a = [1,2,3,4,5]
b = [1,3,5,6]
c = a and b
print c
实际输出:[1,3,5,6]
期望输出:[1,3,5]
如何在两个列表上执行布尔AND操作(列表交集)?
a = [1,2,3,4,5]
b = [1,3,5,6]
c = a and b
print c
实际输出:[1,3,5,6]
期望输出:[1,3,5]
如何在两个列表上执行布尔AND操作(列表交集)?
如果顺序不重要且您不需要担心重复项,则可以使用集合交集:
>>> a = [1,2,3,4,5]
>>> b = [1,3,5,6]
>>> list(set(a) & set(b))
[1, 3, 5]
a = [1,1,2,3,4,5]
且 b = [1,1,3,5,6]
,则它们的交集是 [1,1,3,5]
,但是使用上述方法只会得到一个 1
,即 [1, 3, 5]
。那么正确的做法是什么呢? - Nitish Kumar Palintersection
通常被理解为基于集合的操作。您需要找一个稍微不同的操作-您可能需要通过对每个列表进行排序并合并结果来手动完成操作-在合并中保留重复项。 - WestCoastProjects对于我来说,使用列表推导式是相当明显的选择。不确定性能如何,但至少这些东西保持为列表。
[x for x in a if x in b]
或者说,“如果X的值在B中,则所有在A中的X值”。
b
设为一个集合,就能达到 O(n) 的时间复杂度。 - jcchuks如果您将两个列表中较大的那个转换为集合,您可以使用intersection()
函数获取该集合与任何可迭代对象的交集:
a = [1,2,3,4,5]
b = [1,3,5,6]
set(a).intersection(b)
list(set(a) & set(b))
有什么不同吗? - user1767754将较大的集合转换为集合:
_auxset = set(a)
那么,
c = [x for x in b if x in _auxset]
以下代码可以实现你的需求(保留b
列表的顺序,但不能同时保留a
的顺序),而且速度非常快。在列表推导式中使用条件if x in a
也可以实现此目的,并避免构建_auxset
,但对于长度较长的列表,这样的做法会明显变慢。
如果你希望结果按照顺序排序,而不是保留任何一个列表的顺序,一种更简洁的方法可能是:
c = sorted(set(a).intersection(b))
#!/usr/bin/env python
''' Time list- vs set-based list intersection
See https://dev59.com/B3A65IYBdhLWcg3wqQg8
Written by PM 2Ring 2015.10.16
'''
from __future__ import print_function, division
from timeit import Timer
setup = 'from __main__ import a, b'
cmd_lista = '[u for u in a if u in b]'
cmd_listb = '[u for u in b if u in a]'
cmd_lcsa = 'sa=set(a);[u for u in b if u in sa]'
cmd_seta = 'list(set(a).intersection(b))'
cmd_setb = 'list(set(b).intersection(a))'
reps = 3
loops = 50000
def do_timing(heading, cmd, setup):
t = Timer(cmd, setup)
r = t.repeat(reps, loops)
r.sort()
print(heading, r)
return r[0]
m = 10
nums = list(range(6 * m))
for n in range(1, m + 1):
a = nums[:6*n:2]
b = nums[:6*n:3]
print('\nn =', n, len(a), len(b))
#print('\nn = %d\n%s %d\n%s %d' % (n, a, len(a), b, len(b)))
la = do_timing('lista', cmd_lista, setup)
lb = do_timing('listb', cmd_listb, setup)
lc = do_timing('lcsa ', cmd_lcsa, setup)
sa = do_timing('seta ', cmd_seta, setup)
sb = do_timing('setb ', cmd_setb, setup)
print(la/sa, lb/sa, lc/sa, la/sb, lb/sb, lc/sb)
输出
n = 1 3 2
lista [0.082171916961669922, 0.082588911056518555, 0.0898590087890625]
listb [0.069530963897705078, 0.070394992828369141, 0.075379848480224609]
lcsa [0.11858987808227539, 0.1188349723815918, 0.12825107574462891]
seta [0.26900982856750488, 0.26902294158935547, 0.27298116683959961]
setb [0.27218389511108398, 0.27459001541137695, 0.34307217597961426]
0.305460649521 0.258469975867 0.440838458259 0.301898526833 0.255455833892 0.435697630214
n = 2 6 4
lista [0.15915989875793457, 0.16000485420227051, 0.16551494598388672]
listb [0.13000702857971191, 0.13060092926025391, 0.13543915748596191]
lcsa [0.18650484085083008, 0.18742108345031738, 0.19513416290283203]
seta [0.33592700958251953, 0.34001994132995605, 0.34146714210510254]
setb [0.29436492919921875, 0.2953648567199707, 0.30039691925048828]
0.473793098554 0.387009751735 0.555194537893 0.540689066428 0.441652573672 0.633583767462
n = 3 9 6
lista [0.27657914161682129, 0.28098297119140625, 0.28311991691589355]
listb [0.21585917472839355, 0.21679902076721191, 0.22272896766662598]
lcsa [0.22559309005737305, 0.2271728515625, 0.2323150634765625]
seta [0.36382699012756348, 0.36453008651733398, 0.36750602722167969]
setb [0.34979605674743652, 0.35533690452575684, 0.36164689064025879]
0.760194128313 0.59330170819 0.62005595016 0.790686848184 0.61710008036 0.644927481902
n = 4 12 8
lista [0.39616990089416504, 0.39746403694152832, 0.41129183769226074]
listb [0.33485794067382812, 0.33914685249328613, 0.37850618362426758]
lcsa [0.27405810356140137, 0.2745978832244873, 0.28249192237854004]
seta [0.39211201667785645, 0.39234519004821777, 0.39317893981933594]
setb [0.36988520622253418, 0.37011313438415527, 0.37571001052856445]
1.01034878821 0.85398540833 0.698928091731 1.07106176249 0.905302334456 0.740927452493
n = 5 15 10
lista [0.56792402267456055, 0.57422614097595215, 0.57740211486816406]
listb [0.47309303283691406, 0.47619009017944336, 0.47628307342529297]
lcsa [0.32805585861206055, 0.32813096046447754, 0.3349759578704834]
seta [0.40036201477050781, 0.40322518348693848, 0.40548801422119141]
setb [0.39103078842163086, 0.39722800254821777, 0.43811702728271484]
1.41852623806 1.18166313332 0.819398061028 1.45237674242 1.20986133789 0.838951479847
n = 6 18 12
lista [0.77897095680236816, 0.78187918663024902, 0.78467702865600586]
listb [0.629547119140625, 0.63210701942443848, 0.63321495056152344]
lcsa [0.36563992500305176, 0.36638498306274414, 0.38175487518310547]
seta [0.46695613861083984, 0.46992206573486328, 0.47583580017089844]
setb [0.47616910934448242, 0.47661614418029785, 0.4850609302520752]
1.66818870637 1.34819326075 0.783028414812 1.63591241329 1.32210827369 0.767878297495
n = 7 21 14
lista [0.9703209400177002, 0.9734041690826416, 1.0182771682739258]
listb [0.82394003868103027, 0.82625699043273926, 0.82796716690063477]
lcsa [0.40975093841552734, 0.41210508346557617, 0.42286920547485352]
seta [0.5086359977722168, 0.50968098640441895, 0.51014018058776855]
setb [0.48688101768493652, 0.4879908561706543, 0.49204087257385254]
1.90769222837 1.61990115188 0.805587768483 1.99293236904 1.69228211566 0.841583309951
n = 8 24 16
lista [1.204819917678833, 1.2206029891967773, 1.258256196975708]
listb [1.014998197555542, 1.0206191539764404, 1.0343101024627686]
lcsa [0.50966787338256836, 0.51018595695495605, 0.51319599151611328]
seta [0.50310111045837402, 0.50556015968322754, 0.51335406303405762]
setb [0.51472997665405273, 0.51948785781860352, 0.52113485336303711]
2.39478683834 2.01748351664 1.01305257092 2.34068341135 1.97190418975 0.990165516871
n = 9 27 18
lista [1.511646032333374, 1.5133969783782959, 1.5639569759368896]
listb [1.2461750507354736, 1.254518985748291, 1.2613379955291748]
lcsa [0.5565330982208252, 0.56119203567504883, 0.56451296806335449]
seta [0.5966339111328125, 0.60275578498840332, 0.64791703224182129]
setb [0.54694414138793945, 0.5508568286895752, 0.55375313758850098]
2.53362406013 2.08867620074 0.932788243907 2.76380331728 2.27843203069 1.01753187594
n = 10 30 20
lista [1.7777848243713379, 2.1453688144683838, 2.4085969924926758]
listb [1.5070111751556396, 1.5202279090881348, 1.5779800415039062]
lcsa [0.5954139232635498, 0.59703707695007324, 0.60746097564697266]
seta [0.61563014984130859, 0.62125110626220703, 0.62354087829589844]
setb [0.56723213195800781, 0.57257509231567383, 0.57460403442382812]
2.88774814689 2.44791645689 0.967161734066 3.13413984189 2.6567803378 1.04968299523
这段文本是在一台2GHz单核机器上使用2GB内存运行Python 2.6.6,运行的Linux系统是Debian(同时Firefox在后台运行)生成的。
这些数字只是一个大概指南,因为各种算法的实际速度受源列表中两个列表重复元素比例的影响是不同的。
可以使用filter
和lambda
运算符实现一种函数式的方法。
list1 = [1,2,3,4,5,6]
list2 = [2,4,6,9,10]
>>> list(filter(lambda x:x in list1, list2))
[2, 4, 6]
编辑:它过滤掉了在list1和list中都存在的x,使用set difference也可以实现:
>>> list(filter(lambda x:x not in list1, list2))
[9,10]
编辑2:python3的filter
函数返回一个过滤器对象,使用list
封装该对象可以得到输出列表。
list(filter(lambda x:x in list1, list2))
将其作为列表获取。 - Adrian Wa = [1,2,3,4,5]
b = [1,3,5,6]
c = list(set(a).intersection(set(b)))
应该可以很好地工作。如果可能的话,使用集合而不是列表以避免所有这些类型转换!
您还可以使用numpy.intersect1d(ar1, ar2)
函数。
该函数返回两个数组中都存在的唯一且已排序的值。
这种方法可以获取两个列表的交集,并获取它们共有的重复元素。
>>> from collections import Counter
>>> a = Counter([1,2,3,4,5])
>>> b = Counter([1,3,5,6])
>>> a &= b
>>> list(a.elements())
[1, 3, 5]
map
很方便:>>> lists = [[1, 2, 3], [2, 3, 4], [2, 3, 5]]
>>> set(lists.pop()).intersection(*map(set, lists))
{2, 3}
可以适用于类似的可迭代对象:
>>> lists = ['ash', 'nazg']
>>> set(lists.pop()).intersection(*map(set, lists))
{'a'}
pop
会在列表为空时引发错误,因此您可能需要将其封装在函数中:
def intersect_lists(lists):
try:
return set(lists.pop()).intersection(*map(set, lists))
except IndexError: # pop from empty list
return set()
a and b
的工作方式如文档中提到的那样:“表达式x and y
首先评估x
;如果x
为false,则返回其值;否则,评估y
并返回其结果值。” - Tadeck