Python 是否有一个numpy内置项来拒绝列表中的异常值
是否有一个numpy内置来执行以下操作?也就是说,获取一个列表Python 是否有一个numpy内置项来拒绝列表中的异常值,python,numpy,Python,Numpy,是否有一个numpy内置来执行以下操作?也就是说,获取一个列表d,并返回一个列表filtered\u d,其中根据d中的点的某些假定分布删除了任何外围元素 import numpy as np def reject_outliers(data): m = 2 u = np.mean(data) s = np.std(data) filtered = [e for e in data if (u - 2 * s < e < u + 2 * s)]
d
,并返回一个列表filtered\u d
,其中根据d
中的点的某些假定分布删除了任何外围元素
import numpy as np
def reject_outliers(data):
m = 2
u = np.mean(data)
s = np.std(data)
filtered = [e for e in data if (u - 2 * s < e < u + 2 * s)]
return filtered
>>> d = [2,4,5,1,6,5,40]
>>> filtered_d = reject_outliers(d)
>>> print filtered_d
[2,4,5,1,6,5]
将numpy导入为np
def拒绝_异常值(数据):
m=2
u=np.平均值(数据)
s=np.std(数据)
过滤=[e表示数据中的e,如果(u-2*s>>d=[2,4,5,1,6,5,40]
>>>过滤的\u d=拒绝异常值(d)
>>>打印过滤的
[2,4,5,1,6,5]
我之所以说“类似”,是因为函数可能允许这些分布中的不同分布(泊松分布、高斯分布等)和不同的异常值阈值(如我在这里使用的
m
)。此方法与您的方法几乎相同,只是更多的numpy(也仅适用于numpy数组):
def拒绝_异常值(数据,m=2):
返回数据[abs(数据-np.平均值(数据))
处理异常值时,重要的一点是,应尽量使用稳健的估计值。分布的平均值会因异常值而产生偏差,但中位数会小得多
基于eumiro的答案:
def reject_outliers(data, m = 2.):
d = np.abs(data - np.median(data))
mdev = np.median(d)
s = d/mdev if mdev else 0.
return data[s<m]
def拒绝_异常值(数据,m=2):
d=np.abs(数据-np.median(数据))
mdev=np.中值(d)
如果mdev为0,则s=d/mdev。
返回数据[s建立在Benjamin的基础上,使用pandas.Series
,并替换:
def拒绝异常值(sr,iq范围=0.5):
pcnt=(1-智商范围)/2
qlow,中位数,qhigh=sr.dropna().分位数([pcnt,0.50,1-pcnt])
iqr=qhigh-qlow
返回sr[(sr-中值).abs()另一种方法是对标准偏差进行稳健估计(假设高斯统计)。查找在线计算器,我发现90%的百分位对应于1.2815σ,95%的百分位对应于1.645σ()
举个简单的例子:
import numpy as np
# Create some random numbers
x = np.random.normal(5, 2, 1000)
# Calculate the statistics
print("Mean= ", np.mean(x))
print("Median= ", np.median(x))
print("Max/Min=", x.max(), " ", x.min())
print("StdDev=", np.std(x))
print("90th Percentile", np.percentile(x, 90))
# Add a few large points
x[10] += 1000
x[20] += 2000
x[30] += 1500
# Recalculate the statistics
print()
print("Mean= ", np.mean(x))
print("Median= ", np.median(x))
print("Max/Min=", x.max(), " ", x.min())
print("StdDev=", np.std(x))
print("90th Percentile", np.percentile(x, 90))
# Measure the percentile intervals and then estimate Standard Deviation of the distribution, both from median to the 90th percentile and from the 10th to 90th percentile
p90 = np.percentile(x, 90)
p10 = np.percentile(x, 10)
p50 = np.median(x)
# p50 to p90 is 1.2815 sigma
rSig = (p90-p50)/1.2815
print("Robust Sigma=", rSig)
rSig = (p90-p10)/(2*1.2815)
print("Robust Sigma=", rSig)
我得到的结果是:
Mean= 4.99760520022
Median= 4.95395274981
Max/Min= 11.1226494654 -2.15388472011
Sigma= 1.976629928
90th Percentile 7.52065379649
Mean= 9.64760520022
Median= 4.95667658782
Max/Min= 2205.43861943 -2.15388472011
Sigma= 88.6263902244
90th Percentile 7.60646688694
Robust Sigma= 2.06772555531
Robust Sigma= 1.99878292462
这接近于预期值2
如果我们要删除高于/低于5个标准偏差的点(1000个点,我们希望1个值>3个标准偏差):
我不知道哪种方法更有效/更稳健当距离中位数的中位数为0时,Benjamin Bannier的答案会产生一个通过,因此我发现这个修改版本对于下面示例中给出的情况更有用
def reject_outliers_2(data, m=2.):
d = np.abs(data - np.median(data))
mdev = np.median(d)
s = d / (mdev if mdev else 1.)
return data[s < m]
给出:
[[10, 10, 10, 17, 10, 10]] # 17 is not filtered
[10, 10, 10, 10, 10] # 17 is filtered (it's distance, 7, is greater than m)
我想做一些类似的事情,除了将数字设置为NaN,而不是从数据中删除它,因为如果删除它,则会更改长度,这可能会导致打印混乱(即,如果只从表中的一列中删除异常值,但需要它与其他列保持相同,以便可以相互打印)
为此,我使用了:
我想在这个答案中提供两种方法,基于“z分数”的解决方案和基于“IQR”的解决方案
此答案中提供的代码适用于单个dimnumpy
阵列和多个numpy
阵列
让我们先导入一些模块
import collections
import numpy as np
import scipy.stats as stat
from scipy.stats import iqr
基于z分数的方法
此方法将测试数字是否超出三个标准差。根据此规则,如果值为异常值,则该方法将返回true,如果不是,则返回false
def sd_outlier(x, axis = None, bar = 3, side = 'both'):
assert side in ['gt', 'lt', 'both'], 'Side should be `gt`, `lt` or `both`.'
d_z = stat.zscore(x, axis = axis)
if side == 'gt':
return d_z > bar
elif side == 'lt':
return d_z < -bar
elif side == 'both':
return np.abs(d_z) > bar
最后,如果要过滤掉异常值,请使用numpy
选择器
祝你愉快。考虑一下,当你的标准偏差由于巨大的异常值而变得非常大时,上述所有方法都会失败
(Simalar表示平均计算失败,应该计算中位数。不过,平均值“更容易出现stdDv这样的错误”。)
您可以尝试迭代应用算法,或者使用四分位数范围进行过滤:
(此处“因子”与n*西格玛范围相关,但仅当数据服从高斯分布时)
将numpy导入为np
def SORTOUTLIERS(数据输入,系数):
quant3,quant1=np.百分位(数据输入[75,25])
iqr=量程3-量程1
iqrSigma=iqr/1.34896
medData=np.中值(数据单位)
dataOut=[x代表数据输入if中的x((x>medData-因子*iqrSigma)和(x
如果您想获取异常值的索引位置,idx\U列表将返回它
def reject_outliers(data, m = 2.):
d = np.abs(data - np.median(data))
mdev = np.median(d)
s = d/mdev if mdev else 0.
data_range = np.arange(len(data))
idx_list = data_range[s>=m]
return data[s<m], idx_list
data_points = np.array([8, 10, 35, 17, 73, 77])
print(reject_outliers(data_points))
after rejection: [ 8 10 35 17], index positions of outliers: [4 5]
def拒绝_异常值(数据,m=2):
d=np.abs(数据-np.median(数据))
mdev=np.中值(d)
如果mdev为0,则s=d/mdev。
数据范围=np.arange(len(数据))
idx\u列表=数据范围[s>=m]
返回一组图像的数据[s(每个图像有3个维度),其中我想拒绝我使用的每个像素的异常值:
mean = np.mean(imgs, axis=0)
std = np.std(imgs, axis=0)
mask = np.greater(0.5 * std + 1, np.abs(imgs - mean))
masked = np.multiply(imgs, mask)
然后可以计算平均值:
masked_mean = np.divide(np.sum(masked, axis=0), np.sum(mask, axis=0))
(我将其用于背景减法)在这里,我在x
中找到异常值,并用它们周围的一个点窗口(win
)的中位数替换它们(从Benjamin Bannier处获取中位数偏差)
def outlier\u平滑器(x,m=3,win=3,plots=False):
''发现x中的异常值,点>m*mdev(x)[mdev:中值偏差]
并将其替换为其周围的赢点数中值“
x_corr=np.拷贝(x)
d=np.abs(x-np.median(x))
mdev=np.中值(d)
idxs_异常值=np.非零(d>m*mdev)[0]
对于idxs_中的i异常值:
如果i-win<0:
x_corr[i]=np.median(np.append(x[0:i],x[i+1:i+win+1]))
如果i+win+1>len(x):
x_corr[i]=np.median(np.append(x[i-win:i],x[i+1:len(x)])
其他:
x_corr[i]=np.中值(np.追加(x[i-win:i],x[i+1:i+win+1]))
如果绘制:
plt.figure('异常值更平滑',clear=True)
plt.绘图(x,label='orig',lw=5)
plt.plot(idxs_异常值,x[idxs_异常值],'ro',label='outliers')
plt.绘图(x_corr,'-o',label='corrected')
plt。
def reject_outliers(data, m=2):
stdev = np.std(data)
mean = np.mean(data)
maskMin = mean - stdev * m
maskMax = mean + stdev * m
mask = np.ma.masked_outside(data, maskMin, maskMax)
print('Masking values outside of {} and {}'.format(maskMin, maskMax))
return mask
import collections
import numpy as np
import scipy.stats as stat
from scipy.stats import iqr
def sd_outlier(x, axis = None, bar = 3, side = 'both'):
assert side in ['gt', 'lt', 'both'], 'Side should be `gt`, `lt` or `both`.'
d_z = stat.zscore(x, axis = axis)
if side == 'gt':
return d_z > bar
elif side == 'lt':
return d_z < -bar
elif side == 'both':
return np.abs(d_z) > bar
def q1(x, axis = None):
return np.percentile(x, 25, axis = axis)
def q3(x, axis = None):
return np.percentile(x, 75, axis = axis)
def iqr_outlier(x, axis = None, bar = 1.5, side = 'both'):
assert side in ['gt', 'lt', 'both'], 'Side should be `gt`, `lt` or `both`.'
d_iqr = iqr(x, axis = axis)
d_q1 = q1(x, axis = axis)
d_q3 = q3(x, axis = axis)
iqr_distance = np.multiply(d_iqr, bar)
stat_shape = list(x.shape)
if isinstance(axis, collections.Iterable):
for single_axis in axis:
stat_shape[single_axis] = 1
else:
stat_shape[axis] = 1
if side in ['gt', 'both']:
upper_range = d_q3 + iqr_distance
upper_outlier = np.greater(x - upper_range.reshape(stat_shape), 0)
if side in ['lt', 'both']:
lower_range = d_q1 - iqr_distance
lower_outlier = np.less(x - lower_range.reshape(stat_shape), 0)
if side == 'gt':
return upper_outlier
if side == 'lt':
return lower_outlier
if side == 'both':
return np.logical_or(upper_outlier, lower_outlier)
import numpy as np
def sortoutOutliers(dataIn,factor):
quant3, quant1 = np.percentile(dataIn, [75 ,25])
iqr = quant3 - quant1
iqrSigma = iqr/1.34896
medData = np.median(dataIn)
dataOut = [ x for x in dataIn if ( (x > medData - factor* iqrSigma) and (x < medData + factor* iqrSigma) ) ]
return(dataOut)
def reject_outliers(data, m = 2.):
d = np.abs(data - np.median(data))
mdev = np.median(d)
s = d/mdev if mdev else 0.
data_range = np.arange(len(data))
idx_list = data_range[s>=m]
return data[s<m], idx_list
data_points = np.array([8, 10, 35, 17, 73, 77])
print(reject_outliers(data_points))
after rejection: [ 8 10 35 17], index positions of outliers: [4 5]
mean = np.mean(imgs, axis=0)
std = np.std(imgs, axis=0)
mask = np.greater(0.5 * std + 1, np.abs(imgs - mean))
masked = np.multiply(imgs, mask)
masked_mean = np.divide(np.sum(masked, axis=0), np.sum(mask, axis=0))
def outlier_smoother(x, m=3, win=3, plots=False):
''' finds outliers in x, points > m*mdev(x) [mdev:median deviation]
and replaces them with the median of win points around them '''
x_corr = np.copy(x)
d = np.abs(x - np.median(x))
mdev = np.median(d)
idxs_outliers = np.nonzero(d > m*mdev)[0]
for i in idxs_outliers:
if i-win < 0:
x_corr[i] = np.median(np.append(x[0:i], x[i+1:i+win+1]))
elif i+win+1 > len(x):
x_corr[i] = np.median(np.append(x[i-win:i], x[i+1:len(x)]))
else:
x_corr[i] = np.median(np.append(x[i-win:i], x[i+1:i+win+1]))
if plots:
plt.figure('outlier_smoother', clear=True)
plt.plot(x, label='orig.', lw=5)
plt.plot(idxs_outliers, x[idxs_outliers], 'ro', label='outliers')
plt.plot(x_corr, '-o', label='corrected')
plt.legend()
return x_corr