ipta = raw_input("Word: ") ## asks for input
words = [] ## creates list

while ipta: ## while loop to ask for input and append in list
  words.append(ipta)
  ipta = raw_input("Word: ")
  words.append(ipta)
#Create a set, sets do not have repeats
unique_words = set(words)

print "There are " +  str(len(unique_words)) + " unique words!"

尽管使用set是最简单的方法，但也可以使用dict和some_dict.has(key)来填充仅使用唯一键和值的字典。

假设你已经用用户的输入填充了words[]，创建一个字典，将列表中唯一的单词映射到一个数字:

word_map = {}
i = 1
for j in range(len(words)):
    if not word_map.has_key(words[j]):
        word_map[words[j]] = i
        i += 1                                                             
num_unique_words = len(new_map) # or num_unique_words = i, however you prefer

这是我自己的版本

def unique_elements():
    elem_list = []
    dict_unique_word = {}
    for i in range(5):# say you want to check for unique words from five given words
        word_input = input('enter element: ')
        elem_list.append(word_input)
        if word_input not in dict_unique_word:
            dict_unique_word[word_input] = 1
        else:
            dict_unique_word[word_input] += 1
    return elem_list, dict_unique_word
result_1, result_2 = unique_elements() 
# result_1 holds the list of all inputted elements
# result_2 contains unique words with their count
print(result_2)

你可以使用一个set来删除重复的元素，然后使用len函数来计数set中的元素:

len(set(new_words))

Values, counts = np。独特的(话说,return_counts = True)

更详细地

import numpy as np

words = ['b', 'a', 'a', 'c', 'c', 'c']
values, counts = np.unique(words, return_counts=True)

函数numpy。Unique返回输入列表中已排序的唯一元素及其计数:

['a', 'b', 'c']
[2, 1, 3]

aa="XXYYYSBAA"
bb=dict(zip(list(aa),[list(aa).count(i) for i in list(aa)]))
print(bb)
# output:
# {'X': 2, 'Y': 3, 'S': 1, 'B': 1, 'A': 2}