计算元素在列表中的出现次数

Question

我有一个元素列表：

l = ['a', 'b', 'b', 'a', 'c', 'd', 'd', 'c']

我想生成一个元素出现的列表，所以对于每个元素，它都会计算它出现的次数，直到这一点：

这样我会得到：

elements_occurence: [1, 1, 2, 2, 1, 1, 2, 2]

我试过了：

occurrences = {}
for i in l:
if i in occurrences:
occurrences[i] += 1
else:
occurrences[i] = 1

但它给了我这样的结果：

{'a': 2, 'b': 2, 'c': 2, 'd': 2}

Answer 1

您当前正在检查dict的最终结果；保存"迭代时的中间值，您将得到您的结果：

occurrences = {}
elements_occurence = []
for i in l:
if i in occurrences:
occurrences[i] += 1
else:
occurrences[i] = 1
elements_occurence.append(occurrences[i])
print(elements_occurence)
print(occurrences)

提供：

[1, 1, 2, 2, 1, 1, 2, 2]
{'a': 2, 'b': 2, 'c': 2, 'd': 2}

---

提高计数的几种方法(按"蟒蛇">的升序(：

1. 使用setdefault:

   for i in l:
   occurrences[i] = occurrences.setdefault(i, 0) + 1
   

2. 使用get:

   for i in l:
   occurrences[i] = occurrences.get(i, 0) + 1
   

3. 使用defaultdict:

   from collections import defaultdict
   occurrences = defaultdict(int)
   for i in l:
   occurrences[i] += 1