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我正在尝试从带有日期的 GPS 高程位移散点图中过滤掉一些异常值

我正在尝试使用 df.rolling 计算每个窗口的中值和标准偏差，然后如果它大于 3 个标准偏差则删除该点。

但是，我想不出一种方法来遍历列并比较滚动计算的中值。

这是我到目前为止的代码

import pandas as pd
import numpy as np

def median_filter(df, window):
    cnt = 0
    median = df['b'].rolling(window).median()
    std = df['b'].rolling(window).std()
    for row in df.b:
      #compare each value to its median




df = pd.DataFrame(np.random.randint(0,100,size=(100,2)), columns = ['a', 'b'])

median_filter(df, 10)

如何循环并比较每个点并将其删除？

4

3 回答 3

17

只需过滤数据框

df['median']= df['b'].rolling(window).median()
df['std'] = df['b'].rolling(window).std()

#filter setup
df = df[(df.b <= df['median']+3*df['std']) & (df.b >= df['median']-3*df['std'])]
于 2017-10-27T03:55:23.603 回答
0

这是我对创建中值滤波器的看法：

def median_filter(num_std=3):
    def _median_filter(x):
        _median = np.median(x)
        _std = np.std(x)
        s = x[-1]
        return s if s >= _median - num_std * _std and s <= _median + num_std * _std else np.nan
    return _median_filter

df.y.rolling(window).apply(median_filter(num_std=3), raw=True)
于 2019-10-22T14:49:38.493 回答
0

很可能有一种更熊猫的方式来做到这一点 - 这有点像黑客，依靠一种将原始 df 的索引映射到每个滚动窗口的手动方式。（我选择了 6 号）。直到第 6 行的记录与第一个窗口相关联；第 7 行是第二个窗口，依此类推。

n = 100
df = pd.DataFrame(np.random.randint(0,n,size=(n,2)), columns = ['a','b'])

## set window size
window=6
std = 1  # I set it at just 1; with real data and larger windows, can be larger

## create df with rolling stats, upper and lower bounds
bounds = pd.DataFrame({'median':df['b'].rolling(window).median(),
'std':df['b'].rolling(window).std()})

bounds['upper']=bounds['median']+bounds['std']*std
bounds['lower']=bounds['median']-bounds['std']*std

## here, we set an identifier for each window which maps to the original df
## the first six rows are the first window; then each additional row is a new window
bounds['window_id']=np.append(np.zeros(window),np.arange(1,n-window+1))

## then we can assign the original 'b' value back to the bounds df
bounds['b']=df['b']

## and finally, keep only rows where b falls within the desired bounds
bounds.loc[bounds.eval("lower<b<upper")]
于 2017-10-26T23:13:59.737 回答