python find duplicate objects in a list

Question

I have a list

l = [object0, object1, object2, object3, object4....object499]

maximum length is never more than 500

All objects have attribute x, y, z, a

I have to modify object.y if duplicates in object.x as:

if object9.x == object10.x:
  object9.y = object9.z * object9.a/1000
  object10.y = object10.y - object9.z * object9.a/1000

It is guaranteed that duplicates will be in consecutive objects.

There may be more than 2 duplicates ie for example

object12.x == object13.x == object14.x

so modification will proceed in the same manner of the third duplicate based on the value of modified 2nd duplicate.

I have written a loop to do it, but was thinking if there is any pythonic/faster way of doing it. I am using python3.7

EDIT:

tag = None
for i, o in enumerate(l):
  if tag is None:
     x_a = o.x
  elif x_a == o.x #duplicate found
    temp = o.y
    c_over = 0
    c_value = o[i-1].z * o[i-1].a/1000
    if c_value < o[i-1].y:
       o[i-1].y = c_value
       c_over = temp - c_value #carry over value
    o[i] = c_over # either zero of carry over value
  x_a = o.x

Could you show your existing loop so others may be able to suggest how to improve it, or otherwise suggest alternative methods? — Jon Clements
– Jon Clements, Commented Mar 13, 2020 at 10:17
You want itertools.groupby, with operator.attrgetter('x') as the key function. — ShadowRanger
– ShadowRanger, Commented Mar 13, 2020 at 10:21
You can iterate on couples of your list with for (prev, current) in zip(l, l[1:]): — rolf82
– rolf82, Commented Mar 13, 2020 at 10:23
Right now enumerating on list and if duplicate found then modify the previous element. posting the code of the loop — Shh
– Shh, Commented Mar 13, 2020 at 10:32

a_guest · Accepted Answer · 2020-03-13 10:31:59Z

3

Since you're modifying attributes of your objects you can loop over consecutive pairs via zip:

for a, b in zip(l, l[1:]):
    if a.x == b.x:
        a.y = a.z * a.a/1000
        b.y = b.y - a.y

answered Mar 13, 2020 at 10:31

a_guest

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4 Comments

Ma0 Over a year ago

what makes it a bit more tricky than that is that there might be a c for which b.x == c.x but since you are modifying b before checking c you are missing it.

a_guest Over a year ago

@Ev.Kounis b.x is never modified so nothing is missed here. x is used for the check and y gets modified.

Ma0 Over a year ago

You are right. There is a bit of ambiguity there in the question come to think of it a bit harder. It is not clear to me if the modification takes place from the beginning or the end of the group. Assuming a triplet of (a, b, c), the modification of c will take place with the already modified value of b in both your and my answer.

a_guest Over a year ago

@Ev.Kounis That's explicitly required by the OP: "[...] so modification will proceed in the same manner of the third duplicate based on the value of modified 2nd duplicate."

Ma0 · Accepted Answer · 2020-03-13 12:03:06Z

3

@Shadow Ranger's comment is definitely the way to go here. I did not test this any but it should do the trick

import itertools
import operator


l = [object0, object1, object2, object3, object4]

for k, g in itertools.groupby(l, key=operator.attrgetter('x')):
    l = list(g)
    for o1, o2, in zip(l, l[1:]):  # Note 1, 2
        o1.y = o1.z * o1.a / 1000
        o2.y = o2.z - o1.y

Notes:

assuming you modify the groups from beginning to the end. If you want to do it the other way round, you have to replace zip(l, l[1:]) with zip(l[len(l)-2::-1], l[::-1])
kudos @ShadowRanger for pointing out that the if check is not needed (handled by the zip)

edited Mar 13, 2020 at 12:03

answered Mar 13, 2020 at 10:32

Ma0

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1 Comment

ShadowRanger Over a year ago

Note: You don't need to check the length (the zip will just terminate immediately if l[1:] is empty), and if you're running forwards, you could use itertools.tee, to avoid listifying at all (which makes it completely lazy; no realizing elements before you need them). g, g2 = itertools.tee(g), next(g2), for o1, o2 in zip(g, g2):

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