Group array by the first element value Python

Question

i have an Array arr = [title, fileurl] so when i do print arr it goes like this:

['name1', 'url1']
['name1', 'url2']
['name1', 'url3']
['name2', 'url1']
['name2', 'url2']
['name3', 'url1']

I would like to group these array by the first element, it means I would like to have:

['name1', 'url1', 'url2', 'url3']
['name2', 'url1', 'url2']
['name3', 'url1']

My code:

for final in posterlink:
    pagesourcec = requests.get(final)
    soupc = BeautifulSoup(pagesourcec.text, "html.parser")
    strc = soupc.findAll("iframe", attrs={"id": "myframe"})
    title = soupb.find("li",{"class": "breadcrumb-item active"}).get_text()
    for embedlink in strc:
        fff = embedlink.get('data-src')
        arr = [title, fff]
        print arr

Do you really need an array? Would not it be better with a dictionary instead? so you could have at the end {"name1": ["url1", "url2", "url3"], "name2": ["url1", "url2"]} — Manuel Sánchez
– Manuel Sánchez, Commented Nov 21, 2018 at 11:22
sorry it's variable title = soupb.find("li",{"class": "breadcrumb-item active"}).get_text() — Lalmi Mohammed Issam
– Lalmi Mohammed Issam, Commented Nov 21, 2018 at 11:24

Silver · Accepted Answer · 2018-11-22 07:47:59Z

4

You can do this:

from collections import defaultdict as ddict

group = ddict(list)

for name, url in arr:
  group[name].append(url)

And if you absolutely want it as a list of lists, you can then follow up with this:

group = [[name, *urls] for name, urls in group.items()]

Edit: It's important to note that the above line works with python 3, which is what you should be using anyways. However, for the sake of completeness if you're using python 2.7, then use this:

group = [[name] + urls for name, urls in group.items()]

edited Nov 22, 2018 at 7:47

answered Nov 21, 2018 at 11:23

Silver

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Comments

Rohit-Pandey · Accepted Answer · 2018-11-21 11:43:03Z

-2

Try This:

a = [['name1', 'url1'],
 ['name1', 'url2'],
 ['name1', 'url3'],
 ['name2', 'url1'],
 ['name2', 'url2'],
 ['name3', 'url1']]
d = {}
for elem in a:
    if elem[0] not in d:
        d[elem[0]] = []
    d[elem[0]].append(elem[1:])

Output:

{'name1': [['url1'], ['url2'], ['url3']], 'name2': [['url1'], ['url2']], 'name3': [['url1']]}

edited Nov 21, 2018 at 11:43

answered Nov 21, 2018 at 11:28

Rohit-Pandey

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6 Comments

Andersson Over a year ago

You should not use dict as variable name

Rohit-Pandey Over a year ago

Now @Andersson.

Andersson Over a year ago

Now the dict name is OK, but it still not clear what your code should do... What is a? Is it list of lists? Also the output not really looks like the one that OP is looking for

Rohit-Pandey Over a year ago

Now Fully Updated with a. btw a is lists of list.

Andersson Over a year ago

IMO instead of re-formatting data it's better to get data in correct format initially... Also your key values are not lists of elements, but lists of lists with single element - still doesn't look like OP requirements met...

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