counts[word]=counts.get(word,0)+1_counts.get(word, 0) + 1

注：
counts[word] = counts.get(word,0)+1
是对进行计数word出现的频率进行统计，
当word不在words时，返回值是0，
当word在words中时，返回+1，
以此进行累计计数。

txt="a b c d a b c d a b c d"
print(txt)#a b c d a b c d a b c d
words=txt.split()#将每个字符按照空格分开
#就是将一个字符串分隔成多个字符串组成的列表(重复 有序)
print(words)#['a', 'b', 'c', 'd', 'a', 'b', 'c', 'd', 'a', 'b', 'c', 'd']

counts={}#新建1个字典
for word in words:#['a', 'b', 'c', 'd', 'a', 'b', 'c', 'd', 'a', 'b', 'c', 'd']
    counts[word]=counts.get(word,0)+1
    #print(counts.items())  
    print(list(counts.items()))#去除dict_items()包裹
    
#a b c d a b c d a b c d
#['a', 'b', 'c', 'd', 'a', 'b', 'c', 'd', 'a', 'b', 'c', 'd']
#[('a', 1)]
#[('a', 1), ('b', 1)]
#[('a', 1), ('b', 1), ('c', 1)]
#[('a', 1), ('b', 1), ('c', 1), ('d', 1)]
#[('a', 2), ('b', 1), ('c', 1), ('d', 1)]
#[('a', 2), ('b', 2), ('c', 1), ('d', 1)]
#[('a', 2), ('b', 2), ('c', 2), ('d', 1)]
#[('a', 2), ('b', 2), ('c', 2), ('d', 2)]
#[('a', 3), ('b', 2), ('c', 2), ('d', 2)]
#[('a', 3), ('b', 3), ('c', 2), ('d', 2)]
#[('a', 3), ('b', 3), ('c', 3), ('d', 2)]
#[('a', 3), ('b', 3), ('c', 3), ('d', 3)]
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counts[word]=counts.get(word,0)+1
一开始见到字典内的key的值:“a”,“b”,“c”,“d”，
因为之前没有遇到过，
所以就会给赋值后面的0.
对于前四个key:“a”,“b”,“c”,“d”
{"a":1,"b":1,"c":1,"d":1}
而对于第二轮的"a",“b”,“c”,“d”,
因为不是第一次见到了
所以不会赋给后面的值0,而是在他们key对应的value基础上+1
{"a":2,"b":2,"c":2,"d":2}
后面几轮的同理