返回零（计数）作为 Python 词典中的值

Question

I have a code which works pretty well if not returning keys with zero counts as values. I am still new to Python, so this may be a matter of a simple tweek, but don't know how So please how can I return 0 as a value of a key,if there are 0 counts.

def count_types(string):
    dictionary = {}

   for char in string:
        if 'Z' >= char >= 'A':
            dictionary.setdefault("upper", 0)
            dictionary["upper"] += 1
        elif 'z' >= char >= 'a':
            dictionary.setdefault("lower", 0)
            dictionary["lower"] += 1
        elif char == ' ':
            dictionary.setdefault("space", 0)
            dictionary["space"] += 1
        elif '9' >= char >= '0':
            dictionary.setdefault("numeral", 0)
            dictionary["numeral"] += 1
        else:
            dictionary.setdefault("punctuation", 0)
            dictionary["punctuation"] += 1
  return dictionary

print(count_types("aabbccc")) 

Output is:

{'lower': 7}

Desired output is:

{'lower': 7, 'upper': 0, 'punctuation': 0, 'space': 0, 'numeral': 0}

Answer 1

Try like this. Initialize the dictionary earlier

def count_types(string):
    dictionary = {'lower':0, 'upper':0, 'space':0, 'numeral':0, 'punctuation':0}
    
    for char in string:
        if 'Z' >= char >= 'A':
            dictionary.setdefault("upper", 0)
            dictionary["upper"] += 1
        elif 'z' >= char >= 'a':
            dictionary.setdefault("lower", 0)
            dictionary["lower"] += 1
        elif char == ' ':
            dictionary.setdefault("space", 0)
            dictionary["space"] += 1
        elif '9' >= char >= '0':
            dictionary.setdefault("numeral", 0)
            dictionary["numeral"] += 1
        else:
            dictionary.setdefault("punctuation", 0)
            dictionary["punctuation"] += 1
    return dictionary

print(count_types("aabbccc")) 

output

{'lower': 7, 'upper': 0, 'space': 0, 'numeral': 0, 'punctuation': 0}
> 

Answer 2

The bug in your code is that if there are no numerical chars in the string for example, the code in the if statement never executes so the key will never get inserted to the dict.

What I suggest is preparing the dict in adavance with 0 as a default value, and then you can remove all of the defaultdict usage and make your code much shorter:

def count_chars(string):
    dictionary = {'lower': 0, 'upper': 0, 'space': 0, 'numeral': 0, 'punctuation': 0}
    
    for char in string:
        if 'Z' >= char >= 'A':
            dictionary["upper"] += 1
        elif 'z' >= char >= 'a':
            dictionary["lower"] += 1
        elif char == ' ':
            dictionary["space"] += 1
        elif '9' >= char >= '0':
            dictionary["numeral"] += 1
        else:
            dictionary["punctuation"] += 1
    return dictionary

print(count_chars("testtest")) 

Answer 3

I see other answers But you can use import string and use string.ascii_uppercase or string.ascii_lowercase or ... in if condition like below:

import string
 
def count_types(chars):
    dictionary = {'lower':0, 'upper':0, 'space':0, 'numeral':0, 'punctuation':0}
    for char in chars:
        if char in string.ascii_uppercase   : dictionary["upper"] += 1
        elif char in string.ascii_lowercase : dictionary["lower"] += 1
        elif char == ' '                    : dictionary["space"] += 1
        elif char in string.digits          : dictionary["numeral"] += 1
        else                                : dictionary["punctuation"] += 1
    return dictionary
print(count_types("a bBC123"))

Output:

{'lower': 2, 'upper': 2, 'space': 1, 'numeral': 3, 'punctuation': 0}