Question

a = [1, 2, 9, 5, 1]
b = [9, 8, 7, 6, 5]

我想计算两个名单之间的重复数。因此,我想再回一个数字2,因为9和5是两个名单的共同点。

我曾尝试过这样的事情,但却没有做很多工作。

def filter_(x, y):
    count = 0
    for num in y:
        if num in x:
            count += 1
            return count

Answer 1

更短的方式和更好:

>>> a = [1, 2, 9, 5, 1]
>>> b = [9, 8, 7, 6, 5]
>>> len(set(a) & set(b))     # & is intersection - elements common to both
2

Why your code doesn t work:

>>> def filter_(x, y):
...     count = 0
...     for num in y:
...             if num in x:
...                     count += 1
...     return count
... 
>>> filter_(a, b)
2

页: 1

Answer 2

您可使用set.intersection :

>>> set(a).intersection(set(b)) # or just: set(a).intersection(b)
set([9, 5])

或者,对于交叉点的长度:

>>> len(set(a).intersection(set(b)))
2

或更简洁:

>>> len(set(a) & set(b))
2

Answer 3

如果你想看到多彩的条目,基于解决办法就会失败;你们需要像样的东西。

from collections import Counter

def numDups(a, b):
    if len(a)>len(b):
        a,b = b,a

    a_count = Counter(a)
    b_count = Counter(b)

    return sum(min(b_count[ak], av) for ak,av in a_count.iteritems())

之后

numDups([1,1,2,3], [1,1,1,1,1])

回返 2. 本比额表的运行时间为O(n+m)。

此外,您的初步解决办法

for num in y:
    if num in x:
        count += 1

错误——适用于[1,2,3,3]和[1,1,1,1,1,1,1,3],你的代码将退回3或6,neither,其中正确(回答应为2)。

Answer 4

将其转换成<条码>。

 len(set(a).intersection(set(b)))

Answer 5

以下解决办法还涉及清单中的重复内容:

from collections import Counter

def number_of_duplicates(list_a, list_b):
    count_a = Counter(list_a)
    count_b = Counter(list_b)

    common_keys = set(count_a.keys()).intersection(count_b.keys())
    return sum(min(count_a[key], count_b[key]) for key in common_keys)

number_of_duplicates ([1, 2, 2, 2, 3], [1, 2, 2, 4] results in the expected 3.





请注意,@Hugh bothwell也提供了类似的解决办法,但是,如果某一要素只包含在较短的清单中,则有时会扔下<条码>。

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