如果订单不重要且您不需要担心重复,那么您可以使用set intersection:
>>> a = [1,2,3,4,5] >>> b = [1,3,5,6] >>> list(set(a) & set(b)) [1, 3, 5]
如果`a = [1,1,2,3,4,5]`和`b = [1,1,3,5,6]`那么交集是`[1,1,3,5]`但是通过上面的方法,它只会产生一个`1`,即`[1,3,5]`那么它的写入方式是什么呢？ (8认同)
@NItishKumarPal`intersection`通常被认为是基于集合的。您正在寻找稍有不同的动物-您可能需要通过对每个列表进行排序并合并结果来手动进行操作-并在合并中保留复制。 (4认同)
Lodewijk.. 53
使用列表推导对我来说非常明显.不确定性能,但至少事情保持清单.
[x for x in a if x in b]
或者"如果X值在B中,则A中的所有x值".
如果订单不重要且您不需要担心重复,那么您可以使用set intersection:
>>> a = [1,2,3,4,5] >>> b = [1,3,5,6] >>> list(set(a) & set(b)) [1, 3, 5]
使用列表推导对我来说非常明显.不确定性能,但至少事情保持清单.
[x for x in a if x in b]
或者"如果X值在B中,则A中的所有x值".
如果将两个列表中较大的一个转换为一个集合,则可以使用以下命令获取该集合与任何iterable的交集intersection():
a = [1,2,3,4,5] b = [1,3,5,6] set(a).intersection(b)
制作一个较大的一个:
_auxset = set(a)
然后,
c = [x for x in b if x in _auxset]
会做你想做的事(保留b的顺序,而不是a- 不一定能保留两者)并快速完成.(使用if x in a列表理解中的条件也可以工作,并且避免构建的需要_auxset,但不幸的是,对于相当长的列表,它会慢得多).
如果您希望对结果进行排序,而不是保留列表的排序,则可以采用更简洁的方式:
c = sorted(set(a).intersection(b))
下面是一些Python 2/Python 3代码,它为基于列表和基于集合的方法生成定时信息,以查找两个列表的交集.
纯列表推导算法是O(n ^ 2),因为in在列表上是线性搜索.基于集合的算法是O(n),因为集合搜索是O(1),集合创建是O(n)(并且将集合转换为列表也是O(n)).因此,对于足够大的n,基于集合的算法更快,但是对于小n,创建集合的开销使得它们比纯列表comp算法慢.
#!/usr/bin/env python
''' Time list- vs set-based list intersection
See http://stackoverflow.com/q/3697432/4014959
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)
产量
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
使用2GHz单核机器生成2GB内存,运行Python 2.6.6,在Debian风格的Linux上运行(Firefox在后台运行).
这些数字只是一个粗略的指导,因为各种算法的实际速度受两个源列表中元素的比例的不同影响.
a = [1,2,3,4,5] b = [1,3,5,6] c = list(set(a).intersection(set(b)))
应该像梦一样工作.并且,如果可以,使用集合而不是列表来避免所有这种类型的变化!
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