在numpy数组求和中将nan视为零,除了所有数组中的nan

Question

我有两个numpy数组NS,EW总结.他们每个人在不同的位置都有缺失值,比如

NS = array([[  1.,   2.,  nan],
       [  4.,   5.,  nan],
       [  6.,  nan,  nan]])
EW = array([[  1.,   2.,  nan],
       [  4.,  nan,  nan],
       [  6.,  nan,   9.]]

如何以numpy方式执行求和操作,如果一个阵列在一个位置具有nan,则将nan视为零,并且如果两个阵列在同一位置具有nan,则保持nan.

我期望看到的结果是

SUM = array([[  2.,   4.,  nan],
           [  8.,  5.,  nan],
           [  12.,  nan,   9.]])

当我尝试

SUM=np.add(NS,EW)

它给了我

SUM=array([[  2.,   4.,  nan],
       [  8.,  nan,  nan],
       [ 12.,  nan,  nan]])

当我尝试

SUM = np.nansum(np.dstack((NS,EW)),2)

它给了我

SUM=array([[  2.,   4.,   0.],
       [  8.,   5.,   0.],
       [ 12.,   0.,   9.]])

当然,我可以通过进行元素级操作来实现我的目标,

for i in range(np.size(NS,0)):
    for j in range(np.size(NS,1)):
        if np.isnan(NS[i,j]) and np.isnan(EW[i,j]):
            SUM[i,j] = np.nan
        elif np.isnan(NS[i,j]):
            SUM[i,j] = EW[i,j]
        elif np.isnan(EW[i,j]):
            SUM[i,j] = NS[i,j]
        else:
            SUM[i,j] = NS[i,j]+EW[i,j]

但它很慢.所以我正在寻找一个更加笨拙的解决方案来解决这个问题.

提前感谢您的帮助!

Answer 1

方法#1: 一种方法np.where-

def sum_nan_arrays(a,b):
    ma = np.isnan(a)
    mb = np.isnan(b)
    return np.where(ma&mb, np.nan, np.where(ma,0,a) + np.where(mb,0,b))

样品运行 -

In [43]: NS
Out[43]: 
array([[  1.,   2.,  nan],
       [  4.,   5.,  nan],
       [  6.,  nan,  nan]])

In [44]: EW
Out[44]: 
array([[  1.,   2.,  nan],
       [  4.,  nan,  nan],
       [  6.,  nan,   9.]])

In [45]: sum_nan_arrays(NS, EW)
Out[45]: 
array([[  2.,   4.,  nan],
       [  8.,   5.,  nan],
       [ 12.,  nan,   9.]])

方法#2:可能是一个更快的混合boolean-indexing-

def sum_nan_arrays_v2(a,b):
    ma = np.isnan(a)
    mb = np.isnan(b)
    m_keep_a = ~ma & mb
    m_keep_b = ma & ~mb
    out = a + b
    out[m_keep_a] = a[m_keep_a]
    out[m_keep_b] = b[m_keep_b]
    return out

运行时测试 -

In [140]: # Setup input arrays with 4/9 ratio of NaNs (same as in the question)
     ...: a = np.random.rand(3000,3000)
     ...: b = np.random.rand(3000,3000)
     ...: a.ravel()[np.random.choice(range(a.size), size=4000000, replace=0)] = np.nan
     ...: b.ravel()[np.random.choice(range(b.size), size=4000000, replace=0)] = np.nan
     ...: 

In [141]: np.nanmax(np.abs(sum_nan_arrays(a, b) - sum_nan_arrays_v2(a, b))) # Verify
Out[141]: 0.0

In [142]: %timeit sum_nan_arrays(a, b)
10 loops, best of 3: 141 ms per loop

In [143]: %timeit sum_nan_arrays_v2(a, b)
10 loops, best of 3: 177 ms per loop

In [144]: # Setup input arrays with lesser NaNs
     ...: a = np.random.rand(3000,3000)
     ...: b = np.random.rand(3000,3000)
     ...: a.ravel()[np.random.choice(range(a.size), size=4000, replace=0)] = np.nan
     ...: b.ravel()[np.random.choice(range(b.size), size=4000, replace=0)] = np.nan
     ...: 

In [145]: np.nanmax(np.abs(sum_nan_arrays(a, b) - sum_nan_arrays_v2(a, b))) # Verify
Out[145]: 0.0

In [146]: %timeit sum_nan_arrays(a, b)
10 loops, best of 3: 69.6 ms per loop

In [147]: %timeit sum_nan_arrays_v2(a, b)
10 loops, best of 3: 38 ms per loop