sammtt - 3 months ago 6x

Python Question

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

Answer

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]
```

Source (Stackoverflow)

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