问题:检测并排除熊猫数据框中的异常值

我有一个只有几列的熊猫数据框。

现在我知道某些行是基于某个列值的离群值。

例如

“ Vol”列的所有值都在周围,12xx而一个值是4000(离群值)。

现在,我想排除具有Vol此类列的行。

因此,从本质上讲,我需要在数据帧上放置一个过滤器,以便我们选择某一列的值在均值例如3个标准差以内的所有行。

有什么优雅的方法可以做到这一点?

I have a pandas data frame with few columns.

Now I know that certain rows are outliers based on a certain column value.

For instance

column ‘Vol’ has all values around 12xx and one value is 4000 (outlier).

Now I would like to exclude those rows that have Vol column like this.

So, essentially I need to put a filter on the data frame such that we select all rows where the values of a certain column are within, say, 3 standard deviations from mean.

What is an elegant way to achieve this?


回答 0

如果您的数据框中有多个列,并且想要删除至少一列中具有异常值的所有行,则以下表达式可以一次性完成。

df = pd.DataFrame(np.random.randn(100, 3))

from scipy import stats
df[(np.abs(stats.zscore(df)) < 3).all(axis=1)]

描述:

  • 对于每列,首先要计算列中每个值相对于列均值和标准差的Z分数。
  • 然后取Z分数的绝对值,因为方向无关紧要,只有方向低于阈值时才行。
  • all(axis = 1)确保对于每一行,所有列均满足约束。
  • 最后,此条件的结果用于索引数据帧。

If you have multiple columns in your dataframe and would like to remove all rows that have outliers in at least one column, the following expression would do that in one shot.

df = pd.DataFrame(np.random.randn(100, 3))

from scipy import stats
df[(np.abs(stats.zscore(df)) < 3).all(axis=1)]

description:

  • For each column, first it computes the Z-score of each value in the column, relative to the column mean and standard deviation.
  • Then is takes the absolute of Z-score because the direction does not matter, only if it is below the threshold.
  • all(axis=1) ensures that for each row, all column satisfy the constraint.
  • Finally, result of this condition is used to index the dataframe.

回答 1

boolean就像在索引中那样使用索引numpy.array

df = pd.DataFrame({'Data':np.random.normal(size=200)})
# example dataset of normally distributed data. 

df[np.abs(df.Data-df.Data.mean()) <= (3*df.Data.std())]
# keep only the ones that are within +3 to -3 standard deviations in the column 'Data'.

df[~(np.abs(df.Data-df.Data.mean()) > (3*df.Data.std()))]
# or if you prefer the other way around

对于系列,它类似于:

S = pd.Series(np.random.normal(size=200))
S[~((S-S.mean()).abs() > 3*S.std())]

Use boolean indexing as you would do in numpy.array

df = pd.DataFrame({'Data':np.random.normal(size=200)})
# example dataset of normally distributed data. 

df[np.abs(df.Data-df.Data.mean()) <= (3*df.Data.std())]
# keep only the ones that are within +3 to -3 standard deviations in the column 'Data'.

df[~(np.abs(df.Data-df.Data.mean()) > (3*df.Data.std()))]
# or if you prefer the other way around

For a series it is similar:

S = pd.Series(np.random.normal(size=200))
S[~((S-S.mean()).abs() > 3*S.std())]

回答 2

对于每个dataframe列,您可以使用以下方法获得分位数:

q = df["col"].quantile(0.99)

然后过滤:

df[df["col"] < q]

如果需要删除上下限离群值,则将条件与AND语句结合使用:

q_low = df["col"].quantile(0.01)
q_hi  = df["col"].quantile(0.99)

df_filtered = df[(df["col"] < q_hi) & (df["col"] > q_low)]

For each of your dataframe column, you could get quantile with:

q = df["col"].quantile(0.99)

and then filter with:

df[df["col"] < q]

If one need to remove lower and upper outliers, combine condition with an AND statement:

q_low = df["col"].quantile(0.01)
q_hi  = df["col"].quantile(0.99)

df_filtered = df[(df["col"] < q_hi) & (df["col"] > q_low)]

回答 3

此答案与@tanemaki提供的答案相似,但使用的是lambda表达式而不是scipy stats

df = pd.DataFrame(np.random.randn(100, 3), columns=list('ABC'))

df[df.apply(lambda x: np.abs(x - x.mean()) / x.std() < 3).all(axis=1)]

要过滤只有一个列(例如“ B”)在三个标准差以内的DataFrame:

df[((df.B - df.B.mean()) / df.B.std()).abs() < 3]

有关如何在滚动基础上应用此z评分的信息,请参见此处:将滚动z评分应用于熊猫数据框

This answer is similar to that provided by @tanemaki, but uses a lambda expression instead of scipy stats.

df = pd.DataFrame(np.random.randn(100, 3), columns=list('ABC'))

df[df.apply(lambda x: np.abs(x - x.mean()) / x.std() < 3).all(axis=1)]

To filter the DataFrame where only ONE column (e.g. ‘B’) is within three standard deviations:

df[((df.B - df.B.mean()) / df.B.std()).abs() < 3]

See here for how to apply this z-score on a rolling basis: Rolling Z-score applied to pandas dataframe


回答 4

#------------------------------------------------------------------------------
# accept a dataframe, remove outliers, return cleaned data in a new dataframe
# see http://www.itl.nist.gov/div898/handbook/prc/section1/prc16.htm
#------------------------------------------------------------------------------
def remove_outlier(df_in, col_name):
    q1 = df_in[col_name].quantile(0.25)
    q3 = df_in[col_name].quantile(0.75)
    iqr = q3-q1 #Interquartile range
    fence_low  = q1-1.5*iqr
    fence_high = q3+1.5*iqr
    df_out = df_in.loc[(df_in[col_name] > fence_low) & (df_in[col_name] < fence_high)]
    return df_out
#------------------------------------------------------------------------------
# accept a dataframe, remove outliers, return cleaned data in a new dataframe
# see http://www.itl.nist.gov/div898/handbook/prc/section1/prc16.htm
#------------------------------------------------------------------------------
def remove_outlier(df_in, col_name):
    q1 = df_in[col_name].quantile(0.25)
    q3 = df_in[col_name].quantile(0.75)
    iqr = q3-q1 #Interquartile range
    fence_low  = q1-1.5*iqr
    fence_high = q3+1.5*iqr
    df_out = df_in.loc[(df_in[col_name] > fence_low) & (df_in[col_name] < fence_high)]
    return df_out

回答 5

对于数据框中的每个系列,您可以使用betweenquantile删除异常值。

x = pd.Series(np.random.normal(size=200)) # with outliers
x = x[x.between(x.quantile(.25), x.quantile(.75))] # without outliers

For each series in the dataframe, you could use between and quantile to remove outliers.

x = pd.Series(np.random.normal(size=200)) # with outliers
x = x[x.between(x.quantile(.25), x.quantile(.75))] # without outliers

回答 6

由于我还没有看到涉及数字非数字属性的答案,因此这里是一个补充性答案。

您可能只想将离群值放在数字属性上(分类变量几乎不可能是离群值)。

功能定义

我还扩展了@tanemaki的建议,以在还存在非数字属性时处理数据:

from scipy import stats

def drop_numerical_outliers(df, z_thresh=3):
    # Constrains will contain `True` or `False` depending on if it is a value below the threshold.
    constrains = df.select_dtypes(include=[np.number]) \
        .apply(lambda x: np.abs(stats.zscore(x)) < z_thresh, reduce=False) \
        .all(axis=1)
    # Drop (inplace) values set to be rejected
    df.drop(df.index[~constrains], inplace=True)

用法

drop_numerical_outliers(df)

想象一个数据集df,其中包含有关房屋的一些值:胡同,土地轮廓,售价,…例如:数据文档

首先,您要可视化散点图上的数据(z分数Thresh = 3):

# Plot data before dropping those greater than z-score 3. 
# The scatterAreaVsPrice function's definition has been removed for readability's sake.
scatterAreaVsPrice(df)

在此之前-Gr Liv Area与SalePrice

# Drop the outliers on every attributes
drop_numerical_outliers(train_df)

# Plot the result. All outliers were dropped. Note that the red points are not
# the same outliers from the first plot, but the new computed outliers based on the new data-frame.
scatterAreaVsPrice(train_df)

售后-Gr Liv地区对售价

Since I haven’t seen an answer that deal with numerical and non-numerical attributes, here is a complement answer.

You might want to drop the outliers only on numerical attributes (categorical variables can hardly be outliers).

Function definition

I have extended @tanemaki’s suggestion to handle data when non-numeric attributes are also present:

from scipy import stats

def drop_numerical_outliers(df, z_thresh=3):
    # Constrains will contain `True` or `False` depending on if it is a value below the threshold.
    constrains = df.select_dtypes(include=[np.number]) \
        .apply(lambda x: np.abs(stats.zscore(x)) < z_thresh, reduce=False) \
        .all(axis=1)
    # Drop (inplace) values set to be rejected
    df.drop(df.index[~constrains], inplace=True)

Usage

drop_numerical_outliers(df)

Example

Imagine a dataset df with some values about houses: alley, land contour, sale price, … E.g: Data Documentation

First, you want to visualise the data on a scatter graph (with z-score Thresh=3):

# Plot data before dropping those greater than z-score 3. 
# The scatterAreaVsPrice function's definition has been removed for readability's sake.
scatterAreaVsPrice(df)

Before - Gr Liv Area Versus SalePrice

# Drop the outliers on every attributes
drop_numerical_outliers(train_df)

# Plot the result. All outliers were dropped. Note that the red points are not
# the same outliers from the first plot, but the new computed outliers based on the new data-frame.
scatterAreaVsPrice(train_df)

After - Gr Liv Area Versus SalePrice


回答 7

scipy.stats有方法trim1()trimboth()可以根据排名和删除值的引入百分比将异常值切成一行。

scipy.stats has methods trim1() and trimboth() to cut the outliers out in a single row, according to the ranking and an introduced percentage of removed values.


回答 8

另一种选择是转换数据,以减轻异常值的影响。您可以通过取消存储数据来做到这一点。

import pandas as pd
from scipy.stats import mstats
%matplotlib inline

test_data = pd.Series(range(30))
test_data.plot()

原始数据

# Truncate values to the 5th and 95th percentiles
transformed_test_data = pd.Series(mstats.winsorize(test_data, limits=[0.05, 0.05])) 
transformed_test_data.plot()

Winsorized数据

Another option is to transform your data so that the effect of outliers is mitigated. You can do this by winsorizing your data.

import pandas as pd
from scipy.stats import mstats
%matplotlib inline

test_data = pd.Series(range(30))
test_data.plot()

Original data

# Truncate values to the 5th and 95th percentiles
transformed_test_data = pd.Series(mstats.winsorize(test_data, limits=[0.05, 0.05])) 
transformed_test_data.plot()

Winsorized data


回答 9

如果您喜欢方法链接,则可以为所有数字列获取布尔条件,如下所示:

df.sub(df.mean()).div(df.std()).abs().lt(3)

每列的每个值将True/False根据其与平均值之间的距离是否小于三个标准偏差进行转换。

If you like method chaining, you can get your boolean condition for all numeric columns like this:

df.sub(df.mean()).div(df.std()).abs().lt(3)

Each value of each column will be converted to True/False based on whether its less than three standard deviations away from the mean or not.


回答 10

您可以使用布尔掩码:

import pandas as pd

def remove_outliers(df, q=0.05):
    upper = df.quantile(1-q)
    lower = df.quantile(q)
    mask = (df < upper) & (df > lower)
    return mask

t = pd.DataFrame({'train': [1,1,2,3,4,5,6,7,8,9,9],
                  'y': [1,0,0,1,1,0,0,1,1,1,0]})

mask = remove_outliers(t['train'], 0.1)

print(t[mask])

输出:

   train  y
2      2  0
3      3  1
4      4  1
5      5  0
6      6  0
7      7  1
8      8  1

You can use boolean mask:

import pandas as pd

def remove_outliers(df, q=0.05):
    upper = df.quantile(1-q)
    lower = df.quantile(q)
    mask = (df < upper) & (df > lower)
    return mask

t = pd.DataFrame({'train': [1,1,2,3,4,5,6,7,8,9,9],
                  'y': [1,0,0,1,1,0,0,1,1,1,0]})

mask = remove_outliers(t['train'], 0.1)

print(t[mask])

output:

   train  y
2      2  0
3      3  1
4      4  1
5      5  0
6      6  0
7      7  1
8      8  1

回答 11

由于我正处于数据科学之旅的早期阶段,因此我使用以下代码处理异常值。

#Outlier Treatment

def outlier_detect(df):
    for i in df.describe().columns:
        Q1=df.describe().at['25%',i]
        Q3=df.describe().at['75%',i]
        IQR=Q3 - Q1
        LTV=Q1 - 1.5 * IQR
        UTV=Q3 + 1.5 * IQR
        x=np.array(df[i])
        p=[]
        for j in x:
            if j < LTV or j>UTV:
                p.append(df[i].median())
            else:
                p.append(j)
        df[i]=p
    return df

Since I am in a very early stage of my data science journey, I am treating outliers with the code below.

#Outlier Treatment

def outlier_detect(df):
    for i in df.describe().columns:
        Q1=df.describe().at['25%',i]
        Q3=df.describe().at['75%',i]
        IQR=Q3 - Q1
        LTV=Q1 - 1.5 * IQR
        UTV=Q3 + 1.5 * IQR
        x=np.array(df[i])
        p=[]
        for j in x:
            if j < LTV or j>UTV:
                p.append(df[i].median())
            else:
                p.append(j)
        df[i]=p
    return df

回答 12

获得第98和第2个百分位数作为离群值的限制

upper_limit = np.percentile(X_train.logerror.values, 98) 
lower_limit = np.percentile(X_train.logerror.values, 2) # Filter the outliers from the dataframe
data[‘target’].loc[X_train[‘target’]>upper_limit] = upper_limit data[‘target’].loc[X_train[‘target’]<lower_limit] = lower_limit

Get the 98th and 2nd percentile as the limits of our outliers

upper_limit = np.percentile(X_train.logerror.values, 98) 
lower_limit = np.percentile(X_train.logerror.values, 2) # Filter the outliers from the dataframe
data[‘target’].loc[X_train[‘target’]>upper_limit] = upper_limit data[‘target’].loc[X_train[‘target’]<lower_limit] = lower_limit

回答 13

包含数据和2个组的完整示例如下:

进口:

from StringIO import StringIO
import pandas as pd
#pandas config
pd.set_option('display.max_rows', 20)

具有2组的数据示例:G1:组1。G2:组2:

TESTDATA = StringIO("""G1;G2;Value
1;A;1.6
1;A;5.1
1;A;7.1
1;A;8.1

1;B;21.1
1;B;22.1
1;B;24.1
1;B;30.6

2;A;40.6
2;A;51.1
2;A;52.1
2;A;60.6

2;B;80.1
2;B;70.6
2;B;90.6
2;B;85.1
""")

将文本数据读取到pandas数据框:

df = pd.read_csv(TESTDATA, sep=";")

使用标准偏差定义离群值

stds = 1.0
outliers = df[['G1', 'G2', 'Value']].groupby(['G1','G2']).transform(
           lambda group: (group - group.mean()).abs().div(group.std())) > stds

定义过滤后的数据值和离群值:

dfv = df[outliers.Value == False]
dfo = df[outliers.Value == True]

打印结果:

print '\n'*5, 'All values with decimal 1 are non-outliers. In the other hand, all values with 6 in the decimal are.'
print '\nDef DATA:\n%s\n\nFiltred Values with %s stds:\n%s\n\nOutliers:\n%s' %(df, stds, dfv, dfo)

a full example with data and 2 groups follows:

Imports:

from StringIO import StringIO
import pandas as pd
#pandas config
pd.set_option('display.max_rows', 20)

Data example with 2 groups: G1:Group 1. G2: Group 2:

TESTDATA = StringIO("""G1;G2;Value
1;A;1.6
1;A;5.1
1;A;7.1
1;A;8.1

1;B;21.1
1;B;22.1
1;B;24.1
1;B;30.6

2;A;40.6
2;A;51.1
2;A;52.1
2;A;60.6

2;B;80.1
2;B;70.6
2;B;90.6
2;B;85.1
""")

Read text data to pandas dataframe:

df = pd.read_csv(TESTDATA, sep=";")

Define the outliers using standard deviations

stds = 1.0
outliers = df[['G1', 'G2', 'Value']].groupby(['G1','G2']).transform(
           lambda group: (group - group.mean()).abs().div(group.std())) > stds

Define filtered data values and the outliers:

dfv = df[outliers.Value == False]
dfo = df[outliers.Value == True]

Print the result:

print '\n'*5, 'All values with decimal 1 are non-outliers. In the other hand, all values with 6 in the decimal are.'
print '\nDef DATA:\n%s\n\nFiltred Values with %s stds:\n%s\n\nOutliers:\n%s' %(df, stds, dfv, dfo)

回答 14

我删除异常值的功能

def drop_outliers(df, field_name):
    distance = 1.5 * (np.percentile(df[field_name], 75) - np.percentile(df[field_name], 25))
    df.drop(df[df[field_name] > distance + np.percentile(df[field_name], 75)].index, inplace=True)
    df.drop(df[df[field_name] < np.percentile(df[field_name], 25) - distance].index, inplace=True)

My function for dropping outliers

def drop_outliers(df, field_name):
    distance = 1.5 * (np.percentile(df[field_name], 75) - np.percentile(df[field_name], 25))
    df.drop(df[df[field_name] > distance + np.percentile(df[field_name], 75)].index, inplace=True)
    df.drop(df[df[field_name] < np.percentile(df[field_name], 25) - distance].index, inplace=True)

回答 15

我更喜欢剪辑而不是放下。下面的内容将在第2个和第98个百分位处固定。

df_list = list(df)
minPercentile = 0.02
maxPercentile = 0.98

for _ in range(numCols):
    df[df_list[_]] = df[df_list[_]].clip((df[df_list[_]].quantile(minPercentile)),(df[df_list[_]].quantile(maxPercentile)))

I prefer to clip rather than drop. the following will clip inplace at the 2nd and 98th pecentiles.

df_list = list(df)
minPercentile = 0.02
maxPercentile = 0.98

for _ in range(numCols):
    df[df_list[_]] = df[df_list[_]].clip((df[df_list[_]].quantile(minPercentile)),(df[df_list[_]].quantile(maxPercentile)))

回答 16

我认为从统计上删除和删除异常值是错误的。它使数据与原始数据不同。也使数据不均匀地成形,因此最好的方法是通过对数据进行对数变换来减少或避免离群值的影响。这为我工作:

np.log(data.iloc[:, :])

Deleting and dropping outliers I believe is wrong statistically. It makes the data different from original data. Also makes data unequally shaped and hence best way is to reduce or avoid the effect of outliers by log transform the data. This worked for me:

np.log(data.iloc[:, :])

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