假设我有一个numpy数组x = [5, 2, 3, 1, 4, 5],y = ['f', 'o', 'o', 'b', 'a', 'r'].我想选择y对应于x大于1且小于5的元素的元素.
我试过了
x = array([5, 2, 3, 1, 4, 5])
y = array(['f','o','o','b','a','r'])
output = y[x > 1 & x < 5] # desired output is ['o','o','a']
但这不起作用.我该怎么做？
jfs*_*jfs 189
如果添加括号,则表达式有效:
>>> y[(1 < x) & (x < 5)]
array(['o', 'o', 'a'],
dtype='|S1')
Mar*_*ski 31
IMO OP实际上并不想要np.bitwise_and()(又名&)但实际上想要,np.logical_and()因为他们正在比较逻辑值,例如True和False- 请参阅逻辑与按位的 SO帖子以查看差异.
>>> x = array([5, 2, 3, 1, 4, 5])
>>> y = array(['f','o','o','b','a','r'])
>>> output = y[np.logical_and(x > 1, x < 5)] # desired output is ['o','o','a']
>>> output
array(['o', 'o', 'a'],
dtype='|S1')
这样做的等效方法是np.all()通过axis适当地设置参数.
>>> output = y[np.all([x > 1, x < 5], axis=0)] # desired output is ['o','o','a']
>>> output
array(['o', 'o', 'a'],
dtype='|S1')
按数字:
>>> %timeit (a < b) & (b < c)
The slowest run took 32.97 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 1.15 µs per loop
>>> %timeit np.logical_and(a < b, b < c)
The slowest run took 32.59 times longer than the fastest. This could mean that an intermediate result is being cached.
1000000 loops, best of 3: 1.17 µs per loop
>>> %timeit np.all([a < b, b < c], 0)
The slowest run took 67.47 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 5.06 µs per loop
所以使用np.all()比较慢,但&和logical_and大致相同.
Goo*_*ill 19
在@JF Sebastian和@Mark Mikofski的答案中添加一个细节:
如果想要获得相应的索引(而不是数组的实际值),以下代码将执行:
为了满足多个(所有)条件:
select_indices = np.where( np.logical_and( x > 1, x < 5) )[0] # 1 < x <5
为了满足多个(或)条件:
select_indices = np.where( np.logical_or( x < 1, x > 5 ) )[0] # x <1 or x >5
小智 5
我喜欢用np.vectorize这些任务.考虑以下:
>>> # Arrays
>>> x = np.array([5, 2, 3, 1, 4, 5])
>>> y = np.array(['f','o','o','b','a','r'])
>>> # Function containing the constraints
>>> func = np.vectorize(lambda t: t>1 and t<5)
>>> # Call function on x
>>> y[func(x)]
>>> array(['o', 'o', 'a'], dtype='<U1')
优点是您可以在矢量化函数中添加更多类型的约束.
希望能帮助到你.