Question

我有三个相同大小的数组（可以很长到5000）。我必须在前两个数组中查找一对值（这将始终是唯一的），例如，（2,3），并相应地从同一索引处的第三个数组中获取值。执行此操作的最快方法是什么，或者为此创建的任何简单的python内置库？解决该问题的最简单方法是：

a = [1,1,1,2,2,2,3,3,3,4,4,4]
b = [2,3,4,3,4,5,7,3,2,1,8,9]
c = [4,5,6,13,4,8,80,4,2,3,7,11]
for i in range(0, len(a)):
  if a[i] == 2 and b[i] == 3:
    fetch = c[i]

Answer 1

找到索引并使用它：

>>> c[zip(a, b).index((2, 3))]
13

或者准备一张dict来看看这对：

>>> dict(zip(zip(a, b), c))[2, 3]
13

如果你想查找很多对，而不仅仅是一对，这会更快。并且您可以使用其get，以防该对可能不存在。

Answer 2

您可以在所有列表中使用next()和zip()的生成器表达式：

>>> next((z for x, y, z in zip(a, b, c) if (x, y) == (2, 3)), 'None found')
13

Answer 3

使用Python 2.7.14对使用长度10000的列表进行基准测试，使用类似于所示的列表并搜索所有现有对，

 2.380 seconds rajat      # The original
 1.712 seconds rajat2     #   Precomputed range
 1.843 seconds rajat3     #   xrange instead of range
 5.243 seconds stefan1    # zip(a, b).index
 0.954 seconds stefan1b   #   Precomputed zip(a, b).index
16.108 seconds stefan2    # dict
 0.002 seconds stefan2b   #   Precomputed dict
10.782 seconds chris      # next(generator)
 6.728 seconds chris2     #   bit optimized
 1.754 seconds chris3     #   Precomputed zip(a, b, c)

代码：

from timeit import timeit

b = range(100) * 100
a = sorted(b)
c = range(10000)

#-------------------------------------------------------------------------

def rajat(A, B):
    'The original'
    for i in range(0, len(a)):
        if a[i] == A and b[i] == B:
            return c[i]

def rajat2(A, B, r=range(0, len(a))):
    '  Precomputed range'
    for i in r:
        if a[i] == A and b[i] == B:
            return c[i]

def rajat3(A, B):
    '  xrange instead of range'
    for i in xrange(0, len(a)):
        if a[i] == A and b[i] == B:
            return c[i]

def stefan1(A, B):
    'zip(a, b).index'
    return c[zip(a, b).index((A, B))]

def stefan1b(A, B, index=zip(a, b).index):
    '  Precomputed zip(a, b).index'
    return c[index((A, B))]

def stefan2(A, B):
    'dict'
    return dict(zip(zip(a, b), c))[A, B]

def stefan2b(A, B, d=dict(zip(zip(a, b), c))):
    '  Precomputed dict'
    return d[A, B]

def chris(A, B):
    'next(generator)'
    return next((z for x, y, z in zip(a, b, c) if (x, y) == (A, B)), 'None found')

def chris2(A, B):
    '  bit optimized'
    return next((z for x, y, z in zip(a, b, c) if x == A and y == B), 'None found')

def chris3(A, B, abc=zip(a, b, c)):
    '  Precomputed zip(a, b, c)'
    return next((z for x, y, z in abc if x == A and y == B), 'None found')

#-------------------------------------------------------------------------

ab = zip(a, b)
def test():
    for A, B in ab:
        func(A, B)

for func in rajat, rajat2, rajat3, stefan1, stefan1b, stefan2, stefan2b, chris, chris2, chris3:
    t = timeit(test, number=1)
    print '%6.3f seconds %-10s # %s' % (t, func.__name__, func.__doc__)

从python

3 个答案: