标签归档:statsmodels

使用Pandas Data Frame运行OLS回归

问题:使用Pandas Data Frame运行OLS回归

我有一个pandas数据框,我希望能够从B和C列中的值预测A列的值。这是一个玩具示例:

import pandas as pd
df = pd.DataFrame({"A": [10,20,30,40,50], 
                   "B": [20, 30, 10, 40, 50], 
                   "C": [32, 234, 23, 23, 42523]})

理想情况下,我会有类似的东西,ols(A ~ B + C, data = df)但是当我查看算法库中的示例时,看起来好像scikit-learn是用行而不是列的列表将数据提供给模型。这将要求我将数据重新格式化为列表内的列表,这似乎首先使使用熊猫的目的遭到了破坏。在熊猫数据框中的数据上运行OLS回归(或更通用的任何机器学习算法)的最有效方法是什么?

I have a pandas data frame and I would like to able to predict the values of column A from the values in columns B and C. Here is a toy example:

import pandas as pd
df = pd.DataFrame({"A": [10,20,30,40,50], 
                   "B": [20, 30, 10, 40, 50], 
                   "C": [32, 234, 23, 23, 42523]})

Ideally, I would have something like ols(A ~ B + C, data = df) but when I look at the examples from algorithm libraries like scikit-learn it appears to feed the data to the model with a list of rows instead of columns. This would require me to reformat the data into lists inside lists, which seems to defeat the purpose of using pandas in the first place. What is the most pythonic way to run an OLS regression (or any machine learning algorithm more generally) on data in a pandas data frame?


回答 0

我认为您可以使用statsmodels包几乎完成您认为理想的事情,该包是0.20.0版pandas之前的“可选依赖项pandas”之一(在中有一些用途pandas.stats。)

>>> import pandas as pd
>>> import statsmodels.formula.api as sm
>>> df = pd.DataFrame({"A": [10,20,30,40,50], "B": [20, 30, 10, 40, 50], "C": [32, 234, 23, 23, 42523]})
>>> result = sm.ols(formula="A ~ B + C", data=df).fit()
>>> print(result.params)
Intercept    14.952480
B             0.401182
C             0.000352
dtype: float64
>>> print(result.summary())
                            OLS Regression Results                            
==============================================================================
Dep. Variable:                      A   R-squared:                       0.579
Model:                            OLS   Adj. R-squared:                  0.158
Method:                 Least Squares   F-statistic:                     1.375
Date:                Thu, 14 Nov 2013   Prob (F-statistic):              0.421
Time:                        20:04:30   Log-Likelihood:                -18.178
No. Observations:                   5   AIC:                             42.36
Df Residuals:                       2   BIC:                             41.19
Df Model:                           2                                         
==============================================================================
                 coef    std err          t      P>|t|      [95.0% Conf. Int.]
------------------------------------------------------------------------------
Intercept     14.9525     17.764      0.842      0.489       -61.481    91.386
B              0.4012      0.650      0.617      0.600        -2.394     3.197
C              0.0004      0.001      0.650      0.583        -0.002     0.003
==============================================================================
Omnibus:                          nan   Durbin-Watson:                   1.061
Prob(Omnibus):                    nan   Jarque-Bera (JB):                0.498
Skew:                          -0.123   Prob(JB):                        0.780
Kurtosis:                       1.474   Cond. No.                     5.21e+04
==============================================================================

Warnings:
[1] The condition number is large, 5.21e+04. This might indicate that there are
strong multicollinearity or other numerical problems.

I think you can almost do exactly what you thought would be ideal, using the statsmodels package which was one of pandas‘ optional dependencies before pandas‘ version 0.20.0 (it was used for a few things in pandas.stats.)

>>> import pandas as pd
>>> import statsmodels.formula.api as sm
>>> df = pd.DataFrame({"A": [10,20,30,40,50], "B": [20, 30, 10, 40, 50], "C": [32, 234, 23, 23, 42523]})
>>> result = sm.ols(formula="A ~ B + C", data=df).fit()
>>> print(result.params)
Intercept    14.952480
B             0.401182
C             0.000352
dtype: float64
>>> print(result.summary())
                            OLS Regression Results                            
==============================================================================
Dep. Variable:                      A   R-squared:                       0.579
Model:                            OLS   Adj. R-squared:                  0.158
Method:                 Least Squares   F-statistic:                     1.375
Date:                Thu, 14 Nov 2013   Prob (F-statistic):              0.421
Time:                        20:04:30   Log-Likelihood:                -18.178
No. Observations:                   5   AIC:                             42.36
Df Residuals:                       2   BIC:                             41.19
Df Model:                           2                                         
==============================================================================
                 coef    std err          t      P>|t|      [95.0% Conf. Int.]
------------------------------------------------------------------------------
Intercept     14.9525     17.764      0.842      0.489       -61.481    91.386
B              0.4012      0.650      0.617      0.600        -2.394     3.197
C              0.0004      0.001      0.650      0.583        -0.002     0.003
==============================================================================
Omnibus:                          nan   Durbin-Watson:                   1.061
Prob(Omnibus):                    nan   Jarque-Bera (JB):                0.498
Skew:                          -0.123   Prob(JB):                        0.780
Kurtosis:                       1.474   Cond. No.                     5.21e+04
==============================================================================

Warnings:
[1] The condition number is large, 5.21e+04. This might indicate that there are
strong multicollinearity or other numerical problems.

回答 1

注意: pandas.stats 已被 0.20.0 删除


可以使用pandas.stats.ols

>>> from pandas.stats.api import ols
>>> df = pd.DataFrame({"A": [10,20,30,40,50], "B": [20, 30, 10, 40, 50], "C": [32, 234, 23, 23, 42523]})
>>> res = ols(y=df['A'], x=df[['B','C']])
>>> res
-------------------------Summary of Regression Analysis-------------------------

Formula: Y ~ <B> + <C> + <intercept>

Number of Observations:         5
Number of Degrees of Freedom:   3

R-squared:         0.5789
Adj R-squared:     0.1577

Rmse:             14.5108

F-stat (2, 2):     1.3746, p-value:     0.4211

Degrees of Freedom: model 2, resid 2

-----------------------Summary of Estimated Coefficients------------------------
      Variable       Coef    Std Err     t-stat    p-value    CI 2.5%   CI 97.5%
--------------------------------------------------------------------------------
             B     0.4012     0.6497       0.62     0.5999    -0.8723     1.6746
             C     0.0004     0.0005       0.65     0.5826    -0.0007     0.0014
     intercept    14.9525    17.7643       0.84     0.4886   -19.8655    49.7705
---------------------------------End of Summary---------------------------------

请注意,您需要statsmodels安装软件包,该软件包在内部使用pandas.stats.ols

Note: pandas.stats has been removed with 0.20.0


It’s possible to do this with pandas.stats.ols:

>>> from pandas.stats.api import ols
>>> df = pd.DataFrame({"A": [10,20,30,40,50], "B": [20, 30, 10, 40, 50], "C": [32, 234, 23, 23, 42523]})
>>> res = ols(y=df['A'], x=df[['B','C']])
>>> res
-------------------------Summary of Regression Analysis-------------------------

Formula: Y ~ <B> + <C> + <intercept>

Number of Observations:         5
Number of Degrees of Freedom:   3

R-squared:         0.5789
Adj R-squared:     0.1577

Rmse:             14.5108

F-stat (2, 2):     1.3746, p-value:     0.4211

Degrees of Freedom: model 2, resid 2

-----------------------Summary of Estimated Coefficients------------------------
      Variable       Coef    Std Err     t-stat    p-value    CI 2.5%   CI 97.5%
--------------------------------------------------------------------------------
             B     0.4012     0.6497       0.62     0.5999    -0.8723     1.6746
             C     0.0004     0.0005       0.65     0.5826    -0.0007     0.0014
     intercept    14.9525    17.7643       0.84     0.4886   -19.8655    49.7705
---------------------------------End of Summary---------------------------------

Note that you need to have statsmodels package installed, it is used internally by the pandas.stats.ols function.


回答 2

我不知道这是否是新的sklearn还是pandas,但我能直接传递数据帧sklearn没有数据帧转换为numpy的阵列或任何其它数据类型。

from sklearn import linear_model

reg = linear_model.LinearRegression()
reg.fit(df[['B', 'C']], df['A'])

>>> reg.coef_
array([  4.01182386e-01,   3.51587361e-04])

I don’t know if this is new in sklearn or pandas, but I’m able to pass the data frame directly to sklearn without converting the data frame to a numpy array or any other data types.

from sklearn import linear_model

reg = linear_model.LinearRegression()
reg.fit(df[['B', 'C']], df['A'])

>>> reg.coef_
array([  4.01182386e-01,   3.51587361e-04])

回答 3

这将要求我将数据重新格式化为列表内的列表,这似乎首先使使用熊猫的目的无法实现。

不,不是,只是转换为NumPy数组:

>>> data = np.asarray(df)

这会花费固定的时间,因为它只会创建数据视图。然后将其提供给scikit-learn:

>>> from sklearn.linear_model import LinearRegression
>>> lr = LinearRegression()
>>> X, y = data[:, 1:], data[:, 0]
>>> lr.fit(X, y)
LinearRegression(copy_X=True, fit_intercept=True, normalize=False)
>>> lr.coef_
array([  4.01182386e-01,   3.51587361e-04])
>>> lr.intercept_
14.952479503953672

This would require me to reformat the data into lists inside lists, which seems to defeat the purpose of using pandas in the first place.

No it doesn’t, just convert to a NumPy array:

>>> data = np.asarray(df)

This takes constant time because it just creates a view on your data. Then feed it to scikit-learn:

>>> from sklearn.linear_model import LinearRegression
>>> lr = LinearRegression()
>>> X, y = data[:, 1:], data[:, 0]
>>> lr.fit(X, y)
LinearRegression(copy_X=True, fit_intercept=True, normalize=False)
>>> lr.coef_
array([  4.01182386e-01,   3.51587361e-04])
>>> lr.intercept_
14.952479503953672

回答 4

Statsmodels可使用直接引用熊猫数据框的列引用来构建OLS模型

简短而甜美:

model = sm.OLS(df[y], df[x]).fit()


代码详细信息和回归摘要:

# imports
import pandas as pd
import statsmodels.api as sm
import numpy as np

# data
np.random.seed(123)
df = pd.DataFrame(np.random.randint(0,100,size=(100, 3)), columns=list('ABC'))

# assign dependent and independent / explanatory variables
variables = list(df.columns)
y = 'A'
x = [var for var in variables if var not in y ]

# Ordinary least squares regression
model_Simple = sm.OLS(df[y], df[x]).fit()

# Add a constant term like so:
model = sm.OLS(df[y], sm.add_constant(df[x])).fit()

model.summary()

输出:

                            OLS Regression Results                            
==============================================================================
Dep. Variable:                      A   R-squared:                       0.019
Model:                            OLS   Adj. R-squared:                 -0.001
Method:                 Least Squares   F-statistic:                    0.9409
Date:                Thu, 14 Feb 2019   Prob (F-statistic):              0.394
Time:                        08:35:04   Log-Likelihood:                -484.49
No. Observations:                 100   AIC:                             975.0
Df Residuals:                      97   BIC:                             982.8
Df Model:                           2                                         
Covariance Type:            nonrobust                                         
==============================================================================
                 coef    std err          t      P>|t|      [0.025      0.975]
------------------------------------------------------------------------------
const         43.4801      8.809      4.936      0.000      25.996      60.964
B              0.1241      0.105      1.188      0.238      -0.083       0.332
C             -0.0752      0.110     -0.681      0.497      -0.294       0.144
==============================================================================
Omnibus:                       50.990   Durbin-Watson:                   2.013
Prob(Omnibus):                  0.000   Jarque-Bera (JB):                6.905
Skew:                           0.032   Prob(JB):                       0.0317
Kurtosis:                       1.714   Cond. No.                         231.
==============================================================================

如何直接获得R平方,系数和p值:

# commands:
model.params
model.pvalues
model.rsquared

# demo:
In[1]: 
model.params
Out[1]:
const    43.480106
B         0.124130
C        -0.075156
dtype: float64

In[2]: 
model.pvalues
Out[2]: 
const    0.000003
B        0.237924
C        0.497400
dtype: float64

Out[3]:
model.rsquared
Out[2]:
0.0190

Statsmodels kan build an OLS model with column references directly to a pandas dataframe.

Short and sweet:

model = sm.OLS(df[y], df[x]).fit()


Code details and regression summary:

# imports
import pandas as pd
import statsmodels.api as sm
import numpy as np

# data
np.random.seed(123)
df = pd.DataFrame(np.random.randint(0,100,size=(100, 3)), columns=list('ABC'))

# assign dependent and independent / explanatory variables
variables = list(df.columns)
y = 'A'
x = [var for var in variables if var not in y ]

# Ordinary least squares regression
model_Simple = sm.OLS(df[y], df[x]).fit()

# Add a constant term like so:
model = sm.OLS(df[y], sm.add_constant(df[x])).fit()

model.summary()

Output:

                            OLS Regression Results                            
==============================================================================
Dep. Variable:                      A   R-squared:                       0.019
Model:                            OLS   Adj. R-squared:                 -0.001
Method:                 Least Squares   F-statistic:                    0.9409
Date:                Thu, 14 Feb 2019   Prob (F-statistic):              0.394
Time:                        08:35:04   Log-Likelihood:                -484.49
No. Observations:                 100   AIC:                             975.0
Df Residuals:                      97   BIC:                             982.8
Df Model:                           2                                         
Covariance Type:            nonrobust                                         
==============================================================================
                 coef    std err          t      P>|t|      [0.025      0.975]
------------------------------------------------------------------------------
const         43.4801      8.809      4.936      0.000      25.996      60.964
B              0.1241      0.105      1.188      0.238      -0.083       0.332
C             -0.0752      0.110     -0.681      0.497      -0.294       0.144
==============================================================================
Omnibus:                       50.990   Durbin-Watson:                   2.013
Prob(Omnibus):                  0.000   Jarque-Bera (JB):                6.905
Skew:                           0.032   Prob(JB):                       0.0317
Kurtosis:                       1.714   Cond. No.                         231.
==============================================================================

How to directly get R-squared, Coefficients and p-value:

# commands:
model.params
model.pvalues
model.rsquared

# demo:
In[1]: 
model.params
Out[1]:
const    43.480106
B         0.124130
C        -0.075156
dtype: float64

In[2]: 
model.pvalues
Out[2]: 
const    0.000003
B        0.237924
C        0.497400
dtype: float64

Out[3]:
model.rsquared
Out[2]:
0.0190

如何遍历熊猫数据框的列以运行回归

问题:如何遍历熊猫数据框的列以运行回归

我敢肯定这很简单,但是作为python的完整新手,我在弄清楚如何遍历pandas数据帧中的变量并对每个变量进行回归时都遇到了麻烦。

这是我在做什么:

all_data = {}
for ticker in ['FIUIX', 'FSAIX', 'FSAVX', 'FSTMX']:
    all_data[ticker] = web.get_data_yahoo(ticker, '1/1/2010', '1/1/2015')

prices = DataFrame({tic: data['Adj Close'] for tic, data in all_data.iteritems()})  
returns = prices.pct_change()

我知道我可以像这样进行回归:

regs = sm.OLS(returns.FIUIX,returns.FSTMX).fit()

但是假设我要对数据框中的每一列执行此操作。特别是,我想在FSTMX上还原FIUIX,然后在FSTMX上还原FSAIX,然后在FSTMX上还原FSAVX。每次回归后,我想存储残差。

我尝试了以下各种版本,但语法一定有误:

resids = {}
for k in returns.keys():
    reg = sm.OLS(returns[k],returns.FSTMX).fit()
    resids[k] = reg.resid

我认为问题是我不知道如何按键引用return列,所以returns[k]可能是错误的。

任何有关最佳方法的指导将不胜感激。也许我缺少一种常见的熊猫方法。

I’m sure this is simple, but as a complete newbie to python, I’m having trouble figuring out how to iterate over variables in a pandas dataframe and run a regression with each.

Here’s what I’m doing:

all_data = {}
for ticker in ['FIUIX', 'FSAIX', 'FSAVX', 'FSTMX']:
    all_data[ticker] = web.get_data_yahoo(ticker, '1/1/2010', '1/1/2015')

prices = DataFrame({tic: data['Adj Close'] for tic, data in all_data.iteritems()})  
returns = prices.pct_change()

I know I can run a regression like this:

regs = sm.OLS(returns.FIUIX,returns.FSTMX).fit()

but suppose I want to do this for each column in the dataframe. In particular, I want to regress FIUIX on FSTMX, and then FSAIX on FSTMX, and then FSAVX on FSTMX. After each regression I want to store the residuals.

I’ve tried various versions of the following, but I must be getting the syntax wrong:

resids = {}
for k in returns.keys():
    reg = sm.OLS(returns[k],returns.FSTMX).fit()
    resids[k] = reg.resid

I think the problem is I don’t know how to refer to the returns column by key, so returns[k] is probably wrong.

Any guidance on the best way to do this would be much appreciated. Perhaps there’s a common pandas approach I’m missing.


回答 0

for column in df:
    print(df[column])
for column in df:
    print(df[column])

回答 1

您可以使用iteritems()

for name, values in df.iteritems():
    print('{name}: {value}'.format(name=name, value=values[0]))

You can use iteritems():

for name, values in df.iteritems():
    print('{name}: {value}'.format(name=name, value=values[0]))

回答 2

这个答案是要遍历DF中的选定列以及所有列。

df.columns给出包含DF中所有列名称的列表。现在,如果要遍历所有列,则不是很有帮助。但是,当您只想遍历所选列时,它会派上用场。

我们可以根据需要轻松使用Python的列表切片对df.columns进行切片。例如,要遍历除第一列之外的所有列,我们可以这样做:

for column in df.columns[1:]:
    print(df[column])

类似于以相反的顺序遍历所有列,我们可以执行以下操作:

for column in df.columns[::-1]:
    print(df[column])

我们可以使用这种技术以许多很酷的方式遍历所有列。还请记住,您可以使用以下命令轻松获取所有列的索引:

for ind, column in enumerate(df.columns):
    print(ind, column)

This answer is to iterate over selected columns as well as all columns in a DF.

df.columns gives a list containing all the columns’ names in the DF. Now that isn’t very helpful if you want to iterate over all the columns. But it comes in handy when you want to iterate over columns of your choosing only.

We can use Python’s list slicing easily to slice df.columns according to our needs. For eg, to iterate over all columns but the first one, we can do:

for column in df.columns[1:]:
    print(df[column])

Similarly to iterate over all the columns in reversed order, we can do:

for column in df.columns[::-1]:
    print(df[column])

We can iterate over all the columns in a lot of cool ways using this technique. Also remember that you can get the indices of all columns easily using:

for ind, column in enumerate(df.columns):
    print(ind, column)

回答 3

您可以使用来按位置索引数据框列ix

df1.ix[:,1]

例如,这将返回第一列。(0为索引)

df1.ix[0,]

这将返回第一行。

df1.ix[:,1]

这将是第0行与第1列的交集处的值:

df1.ix[0,1]

等等。因此,您可以enumerate() returns.keys():并使用数字来索引数据框。

You can index dataframe columns by the position using ix.

df1.ix[:,1]

This returns the first column for example. (0 would be the index)

df1.ix[0,]

This returns the first row.

df1.ix[:,1]

This would be the value at the intersection of row 0 and column 1:

df1.ix[0,1]

and so on. So you can enumerate() returns.keys(): and use the number to index the dataframe.


回答 4

一种解决方法是对进行转置DataFrame并在行上进行迭代。

for column_name, column in df.transpose().iterrows():
    print column_name

A workaround is to transpose the DataFrame and iterate over the rows.

for column_name, column in df.transpose().iterrows():
    print column_name

回答 5

使用列表推导,您可以获得所有列名(标题):

[column for column in df]

Using list comprehension, you can get all the columns names (header):

[column for column in df]


回答 6

根据接受的答案,是否还需要与各列相对应的索引

for i, column in enumerate(df):
    print i, df[column]

上面的df[column]类型是Series,可以简单地转换为numpy ndarrays:

for i, column in enumerate(df):
    print i, np.asarray(df[column])

Based on the accepted answer, if an index corresponding to each column is also desired:

for i, column in enumerate(df):
    print i, df[column]

The above df[column] type is Series, which can simply be converted into numpy ndarrays:

for i, column in enumerate(df):
    print i, np.asarray(df[column])

回答 7

我来晚了,但是这是我的方法。步骤:

  1. 创建所有列的列表
  2. 使用itertools进行x组合
  3. 将每个结果R平方值与排除的列列表一起附加到结果数据帧
  4. 以R平方的降序对结果DF排序,以找出最合适的DF。

这是我在DataFrame上使用的称为的代码aft_tmt。随意推断您的用例。

import pandas as pd
# setting options to print without truncating output
pd.set_option('display.max_columns', None)
pd.set_option('display.max_colwidth', None)

import statsmodels.formula.api as smf
import itertools

# This section gets the column names of the DF and removes some columns which I don't want to use as predictors.
itercols = aft_tmt.columns.tolist()
itercols.remove("sc97")
itercols.remove("sc")
itercols.remove("grc")
itercols.remove("grc97")
print itercols
len(itercols)

# results DF
regression_res = pd.DataFrame(columns = ["Rsq", "predictors", "excluded"])

# excluded cols
exc = []

# change 9 to the number of columns you want to combine from N columns.
#Possibly run an outer loop from 0 to N/2?
for x in itertools.combinations(itercols, 9):
    lmstr = "+".join(x)
    m = smf.ols(formula = "sc ~ " + lmstr, data = aft_tmt)
    f = m.fit()
    exc = [item for item in x if item not in itercols]
    regression_res = regression_res.append(pd.DataFrame([[f.rsquared, lmstr, "+".join([y for y in itercols if y not in list(x)])]], columns = ["Rsq", "predictors", "excluded"]))

regression_res.sort_values(by="Rsq", ascending = False)

I’m a bit late but here’s how I did this. The steps:

  1. Create a list of all columns
  2. Use itertools to take x combinations
  3. Append each result R squared value to a result dataframe along with excluded column list
  4. Sort the result DF in descending order of R squared to see which is the best fit.

This is the code I used on DataFrame called aft_tmt. Feel free to extrapolate to your use case..

import pandas as pd
# setting options to print without truncating output
pd.set_option('display.max_columns', None)
pd.set_option('display.max_colwidth', None)

import statsmodels.formula.api as smf
import itertools

# This section gets the column names of the DF and removes some columns which I don't want to use as predictors.
itercols = aft_tmt.columns.tolist()
itercols.remove("sc97")
itercols.remove("sc")
itercols.remove("grc")
itercols.remove("grc97")
print itercols
len(itercols)

# results DF
regression_res = pd.DataFrame(columns = ["Rsq", "predictors", "excluded"])

# excluded cols
exc = []

# change 9 to the number of columns you want to combine from N columns.
#Possibly run an outer loop from 0 to N/2?
for x in itertools.combinations(itercols, 9):
    lmstr = "+".join(x)
    m = smf.ols(formula = "sc ~ " + lmstr, data = aft_tmt)
    f = m.fit()
    exc = [item for item in x if item not in itercols]
    regression_res = regression_res.append(pd.DataFrame([[f.rsquared, lmstr, "+".join([y for y in itercols if y not in list(x)])]], columns = ["Rsq", "predictors", "excluded"]))

regression_res.sort_values(by="Rsq", ascending = False)