比较2个列表并返回List2中存在的List1元素中的事件计数后返回列表2

Question

如果我想返回list1中的值的列表与list2相比，该怎么办？

list1 = [1, 2, 3, 4, 5]

list2 = [5, 6, 7, 8, 9]

我期望0出现1，0发生2，0发生3，0发生4，1出现5；要拥有下面的新列表，

new_list = [1, 0, 0, 0, 0]

请参见下文。我在做什么错？

from collections import Counter
list1 = [1, 2, 3, 4, 5]
list2 = [5, 6, 7, 8, 9]

def matchingStrings(list1, list2):
    count_all=Counter(list1)
    counts= {x: count_all[x] for x in list2 if x in list1 }
    list_output=list(counts.values())
    print(list_output)
    return list_output
    # Write your code here


if __name__ == '__main__':
    matchingStrings(list1,list2)

输出

预期输出

[1, 0, 0, 0, 0]

Answer 1

直接将一组需要值传递给counter然后在list1上迭代以获取其计数或0

def matchingStrings(list1, list2):
    counts = Counter(list1)
    return [counts.get(value, 0) for value in list2]

print(matchingStrings(list1, list2))  # [1, 0, 0, 0, 0]

---

基准的Counter> Counter vs list.count

from collections import Counter, defaultdict
from datetime import datetime
import numpy as np

def matchingStrings(list1, list2):
    counts = Counter(list1)
    return [counts.get(value, 0) for value in list2]

def matchingStrings2(list1, list2):
    return [list1.count(a) for a in list2]

if __name__ == '__main__':
    nb = 5000
    times = defaultdict(list)
    for i in range(10):
        list1 = list(np.random.randint(0, 100, nb))
        list2 = list(np.random.randint(0, 100, nb))

        s = datetime.now()
        x1 = matchingStrings(list1, list2)
        times["counter"].append(datetime.now() - s)

        s = datetime.now()
        x2 = matchingStrings2(list1, list2)
        times["list"].append(datetime.now() - s)

    print(np.mean(times['list']) / np.mean(times['counter']))

    for key, values in times.items():
        print(f"{key:7s} => {np.mean(values)}")

计数器解决方案的复杂性是2n，而list.count.count解决方案是n²

这是一个几个数字

nb = 1000                  # lists size
39.18844022169438          # counter 50 times faster
counter => 0:00:00.001263
list    => 0:00:00.049495

nb = 5000                  # lists size
481.512173128945           # counter 500 times faster
counter => 0:00:00.003327
list    => 0:00:01.601991

nb = 10000                 # lists size
1104.0679151061174         # counter 1000 times faster
counter => 0:00:00.004005
list    => 0:00:04.421792

Answer 2

一个小的更正修复了：

from collections import Counter
list1 = [1, 2, 3, 4, 5]
list2 = [5, 6, 7, 8, 9]

def matchingStrings(list1, list2):
    count_all=Counter(list2)
    counts= {x: count_all[x] for x in list1 }
    list_output=list(counts.values())
    print(list_output)
    return list_output
    # Write your code here


if __name__ == '__main__':
    matchingStrings(list1,list2)

Answer 3

尝试使用Count方法。

list1 = [1, 2, 3, 4, 5]

list2 = [5, 6, 7, 8, 9]
def matchingStrings(list1, list2):
    new_list = [list1.count(a) for a in list2]
    print(new_list)
    return new_list


if __name__ == '__main__':
    matchingStrings(list1,list2)

输出：

[1, 0, 0, 0, 0]