Python : List of dict, if exists increment a dict value, if not append a new dict

Question

I would like do something like that.

list_of_urls = ['http://www.google.fr/', 'http://www.google.fr/', 
                'http://www.google.cn/', 'http://www.google.com/', 
                'http://www.google.fr/', 'http://www.google.fr/', 
                'http://www.google.fr/', 'http://www.google.com/', 
                'http://www.google.fr/', 'http://www.google.com/', 
                'http://www.google.cn/']

urls = [{'url': 'http://www.google.fr/', 'nbr': 1}]

for url in list_of_urls:
    if url in [f['url'] for f in urls]:
         urls[??]['nbr'] += 1
    else:
         urls.append({'url': url, 'nbr': 1})

How can I do ? I don't know if I should take the tuple to edit it or figure out the tuple indices?

Any help ?

Stonecutter

Answer 1

That is a very strange way to organize things. If you stored in a dictionary, this is easy:

# This example should work in any version of Python.
# urls_d will contain URL keys, with counts as values, like: {'http://www.google.fr/' : 1 }
urls_d = {}
for url in list_of_urls:
    if not url in urls_d:
        urls_d[url] = 1
    else:
        urls_d[url] += 1

This code for updating a dictionary of counts is a common "pattern" in Python. It is so common that there is a special data structure, defaultdict, created just to make this even easier:

from collections import defaultdict  # available in Python 2.5 and newer

urls_d = defaultdict(int)
for url in list_of_urls:
    urls_d[url] += 1

If you access the defaultdict using a key, and the key is not already in the defaultdict, the key is automatically added with a default value. The defaultdict takes the callable you passed in, and calls it to get the default value. In this case, we passed in class int; when Python calls int() it returns a zero value. So, the first time you reference a URL, its count is initialized to zero, and then you add one to the count.

But a dictionary full of counts is also a common pattern, so Python provides a ready-to-use class: containers.Counter You just create a Counter instance by calling the class, passing in any iterable; it builds a dictionary where the keys are values from the iterable, and the values are counts of how many times the key appeared in the iterable. The above example then becomes:

from collections import Counter  # available in Python 2.7 and newer

urls_d = Counter(list_of_urls)

If you really need to do it the way you showed, the easiest and fastest way would be to use any one of these three examples, and then build the one you need.

from collections import defaultdict  # available in Python 2.5 and newer

urls_d = defaultdict(int)
for url in list_of_urls:
    urls_d[url] += 1

urls = [{"url": key, "nbr": value} for key, value in urls_d.items()]

If you are using Python 2.7 or newer you can do it in a one-liner:

from collections import Counter

urls = [{"url": key, "nbr": value} for key, value in Counter(list_of_urls).items()]

Menedez

I do like that to send it to a django template so I can do : `{% for u in urls %} {{ u.url }} : {{ u.nbr }}{% endfor %} – Stonecutter

You can still do {% for url, nbr in urls.items %}{{ url }} : {{ nbr }}{% endfor %} – Whorled

Nice! defaultdict(lambda: 0) might a bit easier to understand. – Oconnell

@Oconnell defaultdict(int) is an idiom that Python devs should be familiar with as they are likely to see it in other peoples' code. – Menedez

Answer 2

Using the default works, but so does:

urls[url] = urls.get(url, 0) + 1

using .get, you can get a default return if it doesn't exist. By default it's None, but in the case I sent you, it would be 0.

Metronymic

Actually I think this is the best answer, since it is agnostic on the given dictionary, which is a huge bonus imo. – Streaky

This is a nice clean solution. – Citole

This should be the answer. Efficient, clean and to the point !! I hope stackoverflow allows the community to decide the answer along with the question poster. – Vespucci

Really like this answer just not working if the key is None ^^ Or well... Needs some more steps... – Rrhoea

Definitely good answer - should be the actual solution. – Alum

Much slower than defaultdict(). – Etesian

Answer 3

Use defaultdict:

from collections import defaultdict

urls = defaultdict(int)

for url in list_of_urls:
    urls[url] += 1

Alexio

Answer 4

This always works fine for me:

for url in list_of_urls:
    urls.setdefault(url, 0)
    urls[url] += 1

Gerber

Answer 5

Except for the first time, each time a word is seen the if statement's test fails. If you are counting a large number of words, many will probably occur multiple times. In a situation where the initialization of a value is only going to occur once and the augmentation of that value will occur many times it is cheaper to use a try statement:

urls_d = {}
for url in list_of_urls:
    try:
        urls_d[url] += 1
    except KeyError:
        urls_d[url] = 1

you can read more about this: https://wiki.python.org/moin/PythonSpeed/PerformanceTips

Snippy

Answer 6

To do it exactly your way? You could use the for...else structure

for url in list_of_urls:
    for url_dict in urls:
        if url_dict['url'] == url:
            url_dict['nbr'] += 1
            break
    else:
        urls.append(dict(url=url, nbr=1))

But it is quite inelegant. Do you really have to store the visited urls as a LIST? If you sort it as a dict, indexed by url string, for example, it would be way cleaner:

urls = {'http://www.google.fr/': dict(url='http://www.google.fr/', nbr=1)}

for url in list_of_urls:
    if url in urls:
        urls[url]['nbr'] += 1
    else:
        urls[url] = dict(url=url, nbr=1)

A few things to note in that second example:

• see how using a dict for urls removes the need for going through the whole urls list when testing for one single url. This approach will be faster.
• Using dict( ) instead of braces makes your code shorter
• using list_of_urls, urls and url as variable names make the code quite hard to parse. It's better to find something clearer, such as urls_to_visit, urls_already_visited and current_url. I know, it's longer. But it's clearer.

And of course I'm assuming that dict(url='http://www.google.fr', nbr=1) is a simplification of your own data structure, because otherwise, urls could simply be:

urls = {'http://www.google.fr':1}

for url in list_of_urls:
    if url in urls:
        urls[url] += 1
    else:
        urls[url] = 1

Which can get very elegant with the defaultdict stance:

urls = collections.defaultdict(int)
for url in list_of_urls:
    urls[url] += 1

Farthingale

The second version is good since I can convert the dict as a list after. – Stonecutter

Answer 7

It can be done with standard dictionary as well.

urls = {}

for url in list_of_urls:
    urls[url] = urls.get(url, 0) + 1

Beanery