Question

假设我有一个100000 x 100的矩阵

import numpy as np

mat = np.random.randint(2, size=(100000,100))

我希望遍历此矩阵，如果每个row完全包含1或0，则希望将state变量更改为该值。如果状态未更改，则希望将整个row的值设置为state。 state的初始值为0。

天真地在for循环中，可以按以下步骤完成

state = 0

for row in mat:
    if set(row) == {1}:
        state = 1
    elif set(row) == {0}:
        state = 0
    else:
        row[:] = state

但是，当矩阵的大小增加时，这将花费不切实际的时间。有人可以向我指出如何利用numpy来向量化此循环并加快循环速度的方向吗？

所以对于示例输入

array([[0, 1, 0],
       [0, 0, 1],
       [1, 1, 1],
       [0, 0, 1],
       [0, 0, 1]])

在这种情况下，预期输出为

array([[0, 0, 0],
       [0, 0, 0],
       [1, 1, 1],
       [1, 1, 1],
       [1, 1, 1]])

Answer 1

方法1：NumPy矢量化

这是矢量化的-

def check_all(a, state): # a is input matrix/array
    # Get zeros and ones all masks
    zm = (a==0).all(1)
    om = (a==1).all(1)

    # "Attach" boundaries with False values at the start of these masks.
    # These will be used to detect rising edges (as indices) on these masks.
    zma = np.r_[False,zm]
    oma = np.r_[False,om]

    omi = np.flatnonzero(oma[:-1] < oma[1:])
    zmi = np.flatnonzero(zma[:-1] < zma[1:])

    # Group the indices and the signatures (values as 1s and -1s)
    ai = np.r_[omi,zmi]
    av = np.r_[np.ones(len(omi),dtype=int),-np.ones(len(zmi),dtype=int)]

    # Sort the grouped-indices, thus we would know the positions
    # of these group starts. Then index into the signatures/values
    # and indices with those, giving us the information on how these signatures
    # occur through the length of the input
    sidx = ai.argsort()
    val,aidx = av[sidx],ai[sidx]

    # The identical consecutive signatures are to be removed
    mask = np.r_[True,val[:-1]!=val[1:]]
    v,i = val[mask],aidx[mask]

    # Also, note that we are assigning all 1s as +1 signature and all 0s as -1
    # So, in case the starting signature is a 0, assign a value of 0
    if v[0]==-1:
        v[0] = 0

    # Initialize 1D o/p array, which stores the signatures as +1s and -1s.
    # The bigger level idea is that performing cumsum at the end would give us the
    # desired 1D output
    out1d = np.zeros(len(a),dtype=a.dtype)

    # Assign the values at i positions
    out1d[i] = v

    # Finally cumsum to get desired output
    out1dc = out1d.cumsum()

    # Correct the starting positions based on starting state value
    out1dc[:i[0]] = state

    # Convert to 2D view for mem. and perf. efficiency
    out = np.broadcast_to(out1dc[:,None],a.shape)
    return out

方法2：基于Numba

这是另一个基于numba的内存和perf。效率-

@njit(parallel=True)
def func1(zm, om, out, start_state, cur_state):
    # This outputs 1D version of required output.

    # Start off with the starting given state
    newval = start_state

    # Loop through zipped zeros-all and ones-all masks and in essence do :
    # Switch between zeros and ones based on whether the other ones
    # are occuring through or not, prior to the current state
    for i,(z,o) in enumerate(zip(zm,om)):
        if z and cur_state:
            cur_state = ~cur_state
            newval = 0
        if o and ~cur_state:
            cur_state = ~cur_state
            newval = 1
        out[i] = newval
    return out

def check_all_numba(a, state):
    # Get zeros and ones all masks
    zm = (a==0).all(1)
    om = (a==1).all(1)

    # Decide the starting state
    cur_state = zm.argmax() < om.argmax()

    # Initialize 1D o/p array with given state values
    out1d = np.full(len(a), fill_value=state)
    func1(zm, om, out1d, state, cur_state)

    # Broadcast into the 2D view for memory and perf. efficiency
    return np.broadcast_to(out1d[:,None],a.shape)

Answer 2

您可以利用np.accumulate来做到这一点而无需循环：

R = 5 # 100000
C = 3 # 100

mat   = np.random.randint(2, size=(R,C))
print(mat) # original matrix

state    = np.zeros((1,C))                        # or np.ones((1,C))
mat      = np.concatenate([state,mat])            # insert state row
zRows    = np.isin(np.sum(mat,1),[0,C])           # all zeroes or all ones
iRows    = np.arange(R+1) * zRows.astype(np.int)  # base indexes
mat      = mat[np.maximum.accumulate(iRows)][1:]  # indirection, remove state
print(mat) # modified

#original
[[0 0 1]
 [1 1 1]
 [1 0 1]
 [0 0 0]
 [1 0 1]]

# modified
[[0 0 0]
 [1 1 1]
 [1 1 1]
 [0 0 0]
 [0 0 0]]

它的工作方式是为需要更改的行准备一个间接数组。这是通过np.arange行索引完成的，其中将需要替换的索引设置为零。累积最大索引会将每个替换行映射到它之前的全零或全一行。

例如：

  [ 0, 1, 2, 3, 4, 5 ] # row indexes
  [ 0, 1, 0, 0, 1, 0 ] # rows that are all zeroes or all ones (zRows)
  [ 0, 1, 0, 0, 4, 0 ] # multiplied (iRows)
  [ 0, 1, 1, 1, 4, 4 ] # np.maximum.accumulate

这为我们提供了应从中获取行内容的索引列表。

状态由执行操作之前在矩阵的开头插入并在之后删除的额外行表示。

对于很小的矩阵（5x3），此解决方案的速度会稍慢一些，但对于较大的矩阵（100000x100：0.7秒vs 14秒），它可以使速度提高20倍。

Answer 3

这是一个简单快速的numpy方法：

import numpy as np

def pp():
    m,n = a.shape
    A = a.sum(axis=1)    
    A = np.where((A==0)|(A==n))[0]
    if not A.size:
        return np.ones_like(a) if state else np.zeros_like(a)
    st = np.concatenate([np.arange(A[0]!=0), A, [m]])
    v = a[st[:-1],0]
    if A[0]:
        v[0] = state
    return np.broadcast_to(v.repeat(st[1:]-st[:-1])[:,None],(m,n))

我以此为准

state=0
a = (np.random.random((100000,100))<np.random.random((100000,1))).astype(int)

简单的测试用例：

0.8655898020006134   # me
4.089095343002555    # Alain T.
2.2958932030014694   # Divakar 1
2.2178015549980046   # & 2

如何通过矩阵numpy向量化循环

3 个答案: