如何列出另一个列表中出现 integer 的时间？

Question

I wrote some code (Python) that prints that expresses a number in prime numbers. Here is the code that I have:

n = 24
lst=[]
i = 2
while n>1:
    if n%i==0:
      lst.append(i)
      n=n/i
      i-=1
    i+=1  
print(lst)

For example when my n is 24 , the output from the list is [2,2,2,3] , because 2*2*2*3=24 . Now I want to make a list that gives how many times a number occurs in my list. So for this example I want that my output is: [3,1] , because 2 occurs 3 times in my list en 3 occurs one time in my list. I have no idea how I can do this. When I uses the function count(), it prints [3,3,3,1] , because it checks the other list and checks for each item in it how often it exists, so also for the second and third item, which is the same as the first. Does anyone know how my code must look in order to have the output [3,1] ?

Answer 1

There's a datatype called a Counter in the standard library. Given an input iterable, it will output a dictionary of { value: count of value } pairs. You can use it as follows (output is from the ipython REPL ):

In [1]: from collections import Counter

In [2]: primes = [2,2,2,3]

In [3]: counts = Counter(primes)

In [4]: counts
Out[4]: Counter({2: 3, 3: 1})

In [5]: counts.values()
Out[5]: dict_values([3, 1])

In [6]: list(counts.values())
Out[6]: [3, 1]

For illustrative purposes, let's also construct this manually using a defaultdict :

In [7]: from collections import defaultdict

In [8]: counts = defaultdict(int)

In [9]: for p in primes:
   ...:     counts[p] += 1
   ...:

In [10]: counts
Out[10]: defaultdict(int, {2: 3, 3: 1})

In [11]: counts.values()
Out[11]: dict_values([3, 1])

In [12]: list(counts.values())
Out[12]: [3, 1]

Answer 2

Use collections.Counter

from collections import Counter

lst = [2,2,2,3]
c = Counter(lst) # Counter({2: 3, 3: 1})

To get the count of a value:

c[2]  # count of 2 is 3