python - 如果函数在匹配来自不同列表的两个用户 ID 时执行函数？

Question

我正在寻找一些帮助添加到我当前的代码中，我有两个用户名列表，这些列表看起来像这样：

Fishermen A:

George
Tom
Joel
Tom
Lance


Fishermen B:

George
Tom
Tom

我想要做的基本上是如果用户名出现在 Fisherman A 列表和 Fisherman B 列表中，然后计算它在两个列表中出现的次数。所以在这种情况下，代码将列出汤姆 4 次和乔治 2 次，否则什么也不做。我是编码的相对新手，因此将不胜感激任何评论和帮助。

Answer 1

fishermanA = ['George', 'Tom', 'Joel', 'Tom', 'Lance']
fishermanB = ['George', 'Tom', 'Tom']
a_set = set(fishermanA)
b_set = set(fishermanB)
inter = a_set.intersection(b_set)
for i in inter:
    print(i, fishermanA.count(i) + fishermanB.count(i))


Output:
('George', 2)
('Tom', 4)

Answer 2

看起来像一份工作collections.Counter：

>>> from collections import Counter
>>> l1 = ['George', 'Tom', 'Joel', 'Tom', 'Lance']
>>> l2 = ['George', 'Tom', 'Tom']
>>> Counter(filter((set(l1) & set(l2)).__contains__, l1 + l2))
Counter({'Tom': 4, 'George': 2})

Answer 3

这个怎么样：

from collections import defaultdict

fmenA = [
    "George",
    "Tom",
    "Joel",
    "Tom",
    "Lance",
]

fmenB = [
    "George",
    "Tom",
    "Tom",
]

countsA = defaultdict(int)
countsB = defaultdict(int) 

for name in fmenA:
    countsA[name] += 1

for name in fmenB:
    countsB[name] += 1

print {
         name: countsA[name] + countsB[name] 
         for name in countsA if name in countsB
      }


--output:--
{'George': 2, 'Tom': 4}



#The following data is highly skewed against count()

print len(string.printable)   #-->100

fmenA = list(string.printable)[:10]
fmenB = list(string.printable)[:10]
--------------------------------------
2.14819002151   defaultdict
1.860476017     count()
3.48084497452   Counter_arshajii
5.76169896126   Counter_jpmc26



fmenA = list(string.printable)[:20]
fmenB = list(string.printable)[:20]
--------------------------------------
3.87321305275    defaultdict
4.63102507591    count()
5.21796107292    Counter_arshajii
8.44607114792    Counter_jpmc26


fmenA = list(string.printable)[:40]
fmenB = list(string.printable)[:40]
--------------------------------------
7.59739494324         defaultdict
13.643941879272461    count()
9.3110909462          Counter_arshajii
15.3523819447         Counter_jpmc26



fmenA = list(string.printable)
fmenB = list(string.printable)
-------------------------------
18.7256119251    defaultdict
80.9080910683    count()
22.0006680489    Counter_arshajii
37.6448471546    Counter_jpmc26



import timeit

setup ="""
from collections import defaultdict
from collections import Counter
import string

fmenA = list(string.printable) 
fmenB = list(string.printable) 
"""

s1 = """
countsA = defaultdict(int)
countsB = defaultdict(int) 

for name in fmenA:
    countsA[name] += 1

for name in fmenB:
    countsB[name] += 1

{
    name: (countsA[name] + countsB[name]) 
    for name in countsA if name in countsB
}

"""

s2 = """
a_set = set(fmenA)
b_set = set(fmenB)
inter = a_set.intersection(b_set)

{
    name: fmenA.count(name) + fmenB.count(name) 
    for name in inter
}
"""

s3 = """
Counter(filter((set(fmenA) & set(fmenB)).__contains__, fmenA + fmenB))
"""

s4 = """
a = Counter(fmenA)
b = Counter(fmenB)
{k: a[k] + b[k] for k in a if a[k] > 0 and b[k] > 0}
"""

t = timeit.Timer(stmt=s1, setup=setup)
print(t.timeit(number=100000))
t = timeit.Timer(stmt=s2, setup=setup)
print(t.timeit(number=100000))
t = timeit.Timer(stmt=s3, setup=setup)
print(t.timeit(number=100000))
t = timeit.Timer(stmt=s4, setup=setup)
print(t.timeit(number=100000))

Answer 4

我同意 arshaji 这Counter是要走的路，但在我看来，创建额外set的 s 和直接访问魔法方法是不必要的。

>>> from collections import Counter
>>> l1 = ['George', 'Tom', 'Joel', 'Tom', 'Lance']
>>> l2 = ['George', 'Tom', 'Tom']
>>> a = Counter(l1)
>>> b = Counter(l2)
>>> counts = {k: a[k] + b[k] for k in a if a[k] > 0 and b[k] > 0}
>>> counts
{'George': 2, 'Tom': 4}
>>> for k in counts:
...     print str(k) + ': ' + str(counts[k])
...
George: 2
Tom: 4

请注意，我们只迭代 one 中的键是可以的Counter。关键必须在两个列表中才能让我们关心它，所以如果它在两个列表中，它将在Counter我们迭代中。

与 Ankur Ankan 的解决方案相比，优势在于大型列表的效率。Ankur Ankan 的解决方案对每个公共元素的两个完整列表进行迭代。Counteronly 遍历每个列表一次，然后遍历一个Counter。对于大型列表和大量公共元素，性能差异会非常大。对于小型列表，性能影响可以忽略不计。