how can I get a dictionary to display 0 when something appears no times?-CodePudding

I want to search a list of strings for the letter a and create a dictionary that shows how many times it appeared. What I have now does this except when a doesn't appear it returns an empty dictionary. I want it to return {a: 0}. Here's what I have now in a function where its indented properly. what can i change in what i have now so that the created dictionary is {a:0}

list_strings = (["this","sentence"]
result ={}
for letter in list_strings:
    list_strings = list_strings.lower()
    for letter in letters:
        if letter == "a":
            if letter not in result:
                result[letter] = 0
            result[letter]=result[letter]   1
return result

CodePudding user response：

You can put {'a': 0} in the result ahead of time:

result = {'a': 0}

And then do the next loop If you want to count every letter in list of words, the defaultdict is useful.

from collections import defaultdict

list_strings = ["this", "sentence", "apple"]
result = defaultdict(int)
result.update({'a': 0})
for word in list_strings:
    for letter in word:
        result[letter]  = 1

print(result)
print(result.get('t', 0))
print(result.get('a', 0))

After that, you can take value by function: get, the second parameter is optional, if element not in dictionary, get will return the second parameter.

CodePudding user response：

There are some errors in the posted code.

Guess what you need is to count the number of "a"s in all the strings in list_strings and store it in a dictionary. If there are no "a"s you need to be dictionary value for "a" to be 0.

You can initialize dictionary value "a" to zero at the beginning. I have corrected errors and do this in the below code.

list_strings = (["this","sentence"])
result ={}
result["a"] = 0
for string in list_strings:
    string = string.lower()
    for letter in string:
        if letter == "a":
            result[letter]=result[letter]   1


print(result)

If you need to count other characters you can create the initial dictionary as follows.

d1 = dict.fromkeys(string.ascii_lowercase, 0)

CodePudding user response：

You could collections.Counter to count all the letters and then use the get function on the Counter dictionary (to return 0 as the default)

>>> from collections import Counter
>>> list_strings = ["this","sentence"]
>>> c = Counter()
>>> for string in list_strings:
    c.update(string)

    
>>> c
Counter({'e': 3, 't': 2, 's': 2, 'n': 2, 'h': 1, 'i': 1, 'c': 1})
>>> c.get('a', 0)
0