CODEWITHSUNDEEP
pythonlistdictionary
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Reputation: 105

Count the number of words in dictionary from a list

Given a list:

lst = ['apple', 'orange', 'pears', 'pears', 'banana']

and a dictionary

dict = {'orange': 4, 'apple':2, 'pears': 1}

if a string from the list already exist in dict update the value else add a new key and its counting.

result:

dict = {'orange' = 5, 'apple':3, 'pears':3, 'banana':1}

I tried:

count = 0
for string on lst:
    if string in dict.keys():
        for num in dict:
            count = count + num
            num = count

I don't know how to continue

Upvotes: 0

Views: 2651

Answers (4)

karsco
karsco

Reputation: 1

Before extending, there is a typo in line 2, which should be for string in list.

This is my suggested solution. As you iterate through the list, check each entry to see if it is a key in the dictionary (as you have already done). If it is, the dict[string] will be the number value paired with that key and you can add one to it. If not, then you can add the string as a new key with a value of 1.

# original data
lst = ['apple', 'orange', 'pears', 'pears', 'banana']  
dict = {'orange': 4, 'apple':2, 'pears': 1}

# iterate through lst and add 1 to each corresponding key value
for string in lst:
    if string in dict.keys():
        #  increment count for a found key 
        #  which can be accessed in dict[string] - no need for num
        count = int(dict[string])
        dict[string] = count + 1
    else:
        # add new key and a count of 1 to dict
        dict[string] = 1

Upvotes: 0

Rahul Chawla
Rahul Chawla

Reputation: 1078

This can easily be done with simple list looping and dict.get method, though other efficient method exist.

lst = ['apple', 'orange', 'pears', 'pears', 'banana']
dict = {'orange': 4, 'apple':2, 'pears': 1}

for st in lst:
     dict[st] = dict.get(st,0)+1

dict
{'orange': 5, 'apple': 3, 'pears': 3, 'banana': 1}

Upvotes: 1

Luca Giorgi
Luca Giorgi

Reputation: 1010

Your answer was almost correct:

for string in lst:
    if string in dict.keys():
        dict[string] += 1
    else:
        dict[string] = 1

This is assuming that a string you haven't seen yet starts with a value of 1, which seems to be the case given your output.

You could also remove the .keys(), as python will automatically check in the keys for the values you are looping on, hence:

for string in lst:
    if string in dict:
        dict[string] += 1
    else:
        dict[string] = 1

Upvotes: 1

Cory Kramer
Cory Kramer

Reputation: 117856

You can use collections.Counter

from collections import Counter

>>> lst = ['apple', 'orange', 'pears', 'pears', 'banana']
>>> d = {'orange': 4, 'apple':2, 'pears': 1}

>>> count = Counter(d)
>>> count
Counter({'orange': 4, 'apple': 2, 'pears': 1})
>>> count += Counter(lst)
>>> count
Counter({'orange': 5, 'pears': 3, 'apple': 3, 'banana': 1})

Upvotes: 1

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