[英]Intersection of two lists including duplicates?
>>> a = [1,1,1,2,3,4,4]
>>> b = [1,1,2,3,3,3,4]
[1,1,2,3,4]
Please note this is not the same question as this: Python intersection of two lists keeping duplicates Because even though there are three 1s in list a, there are only two in list b so the result should only have two.请注意,这与以下问题不同: Python 两个列表保持重复的交集因为即使列表 a 中有三个 1,但列表 b 中只有两个,所以结果应该只有两个。
You can use collections.Counter<\/code><\/a> for this, which will provide the lowest count found in either list for each element when you take the intersection.
您可以为此使用
collections.Counter<\/code><\/a> ,它会在您选择交叉点时为每个元素提供在任一列表中找到的最低计数。
from collections import Counter
c = list((Counter(a) & Counter(b)).elements())
Simple with no additional imports and easy to debug :)简单,无需额外导入且易于调试 :)
Disadvantage: The value of list b is changed.缺点:列表b的值改变了。 Work on a copy of b if you don't want to change b.
如果您不想更改 b,请处理 b 的副本。
c = list()
for x in a:
if x in b:
b.remove(x)
c.append(x)
The accepted solution posted using Counter is simple, but I think this approach using a dictionary will work too and can be faster -- even on lists that aren't ordered (that requirement wasn't really mentioned, but at least one of the other solutions assumes that is the case).使用 Counter 发布的公认解决方案很简单,但我认为这种使用字典的方法也可以工作并且可以更快 - 即使在未排序的列表上也是如此(该要求并未真正提及,但至少其中一个)解决方案假设是这种情况)。
a = [1, 1, 1, 2, 3, 4, 4]
b = [1, 1, 2, 3, 3, 3, 4]
def intersect(nums1, nums2):
match = {}
for x in a:
if x in match:
match[x] += 1
else:
match[x] = 1
i = []
for x in b:
if x in match:
i.append(x)
match[x] -= 1
if match[x] == 0:
del match[x]
return i
def intersect2(nums1, nums2):
return list((Counter(nums1) & Counter(nums2)).elements())
timeit intersect(a,b)
100000 loops, best of 3: 3.8 µs per loop
timeit intersect2(a,b)
The slowest run took 4.90 times longer than the fastest. This could mean
that an intermediate result is being cached.
10000 loops, best of 3: 20.4 µs per loop
This should also works.这也应该有效。
a = [1, 1, 1, 2, 3, 4, 4]
b = [1, 1, 2, 3, 3, 3, 4]
c = []
i, j = 0, 0
while i < len(a) and j < len(b):
if a[i] == b[j]:
c.append(a[i])
i += 1
j += 1
elif a[i] > b[j]:
j += 1
else:
i += 1
print(c) # [1, 1, 2, 3, 4]
This should also work:这也应该有效:
def list_intersect(lisA, lisB):
""" Finds the intersection of 2 lists including common duplicates"""
Iset = set(lisA).intersection(set(lisB))
Ilis = []
for i in Iset:
num = min(lisA.count(i), lisB.count(i))
for j in range(num):
Ilis.append(i)
return Ilis
This will do:这将做:
from itertools import chain
list(chain.from_iterable([(val,)*min(a.count(val), b.count(val)) for val in (set(a) & set(b))]))
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