基于雇员流失率数据进行多分类模型训练及阈值调整实践-大数据ML样本集案例实战
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1 数据的预处理分析
from __future__ import division
import pandas as pd
import numpy as np
churn_df = pd.read_csv('churn.csv')
col_names = churn_df.columns.tolist()
print "Column names:"
print col_names
#前六个后六个
to_show = col_names[:6] + col_names[-6:]
print "\nSample data:"
churn_df[to_show].head(6)
2 数据标准化处理
churn_result = churn_df['Churn?']
y = np.where(churn_result == 'True.',1,0)
# We don't need these columns
to_drop = ['State','Area Code','Phone','Churn?']
churn_feat_space = churn_df.drop(to_drop,axis=1)
# 'yes'/'no' has to be converted to boolean values
# NumPy converts these from boolean to 1. and 0. later
yes_no_cols = ["Int'l Plan","VMail Plan"]
churn_feat_space[yes_no_cols] = churn_feat_space[yes_no_cols] == 'yes'
# Pull out features for future use
features = churn_feat_space.columns
X = churn_feat_space.as_matrix().astype(np.float)
# This is important
from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()
X = scaler.fit_transform(X)
print "Feature space holds %d observations and %d features" % X.shape
print "Unique target labels:", np.unique(y)
print X[0]
print len(y[y == 0])
Feature space holds 3333 observations and 17 features
Unique target labels: [0 1]
[ 0.67648946 -0.32758048 1.6170861 1.23488274 1.56676695 0.47664315
1.56703625 -0.07060962 -0.05594035 -0.07042665 0.86674322 -0.46549436
0.86602851 -0.08500823 -0.60119509 -0.0856905 -0.42793202]
2850
3 sklearn多模型封装(已废弃,学思想)
from sklearn.cross_validation import KFold
def run_cv(X,y,clf_class,**kwargs):
# Construct a kfolds object
kf = KFold(len(y),n_folds=5,shuffle=True)
y_pred = y.copy()
# Iterate through folds
for train_index, test_index in kf:
X_train, X_test = X[train_index], X[test_index]
y_train = y[train_index]
# Initialize a classifier with key word arguments
clf = clf_class(**kwargs)
clf.fit(X_train,y_train)
y_pred[test_index] = clf.predict(X_test)
return y_pred
from sklearn.svm import SVC
from sklearn.ensemble import RandomForestClassifier as RF
from sklearn.neighbors import KNeighborsClassifier as KNN
def accuracy(y_true,y_pred):
# NumPy interprets True and False as 1. and 0.
return np.mean(y_true == y_pred)
print "Support vector machines:"
print "%.3f" % accuracy(y, run_cv(X,y,SVC))
print "Random forest:"
print "%.3f" % accuracy(y, run_cv(X,y,RF))
print "K-nearest-neighbors:"
print "%.3f" % accuracy(y, run_cv(X,y,KNN))
Support vector machines:
0.916
Random forest:
0.944
K-nearest-neighbors:
0.893
4 阈值概率调整
def run_prob_cv(X, y, clf_class, **kwargs):
kf = KFold(len(y), n_folds=5, shuffle=True)
y_prob = np.zeros((len(y),2))
for train_index, test_index in kf:
X_train, X_test = X[train_index], X[test_index]
y_train = y[train_index]
clf = clf_class(**kwargs)
clf.fit(X_train,y_train)
# Predict probabilities, not classes
y_prob[test_index] = clf.predict_proba(X_test)
return y_prob
import warnings
warnings.filterwarnings('ignore')
# Use 10 estimators so predictions are all multiples of 0.1
pred_prob = run_prob_cv(X, y, RF, n_estimators=10)
#print pred_prob[0]
pred_churn = pred_prob[:,1]
is_churn = y == 1
# Number of times a predicted probability is assigned to an observation
counts = pd.value_counts(pred_churn)
#print counts
# calculate true probabilities
true_prob = {}
for prob in counts.index:
true_prob[prob] = np.mean(is_churn[pred_churn == prob])
true_prob = pd.Series(true_prob)
# pandas-fu
counts = pd.concat([counts,true_prob], axis=1).reset_index()
counts.columns = ['pred_prob', 'count', 'true_prob']
counts
# 0.7以上流式率达到94%,说明阈值为0.7是合适的,低于0.7不管,高于0.7的都认为是流失的
5 总结
方便复习,整成笔记,内容粗略,勿怪
版权声明:本套技术专栏是作者(秦凯新)平时工作的总结和升华,通过从真实商业环境抽取案例进行总结和分享,并给出商业应用的调优建议和集群环境容量规划等内容,请持续关注本套博客。QQ邮箱地址:[email protected],如有任何学术交流,可随时联系。
秦凯新 于深圳 20181211056