Getting distinct values from from a list comprised of lists containing a comma delimited string

Question

Main list:

data = [
["629-2, text1, 12"],
["629-2, text2, 12"],
["407-3, text9, 6"],
["407-3, text4, 6"],
["000-5, text7, 0"],
["000-5, text6, 0"],
]

I want to get a list comprised of unique lists like so:

data_unique = [
["629-2, text1, 12"],
["407-3, text9, 6"],
["000-5, text6, 0"],
]

I’ve tried using numpy.unique but I need to pare it down further as I need the list to be populated by lists containing a single unique version of the numerical designator in the beginning of the string, ie. 629-2…

I’ve also tried using chain from itertools like this:

def get_unique(data):
    return list(set(chain(*data)))

But that only got me as far as numpy.unique.

Thanks in advance.

Asked By: PersonPr7

||

Source

Answer 1

# Convert the list of lists to a set
data_set = set(tuple(x) for x in data)

# Convert the set back to a list
data_unique = [list(x) for x in data_set]

Answered By: a5zima

Answer 2

I have used recursion to solve the problem!

def get_unique(lst):
        if not lst:
            return []
        if lst[0] in lst[1:]:
            return get_unique(lst[1:])
        else:
            return [lst[0]] + get_unique(lst[1:])

data = [
["629-2, text1, 12"],
["629-2, text2, 12"],
["407-3, text9, 6"],
["407-3, text4, 6"],
["000-5, text7, 0"],
["000-5, text6, 0"],
]
print(get_unique(data))

Here I am storing the last occurrence of the element in list.

Answered By: Punit Choudhary

Answer 3

Code

from itertools import groupby

def get_unique(data):
    def designated_version(item):
        return item[0].split(',')[0]

    return [list(v)[0] 
            for _, v in groupby(sorted(data, 
                                       key = designated_version),
                                designated_version)
           ]

Test

print(get_unique(data))
# Output
[['629-2, text1, 12'], ['407-3, text9, 6'], ['000-5, text7, 0']]

Explanation

Sorts data by designated number (in case not already sorted)
Uses groupby to group by the unique version of the numerical designator of each item in list i.e. lambda item: item[0].split(',')[0]
List comprehension keeps the first item in each grouped list i.e. list(v)[0]

Answered By: DarrylG

Getting distinct values from from a list comprised of lists containing a comma delimited string

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