如何从 Python 的对象列表中删除具有相同“标题”字段的重复项？

Question

I have class with field title . I have also list of objects, but with duplicates.

for i in self.qList:
    print(i.title)

gives:

    War
    Law
    Mummy
    War
    War

I want to remove War from the list, leaving only one.

I tried this:

newlist=[]
for i in self.qList:
    if i.title not in self.qList.title
         newlist.append(i)

but got the error:

AttributeError: 'list' object has no attribute 'title'

Answer 1

The easiest way to do this is keep a set of just the titles, and check against that:

newlist = []
titles = set()
for i in self.qList:
    if i.title not in titles:
         newlist.append(i)
         titles.add(i.title)

Your code might be easier to read with better variable names, too; what is qList , or i ?

---

If you want to go the list comprehension route, you can do something sneaky with lazy and evaluation:

titles = set()
newlist = [i for i in self.qList if i.title not in titles and titles.add(i.title) is None]

I will leave you to puzzle out exactly how this works!

Answer 2

Looks like a list comprehension to me.

newlist = set(i.title for i in self.qList)

if you really wanted a list then just "cast" it back (ok it's not really casting in python but it looks like it)

newlist = list(set(i.title for i in self.qList))

Also, a minor added plus of using a generator expression is that there is no "leftover" i variable that you didn't really need for anything.

EDIT: the above incorrect solution is left as a reminder to myself to read the question first. The "generator based solution" to give the objects would look more like:

new_set = set(i.title for i in self.qList)
new_list = [i for i in self.qList if i.title in new_set]

I suspect that python's optimization of the generator and list comprehension could still make this faster than the loop but accessing the i.title attribute twice is the tradeoff.

I still kind of like it as looking more "pythonic" (performance question pending...)