Question

我有一个4D矩阵，A（形状[251,6,60,141]）包含很多NaN。我想将这个矩阵重塑成另一个矩阵B（形状[73,6,60,141]）。换句话说，在axis = 0中，我想采用不规则间隔步长的numpy.nanmean（）。有效的方法吗？

我希望下面代码中的循环说明我的愿望，但我认为它不起作用，因为它以（似乎）无休止的RuntimeWarnings循环结束：

＆＃34; /opt/anaconda3/lib/python3.4/site-packages/numpy/lib/nanfunctions.py:598：RuntimeWarning：空切片的平均值 warnings.warn（＆＃34;空切片的意思＆＃34;，RuntimeWarning）＆＃34;

import numpy as np

A = np.full(([251,6,60,141]), np.nan)    # Create matrix A full of NaNs
# Assign some random values in random grid boxes in A
A[0, 1, 2, 3] = 4
A[1, 2, 3, 4] = 5
A[2, 3, 4, 5] = 6
A[3, 4, 5, 6] = 7

# Create the 1D array of the number of rows I want to average together in each interval
intvl = [0, 5, 2, 2, 1, 6, 5, 4, 1, 6, 2, 2, 3, 2, 2, 5, 6, 3, 3, 3, 3, 3, 3, 3, 2, 6, 3, 6, 3, 1, 6, 3, 6, 1, 4, 6, 3, 3, 2, 2, 3, 4, 2, 5, 1, 3, 1, 3, 1, 6, 4, 2, 3, 5, 5, 5, 7, 4, 2, 3, 4, 3, 2, 3, 5, 3, 2, 7, 5, 3, 5, 3, 3, 2]
# Sum the intvl array stepwise
intvl_cs = np.cumsum(intvl)

# Loop to perform the interval summation 
B = np.full(([len(intvl),6,60,141]), np.nan)    # Create the matrix B, intially full of NaNs
for b in np.arange(len(intvl)-1):
    for L in np.arange(6):
        for i in np.arange(60):
            for j in np.arange(141):
                B[b,L,i,j] = np.nanmean(A[intvl_cs[b]:intvl_cs[b+1],L,i,j])

Answer 1

为了提高性能，您可以在使用

替换整个循环块时对求和进行矢量化

for b in np.arange(len(intvl)-1):
    B[b, ...] = np.nanmean(A[intvl_cs[b]:intvl_cs[b+1], ...], axis=0)

如果你这样做，那就让它更具可读性

B = np.zeros((len(intvl_cs) - 1, 6, 60, 141))
# no need to initialize to `nan` if we are touching all values anyways

for b, (start, stop) in enumerate(zip(intvl_cs[:-1], intvl_cs[1:])):
    B[b, ...] = np.nanmean(A[start:stop, ...], axis=0)

在我的机器上，计算在几分之一秒内完成。

Python：对轴的选定间隔使用numpy.nanmean（）

1 个答案: