Python:滑动窗口均值,忽略丢失的数据 [英] Python: Sliding windowed mean, ignoring missing data
问题描述
我当前正在尝试处理实验时间序列数据集,该数据集缺少值.我想在处理nan值的同时计算该数据集随时间的滑动窗口均值.对我而言,正确的方法是在每个窗口内计算有限元素的总和,然后将其除以它们的数量.这种非线性迫使我使用非卷积方法来面对这个问题,因此在该过程的这一部分中我遇到了严重的时间瓶颈.作为我要完成的工作的代码示例,我提出以下内容:
I am currently trying to process an experimental timeseries dataset, which has missing values. I would like to calculate the sliding windowed mean of this dataset along time, while handling nan values. The correct way for me to do it is to compute inside each window the sum of the finite elements and divide it with their number. This nonlinearity forces me to use non convolutional methods to face this problem, thus I have a severe time bottleneck in this part of the process. As a code example of what I am trying to accomplish I present the following:
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:
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 ]
可以通过numpy操作产生此结果,而无需使用for循环吗?
Can this result be produced by numpy operations, without using a for loop?
推荐答案
Please note that this would have one extra element on either sides.
If you are working with 2D
data, we can use Scipy's 2D convolution
.
方法-
def original_app(data, win_size):
#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
return result
def numpy_app(data, win_size):
mask = np.isnan(data)
K = np.ones(win_size,dtype=int)
out = np.convolve(np.where(mask,0,data), K)/np.convolve(~mask,K)
return out[1:-1] # Slice out the one-extra elems on sides
样品运行-
In [118]: #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)] = np.nan
...:
In [119]: original_app(data, win_size = 3)
Out[119]:
array([ 0.88356487, 0.86829731, 0.85249541, 0.83776219, nan,
nan, 0.61054015, 0.63111926, 0.63111926, 0.65169837,
0.1857301 , 0.58335324, 0.42088104, 0.5384565 , 0.31027752,
0.40768907, 0.3478563 , 0.34089655, 0.55462903, 0.71784816,
0.93195716, nan, 0.41635575, 0.52211653, 0.65053379,
0.76762282, 0.72888574, 0.35250449, 0.35250449, 0.14500637,
0.06997668, 0.22582318, 0.18621848, 0.36320784, 0.19926647,
0.24506199, 0.09983572, 0.47595439, 0.79792941, 0.5982114 ,
0.42389375, 0.28944089, 0.36246113, 0.48088139, 0.71105449,
0.60234163, 0.40012839, 0.45100475, 0.41768466, 0.41768466])
In [120]: numpy_app(data, win_size = 3)
__main__:36: RuntimeWarning: invalid value encountered in divide
Out[120]:
array([ 0.88356487, 0.86829731, 0.85249541, 0.83776219, nan,
nan, 0.61054015, 0.63111926, 0.63111926, 0.65169837,
0.1857301 , 0.58335324, 0.42088104, 0.5384565 , 0.31027752,
0.40768907, 0.3478563 , 0.34089655, 0.55462903, 0.71784816,
0.93195716, nan, 0.41635575, 0.52211653, 0.65053379,
0.76762282, 0.72888574, 0.35250449, 0.35250449, 0.14500637,
0.06997668, 0.22582318, 0.18621848, 0.36320784, 0.19926647,
0.24506199, 0.09983572, 0.47595439, 0.79792941, 0.5982114 ,
0.42389375, 0.28944089, 0.36246113, 0.48088139, 0.71105449,
0.60234163, 0.40012839, 0.45100475, 0.41768466, 0.41768466])
运行时测试-
In [122]: #Construct sample data
...: n = 50000
...: n_miss = 20000
...: win_size = 3
...: data= np.random.random(n)
...: data[np.random.randint(0,n-1, n_miss)] = np.nan
...:
In [123]: %timeit original_app(data, win_size = 3)
1 loops, best of 3: 1.51 s per loop
In [124]: %timeit numpy_app(data, win_size = 3)
1000 loops, best of 3: 1.09 ms per loop
In [125]: import pandas as pd
# @jdehesa's pandas solution
In [126]: %timeit pd.Series(data).rolling(window=3, min_periods=1).mean()
100 loops, best of 3: 3.34 ms per loop
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