Question

我有很多这些矩阵数组，其中我想用最低索引的非零条目替换0条目。这可以使用for循环轻松完成：

import numpy as np

input_array = np.array([ 0.01561,  0.01561,  0.02039,  0.02039,  0.02776,  0.02776,
  0.03997,  0.,          0.03997,  0.06243,   0.,          0.,       0.0624662,
  0.11105,  0.,          0.,          0.,          0.11105,  0.24986,
  0.,          0.,          0.,          0.,          0.,          0.,
  0.24986])

for i in range(0,len(input_array)) : 
   if input_array[i] == 0 : 
      input_array[i] = input_array[i-1]

有人建议我是否值得努力？

Answer 1

将numpy解决方案应用于：

Most efficient way to forward-fill NaN values in numpy array

def foo2(arr):
    idx=np.where(arr==0,0,np.arange(len(arr)))
    idx=np.maximum.accumulate(idx)
    return arr[idx]

def foo1(arr):
    arr = arr.copy()
    for i in range(len(arr)):
        if arr[i]==0:
            arr[i] = arr[i-1]
    return arr

对于您的测试阵列arr，速度提升是适度的：

In [67]: timeit foo1(arr)
100000 loops, best of 3: 18.1 µs per loop
In [68]: timeit foo2(arr)
The slowest run took 1387.12 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 11.4 µs per loop

但是有了更大的一个，循环随着大小而增加，阵列版本几乎没有变化：

In [69]: arr1=np.concatenate((arr,arr,arr,arr,arr,arr,arr))
In [70]: timeit foo1(arr1)
10000 loops, best of 3: 116 µs per loop
In [71]: timeit foo2(arr1)
The slowest run took 4.16 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 14.6 µs per loop

idx构造的详细信息：

In [72]: idx=np.arange(len(arr))
In [73]: idx[arr==0]=0
In [74]: idx
Out[74]: 
array([ 0,  1,  2,  3,  4,  5,  6,  0,  8,  9,  0,  0, 12, 13,  0,  0,  0, 17, 18,  0,  0,  0,  0,  0,  0, 25])
In [75]: idx=np.maximum.accumulate(idx)
In [76]: idx
Out[76]: 
array([ 0,  1,  2,  3,  4,  5,  6,  6,  8,  9,  9,  9, 12, 13, 13, 13, 13, 17, 18, 18, 18, 18, 18, 18, 18, 25], dtype=int32)

numpy函数可以替换这个for循环以获得速度吗？

1 个答案: