使用字典计算用空格分隔的列表中数字的出现

Question

If I have a list of numbers (ex:- 1 2 3 3 2 2 2 1 3 4 5 3 ) how can I use dictionaries in python to count the occurrences of each number in the list? So the output be something like:

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 

Also if a number occurs only once the output should be "one time".

This is what I have so far:

numbers=input("Enter numbers separated by spaces:-")
count={}
for number in numbers:
    if number in count:
        count[number] = count[number]+1
    else:
        count[number] = 1

print(number)

but my output ends up being the last number, i, input can someone help me?

OK, this is what I have now:

numbers = input("Enter numbers separated by spaces:-") # i.e. '1 2 3 3 2 2 2 1 3 4 5 3'
my_list = list(map(int, numbers.strip().split(' ')))
count = {}
for x in set(my_list):
   count[x] = my_list.count(x)
print(count)
for key, value in count.items():
    if value == 1:
         print('{} occurs one time'.format(key))
    else:
         print('{} occurs {} times'.format(key, value))

This is what I have now and it seems pretty good, if theres any way to make it better please do let me know. Thanks a lot everybody

Answer 1

You're close -- you want print(count) , not print(number) so that you're printing the dictionary.

Incidentally, you can use the Counter class from the collections library to do this for you:

>>> import collections
>>> numbers = input("Enter numbers ").split(' ')
>>> count = Counter(numbers)
>>> print(count)

Answer 2

The set() method returns a list of unique values in an iterable, and the count() method returns the number of occurrences of a particular number a list.

Using these facts, you can solve the problem by doing something like below.

numbers = input("Enter numbers seperated by spaces:-") # i.e. '1 2 3 3 2 2 2 1 3 4 5 3'
my_list = list(map(int, numbers.strip().split(' ')))
count = {}
for x in set(my_list):
    count[x] = my_list.count(x)

for key, value in count.items():
    if value == 1:
        print('{} occurs one time'.format(key))
    else:
        print('{} occurs {} times'.format(key, value))

Answer 3

Try counters:

>>> import collections
>>> number_string = '1 2 3 3 2 2 2 1 3 4 5 3'
>>> number_list = number_string.split(' ') # provide a list split on the spaces
>>> counts = collections.Counter(number_list)
>>> counts
Counter({'2': 4, '3': 4, '1': 2, '4': 1, '5': 1})

You can also count as you go:

>>> counts = collections.Counter()
>>> for l in "GATTACA":
...     counts[l] +=1
... 
>>> counts
Counter({'A': 3, 'T': 2, 'C': 1, 'G': 1})

To print this nicely:

import collections

def print_counts_in_spaced_string(number_string):
    number_list = number_string.split(' ') # provide a list split on the spaces
    counts = collections.Counter(number_list)
    for key, value in counts.items():
        format_string = '{key} occurs {value} {times}'
        if value == 1:
            value, times = 'one', 'time'
        else:
            times = 'times'
        print(format_string.format(key=key, value=value, times=times))

number_string = '1 2 3 3 2 2 2 1 3 4 5 3'
print_counts_in_spaced_string(number_string)

Which prints:

1 occurs 2 times
2 occurs 4 times
3 occurs 4 times
4 occurs one time
5 occurs one time

Answer 4

Your problem is that you start off with an empty dict . This means that your check if number in count will always fail (since count is an empty dictionary).

I would recommend using collections.Counter as others have suggested. But if you're really looking to debug your code, then here's what I would do:

numbers=input("Enter numbers seperated by spaces:-")
count={}
for number in numbers.split(): # ignore all the white-spaces
  if number not in count:
    count[number] = 0
  count[number] += 1
for num,freq in count.items():
  print(num, "occurs", freq, "times")