以下是几种变化,可以计算重复项并忽略b中没有的所有值。"
from collections import Counter
a = [1, 4, 3, 1, 2, 4, 4, 5, 6, 6, 7, 7, 7, 7, 8, 9, 0, 1]
b = [1, 3, 6, 9]
counts = Counter()
for item in b:
counts[item] = 0
for item in a:
if item in b:
counts[item] += 1
print("in 'b' order")
print([(k, v) for k, v in counts.items()])
print("in descending frequency order")
print(counts.most_common())
print("count all occurrences in a of elements that are also in b")
print(sum(counts.values()))
python count_b_in_a.py
in 'b' order
[(1, 3), (3, 1), (6, 2), (9, 1)]
in descending frequency order
[(1, 3), (6, 2), (3, 1), (9, 1)]
count all occurrences in a of elements that are also in b
7
针对您关于性能的评论,以下是Python中扫描列表和扫描集合的比较:
最初的回答:
import datetime
def timestamp():
return datetime.datetime.now()
def time_since(t):
return (timestamp() - t).microseconds // 1000
a = list(range(1000_000))
b = set(a)
iterations = 10
t = timestamp()
for i in range(iterations):
c = 974_152 in a
print("Finished {iterations} iterations of list scan in {duration}ms"
.format(iterations=iterations, duration=time_since(t)))
t = timestamp()
for i in range(iterations):
c = 974_152 in b
print("Finished {iterations} iterations of set scan in {duration}ms"
.format(iterations=iterations, duration=time_since(t)))
python scan.py
Finished 10 iterations of list scan in 248ms
Finished 10 iterations of set scan in 0ms
第一点要注意:Python在这方面也不慢。在旧笔记本电脑上扫描1000万个列表元素只需要1/4秒,这已经很不错了。但它仍然是一个线性扫描。
Python集合类别不同。如果你从
time_since()
中去掉
// 1000
,你会发现Python在不到一微秒的时间内10次扫描了一个100万成员的集合。你会发现其他集合操作也非常快速。无论何时使用Python中的集合,都要使用它们:它们非常棒。
如果你正在考虑将上述代码应用于更大的列表,其中性能很重要,第一件事可能是将
b
转换为一个集合。
len(set(b).intersection(a))
… - Jon Clementslist
调用。 - user2357112