Python 代码返回 no 的总数。项目在同一索引处不同的位置

Question

A=[1,2,3,4,5,6,7,8,9] B=[1,2,3,7,4,6​​,5,8,9]

我必须比较这两个列表并返回没有的计数。使用一行 python 代码的项目不同的位置。

例如： 给定数组的输出应为 4，因为在索引 (3,4,5,6) 处，项目不同。因此，程序应返回 4。

我这样做的方法是使用 for 循环比较每个位置：

count=0
for i in range(0,len(A)):
   if(A[i]==B[i]):
     continue
   else:
     count+=1
print(count)

请帮我写一行python代码。

Answer 1

count = sum(a != b for a, b in zip(A, B))
print(count)

或者只是print sum(a != b for a, b in zip(A, B))

你可以查看zip/lambda/map here，这些工具在python中非常强大和重要..

Here您也可以查看其他使用这些工具的方法。

玩得开心！

Answer 2

有many ways 可以做到这一点。如果你使用 numpy，你可以使用 np.count_nonzero:

>>> a = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9])
>>> b = np.array([1, 2, 3, 7, 4, 6, 5, 8, 9])
>>> a != b
array([False, False, False,  True,  True, False,  True, False, False], dtype=bool)
>>> np.count_nonzero(a != b)
3

请注意，a != b 返回一个包含真假的 数组，具体取决于条件在每个索引处的计算方式。

这是速度比较：

>>> %timeit np.count_nonzero(a != b)
The slowest run took 40.59 times longer than the fastest. This could mean that an intermediate result is being cached.
1000000 loops, best of 3: 752 ns per loop

>>> %timeit sum(i != j for i, j in zip(a, b))
The slowest run took 5.86 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 18.5 µs per loop

缓存掩盖了时间，但40.59 * 0.752 = 30.52µs，而5.86 * 18.5 = 108.41µs，所以numpy的最慢似乎仍然比纯python最慢的运行要快得多。

使用更大的数组会更清楚：

>>> n = 10000
>>> a = np.arange(n)
>>> b = np.arange(n)
>>> k = 50
>>> ids = np.random.randint(0, n, k)
>>> a[ids] = 0
>>> ids = np.random.randint(0, n, k)
>>> b[ids] = 0
>>> %timeit np.count_nonzero(a != b)
The slowest run took 20.50 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 11.5 µs per loop
>>> %timeit sum(i != j for i, j in zip(a, b))
100 loops, best of 3: 15.6 ms per loop

差异更加明显，numpy 最多 235 micro-秒，而纯python 需要15.6 milli-秒平均！