Numpy 2D数组:将所有值更改为NaN右侧

Question

情况

我有一个包含一些nan值的2D Numpy数组.简化示例:

arr = np.array([[3, 5, np.nan, 2, 4],
                [9, 1, 3, 5, 1],
                [8, np.nan, 3, np.nan, 7]])

在控制台输出中看起来像这样:

array([[  3.,   5.,  nan,   2.,   4.],
       [  9.,   1.,   3.,   5.,   1.],
       [  8.,  nan,   3.,  nan,   7.]])

问题

我正在寻找一种将所有值设置在现有nan值右侧的nan好方法.换句话说,我需要将示例数组转换为:

array([[  3.,   5.,  nan,  nan,  nan],
       [  9.,   1.,   3.,   5.,   1.],
       [  8.,  nan,  nan,  nan,  nan]]) 

我知道如何使用循环实现这一点,但我认为只使用Numpy矢量化操作的方法会更有效.有没有人可以帮我找到这样的方法？

Answer 1

cumsum和boolean-indexing- 的一种方法-

arr[np.isnan(arr).cumsum(1)>0] = np.nan

为了性能,最好使用np.maximum.accumulate-

arr[np.maximum.accumulate(np.isnan(arr),axis=1)] = np.nan

使用有点扭曲的另一种方法broadcasting-

n = arr.shape[1]
mask = np.isnan(arr)
idx = mask.argmax(1)
idx[~mask.any(1)] = n
arr[idx[:,None] <= np.arange(n)] = np.nan

样品运行 -

In [96]: arr
Out[96]: 
array([[  3.,   5.,  nan,   2.,   4.],
       [  9.,   1.,   3.,   5.,   1.],
       [  8.,  nan,   3.,  nan,   7.]])

In [97]: arr[np.maximum.accumulate(np.isnan(arr),axis=1)] = np.nan

In [98]: arr
Out[98]: 
array([[  3.,   5.,  nan,  nan,  nan],
       [  9.,   1.,   3.,   5.,   1.],
       [  8.,  nan,  nan,  nan,  nan]])

标杆

方法 -

def func1(arr):
    arr[np.isnan(arr).cumsum(1)>0] = np.nan

def func2(arr):
    arr[np.maximum.accumulate(np.isnan(arr),axis=1)] = np.nan

def func3(arr): # @ MSeifert's suggestion
    mask = np.isnan(arr); 
    accmask = np.cumsum(mask, out=mask, axis=1); 
    arr[accmask] = np.nan

def func4(arr):
    mask = np.isnan(arr); 
    np.maximum.accumulate(mask,axis=1, out = mask)
    arr[mask] = np.nan

def func5(arr):
    n = arr.shape[1]
    mask = np.isnan(arr)
    idx = mask.argmax(1)
    idx[~mask.any(1)] = n
    arr[idx[:,None] <= np.arange(n)] = np.nan

计时 -

In [201]: # Setup inputs
     ...: arr = np.random.rand(5000,5000)
     ...: arr.ravel()[np.random.choice(range(arr.size), 10000, replace=0)] = np.nan
     ...: arr1 = arr.copy()
     ...: arr2 = arr.copy()
     ...: arr3 = arr.copy()
     ...: arr4 = arr.copy()
     ...: arr5 = arr.copy()
     ...: 

In [202]: %timeit func1(arr1)
     ...: %timeit func2(arr2)
     ...: %timeit func3(arr3)
     ...: %timeit func4(arr4)
     ...: %timeit func5(arr5)
     ...: 
10 loops, best of 3: 149 ms per loop
10 loops, best of 3: 90.5 ms per loop
10 loops, best of 3: 88.8 ms per loop
10 loops, best of 3: 88.5 ms per loop
10 loops, best of 3: 75.3 ms per loop

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