Question

除此efficient algorithm for list edits之外，我正在为另一个“循环计算”寻找更有效的算法。这次我有一个矩阵：

grid_z1 = [[1,2,3],
           [4,5,6],
           [7,8,9]]

并且用户可以输入几个参数：目标是，将矩阵内的值更改为次高的参数值（如果矩阵值高于max（参数），则将其更改为nan例如，当用户输入“4”和“7”时，矩阵值“5”将变为“7”（=输入值的下一个最高值）。例如：

h = [2, 7, 4] # user entered this three values
grid_z1 = [[2, 2, 4],
           [4, 7, 7],
           [7, nan, nan]] # this should be my output

此外，我想计算更改为给定值的值的数量。在我的例子中，这应该是[2,2,3] - ＆gt; 2x2,2x4,3x7

h.sort()
h.reverse()

count = [0]*len(h)

for i in range(len(grid_z1)):
    for j in range(len(grid_z1[0])):
        if grid_z1[i][j] > max(h):
            grid_z1[i][j] = float('NaN')
        else:
            for k in range(len(h)-1):
                if grid_z1[i][j] <= h[k] and grid_z1[i][j] > h[k+1]:
                    grid_z1[i][j] = h[k]
                    count[k] += 1
            if grid_z1[i][j] <= min(h):
                grid_z1[i][j] = min(h)
                count[-1] += 1

print grid_z1
print count

但又一次很慢。遗憾的是，我不理解zip方法足以将其用于更复杂的算法。

Answer 1

使用bisect模块：

from bisect import bisect_left
def solve(matrix, param):
    param.sort()             #Sort the params passed by user
    maxx = param[-1]         #Find max
    for row in matrix:
        # If item in the row is greater than `maxx` then use Nan
        # else use bisect_right to get the next highest item from
        # params in O(log N) time.
        yield [float('nan') if item > maxx else
                                 param[bisect_left(param, item)] for item in row]

grid_z1 = [[1,2,3],
           [4,5,6],
           [7,8,9]]
print list(solve(grid_z1, [2, 7, 4]))

输出

[[2, 2, 4], [4, 7, 7], [7, nan, nan]]

Answer 2

for i,row in enumerate(g):
        g[i] = [float('nan') if item > max(h) else min([x for x in h if x >= item]) for item in row]

高效的矩阵计算算法

2 个答案: