在嵌套的字符串列表中查找唯一元素

Question

Similar to the query posted at this URL:

https://stackoverflow.com/questions/54477996/finding-unique-elements-in-nested-list/, 

I have another query.

If I have a list that I have imported from Pandas and I need to get a single list as an output with all the unique elements as

[Ac, Ad, An, Bi, Co, Cr, Dr, Fa, Mu, My, Sc]

Once I have all the unique elements, I want to check the count of each of these elements within the whole list. Can someone advise as to how can I accomplish that?

mylist = df.Abv.str.split().tolist()
mylist
[[‘Ac,Cr,Dr’],
 [‘Ac,Ad,Sc'],
 [‘Ac,Bi,Dr’],
 [‘Ad,Dr,Sc'],
 [‘An,Dr,Fa’],
 [‘Bi,Co,Dr’],
 [‘Dr,Mu’],
 [‘Ac,Co,My’],
 [‘Co,Dr’],
 [‘Ac,Ad,Sc'],
 [‘An,Ac,Ad’],
]

I have tried different things but can't seem to make it work.

Tried to convert it into a string and apply split function on the string, but to no avail.

Answer 1

you can create a dictionary with keys as list value and value as their count

your code may look like this .

mylists = [[‘Ac,Cr,Dr’],
 [‘Ac,Ad,Sc'],
 [‘Ac,Bi,Dr’],
 [‘Ad,Dr,Sc'],
 [‘An,Dr,Fa’],
 [‘Bi,Co,Dr’],
 [‘Dr,Mu’],
 [‘Ac,Co,My’],
 [‘Co,Dr’],
 [‘Ac,Ad,Sc'],
 [‘An,Ac,Ad’],
]
unique = {}

for mylist in mylists:
    for elem in mylist:
        unique[elem] = unique[elem]+1 if elem in unique else 1

unique.keys() will give unique element array and if you want the count of any value you can get this from dictionary eg unique['Ad']

Answer 2

You can use collections.Counter to make a dictionary of the counts of the elements. This will also give you easy access to a list of all unique elements. It looks like you have a list of lists where each sublist contains a ingle string. You will need to split these before you add them to the counter.

from collections import Counter
count = Counter()
mylist = [['Ac,Cr,Dr'],
 ['Ac,Ad,Sc'],
 ['Ac,Bi,Dr'],
 ['Ad,Dr,Sc'],
 ['An,Dr,Fa'],
 ['Bi,Co,Dr'],
 ['Dr,Mu'],
 ['Ac,Co,My'],
 ['Co,Dr'],
 ['Ac,Ad,Sc'],
 ['An,Ac,Ad'],
]

for arr in mylist:
    count.update(arr[0].split(','))

print(count) # dictionary of symbols: counts
print(list(count.keys())) # list of all unique elements

Answer 3

You can do it this way in Python3

    mylist = [['Ac,Cr,Dr'],
    ['Ac,Ad,Sc'],
    ['Ac,Bi,Dr'],
    ['Ad,Dr,Sc'],
    ['An,Dr,Fa'],
    ['Bi,Co,Dr'],
    ['Dr,Mu'],
    ['Ac,Co,My'],
    ['Co,Dr'],
    ['Ac,Ad,Sc'],
    ['An,Ac,Ad'],
    ]

    uniquedict = {}

    for sublist in mylist:
        for item in sublist[0].split(','):
            if item in uniquedict.keys():
                uniquedict[item] += 1
            else:
                uniquedict[item] = 1

    print(uniquedict)
    print(list(uniquedict.keys()))

{'Ac': 6, 'Cr': 1, 'Dr': 7, 'Ad': 4, 'Sc': 3, 'Bi': 2, 'An': 2, 'Fa': 1, 'Co': 3, 'Mu': 1, 'My': 1} ['Ac', 'Cr', 'Dr', 'Ad', 'Sc', 'Bi', 'An', 'Fa', 'Co', 'Mu', 'My']

Answer 4

You can take advantage of the very powerful tools offered by collections , itertools and functools and get a one-line solution.

If your lists contain only one element:

from collections import Counter
from itertools import chain
from functools import partial

if __name__ == '__main__':

    mylist = [
        ['Ac,Cr,Dr'],
        ['Ac,Ad,Sc'],
        ['Ac,Bi,Dr'],
        ['Ad,Dr,Sc'],
        ['An,Dr,Fa'],
        ['Bi,Co,Dr'],
        ['Dr,Mu'],
        ['Ac,Co,My'],
        ['Co,Dr'],
        ['Ac,Ad,Sc'],
        ['An,Ac,Ad'],
     ]

    # if lists contain only one element
    occurrence_count = Counter(chain(*map(lambda x: x[0].split(','), mylist)))

    items = list(occurrence_count.keys())  # items, with no repetitions
    all_items = list(occurrence_count.elements())  # all items
    ac_occurrences = occurrence_count['Ac']  # occurrences of 'Ac'

    print(f"Unique items: {items}")
    print(f"All list elements: {all_items}")
    print(f"Occurrences of 'Ac': {ac_occurrences}")

And this is what you get:

Unique items: ['Ac', 'Cr', 'Dr', 'Ad', 'Sc', 'Bi', 'An', 'Fa', 'Co', 'Mu', 'My']
All list elements: ['Ac', 'Ac', 'Ac', 'Ac', 'Ac', 'Ac', 'Cr', 'Dr', 'Dr', 'Dr', 'Dr', 'Dr', 'Dr', 'Dr', 'Ad', 'Ad', 'Ad', 'Ad', 'Sc', 'Sc', 'Sc', 'Bi', 'Bi', 'An', 'An', 'Fa', 'Co', 'Co', 'Co', 'Mu', 'My']
Occurrences of 'Ac': 6

Otherwise, if your lists have more than one element:

from collections import Counter
from itertools import chain
from functools import partial

if __name__ == '__main__':

    mylist_complex = [
        ['Ac,Cr,Dr', 'Ac,Ad,Sc'],
        ['Ac,Ad,Sc', 'Ac,Bi,Dr'],
        ['Ac,Bi,Dr', 'Ad,Dr,Sc'],
        ['Ad,Dr,Sc', 'An,Dr,Fa'],
        ['An,Dr,Fa', 'Bi,Co,Dr'],
        ['Bi,Co,Dr', 'Dr,Mu'],
        ['Dr,Mu', 'Ac,Co,My'],
        ['Ac,Co,My', 'Co,Dr'],
        ['Co,Dr', 'Ac,Ad,Sc'],
        ['Ac,Ad,Sc', 'An,Ac,Ad'],
        ['An,Ac,Ad', 'Ac,Cr,Dr'],
    ]

    # if lists contain more than one element
    occurrence_count_complex = Counter(chain(*map(lambda x: chain(*map(partial(str.split, sep=','), x)), mylist_complex)))

    items = list(occurrence_count_complex.keys())  # items, with no repetitions
    all_items = list(occurrence_count_complex.elements())  # all items
    ac_occurrences = occurrence_count_complex['Ac']  # occurrences of 'Ac'

    print(f"Unique items: {items}")
    print(f"All list elements: {all_items}")
    print(f"Occurrences of 'Ac': {ac_occurrences}")

And this is what you get in this case:

Unique items: ['Ac', 'Cr', 'Dr', 'Ad', 'Sc', 'Bi', 'An', 'Fa', 'Co', 'Mu', 'My']
All list elements: ['Ac', 'Ac', 'Ac', 'Ac', 'Ac', 'Ac', 'Ac', 'Ac', 'Ac', 'Ac', 'Ac', 'Ac', 'Cr', 'Cr', 'Dr', 'Dr', 'Dr', 'Dr', 'Dr', 'Dr', 'Dr', 'Dr', 'Dr', 'Dr', 'Dr', 'Dr', 'Dr', 'Dr', 'Ad', 'Ad', 'Ad', 'Ad', 'Ad', 'Ad', 'Ad', 'Ad', 'Sc', 'Sc', 'Sc', 'Sc', 'Sc', 'Sc', 'Bi', 'Bi', 'Bi', 'Bi', 'An', 'An', 'An', 'An', 'Fa', 'Fa', 'Co', 'Co', 'Co', 'Co', 'Co', 'Co', 'Mu', 'Mu', 'My', 'My']
Occurrences of 'Ac': 12

Answer 5

Try below:

from itertools import chain

mylist = [['Ac,Cr,Dr'], 
          ['Ac,Ad,Sc'], 
          ['Ac,Bi,Dr'], 
          ['Ad,Dr,Sc'], 
          ['An,Dr,Fa'],   
          ['Bi,Co,Dr'], 
          ['Dr,Mu'], 
          ['Ac,Co,My'], 
          ['Co,Dr'], 
          ['Ac,Ad,Sc'], 
          ['An,Ac,Ad']
         ]

flat_list = list(chain.from_iterable(mylist))

unique_list = set(','.join(flat_list).split(','))