我有一个带有速度
列的 Pandas 数据框,其中偶尔有噪音(数据来自 Garmin,代表运行期间捕获的数据)。
我正在尝试找到一种对相邻点进行平均的方法,但是当我遇到这样的事情时
9.112273445
164.5779550738
84.4553498412
4.231089359
4.3740439706
我陷入了无限循环。
我的算法相当幼稚:
# Get list of indices in which value is great than 6:
idx = z[(z['speed']>=6)].index
while list(idx) != []:
for i in idx:
# check if out of bounds
if i + 1 >= len(z):
z.iloc[i, z.columns.get_indexer(['speed'])] = (z['speed'].ix[i-2] + z['speed'].ix[i-1])/2
elif i - 1 < 0:
z.iloc[i, z.columns.get_indexer(['speed'])] = (z['speed'].ix[i+1] + z['speed'].ix[i+2])/2
else:
z.iloc[i, z.columns.get_indexer(['speed'])] = (z['speed'].ix[i-1] + z['speed'].ix[i+1])/2
idx = z[(z['speed']>=6)].index
当然,问题是当我有两个非常大的相邻值时,就会陷入无限循环。
我正在应用此过滤器(使用汉宁窗)来消除随机噪声:SciPy Cookbook SignalSmooth ,但它没有处理数据中的这些大峰值。
除了丢弃它们或将它们设置为恒定值之外,还有其他简单的方法来处理它们吗?
编辑
我正在测试的值是:
0 NaN
1 3.508394
2 5.097879
3 7.743824
4 9.138245
5 13.315918
6 12.836310
7 12.001393
8 15.815223
9 0.000000
10 16.622944
11 9.061864
12 2.089729
13 2.710874
Name: speed, dtype: float64
最佳答案
如果您想“桥接”大于六的值,您可以这样做:
import numpy as np
# locate outliers and adjacent values
outliers = np.r_[False, (~np.isfinite(data)) | (data > 6), False]
if np.any(outliers):
boundaries = np.where(outliers[:-1] != outliers[1:])[0]
lb = boundaries[::2]
rb = boundaries[1::2]
# special case if leftmost and/or rightmost values are outliers
lv = data[lb-1]
if lb[0] == 0:
lv[0] = data[rb[0]]
rv = data[rb % len(data)]
if rb[-1] == len(data):
rv[-1] = data[lb[-1]-1]
# create fill values; use a bit of trickery to keep it vectorised
lengths = rb-lb
fv = np.repeat((rv-lv)/(lengths+1), lengths)
sw = np.cumsum(lengths[:-1])
fv[sw] += fv[sw-1] - rv[:-1] + lv[1:]
fv[0] += lv[0]
fv = np.cumsum(fv)
# place them
out = data.copy()
out[outliers[1:-1]] = fv
else:
out = data.copy()
关于python - 从 gpx 数据中滤除噪音,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/43083342/