CODEWITHSUNDEEP
pythonlistconditional-statements
sammtt
sammtt

Reputation: 421

Going through a list and checking in Python

I'm currently looking for an algorithm to be able to go through a list such as the following one: [1,1,1,1,2,3,4,5,5,5,3,2] I want, in this example, to be able to select the first "1" as there's a duplicate next to it, and keep going through the list until finding the next number having a duplicate next to it, and then select the last number of this one (ie. "5" in this example). Eventually, make the difference between these 2 numbers (ie. 5-1)

I have this code at the moment:

i=0    

for i in range(len(X)):

    if (X[i] == X[i+1]):
        first_number = X[i]

    elif (X[i] != X[i+1]):
        i+=1

I'd like to add a further condition to my question. Suppose you have the following list: lst=[1,1,1,1,2,3,4,5,5,5,3,3,3,3,2,2,2,4,3] In this case, I'll get the following differences according to your code = lst = [4,-2,-1] and then stops. However, I'd like "4-2" to be added to the list afterwards because "4" is followed by a number less than "4" (thus, going to the opposite direction - up - of what "2" followed "4" were following). I hope this is clear enough. Many thanks

Upvotes: 0

Views: 109

Answers (3)

tobias_k
tobias_k

Reputation: 82889

You can use itertools.groupby to find groups of repeating numbers, then find the difference between the first two of those:

>>> import itertools
>>> lst = [1,1,1,1,2,3,4,5,5,5,3,2]
>>> duplicates = [k for k, g in itertools.groupby(lst) if len(list(g)) > 1] 
>>> duplicates[1] - duplicates[0]
4

Or use duplicates[-1] - duplicates[0] if you want the difference between the first and the last repeated number.

In the more general case, if you want the difference between all pairs of consecutive repeated numbers, you could combine that with zip:

>>> lst = [1,1,1,1,2,3,4,5,5,5,3,3,3,3,2,2,2]
>>> duplicates = [k for k, g in itertools.groupby(lst) if len(list(g)) > 1] 
>>> duplicates
[1, 5, 3, 2]
>>> [x - y for x,y in zip(duplicates, duplicates[1:])]
[-4, 2, 1]

I think now I got what you want: You want the difference between any consecutive "plateaus" in the list, where a plateau is either a repeated value, or a local minimum or maximum. This is a bit more complicated and will take several steps:

>>> lst=[1,1,1,1,2,3,4,5,5,5,3,3,3,3,2,2,2,4,3]
>>> plateaus = [lst[i] for i in range(1, len(lst)-1) if lst[i] == lst[i-1] 
...                                                  or lst[i-1] <= lst[i] >= lst[i+1] 
...                                                  or lst[i-1] >= lst[i] <= lst[i+1]]
>>> condensed = [k for k, g in itertools.groupby(plateaus)]
>>> [y-x for x, y in zip(condensed, condensed[1:])]
[4, -2, -1, 2]

Upvotes: 0

El Stun
El Stun

Reputation: 76

Solution:

def subDupeLimits( aList ):
    dupList = []
    prevX = None
    for x in aList:
        if x == prevX:
            dupList.append(x) # track duplicates
        prevX = x # update previous x

    # return last duplicate minus first
    return dupList[-1] - dupList[0]

# call it
y = subDupeLimits( [1,1,1,1,2,3,4,5,5,5,3,2] )
# y =  4

Upvotes: 1

Moses Koledoye
Moses Koledoye

Reputation: 78546

You can use enumerate with a starting index of 1. Duplicates are detected if the current value is equal to the value at the previous index:

l = [1,1,1,1,2,3,4,5,5,5,3,2]
r = [v for i, v in enumerate(l, 1) if i < len(l) and v == l[i]]

result = r[-1] - r[0]
# 4

The list r is a list of all duplicates. r[-1] is the last item and r[0] is the first.

More trials:

>>> l= [1,1,5,5,5,2,2]
>>> r = [v for i, v in enumerate(l, 1) if i < len(l) and v == l[i]]
>>> r[-1] - r[0]
1

Upvotes: 1

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