For each element in an array, if the element is less than its previous element, increase it till the previous element with one

Question

Suppose I have an array: list1 = [8, 5, 3, 1, 1, 10, 15, 9] Now if the element is less than its previous element, increase it till the previous element with one.

Here:

5 < 8 so 5 should become: 5 + 3 + 1 = 9 i.e (8+1)

3 < 5 so 3 should become: 3 + 2 + 1 = 6 i.e (5+1)

1 < 3 so 1 should become: 1 + 2 + 1 = 4 i.e (3+1)

Now I am able to get the difference between elements if its less than its previous element. But, how to use it in a final list to get an output like this:

finallist = [8, 9, 6, 4, 1, 10, 15, 16]

Also how can I get a final list value of 'k' list in my code? Right now it shows:

[2] [2, 4] [2, 4, 3] [2, 4, 3, 3] [2, 4, 3, 3, 7]

Source code:

list1 = [8, 5, 3, 1, 1, 10, 15, 9]

k = []

def comput(x):
    if i[x] < i[x-1]:
        num = (i[x-1] - i[x]) + 1
        k.append(num)
        print(k)
    return


for i in [list1]:
    for j in range(len(list1)):
        comput(j)

Neil · Accepted Answer · 2017-11-22 10:23:13Z

1

You can use a list comprehension for this. Basically, the following code will check if one is larger than the next. If it is, then it will convert it to the previous+1.

list1 = [8, 5, 3, 1, 1, 10, 15, 9]

k = [list1[0]] + [i if j<=i else j+1 for i,j in zip(list1[1:],list1[:-1])]
cost = [j-i for i,j in zip(list1,k)]
print(k)
print(cost)

Output:

[8, 9, 6, 4, 1, 10, 15, 16]
[0, 4, 3, 3, 0, 0, 0, 7]

edited Nov 22, 2017 at 10:23

answered Nov 22, 2017 at 9:01

Neil

14.3k3 gold badges35 silver badges53 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

Tarak Shah Over a year ago

This is the final list. I also want the cost of increment too. Like 5 got incremented by 4. 3 got incremented by 3. So, how to get that increment values? I got [2, 4, 3, 3, 7] which is correct. How to get it here? Or even better: Can we get a cost list like: [0, 4,3,3,0,0,0,7,] ?

Tarak Shah Over a year ago

Man I just got i :D

Neil Over a year ago

@Tarak I updated my answer. If I helped you, if you could mark as solution / upvote, it would help a lot.

Tarak Shah Over a year ago

Just curious! If I wanted to make increment to my finallist how would I do it? Here final list we got is [8, 9, 6, 4, 1, 10, 15, 16] so applying same logic, can we make final list as [8, 9, 10, 11,12, 13,15,16] ??

Van Peer · Accepted Answer · 2017-11-22 10:20:08Z

0

The following code will create a new list with the required output

l1 = [8, 5, 3, 1, 1, 10, 15, 9]
l = [l1[0]]
c=[0]         # cost / difference list

for i in range(len(l1)-1):
    if l1[i+1] < l1[i]:
        l.append(l1[i]+1)
        c.append(l1[i]+1-l1[i+1])
    else:
        l.append(l1[i+1])
        c.append(0)
print(l)

Output

[8, 9, 6, 4, 1, 10, 15, 16]
[0, 4, 3, 3, 0,  0,  0,  7]

edited Nov 22, 2017 at 10:20

answered Nov 22, 2017 at 9:35

Van Peer

2,1674 gold badges25 silver badges36 bronze badges

5 Comments

Tarak Shah Over a year ago

This is the final list. I also want the cost of increment too. Like 5 got incremented by 4. 3 got incremented by 3.

Tarak Shah Over a year ago

So, how to get that increment values? I got [2, 4, 3, 3, 7] which is correct. How to get it here? Or even better: Can we get a cost list like: [0, 4,3,3,0,0,0,7,2] ?

Van Peer Over a year ago

@TarakShah updated! used a separate list c for differences.

Tarak Shah Over a year ago

Just curious! If I wanted to make increment to my finallist how would I do it? Here final list we got is [8, 9, 6, 4, 1, 10, 15, 16] so applying same logic, can we make final list as [8, 9, 10, 11,12, 13,15,16] ??

Van Peer Over a year ago

simply you can change l1 to [8, 9, 6, 4, 1, 10, 15, 16]. If you want to do this operation twice on the initial l1 then, define a function and run it twice.

Collectives™ on Stack Overflow

For each element in an array, if the element is less than its previous element, increase it till the previous element with one

2 Answers 2

4 Comments

5 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

4 Comments

5 Comments

Your Answer

Sign up or log in

Post as a guest

Related