Question

我想最小化Python中网格上的成本函数。我有两个变量x和y，可以计算为

find . -type f -exec bash -c '
  logcmd() { printf "%q " "$@"; printf "\n"; }  # a more accurate replacement for echo
  for f; do
    logcmd mv "$f" "${f%.*}${0}.${f##*.}"
  done' "$i" {} +

换句话说，网格点（i + 1，j + 1）取决于两个内核x[i+1,j+1], y[i+1,j+1] = f(x[i,j], x[i+1,j], x[i,j+1], foo[i,j], bar[i,j])和foo，并且它是相邻节点（i，j + 1）（i + 1， j）和（i，j）。可以在下面看到一个玩具示例

bar

但是，这是一个非常低效的解决方案。特别是因为N可能会任意大。所以我的问题是：哪种/最有效的计算方法是？使用迭代器（我尝试过np.nditer并没有取得很大的成功），或者使用Numba，或者在Numpy中可以做一些花哨的技巧吗？我已经开始用Numpy研究ufuncs和import numpy as np N = 20 ivec = np.arange(N) jvec = np.arange(N) # Kernels foo = np.sin(ivec[:,None] * jvec[None,:]) bar = np.cos(ivec[:,None] + jvec[None,:]) # We want to find the total cost for traversing over the matrix d = np.zeros((N,N)) # And store the optimal path indices = np.zeros((N,N), "int") for i in range(N-1): for j in range(N-1): # Compute all posibilities for reaching current node dd = [ d[i+1,j] + foo[i,j], d[i,j+1] + bar[i,j], d[i,j] + foo[i,j] * bar[i,j] ] # And find and store the minimim path indices[i+1,j+1] = np.argmin(dd) d[i+1,j+1] = dd[indices[i+1,j+1]] print(d[-1,-1])，但无法立即看到解决方案。

请注意，在ufunc.accumulate中，foo和bar比玩具示例中的要复杂。

Answer 1

如前所述，使用Numba可能是在N很大时使代码更快的最简单方法。

import numba
import numpy as np

@numba.jit(nopython=True)
def dp(foo, bar):
    N = foo.shape[0]

    # We want to find the total cost for traversing over the matrix
    d = np.zeros((N,N))

    # And store the optimal path
    indices = np.zeros((N,N), dtype=np.int64)

    for i in range(N-1):
        for j in range(N-1):

            # Compute all posibilities for reaching current node
            dd = np.array([
                d[i+1,j] + foo[i,j],
                d[i,j+1] + bar[i,j],
                d[i,j] + foo[i,j] * bar[i,j]
            ])

            # And find and store the minimim path
            indices[i+1,j+1] = np.argmin(dd)
            d[i+1,j+1] = dd[indices[i+1,j+1]]
    return d[-1,-1]

Python中的动态编程网格

1 个答案: