Python 使用管道时MSE错误
我试图从我搜集的数据集中预测一些价格。我从未为此使用过Python(我通常使用Python 使用管道时MSE错误,python,scikit-learn,svm,Python,Scikit Learn,Svm,我试图从我搜集的数据集中预测一些价格。我从未为此使用过Python(我通常使用tidyverse,但这次我想探索管道)。 下面是代码片段: import pandas as pd from sklearn.compose import ColumnTransformer from sklearn.preprocessing import StandardScaler from sklearn.preprocessing import OneHotEncoder from sklearn.pipe
tidyverse
,但这次我想探索管道
)。
下面是代码片段:
import pandas as pd
from sklearn.compose import ColumnTransformer
from sklearn.preprocessing import StandardScaler
from sklearn.preprocessing import OneHotEncoder
from sklearn.pipeline import Pipeline
from sklearn.svm import SVR
from sklearn.model_selection import train_test_split
from sklearn.model_selection import GridSearchCV
import numpy as np
df = pd.read_csv("https://raw.githubusercontent.com/norhther/idealista/main/idealistaBCN.csv")
df.drop("info", axis = 1, inplace = True)
df["floor"].fillna(1, inplace=True)
df.drop("neigh", axis = 1, inplace = True)
df.dropna(inplace = True)
df = df[df["habs"] < 11]
X = df.drop("price", axis = 1)
y = df["price"]
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size = 0.2, random_state = 42)
ct = ColumnTransformer(
[("standardScaler", StandardScaler(), ["habs", "m2", "floor"]),
("onehot", OneHotEncoder(), ["type"]
)], remainder="passthrough")
pipe = Pipeline(steps = [("Transformer", ct),
("svr", SVR())])
param_grid = {
"svr__kernel" : ['linear', 'poly', 'rbf', 'sigmoid'],
"svr__degree" : range(3,6),
"svr__gamma" : ['scale', 'auto'],
"svr__coef0" : np.linspace(0.01, 1, 2)
}
search = GridSearchCV(pipe, param_grid, scoring = ['neg_mean_squared_error'], refit='neg_mean_squared_error')
search.fit(X_train, y_train)
print(search.best_score_)
pipe = Pipeline(steps = [("Transformer", ct),
("svr", SVR(coef0 = search.best_params_["svr__coef0"],
degree = search.best_params_["svr__degree"],
kernel =
search.best_params_["svr__kernel"]))])
from sklearn.metrics import mean_squared_error
pipe.fit(X_train, y_train)
preds = pipe.predict(X_train)
mean_squared_error(preds, y_train)
将熊猫作为pd导入
从sklearn.compose导入ColumnTransformer
从sklearn.preprocessing导入StandardScaler
从sklearn.preprocessing导入OneHotEncoder
从sklearn.pipeline导入管道
从sklearn.svm导入SVR
从sklearn.model\u选择导入列车\u测试\u拆分
从sklearn.model_选择导入GridSearchCV
将numpy作为np导入
df=pd.read_csv(“https://raw.githubusercontent.com/norhther/idealista/main/idealistaBCN.csv")
落差(“信息”,轴=1,原地=真)
df[“地板”].fillna(1,原地=真)
落差(“嘶鸣”,轴=1,原地=真)
df.dropna(就地=真)
df=df[df[“habs”]<11]
X=df.下降(“价格”,轴=1)
y=df[“价格”]
X_序列,X_测试,y_序列,y_测试=序列测试分割(X,y,测试大小=0.2,随机状态=42)
ct=柱状变压器(
[(“standardScaler”,standardScaler(),[“habs”,“m2”,“地板]),
(“onehot”,OneHotEncoder(),[“type”]
)],余数=“传递”)
管道=管道(步骤=[(“变压器”,ct),
(“svr”,svr())]
参数网格={
“svr__核”:[‘线性’、‘多边形’、‘rbf’、‘S形’],
“svr___度”:范围(3,6),
“svr__γ”:[“缩放”,“自动”],
“svr__coef0”:np.linspace(0.01,1,2)
}
search=GridSearchCV(管道、参数网格、评分=['neg_mean_squared_error'],refit='neg_mean_squared_error')
search.fit(X_-train,y_-train)
打印(搜索最佳分数)
管道=管道(步骤=[(“变压器”,ct),
(“svr”,svr(coef0=search.best_params)[“svr__coef0”],
度=搜索。最佳参数[“svr度”],
内核=
search.best_params(svr___内核]]))
从sklearn.metrics导入均方误差
管道安装(X_系列、y_系列)
preds=管道预测(X_列)
均方误差(preds,y列)
和搜索。这里的最佳得分是-443829697806.1671
,MSE
是60895397916.3896
我想我搞砸了一些东西,可能是变压器,但我不能完全确定。我想这是一个经过检验的MSE
。我用tidymodels
做了一个非常类似的方法,得到了更好的结果。
所以在这里我想知道变压器是否有问题,或者只是模型太差。原因是参数中没有包含C,需要覆盖整个Cs范围才能进行拟合。如果我们使用默认的C=1进行拟合,您可以看到问题出在哪里:
import matplotlib.pyplot as plt
o = pipe.named_steps["Transformer"].fit_transform(X_train)
mdl = SVR(C=1)
mdl.fit(o,y_train)
plt.scatter(mdl.predict(o),y_train)
有些价格值是平均值的10倍(1e7与5e5的中值)。如果使用mse或r^2,这些值将在很大程度上取决于这些极值。因此,我们需要更密切地跟踪数据,这由C决定,您可以选择C。我们尝试一个范围:
ct = ColumnTransformer(
[("standardScaler", StandardScaler(), ["habs", "m2", "floor"]),
("onehot", OneHotEncoder(), ["type"]
)], remainder="passthrough")
pipe = Pipeline(steps = [("Transformer", ct),
("svr", SVR())])
#, 'poly', 'rbf', 'sigmoid'
param_grid = {
"svr__kernel" : ['rbf'],
"svr__gamma" : ['auto'],
"svr__coef0" : [1,2],
"svr__C" : [1e-03,1e-01,1e1,1e3,1e5,1e7]
}
search = GridSearchCV(pipe, param_grid, scoring = ['neg_mean_squared_error'],
refit='neg_mean_squared_error')
search.fit(X_train, y_train)
print(search.best_score_)
-132061065775.25969
你们的y值很高,MSE值将在y值的方差范围内,所以若我们检查:
y_train.var()
545423126823.4545
132061065775.25969 / y_train.var()
0.24212590057261346
这很好,你将MSE减少到方差的25%左右。我们可以用测试数据来检查这一点,我想在这种情况下,C值非常好是非常幸运的:
from sklearn.metrics import mean_squared_error
o = pipe.named_steps["Transformer"].fit_transform(X_train)
mdl = SVR(C=10000000.0, coef0=1, gamma='auto')
mdl.fit(o,y_train)
o_test = pipe.named_steps["Transformer"].fit_transform(X_test)
pred = mdl.predict(o_test)
print( mean_squared_error(pred,y_test) , mean_squared_error(pred,y_test)/y_test.var())
plt.scatter(mdl.predict(o_test),y_test)
谢谢你的回答。一个问题,在训练方差中看到MSE
意味着什么而不是在常规值中?你的目标值越大,你的MSE就越大?