithm, and it now just remains to turn it into code.

Aladdin

Mine was

To first put those elements which are not Same in both lists

And then put those which are equal

Guess my algorithm is right?

John Rennie

I'm not sure how your algorithm works. Your description of it is a bit unclear to me.

Aladdin

First take a list then put those elements which are not in other list

Take the other list and put those elements which are not in other list

Finally take the elements which are common

@JohnRennie is this ok

John Rennie

Ah, I see. Yes that will work though it seems a bit inefficient because you are having to do lots of checking of whether elements are in lists or not.

Aladdin

Sep 16, 2019 16:42

Figured so. But just for practice I am gonna apply both mine and urs

John Rennie

OK, yes that's good practice.

Aladdin

@JohnRennie little help

John Rennie

@Aladdin yes?

Aladdin

For my code, how do I take the value of index only once for a list if the element it represents is not common

I can't use if statement.... It's taking i multiple times

imgur.com/a/7VikH1M

John Rennie

One of the reasons people like Python is that it makes lots of things easy. In particular to check if a value is in a list you can just do if x in mylist:

So suppose you are getting all the elements of l1 that are not in l2. The way to do this is:

for i in range(len(l1)):
    if not l1[i] in l2:
        c.append(l1[i])

Do you see how that works?

Aladdin

Sep 16, 2019 16:56

Hoe does if not work

Is it similar to if?

John Rennie

Yes. The operator not just converts true to false and false to true.

So:

if 1 == 1:
    print("Equal!")
else:
    print("Not equal!")

That prints "Equal!" which I guess is obvious.

Aladdin

OK

John Rennie

If you stick a not in:

if not 1 == 1:
    print("Equal!")
else:
    print("Not equal!")

Aladdin

Not equal

John Rennie

That will then print "Not equal!.

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