Python-如何返回一个字典，用于计算字符串列表中的出现次数？

Question

I'm trying to make a function that counts occurrences of the first letters of a list of strings and returns them as a dictionary.

For example:

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

dictionary would look like: {b:2, c:1, h:1}

Here is the code I have which iterates but doesn't count properly. That's where I'm getting stuck. How do I update the values to be count?

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 the 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 a value? I need to count None.

Answer 1

Problem with your code is that you seem to iterate on all letters of the word. letter[0] is a substring of the letter (which is a string).

You'd have to do it more simply, no need for a double loop, take each 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 taking a generator comprehension to extract first letter is the best choice and one-liner (with an added condition to filter out empty strings):

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 variant to insert None instead of filtering out empty values would be:

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

Answer 2

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 increases the value of the key by 1 for the already existing key, and if they key has not been found before it creates it with the value 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

Answer 3

Also, what do I need to do if the list contains "None" as a 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))

Answer 4

you need counter:

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

Now, dct stores the number of times every element in lst occurs.

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