Sort list by attribute of list [duplicate]

Question

I know that there are several posts regarding sorts of python lists, but I tried a large bunch of things and didn't accomplish to get what I want.

My code:

    list = []
    things = re.findall('<a class="th tooltip" data-rel=.*? href="(.*?)".*?>   <img src="(.*?)" alt="(.*?)" .*?', content, re.DOTALL)
for url, image, name in things:
    list.append({'url': url, 'image': image, 'name': name})

Now I want to sort this list by name. I found several posts which stated to use list.sort(key=) but I don't know what I should use for the key. Everything I tried resulted in a KeyError.

I'm sorry if I'm duplicating an already solved post, but I can't find the proper solution.

Thanks in advance.

Use a lambda function to map a list item i to i['name'], like so: key=lambda i: i['name'] — Will Vousden
– Will Vousden, Commented Jul 3, 2015 at 10:00
@bereal use operator.attrgetter('name') or it won't work on objects that aren't subscriptable — Patrick
– Patrick, Commented Nov 19, 2017 at 11:51
@Patrick attrgetter won't work on dicts, and that's what we're dealing with in this case. — bereal
– bereal, Commented Nov 19, 2017 at 11:53

Anand S Kumar · Accepted Answer · 2015-07-03 10:03:15Z

3

Use a lambda expression , lambda expression would get the dictionary as parameter, and then you can return back the name element from the dictionary -

lst.sort(key=lambda x: x['name'])

Also, please do not use the list as name of the variable, it will overwrite the built-in list function, which may cause issues when trying to use list(..) function.

answered Jul 3, 2015 at 10:03

Anand S Kumar

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3 Comments

conFusl Over a year ago

wow, thx.. I figured out that I sorted on the wrong place, therefore I got two sublists sorted as output. THX

Patrick Over a year ago

Use key=attrgetter('name') for better performance (import attrgetter from operator)

Anand S Kumar Over a year ago

attrgetter would not work in this , as we are working with dictionaries inside a list. But if you were going to do getattr(x, 'name') or x.name as key , then attrgetter would work . But don't really commend trying to optimize, unless you know this part is the bottleneck of the program.

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