Python list element patterns-CodePudding

I have a VERY simple question.

Honestly, I'm so sorry to ask you this stupid question but I really don't know how to do this.

I have a list like this.

p = [2, 5, 1, 2, 4, 1, 2, 5, 1, 2, 4, 1, 2, 4, 1, 2, 5, 1, 2, 4]

The elements in the list are usually the repetition of either 2, 4, 1 or 2, 5, 1.

Sometimes, there are no 1 at the end of 2, 4 or 2, 5.

I want to put 3 in the new list where the 3 consecutive elements of 2, 4, 1 or 2, 5, 1 are.

2, 4, 1 -> 3
2, 5, 1 -> 3

And I want to put 2 in the new list where the elements of 2, 4 or 2, 5 are.

2, 4 -> 2
2, 5 -> 2

Also, if there is either 2, 4 or 2, 5 at the end of the last 2, 4, 1 or 2, 5, 1, I need to put 3.

[2, 4, 1, 2, 5, 1, 2, 5] ==> [3, 3, 3]

However, if there are more than 2 of either 2, 4 or 2, 5 in a row after 2, 4, 1 or 2, 5, 1, I want to put 2 in the list like below.

[2, 4, 1, 2, 5, 2, 5] ==> [3, 2, 2]

My simple code below doesn't give me 2 at the end of the new list. Why is that?

Actually, I altered my code a lot and I am not getting what I want...

Any help will be GREATLY appreciated!

new_list = []

p = [2, 5, 1, 2, 4, 1, 2, 5, 1, 2, 4, 1, 2, 4, 1, 2, 5, 1, 2, 4] 

#p = [2, 5, 1, 2, 4, 1, 2, 5, 1, 2, 4, 1, 2, 4, 1, 2, 5, 1, 2, 4, 2, 4]
#p = [2, 4]

for i in range(0, len(p)):

    if i <= len(p) - 3:
        
        if p[i] == 2 and p[i 1] >= 4:
                
            if p[i 2] !=1:
                
                new_list.append(2)
                
            elif p[i 2] == 1:
            
                new_list.append(3)

            
print(new_list)

CodePudding user response：

Let's do it step by step :

new_list = []
p = [2, 5, 1, 2, 4, 1, 2, 5, 1, 2, 4, 1, 2, 4, 1, 2, 5, 1, 2, 4]
# p = [2, 4, 1, 2, 5, 1, 2, 5]
# p = [2, 4, 1, 2, 5, 2, 5]


sublists = []

# First we split the list on each 2
for i in p:
    if i == 2:
        sublists.append([2])
    else:
        sublists[-1].append(i)

# Then we add the size of each sublist to the new list
new_list = [len(x) for x in sublists]

# This is the same as:
# for i in sublists:
#     new_list.append(len(i))

# Finally we check the particular case of a 2 at the end of the new list
if len(new_list) > 0 and new_list[-1] == 2:
    new_list[-1] = 3
    if len(new_list) > 1 and new_list[-2] == 2:
        new_list[-1] = 2

print(new_list)

It's often a good idea when you are dealing with multiple elements in a list to split it into sub-lists for more clarity.

I hope you find that useful.

CodePudding user response：

This is my method of doing it:

new_list = []
p = [2, 5, 1, 2, 4, 1, 2, 5, 1, 2, 4, 1, 2, 4, 1, 2, 5, 1, 2, 4] 

while len(p) > 0:
    if p[0] == 2:
        if len(p) >= 3:
            if p[1] == 4 and p[2] == 1:
                new_list.append(3)
                p.pop(0)
                p.pop(0)
                p.pop(0)
            elif p[1] == 4 and p[2] != 1:
                new_list.append(2)
                p.pop(0)
                p.pop(0)
            elif p[1] == 5 and p[2] == 1:
                new_list.append(3)
                p.pop(0)
                p.pop(0)
                p.pop(0)
            elif p[1] == 5 and p[2] != 1:
                new_list.append(2)
                p.pop(0)
                p.pop(0)
        else:
            if p[1] == 4:
                new_list.append(2)
                p.pop(0)
                p.pop(0)
            elif p[1] == 5:
                new_list.append(2)
                p.pop(0)
                p.pop(0)

print(new_list)

Firstly, we being a while loop that works until the list p has 0 elements.

Then, we check if the first element of the list starts with a 2. In this case it does. We then check if the second and third elements are 4 and 1. If they are, we add 3 to the new list, and then remove these 3 terms from the list so that their is no repetition.

We check this for the second and third elements 5 and 1 as well, and we also check if the third element is not 1. If it is not 1, we add a 2.

Then, the code is repeated in a if statement. This is because, when the list only has 2 elements left, the 2, 4 in this case, their would be an index error when we check if p[2] == 1, as their is no 3rd element. Therefore, when their are only 2 elements, we check to see if they are 2, 4 or 2, 5, and if so, add 2 to the list.

The code works as long as the first element of the list is a 2, which it seems like it always is.