목록 내에서 고유 한 값을 계산하는 방법

목록 내에서 고유 한 값을 계산하는 방법

---

그래서 사용자에게 입력을 요청하고 값을 배열 / 목록에 저장하는이 프로그램을 만들려고합니다.
그런 다음 빈 줄을 입력하면 해당 값 중 몇 개가 고유한지 사용자에게 알려줍니다.
나는 이것을 문제 세트가 아닌 실제 삶의 이유로 구축하고 있습니다.

enter: happy
enter: rofl
enter: happy
enter: mpg8
enter: Cpp
enter: Cpp
enter:
There are 4 unique words!

내 코드는 다음과 같습니다.

# ask for input
ipta = raw_input("Word: ")

# create list 
uniquewords = [] 
counter = 0
uniquewords.append(ipta)

a = 0   # loop thingy
# while loop to ask for input and append in list
while ipta: 
  ipta = raw_input("Word: ")
  new_words.append(input1)
  counter = counter + 1

for p in uniquewords:

.. 그게 내가 지금까지 얻은 전부입니다.
목록에서 고유 한 단어 수를 계산하는 방법을 잘 모르겠습니까?
누군가가 해결책을 게시하여 내가 배울 수 있거나 적어도 그것이 얼마나 좋을지 보여줄 수 있다면 감사합니다!

---

또한 collections.Counter 를 사용 하여 코드를 리팩터링합니다.

from collections import Counter

words = ['a', 'b', 'c', 'a']

Counter(words).keys() # equals to list(set(words))
Counter(words).values() # counts the elements' frequency

산출:

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

---

You can use a set to remove duplicates, and then the len function to count the elements in the set:

len(set(new_words))

---

Use a set:

words = ['a', 'b', 'c', 'a']
unique_words = set(words)             # == set(['a', 'b', 'c'])
unique_word_count = len(unique_words) # == 3

Armed with this, your solution could be as simple as:

words = []
ipta = raw_input("Word: ")

while ipta:
  words.append(ipta)
  ipta = raw_input("Word: ")

unique_word_count = len(set(words))

print "There are %d unique words!" % unique_word_count

---

values, counts = np.unique(words, return_counts=True)

---

For ndarray there is a numpy method called unique:

np.unique(array_name)

Examples:

>>> np.unique([1, 1, 2, 2, 3, 3])
array([1, 2, 3])
>>> a = np.array([[1, 1], [2, 3]])
>>> np.unique(a)
array([1, 2, 3])

For a Series there is a function call value_counts():

Series_name.value_counts()

---

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

---

Although a set is the easiest way, you could also use a dict and use some_dict.has(key) to populate a dictionary with only unique keys and values.

Assuming you have already populated words[] with input from the user, create a dict mapping the unique words in the list to a number:

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

---

The following should work. The lambda function filter out the duplicated words.

inputs=[]
input = raw_input("Word: ").strip()
while input:
    inputs.append(input)
    input = raw_input("Word: ").strip()
uniques=reduce(lambda x,y: ((y in x) and x) or x+[y], inputs, [])
print 'There are', len(uniques), 'unique words'

---

I'd use a set myself, but here's yet another way:

uniquewords = []
while True:
    ipta = raw_input("Word: ")
    if ipta == "":
        break
    if not ipta in uniquewords:
        uniquewords.append(ipta)
print "There are", len(uniquewords), "unique words!"

---

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!"

---

Other method by using pandas

import pandas as pd

LIST = ["a","a","c","a","a","v","d"]
counts,values = pd.Series(LIST).value_counts().values, pd.Series(LIST).value_counts().index
df_results = pd.DataFrame(list(zip(values,counts)),columns=["value","count"])

You can then export results in any format you want

---

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}

참고URL : https://stackoverflow.com/questions/12282232/how-do-i-count-unique-values-inside-a-list