Python - Nested groupby statements

Question

Given a list_of_list from a SQL table, I'd like to analyze sets of adjacent rows dividing list_of_list for equal elements of column 2, then for equal element of column 1. For example, given

list_of_list = [[1,1,3,2],
[1,3,3,2],
[1,3,6,2],
[1,3,6,2],
[1,5,5,2],
[1,3,5,2]]

I'd like to get to following output

[1,1,3,2]

[1,3,3,2]

[1,3,6,2]
[1,3,6,2]

[1,5,5,2]

[1,3,5,2]

What's wrong in the following code?

for key1, items1 in groupby(list_of_list, itemgetter(2)):
    for key2, items2 in groupby(items1, itemgetter(1)):
        for i in items2:
            print(i)
            print

it returns the following output

[1,1,3,2]

[1,3,3,2]

[1,3,6,2]

[1,3,6,2]

[1,5,5,2]

[1,3,5,2]

can you give us a sample list_of_list and expected output? Like print(i) and tell us what it's supposed to print. — Bahrom
– Bahrom, Commented Mar 28, 2016 at 15:11
Your code already does that though, after unindenting last print one level — Bahrom
– Bahrom, Commented Mar 28, 2016 at 15:25
Yes, if you unindent the last print, your code will start printing expected output — Bahrom
– Bahrom, Commented Mar 28, 2016 at 15:38

Bahrom · Accepted Answer · 2016-03-28 16:02:04Z

1

As mentioned in the comments, your code works fine if you unindent the last print statement:

for key1, items1 in groupby(list_of_list, itemgetter(2)):
    for key2, items2 in groupby(items1, itemgetter(1)):
        for i in items2:
            print(i)
        print

To get the first element of items2 you can use next like so:

first_of_items2 = next(items2, 'items2 was empty')

answered Mar 28, 2016 at 16:02

Bahrom

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3 Comments

user1403546 Over a year ago

It looks like with "for key, items in groupby(list_of_list, itemgetter(2,1))" I obtain the same effect. Can you confirm that it's a fair way to use it?

Bahrom Over a year ago

Looks alright, you can check the docs if anything. I'll take a look too when I get home tonight. It seems as if it's grouping by one and then the other though, so it should be fine.

Bahrom Over a year ago

Just verified - it generates a break or new group every time the value of the key function changes, so all you need to do is use combined key - as long as your items are sorted by that key you're fine. For more info, see here. Good luck!

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