python - dict.setdefault() 如何统计字符数？

Question

我从 Automate the boring stuff with Python 一书中得到这段代码，但我不明白 setdefault() 方法如何计算唯一字符的数量。

代码:

message = 'It was a bright cold day in April, and the clocks were striking thirteen.'
count = {}
for character in message:
    count.setdefault(character, 0)
    count[character] = count[character] + 1
print(count)

根据本书，setdefault() 方法在字典中搜索键，如果没有找到则更新字典，如果找到则什么都不做。但是我不明白 setdefault 的计数行为以及它是如何完成的？

输出:

{' ': 13, ',': 1, '.': 1, 'A': 1, 'I': 1, 'a': 4, 'c': 3, 'b': 1, 'e': 5, 'd': 3, 'g': 2,
 'i': 6, 'h': 3, 'k': 2, 'l': 3, 'o': 2, 'n': 4, 'p': 1, 's': 3, 'r': 5, 't': 6, 'w': 2, 'y': 1}

请给我解释一下。

Answer 1

在您的示例中，setdefault() 等效于此代码...

if character not in count:
    count[character] = 0

这是一种更好的方式(可以说)做同样的事情:

from collections import defaultdict
message = 'It was a bright cold day in April, and the clocks were striking thirteen.'
count = defaultdict(int)
for character in message:
    count[character] = count[character] + 1
print(count)

之所以有效，是因为默认的 int 是 0。

更好的方法如下:

from collections import Counter
print(Counter(
    'It was a bright cold day in April, '
     'and the clocks were striking thirteen.'))