计算由空格分隔的未指定数量的整数，并使用字典查找python中每个数字的出现

Question

Sample Output

Enter numbers separated by spaces :1 2 3 3 2 2 2 1 3 4 5 3

{'1': 2, '3': 4, '2': 4, '5': 1, '4': 1}

1 occurs 2 times

3 occurs 4 times

2 occurs 4 times

5 occurs one time

4 occurs one time

So I'm a total newbie at python but I was thinking of starting off like this :

d = {}     
user = input("Enter numbers separated by spaces :") 
data = user.split() 

Except every loop i tried kept saying that I cant convert str() to int(), I'd appreciate any help, I've been staring at this problem for a few hours..this is something I tried for when input is string, trying to implement something similar for dictionary

def countdigits (aString):  
  c = 10 * [0]

  for e in aString: 
    c[int(e)] += 1 

  return c 

def main (): 
  n = 0 

  for v in (countdigits(str(input('Enter a string: ')))): 
    if v == 1: 
      print(n, "occurs 1 time")
    elif v!=0:
      print(n, "occurs", v, "times")

    n += 1 

main()

I'd like a similar solution to this, for the ouput given (but using dictionaries)

Answer 1

Try

d = {i:data.count(i) for i in data}

for k,v in d:
    print "{0} occurs {1} times\n".format(k,v)

or like examples from the comments below:

import collections

for a,b in collections.Counter(data).items():
    print "{0} occurs {1} times\n".format(a,b)

Answer 2

I can only guess that you were attempting something like this

>>> user = "1 2 3 3 2 2 2 1 3 4 5 3"
    >>> data = int(user)
    Traceback (most recent call last):
      File "", line 1, in 
    ValueError: invalid literal for int() with base 10: '1 2 3 3 2 2 2 1 3 4 5 3'

Something like this:

data = user.split()
for item in data:
   number = int(item)

should work fine. Note that you probably don't need to convert to int for this problem. Leaving the numbers a str should work just as well

Answer 3

without importing anything

nk="1 2 3 3 2 2 2 1 3 4 5 3"
nk=nk.split()
result={}
for x in nk:
    result.setdefault(x,0)
    result[x]+=1
print result

output

{'1': 2, '3': 4, '2': 4, '5': 1, '4': 1}