当使用另一个列表计算一个列表中元素的出现次数时，有什么更快的方法

Question

如果我有两个列表，List_A和List_B，如果我想从List_A计算List_B中每个元素的出现次数并在新的List_C中解析结果，有什么更快的方法？ 通常Id使用列表理解，但是一旦List_A或List_B中的元素数量增加到100000以上，它就开始花费大量时间

List_A = ['a', 'd','f','h','g','e','f','a','f','h','h','d','b','c','c','a'] List_B = ['a', 'b','c','d','e','f','g','h'] List_C = [List_A.count(x) for x in List_B] List_C #Output: #List_C = [3, 1, 2, 2, 1, 3, 1, 3]

Answer 1

def a():
    List_A = ['a', 'd','f','h','g','e','f','a','f','h','h','d','b','c','c','a']
    List_B = ['a', 'b','c','d','e','f','g','h', 'z']

    count = Counter(List_A)
    List_C = [count.get(x) for x in List_B]

100000个循环，最好为5个：每个循环2.96 usec

---

使用计数器（）并检查无：

def b():
    List_A = ['a', 'd','f','h','g','e','f','a','f','h','h','d','b','c','c','a']
    List_B = ['a', 'b','c','d','e','f','g','h', 'z']

    count = Counter(List_A)

    List_C = []

    for x in List_B:
        val = count.get(x)
        if val != None:
            List_C.append(val)

100000个循环，最好为5个：每个循环3.45 usec

---

不检查“无”值：

def c():
    List_A = ['a', 'd','f','h','g','e','f','a','f','h','h','d','b','c','c','a']
    List_B = ['a', 'b','c','d','e','f','g','h', 'z']

    count = Counter(List_A)

    List_C = []

    for x in List_B:
        List_C.append(count.get(x))

100000个回路，最好为5个：每个回路3.04 usec

---

使用@Mafa的解决方案，如果列表_B中的值未出现在列表_A中，则该解决方案不起作用：

def d():
    List_A = ['a', 'd','f','h','g','e','f','a','f','h','h','d','b','c','c','a']
    List_B = ['a', 'b','c','d','e','f','g','h']
    occ = dict()
    for x in List_A:
        occ.setdefault(x, 0)
        occ[x] += 1
    List_C = [occ[x] for x in List_B]

100000个循环，最好为5个循环：每个循环2.59 usec

---

检查现有值的Mafa解决方案：

def e():
    List_A = ['a', 'd','f','h','g','e','f','a','f','h','h','d','b','c','c','a']
    List_B = ['a', 'b','c','d','e','f','g','h']
    occ = dict()
    for x in List_A:
        occ.setdefault(x, 0)
        occ[x] += 1
    List_C = [occ.get(x, 0) for x in List_B]

100000个循环，最好为5个：每个循环3.1 usec

---

接下来的两个功能由@Alex Waygood提出

def f():
    List_A = ['a', 'd','f','h','g','e','f','a','f','h','h','d','b','c','c','a']
    List_B = ['a', 'b','c','d','e','f','g','h']
    c = Counter(filter(set(List_B).__contains__, List_A))
    List_C = [v for k, v in sorted(c.items())]

50000个循环，最好为5个：每个循环4.45 usec

---

def g():
    List_A = ['a', 'd','f','h','g','e','f','a','f','h','h','d','b','c','c','a']
    List_B = ['a', 'b','c','d','e','f','g','h', 'z']
    c = Counter(filter(List_B.__contains__, List_A))
    List_C = [v for k, v in sorted(c.items())]

50000个循环，最好为5个：每个循环4.78 usec

（不确定是否需要除以2，这里显然不是，因为有50000个循环而不是100000个，如果是这样，我们这里有一个明显的赢家）

Answer 2

使用您的解决方案，您执行的计算数量等于len(List_A) * len(List_B)（您的列表理解）

相反，首先计算发生次数，然后进行列表理解：

List_A = ['a', 'd','f','h','g','e','f','a','f','h','h','d','b','c','c','a']
List_B = ['a', 'b','c','d','e','f','g','h']
occ = dict()
for x in List_A:
    occ.setdefault(x, 0)
    occ[x] += 1
List_C = [occ.get(x, 0) for x in List_B]

通过这种方式，您可以遍历List_A一次和List_B一次

[编辑]

更新了最后一行的列表理解，以解决x不在List_A中的情况（参见@AchilleG的评论）