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Python. How do I return a dictionary that counts occurrences in a list of strings?

I am trying to create a function that takes into account occurrences of the first letters in a list of strings and returns them as a dictionary.

For example:

list = ["banana", "ball", "cat", "hat"]

Vocabulary

will look like this: {b: 2, c: 1, h: 1}

Here is the code I have that is iterating but not properly accounted for. This is where I get stuck. How do I update the values ​​to be calculated?

def count_starts(text):
    new_list=[]
    for word in range(len(text)):
        for letter in text[word]:
            if letter[0]=='':
                new_list.append(None)
            else:
                new_list.append(letter[0])

    new_dict= {x:new_list.count(x) for x in new_list}


    return new_dict

Also how can I avoid an out of range error given the following format:

def count_starts(text):
    import collections
    c=collections.Counter(x[0] for x in text)
    return c

Also, what do I need to do if the list contains "None" as the value? I need to count. No.

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4 answers


The problem with your code is that you seem to be repeating all the letters of that word. letter[0]

- a substring of a letter (which is a string).

You will need to make it easier, no double loop needed, take every first letter of your words:

for word in text:
    if word:  # to filter out empty strings
        first_letter = word[0]

But once again collections.Counter

with the understanding of the generator for extracting the first letter is the best choice and one-liner (with added condition to filter out empty lines):



import collections
c = collections.Counter(x[0] for x in ["banana","ball", "cat", "", "hat"] if x)

c

is now a dict: Counter({'b': 2, 'h': 1, 'c': 1})

one option is to insert None

instead of filtering empty values:

c = collections.Counter(x[0] if x else None for x in ["banana","ball", "cat", "", "hat"])
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my_list=["banana","ball", "cat", "hat"] 

my_dict = dict()

for word in my_list:
   try:
      my_dict[word[0]] += 1
   except KeyError:
      my_dict[word[0]] = 1

This increments the key value by 1 for an already existing key, and if the key was not found before it was created with a value of 1



Alternative:

my_list=["banana","ball", "bubbles", "cat", "hat"] 
my_dict = dict()

for word in my_list:
    if word[0] in my_dict.keys():
        my_dict[word[0]] += 1
    else:
        my_dict[word[0]] = 1
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Also, what do I need to do if the list contains "None" as the value? I need to count None.

removing None

lst_no_Nones = [x for x in lis if x != None]

count None

total_None = (sum(x != None for x in lst))
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you need a counter:

 from collections import Counter
 lst = ["banana","ball", "cat", "hat"]
 dct = Counter(lst)

Now, dct stores the number of times each item in lst occurs.

dct = {'b': 2, 'h': 1, 'c': 1}

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