Python字典：在循环中更改字典键的一个值会更改每个键的每个值[重复]

Question

我有一个字符列表。我想将标点符号从列表中拉出，并使每种标点符号成为字典中的键（因此“。”的键，“！”的键等）然后，我想数一下数字每个标点符号出现在另一个列表中的次数，并计算相应键的值。问题是，我的字典中的每个值都会发生变化，而不仅仅是一个键。

输出应该是这样的，因为有 2 个“.”和 4 个“！”在“标点符号列表”中

{'.': [2], ',': [0], '!': [4], '?': [0]}

但是看起来像这样，因为“！”出现 4 次

{'.': [4], ',': [4], '!': [4], '?': [4]}

# Create a list of characters that we will eventually count
charList = [".","!",".","!","!","!","p","p","p","p","p"]

# Create a list of the punctuation we want a count of
punctuationList = [".",",","!","?"]

# Group each type of punctuation and count the number of times it occurs

dic = dict.fromkeys(punctuationList,[0]) # Turn punctuationList into a dictionary with each punctuation type as a key with a value that is 
                                         # the count of each time the key appears in newList

print (dic)

# Count each punctuation in the dictionary

for theKey in dic: # iterate through each key (each punctuation type)
    counter = 0  # Set the counter at 0
    
    for theChar in charList: # If the key matches the character in the list, then add 1 to the counter
        if theKey == theChar:
            
            counter = counter + 1
            
            dic[theKey][0] = counter # Then change the value of that key to the number of times 
                                                   # that character shows up in the list

print (dic)

Answer 1

dict.fromkeys 为每个键共享相同的值。

你会想要的

dic = {key: [0] for key in punctuationList}

而是为每个键初始化一个单独的列表。 （顺便说一句，确实没有必要将数字包装在列表中。）

也就是说，您的代码可以使用内置的collections.Counter 来实现

from collections import Counter
dic = dict(Counter(ch for ch in charList if ch in punctuationList))