在 Python 中获取唯一单词列表（即，不重复）

Question

I have a list with several words and I want to print the unique words from that list. With "unique" words I mean words that only appear once in my original list. That is, if a word appears twice (or more than twice), then it shouldn't be printed.

Here's my list of words:

my_list = ["door", "table", "door", "chair", "couch", "door", "table", "closet"]

And here's the code I've tried so far:

print(set(my_list))

However, set prints a list with ALL the words, though there aren't duplicates. That is: door, table, chair, couch, closet . However, what I want is a list like chair, couch, closet (because they are the ones that appear only once in the list).

Answer 1

Not a neat way to achieve what you want, but still a good workaround, by using Counter . Counter will count how many times each item appear in a list.

from collections import Counter
my_list = ["door", "table", "door", "chair", "couch", "door", "table", "closet"]
my_list_count = Counter(my_list) # Counter({'door': 3, 'table': 2, 'chair': 1, 'closet': 1, 'couch': 1})

# Unique item have count = 1
print([xx for xx in my_list_count if my_list_count[xx] == 1])
# Results: ['chair', 'closet', 'couch']

Answer 2

You can use something like -

res = [x for x in my_list if my_list.count(x) == 1]

It will return list of elements that occur once.

Answer 3

You can use the Counter for this, which will create a dictionary with each unique word as the key and the respective occurrence count as the corresponding value. Then iterate through the dictionary to find the key with the value of 1 . The sample code is as follows:

from collections import Counter

my_list = ["door", "table", "door", "chair", "couch", "door", "table", "closet"]

count_all = Counter(my_list)
for key, value in count_all.items():
    if 1 == value:
        print key

Answer 4

Here's a short snippet on how you can go about it.

from collections import Counter
count_dict = Counter(my_list)

for k, v in count_dict.items():
    if v == 1:
        print(k)

Answer 5

Here is something you can try

def keep_unique(array):
    counts = {}
    for e in array:
        if e not in counts:
            counts[e] = 0
        counts[e] += 1
    return [k for (k, v) in counts.items() if v == 1]

print(keep_unique(["door", "table", "door", "chair", "couch", "door", "table", "closet"]))

The keep_unique method counts the number of occurrences of each element and then returns only the ones which appear only once.

Answer 6

A solution without Counter could be to sort your list and check if surrounding words are different:

my_list = ["door", "table", "door", "chair", "couch", "door", "table", "closet"]
my_list.sort()

unique = [word for i, word in enumerate(my_list) if my_list[i - 1] != word and my_list[i + 1] != word] 
print(unique)
# ['chair', 'closet', 'couch']