我目前正在处理一组实验时间序列数据,其中存在缺失值。我希望计算该数据集沿时间轴的滑动窗口均值,并处理nan值。我需要在每个窗口内计算有限元素的总和并将其除以它们的数量来完成这项任务。这种非线性计算方式迫使我使用非卷积方法来解决这个问题,因此在处理过程中出现了严重的时间瓶颈。以下是我尝试完成的代码示例:
import numpy as np
#Construct sample data
n = 50
n_miss = 20
win_size = 3
data= np.random.random(50)
data[np.random.randint(0,n-1, n_miss)] = None
#Compute mean
result = np.zeros(data.size)
for count in range(data.size):
part_data = data[max(count - (win_size - 1) / 2, 0): min(count + (win_size + 1) / 2, data.size)]
mask = np.isfinite(part_data)
if np.sum(mask) != 0:
result[count] = np.sum(part_data[mask]) / np.sum(mask)
else:
result[count] = None
print 'Input:\t',data
print 'Output:\t',result
输出结果为:
Input: [ 0.47431791 0.17620835 0.78495647 0.79894688 0.58334064 0.38068788
0.87829696 nan 0.71589171 nan 0.70359557 0.76113969
0.13694387 0.32126573 0.22730891 nan 0.35057169 nan
0.89251851 0.56226354 0.040117 nan 0.37249799 0.77625334
nan nan nan nan 0.63227417 0.92781944
0.99416471 0.81850753 0.35004997 nan 0.80743783 0.60828597
nan 0.01410721 nan nan 0.6976317 nan
0.03875394 0.60924066 0.22998065 nan 0.34476729 0.38090961
nan 0.2021964 ]
Output: [ 0.32526313 0.47849424 0.5867039 0.72241466 0.58765847 0.61410849
0.62949242 0.79709433 0.71589171 0.70974364 0.73236763 0.53389305
0.40644977 0.22850617 0.27428732 0.2889403 0.35057169 0.6215451
0.72739103 0.49829968 0.30119027 0.20630749 0.57437567 0.57437567
0.77625334 nan nan 0.63227417 0.7800468 0.85141944
0.91349722 0.7209074 0.58427875 0.5787439 0.7078619 0.7078619
0.31119659 0.01410721 0.01410721 0.6976317 0.6976317 0.36819282
0.3239973 0.29265842 0.41961066 0.28737397 0.36283845 0.36283845
0.29155301 0.2021964 ]
能否使用numpy操作产生此结果,而不使用for循环?
rolling
函数? - jdehesanumpy/lib/nanfunctions.py
包含了主要和辅助函数,用于在计算时过滤掉nan
值。例如,nanmean
使用arr, mask = _replace_nan(a, 0)
并计算mean=sum(arr)/sum(mask)
。 - hpaulj