Using a dictionary to count the items in a list

Question

Suppose I have a list of items, like:

['apple', 'red', 'apple', 'red', 'red', 'pear']

I want a dictionary that counts how many times each item appears in the list. So for the list above the result should be:

{'apple': 2, 'red': 3, 'pear': 1}

How can I do this simply in Python?

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_{If you are only interested in counting instances of a single element in a list, see How do I count the occurrences of a list item?.}

Answer 1

First create a list of elements to count

elements = [1, 2, 3, 2, 1, 3, 2, 1, 1, 4, 5, 4, 4]

make an empty dictionary, we also do the same with lists

>>> counts = {}

create a for loop that will count for the occurrences of the "key", for each occurrence we add 1

for element in elements:
   if element in counts:
   counts[element] +=1

check whether we encountered the key before or not if so we add 1, if not we use "else" so the new key is added to the dictionary.

>>> else:
>>> counts[element] = 1

Now print counts using 'items() so we could create sequence of the key-value pairs

for element, count in counts.items():
   print(element, ":", count)

here the items() method shows us the key-value pairs, as if we ask: "for element aka in the element load the previous data into 'count' and update it into a sequence of key-value pairs which is what the item() does

CODE WITH NO COMMENTARY:

    elements = [1, 2, 3, 2, 1, 3, 2, 1, 1, 4, 5, 4, 4]
    counts = {}
    for element in elements:
        if element in counts:
            counts[element] += 1
        else:
            counts[element] = 1
    
    for element, count in counts.items():
        print(element, ":", count)

OUTPUT:
    1:4
    2:3
    3:2
    4:3
    5:1

Answer 2

That is an easy answer m8!

def equalizeArray(arr):
    # Counting the frequency of each element in the array
    freq = {}
    for i in arr:
        if i not in freq:
            freq[i] = 1
        else:
            freq[i] += 1
    # Finding the element with the highest frequency
    max_freq = max(freq.values())
    # Calculating the number of deletions required
    for key,value in freq.items():
        if value == max_freq:
            print(key,"been repeated:",value,"times")

Answer 3

mylist = [1,2,1,5,1,1,6,'a','a','b']
result = {}
for i in mylist:
    result[i] = mylist.count(i)
print(result)

Answer 4

If you use Numpy, the unique function can tell you how many times each value appeared by passing return_counts=True:

>>> data = ['apple', 'red', 'apple', 'red', 'red', 'pear']
>>> np.unique(data, return_counts=True)
(array(['apple', 'pear', 'red'], dtype='<U5'), array([2, 1, 3]))

The counts are in the same order as the distinct elements that were found; thus we can use the usual trick to create the desired dictionary (passing the two elements as separate arguments to zip):

>>> dict(zip(*np.unique(data, return_counts=True)))
{'apple': 2, 'pear': 1, 'red': 3}

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If you specifically have a large input Numpy array of small integers, you may get better performance from bincount:

>>> data = np.random.randint(10, size=100)
>>> data
array([1, 0, 0, 3, 3, 4, 2, 4, 4, 0, 4, 8, 7, 4, 4, 8, 7, 0, 0, 2, 4, 2,
       0, 9, 0, 2, 7, 0, 7, 7, 5, 6, 6, 8, 4, 2, 7, 6, 0, 3, 6, 3, 0, 4,
       8, 8, 9, 5, 2, 2, 5, 1, 1, 1, 9, 9, 5, 0, 1, 1, 9, 5, 4, 9, 5, 2,
       7, 3, 9, 0, 1, 4, 9, 1, 1, 5, 4, 7, 5, 0, 3, 5, 1, 9, 4, 8, 8, 9,
       7, 7, 7, 5, 6, 3, 2, 4, 3, 9, 6, 0])
>>> np.bincount(data)
array([14, 10,  9,  8, 14, 10,  6, 11,  7, 11])

The nth value in the output array indicates the number of times that n appeared, so we can create the dictionary if desired using enumerate:

>>> dict(enumerate(np.bincount(data)))
{0: 14, 1: 10, 2: 9, 3: 8, 4: 14, 5: 10, 6: 6, 7: 11, 8: 7, 9: 11}

Answer 5

In 2.7 and 3.1, there is the special Counter (dict subclass) for this purpose.

>>> from collections import Counter
>>> Counter(['apple','red','apple','red','red','pear'])
Counter({'red': 3, 'apple': 2, 'pear': 1})

Answer 6

Simply use list property count\

i = ['apple','red','apple','red','red','pear']
d = {x:i.count(x) for x in i}
print d

output :

{'pear': 1, 'apple': 2, 'red': 3}

Answer 7

I like:

counts = dict()
for i in items:
  counts[i] = counts.get(i, 0) + 1

.get allows you to specify a default value if the key does not exist.

Answer 8

L = ['apple','red','apple','red','red','pear']
d = {}
[d.__setitem__(item,1+d.get(item,0)) for item in L]
print d 

Gives {'pear': 1, 'apple': 2, 'red': 3}

Answer 9

I always thought that for a task that trivial, I wouldn't want to import anything. But i may be wrong, depending on collections.Counter being faster or not.

items = "Whats the simpliest way to add the list items to a dictionary "

stats = {}
for i in items:
    if i in stats:
        stats[i] += 1
    else:
        stats[i] = 1

# bonus
for i in sorted(stats, key=stats.get):
    print("%d×'%s'" % (stats[i], i))

I think this may be preferable to using count(), because it will only go over the iterable once, whereas count may search the entire thing on every iteration. I used this method to parse many megabytes of statistical data and it always was reasonably fast.

Answer 10

>>> L = ['apple','red','apple','red','red','pear']
>>> from collections import defaultdict
>>> d = defaultdict(int)
>>> for i in L:
...   d[i] += 1
>>> d
defaultdict(<type 'int'>, {'pear': 1, 'apple': 2, 'red': 3})