[英]Finding unique elements in nested list of strings
类似于在此URL上发布的查询:
https://stackoverflow.com/questions/54477996/finding-unique-elements-in-nested-list/,
我还有另一个查询。
如果我有一个从Pandas导入的列表,则需要获得一个列表作为所有唯一元素的输出
[Ac, Ad, An, Bi, Co, Cr, Dr, Fa, Mu, My, Sc]
拥有所有唯一元素后,我要检查整个列表中每个元素的计数。 有人可以建议我如何做到吗?
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’],
]
我尝试了不同的方法,但似乎无法使其正常工作。
试图将其转换为字符串并在该字符串上应用拆分功能,但无济于事。
您可以创建一个字典,将键作为列表值,将值作为其计数
您的代码可能如下所示。
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()将给出唯一的元素数组,如果您希望对任何值进行计数,都可以从字典中获取,例如unique ['Ad']
您可以使用collections.Counter
制作元素计数的字典。 这也使您可以轻松访问所有唯一元素的列表。 看起来您有一个列表列表,其中每个子列表都包含一个ingle字符串。 您需要先将其split
然后再将其添加到计数器中。
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
您可以在Python3中这样做
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']
您可以利用collections
, itertools
和functools
提供的功能非常强大的工具,并获得一线解决方案。
如果您的列表仅包含一个元素:
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}")
这就是您得到的:
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
否则,如果您的列表包含多个元素:
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}")
这是在这种情况下的结果:
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
请尝试以下方法:
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(','))
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