在嵌套的字符串列表中查找唯一元素

Question

類似於在此URL上發布的查詢：

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

我還有另一個查詢。

如果我有一個從Pandas導入的列表，則需要獲得一個列表作為所有唯一元素的輸出

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

擁有所有唯一元素后，我要檢查整個列表中每個元素的計數。 有人可以建議我如何做到嗎？

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’],
]

我嘗試了不同的方法，但似乎無法使其正常工作。

試圖將其轉換為字符串並在該字符串上應用拆分功能，但無濟於事。

Answer 1

您可以創建一個字典，將鍵作為列表值，將值作為其計數

您的代碼可能如下所示。

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（）將給出唯一的元素數組，如果您希望對任何值進行計數，都可以從字典中獲取，例如unique ['Ad']

Answer 2

您可以使用collections.Counter制作元素計數的字典。 這也使您可以輕松訪問所有唯一元素的列表。 看起來您有一個列表列表，其中每個子列表都包含一個ingle字符串。 您需要先將其split然后再將其添加到計數器中。

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

您可以在Python3中這樣做

    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'：4，4，'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

您可以利用collections ， itertools和functools提供的功能非常強大的工具，並獲得一線解決方案。

如果您的列表僅包含一個元素：

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}")

這就是您得到的：

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

否則，如果您的列表包含多個元素：

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}")

這是在這種情況下的結果：

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

請嘗試以下方法：

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(','))