我有一个数据集,包含124个特征的分类和数字数据。为了减少其维度,我想删除不相关的功能。但是,为了针对特征选择算法运行数据集,我使用get_dummies对其进行了热编码,这使得特征数量增加到391个。
In[16]:
X_train.columns
Out[16]:
Index([u'port_7', u'port_9', u'port_13', u'port_17', u'port_19', u'port_21',
...
u'os_cpes.1_2', u'os_cpes.1_1'], dtype='object', length=391)
根据结果数据,我可以根据Scikit Learn example运行交叉验证来递归递归功能:
产生:
Cross Validated Score vs Features Graph
鉴于已识别的最佳功能数量为8,如何识别功能名称?我假设我可以将它们提取到一个新的DataFrame中用于分类算法?
[编辑]
的帮助下,我已达到以下目的def column_index(df, query_cols):
cols = df.columns.values
sidx = np.argsort(cols)
return sidx[np.searchsorted(cols, query_cols, sorter = sidx)]
feature_index = []
features = []
column_index(X_dev_train, X_dev_train.columns.values)
for num, i in enumerate(rfecv.get_support(), start=0):
if i == True:
feature_index.append(str(num))
for num, i in enumerate(X_dev_train.columns.values, start=0):
if str(num) in feature_index:
features.append(X_dev_train.columns.values[num])
print("Features Selected: {}\n".format(len(feature_index)))
print("Features Indexes: \n{}\n".format(feature_index))
print("Feature Names: \n{}".format(features))
产生:
Features Selected: 8
Features Indexes:
['5', '6', '20', '26', '27', '28', '67', '98']
Feature Names:
['port_21', 'port_22', 'port_199', 'port_512', 'port_513', 'port_514', 'port_3306', 'port_32768']
鉴于一个热编码引入了多重共线性,我不认为目标列选择是理想的,因为它选择的特征是非编码的连续数据特征。我尝试重新添加目标列未编码,但RFE抛出以下错误,因为数据是分类的:
ValueError: could not convert string to float: Wireless Access Point
我是否需要将多个热编码功能列分组才能充当目标?
[编辑2]
如果我只是对目标列进行LabelEncode,我可以将此目标用作' y'见example again。但是,输出仅将单个要素(目标列)确定为最佳。我想这可能是因为一个热门编码,我应该考虑生成密集数组,如果是这样,它可以针对RFE运行吗?
谢谢,
亚当
答案 0 :(得分:0)
回答我自己的问题,我发现问题与我对热门数据编码的方式有关。最初,我对所有分类列运行了一个热编码,如下所示:
ohe_df = pd.get_dummies(df[df.columns]) # One-hot encode all columns
这引入了大量附加功能。采用不同的方法,在here的帮助下,我修改了编码,以按列/功能为基础对多列进行编码,如下所示:
cf_df = df.select_dtypes(include=[object]) # Get categorical features
nf_df = df.select_dtypes(exclude=[object]) # Get numerical features
ohe_df = nf_df.copy()
for feature in cf_df:
ohe_df[feature] = ohe_df.loc[:,(feature)].str.get_dummies().values.tolist()
产:
ohe_df.head(2) # Only showing a subset of the data
+---+---------------------------------------------------+-----------------+-----------------+-----------------------------------+---------------------------------------------------+
| | os_name | os_family | os_type | os_vendor | os_cpes.0 |
+---+---------------------------------------------------+-----------------+-----------------+-----------------------------------+---------------------------------------------------+
| 0 | [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ... | [0, 1, 0, 0, 0] | [1, 0, 0, 0, 0] | [0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0] | [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, ... |
| 1 | [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ... | [0, 0, 0, 1, 0] | [0, 0, 0, 1, 0] | [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0] | [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ... |
+---+---------------------------------------------------+-----------------+-----------------+-----------------------------------+---------------------------------------------------+
不幸的是,尽管这是我正在寻找的,但它并没有针对RFECV执行。接下来我想也许我可以把所有新功能切片并作为目标传递,但这导致了错误。最后,我意识到我必须遍历所有目标值并从每个目标值中获取最高输出。代码最终看起来像这样:
for num, feature in enumerate(features, start=0):
X = X_dev_train
y = X_dev_train[feature]
# Create the RFE object and compute a cross-validated score.
svc = SVC(kernel="linear")
# The "accuracy" scoring is proportional to the number of correct classifications
# step is the number of features to remove at each iteration
rfecv = RFECV(estimator=svc, step=1, cv=StratifiedKFold(kfold), scoring='accuracy')
try:
rfecv.fit(X, y)
print("Number of observations in each fold: {}".format(len(X)/kfold))
print("Optimal number of features : {}".format(rfecv.n_features_))
g_scores = rfecv.grid_scores_
indices = np.argsort(g_scores)[::-1]
print('Printing RFECV results:')
for num2, f in enumerate(range(X.shape[1]), start=0):
if g_scores[indices[f]] > 0.80:
if num2 < 10:
print("{}. Number of features: {} Grid_Score: {:0.3f}".format(f + 1, indices[f]+1, g_scores[indices[f]]))
print "\nTop features sorted by rank:"
results = sorted(zip(map(lambda x: round(x, 4), rfecv.ranking_), X.columns.values))
for num3, i in enumerate(results, start=0):
if num3 < 10:
print i
# Plot number of features VS. cross-validation scores
plt.rc("figure", figsize=(8, 5))
plt.figure()
plt.xlabel("Number of features selected")
plt.ylabel("CV score (of correct classifications)")
plt.plot(range(1, len(rfecv.grid_scores_) + 1), rfecv.grid_scores_)
plt.show()
except ValueError:
pass
我确定这可能更干净,甚至可以在一张图中绘制,但它对我有用。
干杯,
答案 1 :(得分:0)
您可以这样做:
`
from sklearn.feature_selection import RFE
from sklearn.linear_model import LogisticRegression
model = LogisticRegression()
rfe = RFE(model, 5)
rfe = rfe.fit(X, y)
print(rfe.support_)
print(rfe.ranking_)
f = rfe.get_support(1) #the most important features
X = df[df.columns[f]] # final features`
然后,您可以将X用作神经网络或任何算法的输入