CODEWITHSUNDEEP
python
Dominique
Dominique

Reputation: 475

Python Compare element in a list with the previous one and the next one

Within a list, I want to get rid of elements that are different from the previous and next ones (example: difference greater than 5)

n=[1913, 2048, 2049, 2050, 2052, 2052, 2054, 2055]
[x for x,y in zip(n,n[1:]) if  y-x<5]

It nearly works: it returns: [2048, 2049, 2050, 2052, 2052, 2054]

The point is that the last element is omitted.

Is there a fast and efficient way to get [2048, 2049, 2050, 2052, 2052, 2054, 2055] Thanks in advance Dom

Upvotes: 2

Views: 235

Answers (2)

2rs2ts
2rs2ts

Reputation: 11066

You could fudge it by appending something to your slice which would ensure the last element gets added.

n=[1913, 2048, 2049, 2050, 2052, 2052, 2054, 2055]
>>> [x for x,y in zip(n,n[1:]+[n[-1]]) if y-x<5]
[2048, 2049, 2050, 2052, 2052, 2054, 2055]

You will get a performance boost from using itertools.izip instead of zip, of course. I compared my method to thefourtheye's and with izip they are close:

>>> timeit.repeat(
        stmt="[x for x,y in zip(n,n[1:]+[n[-1]]) if y-x<5]",
        setup="n=[1913, 2048, 2049, 2050, 2052, 2052, 2054, 2055]")
[5.881312771296912, 5.983433510327245, 5.889796803416459]
>>> timeit.repeat(
         stmt="[x for x,y in izip(n,n[1:]+[n[-1]]) if y-x<5]",
         setup="from itertools import izip; n=[1913, 2048, 2049, 2050, 2052, 2052, 2054, 2055]")
[4.871789328236275, 4.895227617064933, 4.80257417537436]
>>> timeit.repeat(
         stmt="[x for x, y in ex_zip(n,n[1:]) if (y or x) - x < 5]",
         setup="from itertools import izip_longest as ex_zip; n=[1913, 2048, 2049, 2050, 2052, 2052, 2054, 2055]")
[4.3260582542764245, 4.375828323146993, 4.177447625285289]

Upvotes: 2

thefourtheye
thefourtheye

Reputation: 239693

zip normally works till the smallest iterable is exhausted. That is why your last value is ignored. We can fix that with itertools.izip_longest, which by default returns None if the shortest iterable is exhausted.

We take the value of x itself, if y is None with the expression y or x.

from itertools import izip_longest as ex_zip
n = [1913, 2048, 2049, 2050, 2052, 2052, 2054, 2055]
print [x for x, y in ex_zip(n,n[1:]) if (y or x) - x < 5]
# [2048, 2049, 2050, 2052, 2052, 2054, 2055]

Upvotes: 2

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