I\'m trying to find a short way to see if any of the following items is in a list, but my first attempt does not work. Besides writing a function to accomplish this, is the
Ah, Tobias you beat me to it. I was thinking of this slight variation on your solution:
>>> a = [1,2,3,4]
>>> b = [2,7]
>>> any(x in a for x in b)
True
In some cases (e.g. unique list elements), set operations can be used.
>>> a=[2,3,4]
>>> set(a) - set([2,3]) != set(a)
True
>>>
Or, using set.isdisjoint(),
>>> not set(a).isdisjoint(set([2,3]))
True
>>> not set(a).isdisjoint(set([5,6]))
False
>>>
I collected several of the solutions mentioned in other answers and in comments, then ran a speed test. not set(a).isdisjoint(b)
turned out the be the fastest, it also did not slowdown much when the result was False
.
Each of the three runs tests a small sample of the possible configurations of a
and b
. The times are in microseconds.
Any with generator and max
2.093 1.997 7.879
Any with generator
0.907 0.692 2.337
Any with list
1.294 1.452 2.137
True in list
1.219 1.348 2.148
Set with &
1.364 1.749 1.412
Set intersection explcit set(b)
1.424 1.787 1.517
Set intersection implicit set(b)
0.964 1.298 0.976
Set isdisjoint explicit set(b)
1.062 1.094 1.241
Set isdisjoint implicit set(b)
0.622 0.621 0.753
import timeit
def printtimes(t):
print '{:.3f}'.format(t/10.0),
setup1 = 'a = range(10); b = range(9,15)'
setup2 = 'a = range(10); b = range(10)'
setup3 = 'a = range(10); b = range(10,20)'
print 'Any with generator and max\n\t',
printtimes(timeit.Timer('any(x in max(a,b,key=len) for x in min(b,a,key=len))',setup=setup1).timeit(10000000))
printtimes(timeit.Timer('any(x in max(a,b,key=len) for x in min(b,a,key=len))',setup=setup2).timeit(10000000))
printtimes(timeit.Timer('any(x in max(a,b,key=len) for x in min(b,a,key=len))',setup=setup3).timeit(10000000))
print
print 'Any with generator\n\t',
printtimes(timeit.Timer('any(i in a for i in b)',setup=setup1).timeit(10000000))
printtimes(timeit.Timer('any(i in a for i in b)',setup=setup2).timeit(10000000))
printtimes(timeit.Timer('any(i in a for i in b)',setup=setup3).timeit(10000000))
print
print 'Any with list\n\t',
printtimes(timeit.Timer('any([i in a for i in b])',setup=setup1).timeit(10000000))
printtimes(timeit.Timer('any([i in a for i in b])',setup=setup2).timeit(10000000))
printtimes(timeit.Timer('any([i in a for i in b])',setup=setup3).timeit(10000000))
print
print 'True in list\n\t',
printtimes(timeit.Timer('True in [i in a for i in b]',setup=setup1).timeit(10000000))
printtimes(timeit.Timer('True in [i in a for i in b]',setup=setup2).timeit(10000000))
printtimes(timeit.Timer('True in [i in a for i in b]',setup=setup3).timeit(10000000))
print
print 'Set with &\n\t',
printtimes(timeit.Timer('bool(set(a) & set(b))',setup=setup1).timeit(10000000))
printtimes(timeit.Timer('bool(set(a) & set(b))',setup=setup2).timeit(10000000))
printtimes(timeit.Timer('bool(set(a) & set(b))',setup=setup3).timeit(10000000))
print
print 'Set intersection explcit set(b)\n\t',
printtimes(timeit.Timer('bool(set(a).intersection(set(b)))',setup=setup1).timeit(10000000))
printtimes(timeit.Timer('bool(set(a).intersection(set(b)))',setup=setup2).timeit(10000000))
printtimes(timeit.Timer('bool(set(a).intersection(set(b)))',setup=setup3).timeit(10000000))
print
print 'Set intersection implicit set(b)\n\t',
printtimes(timeit.Timer('bool(set(a).intersection(b))',setup=setup1).timeit(10000000))
printtimes(timeit.Timer('bool(set(a).intersection(b))',setup=setup2).timeit(10000000))
printtimes(timeit.Timer('bool(set(a).intersection(b))',setup=setup3).timeit(10000000))
print
print 'Set isdisjoint explicit set(b)\n\t',
printtimes(timeit.Timer('not set(a).isdisjoint(set(b))',setup=setup1).timeit(10000000))
printtimes(timeit.Timer('not set(a).isdisjoint(set(b))',setup=setup2).timeit(10000000))
printtimes(timeit.Timer('not set(a).isdisjoint(set(b))',setup=setup3).timeit(10000000))
print
print 'Set isdisjoint implicit set(b)\n\t',
printtimes(timeit.Timer('not set(a).isdisjoint(b)',setup=setup1).timeit(10000000))
printtimes(timeit.Timer('not set(a).isdisjoint(b)',setup=setup1).timeit(10000000))
printtimes(timeit.Timer('not set(a).isdisjoint(b)',setup=setup3).timeit(10000000))
print
Best I could come up with:
any([True for e in (1, 2) if e in a])
Think about what the code actually says!
>>> (1 or 2)
1
>>> (2 or 1)
2
That should probably explain it. :) Python apparently implements "lazy or", which should come as no surprise. It performs it something like this:
def or(x, y):
if x: return x
if y: return y
return False
In the first example, x == 1
and y == 2
. In the second example, it's vice versa. That's why it returns different values depending on the order of them.
1 line without list comprehensions.
>>> any(map(lambda each: each in [2,3,4], [1,2]))
True
>>> any(map(lambda each: each in [2,3,4], [1,5]))
False
>>> any(map(lambda each: each in [2,3,4], [2,4]))
True