泰坦尼克号船员获救数据分析
一、概述
本文分析了泰坦尼克号船员获救的数据集合。数据集包括船员的一些信息（年龄、船舱等级、名字等等）和 是否获救的数据
数据解释
-
PassengerId Survived Pclass Name Sex Age SibSp Parch Ticket Fare Cabin Embarked
-
游客ID 是否被救 船舱等级 名字 性别 年龄 兄弟姐妹数 老人孩子数 票编号 票价 座位号 哪个站登船的
注意：兄弟姐妹数、老人孩子数都是指的是在该船上的统计量
二、流程
1. 使用线性回归构建预测模型
2. 使用逻辑回归构建预测模型
3. 使用决策树构建
4. 使用随机森林
5. 使用集成算法（Ensemble learning）中的聚合多个模型（本文中使用的是随机森林和逻辑回归算法）构建的集合算法模型
三、效果
从上到下，效果呈现出上升趋势
1. 数据导入和预处理
import pandas #ipython notebook
titanic = pandas.read_csv("titanic_train.csv")
titanic.head(5)
print (titanic.describe())
PassengerId Survived Pclass Age SibSp \
count 891.000000 891.000000 891.000000 714.000000 891.000000
mean 446.000000 0.383838 2.308642 29.699118 0.523008
std 257.353842 0.486592 0.836071 14.526497 1.102743
min 1.000000 0.000000 1.000000 0.420000 0.000000
25% 223.500000 0.000000 2.000000 20.125000 0.000000
50% 446.000000 0.000000 3.000000 28.000000 0.000000
75% 668.500000 1.000000 3.000000 38.000000 1.000000
max 891.000000 1.000000 3.000000 80.000000 8.000000
Parch Fare
count 891.000000 891.000000
mean 0.381594 32.204208
std 0.806057 49.693429
min 0.000000 0.000000
25% 0.000000 7.910400
50% 0.000000 14.454200
75% 0.000000 31.000000
max 6.000000 512.329200
titanic.iloc[19:30]
| PassengerId | Survived | Pclass | Name | Sex | Age | SibSp | Parch | Ticket | Fare | Cabin | Embarked | |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 19 | 20 | 1 | 3 | Masselmani, Mrs. Fatima | female | NaN | 0 | 0 | 2649 | 7.2250 | NaN | C |
| 20 | 21 | 0 | 2 | Fynney, Mr. Joseph J | male | 35.0 | 0 | 0 | 239865 | 26.0000 | NaN | S |
| 21 | 22 | 1 | 2 | Beesley, Mr. Lawrence | male | 34.0 | 0 | 0 | 248698 | 13.0000 | D56 | S |
| 22 | 23 | 1 | 3 | McGowan, Miss. Anna "Annie" | female | 15.0 | 0 | 0 | 330923 | 8.0292 | NaN | Q |
| 23 | 24 | 1 | 1 | Sloper, Mr. William Thompson | male | 28.0 | 0 | 0 | 113788 | 35.5000 | A6 | S |
| 24 | 25 | 0 | 3 | Palsson, Miss. Torborg Danira | female | 8.0 | 3 | 1 | 349909 | 21.0750 | NaN | S |
| 25 | 26 | 1 | 3 | Asplund, Mrs. Carl Oscar (Selma Augusta Emilia... | female | 38.0 | 1 | 5 | 347077 | 31.3875 | NaN | S |
| 26 | 27 | 0 | 3 | Emir, Mr. Farred Chehab | male | NaN | 0 | 0 | 2631 | 7.2250 | NaN | C |
| 27 | 28 | 0 | 1 | Fortune, Mr. Charles Alexander | male | 19.0 | 3 | 2 | 19950 | 263.0000 | C23 C25 C27 | S |
| 28 | 29 | 1 | 3 | O'Dwyer, Miss. Ellen "Nellie" | female | NaN | 0 | 0 | 330959 | 7.8792 | NaN | Q |
| 29 | 30 | 0 | 3 | Todoroff, Mr. Lalio | male | NaN | 0 | 0 | 349216 | 7.8958 | NaN | S |
amount_Age_NaN = len(titanic.loc[titanic["Age"].isnull().values,:])
print("Age字段缺失值的个数：",amount_Age_NaN)
Age字段缺失值的个数： 177
1.1 从上表原始数据集中可以看到：对于Age列，有177个年龄值缺失（缺失显示为NaN），需要用fillna函数填充缺失值,这里使用中位数（median）填充
titanic["Age"] = titanic["Age"].fillna(titanic["Age"].median())
print(titanic.describe())
PassengerId Survived Pclass Age SibSp \
count 891.000000 891.000000 891.000000 891.000000 891.000000
mean 446.000000 0.383838 2.308642 29.361582 0.523008
std 257.353842 0.486592 0.836071 13.019697 1.102743
min 1.000000 0.000000 1.000000 0.420000 0.000000
25% 223.500000 0.000000 2.000000 22.000000 0.000000
50% 446.000000 0.000000 3.000000 28.000000 0.000000
75% 668.500000 1.000000 3.000000 35.000000 1.000000
max 891.000000 1.000000 3.000000 80.000000 8.000000
Parch Fare
count 891.000000 891.000000
mean 0.381594 32.204208
std 0.806057 49.693429
min 0.000000 0.000000
25% 0.000000 7.910400
50% 0.000000 14.454200
75% 0.000000 31.000000
max 6.000000 512.329200
1.2 查看某一列属性值有多少种类，并将对应的字符串映射成数字
print(titanic["Sex"].unique())
# Replace all the occurences of male with the number 0.
titanic.loc[titanic["Sex"] == "male", "Sex"] = 0
titanic.loc[titanic["Sex"] == "female", "Sex"] = 1
['male' 'female']
print(titanic["Embarked"].unique())
titanic["Embarked"] = titanic["Embarked"].fillna('S')
titanic.loc[titanic["Embarked"] == "S", "Embarked"] = 0
titanic.loc[titanic["Embarked"] == "C", "Embarked"] = 1
titanic.loc[titanic["Embarked"] == "Q", "Embarked"] = 2
['S' 'C' 'Q' nan]
2. 训练模型
2.1 使用线性回归模型构建分类器
# Import the linear regression class
from sklearn.linear_model import LinearRegression
# Sklearn also has a helper that makes it easy to do cross validation
from sklearn.model_selection import KFold # KFold已经移到了model_selection 模块
# The columns we'll use to predict the target
predictors = ["Pclass", "Sex", "Age", "SibSp", "Parch", "Fare", "Embarked"]
# Initialize our algorithm class
alg = LinearRegression()
# Generate cross validation folds for the titanic dataset. It return the row indices corresponding to train and test.
# We set random_state to ensure we get the same splits every time we run this.
kf = KFold(n_splits=3, random_state=1)
predictions = []
accuracy = []
test_idices = []
# kf.split 会将数据分为n_splits份并返回训练和测试集数据对应的索引，而不是数据本身
# 注意，当Kfold的参数shuffle == false 时，生成的 test值从0开始，例如0,1,2,3.....len(数据集)
for train, test in kf.split(titanic):
# The predictors we're using the train the algorithm. Note how we only take the rows in the train folds.
train_predictors = (titanic[predictors].iloc[train,:])
# The target we're using to train the algorithm.
train_target = titanic["Survived"].iloc[train]
# Training the algorithm using the predictors and target.
alg.fit(train_predictors, train_target)
# We can now make predictions on the test fold
test_predictions = alg.predict(titanic[predictors].iloc[test,:])
predictions.append(test_predictions)
test_idices.append(test)
注意：此时predictions是一个list,里面有三个ndarray，因为交叉验证做了n_splits=3 次
下面的例子表明。Fold函数在当参数Shuffle == False时，生成的test 从 0 依次开始
test_idices
[array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,
13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25,
26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38,
39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51,
52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64,
65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77,
78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90,
91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103,
104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116,
117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129,
130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142,
143, 144, 145, 146, 147, 148, 149, 150, 151, 152, 153, 154, 155,
156, 157, 158, 159, 160, 161, 162, 163, 164, 165, 166, 167, 168,
169, 170, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181,
182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194,
195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207,
208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220,
221, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231, 232, 233,
234, 235, 236, 237, 238, 239, 240, 241, 242, 243, 244, 245, 246,
247, 248, 249, 250, 251, 252, 253, 254, 255, 256, 257, 258, 259,
260, 261, 262, 263, 264, 265, 266, 267, 268, 269, 270, 271, 272,
273, 274, 275, 276, 277, 278, 279, 280, 281, 282, 283, 284, 285,
286, 287, 288, 289, 290, 291, 292, 293, 294, 295, 296]),
array([297, 298, 299, 300, 301, 302, 303, 304, 305, 306, 307, 308, 309,
310, 311, 312, 313, 314, 315, 316, 317, 318, 319, 320, 321, 322,
323, 324, 325, 326, 327, 328, 329, 330, 331, 332, 333, 334, 335,
336, 337, 338, 339, 340, 341, 342, 343, 344, 345, 346, 347, 348,
349, 350, 351, 352, 353, 354, 355, 356, 357, 358, 359, 360, 361,
362, 363, 364, 365, 366, 367, 368, 369, 370, 371, 372, 373, 374,
375, 376, 377, 378, 379, 380, 381, 382, 383, 384, 385, 386, 387,
388, 389, 390, 391, 392, 393, 394, 395, 396, 397, 398, 399, 400,
401, 402, 403, 404, 405, 406, 407, 408, 409, 410, 411, 412, 413,
414, 415, 416, 417, 418, 419, 420, 421, 422, 423, 424, 425, 426,
427, 428, 429, 430, 431, 432, 433, 434, 435, 436, 437, 438, 439,
440, 441, 442, 443, 444, 445, 446, 447, 448, 449, 450, 451, 452,
453, 454, 455, 456, 457, 458, 459, 460, 461, 462, 463, 464, 465,
466, 467, 468, 469, 470, 471, 472, 473, 474, 475, 476, 477, 478,
479, 480, 481, 482, 483, 484, 485, 486, 487, 488, 489, 490, 491,
492, 493, 494, 495, 496, 497, 498, 499, 500, 501, 502, 503, 504,
505, 506, 507, 508, 509, 510, 511, 512, 513, 514, 515, 516, 517,
518, 519, 520, 521, 522, 523, 524, 525, 526, 527, 528, 529, 530,
531, 532, 533, 534, 535, 536, 537, 538, 539, 540, 541, 542, 543,
544, 545, 546, 547, 548, 549, 550, 551, 552, 553, 554, 555, 556,
557, 558, 559, 560, 561, 562, 563, 564, 565, 566, 567, 568, 569,
570, 571, 572, 573, 574, 575, 576, 577, 578, 579, 580, 581, 582,
583, 584, 585, 586, 587, 588, 589, 590, 591, 592, 593]),
array([594, 595, 596, 597, 598, 599, 600, 601, 602, 603, 604, 605, 606,
607, 608, 609, 610, 611, 612, 613, 614, 615, 616, 617, 618, 619,
620, 621, 622, 623, 624, 625, 626, 627, 628, 629, 630, 631, 632,
633, 634, 635, 636, 637, 638, 639, 640, 641, 642, 643, 644, 645,
646, 647, 648, 649, 650, 651, 652, 653, 654, 655, 656, 657, 658,
659, 660, 661, 662, 663, 664, 665, 666, 667, 668, 669, 670, 671,
672, 673, 674, 675, 676, 677, 678, 679, 680, 681, 682, 683, 684,
685, 686, 687, 688, 689, 690, 691, 692, 693, 694, 695, 696, 697,
698, 699, 700, 701, 702, 703, 704, 705, 706, 707, 708, 709, 710,
711, 712, 713, 714, 715, 716, 717, 718, 719, 720, 721, 722, 723,
724, 725, 726, 727, 728, 729, 730, 731, 732, 733, 734, 735, 736,
737, 738, 739, 740, 741, 742, 743, 744, 745, 746, 747, 748, 749,
750, 751, 752, 753, 754, 755, 756, 757, 758, 759, 760, 761, 762,
763, 764, 765, 766, 767, 768, 769, 770, 771, 772, 773, 774, 775,
776, 777, 778, 779, 780, 781, 782, 783, 784, 785, 786, 787, 788,
789, 790, 791, 792, 793, 794, 795, 796, 797, 798, 799, 800, 801,
802, 803, 804, 805, 806, 807, 808, 809, 810, 811, 812, 813, 814,
815, 816, 817, 818, 819, 820, 821, 822, 823, 824, 825, 826, 827,
828, 829, 830, 831, 832, 833, 834, 835, 836, 837, 838, 839, 840,
841, 842, 843, 844, 845, 846, 847, 848, 849, 850, 851, 852, 853,
854, 855, 856, 857, 858, 859, 860, 861, 862, 863, 864, 865, 866,
867, 868, 869, 870, 871, 872, 873, 874, 875, 876, 877, 878, 879,
880, 881, 882, 883, 884, 885, 886, 887, 888, 889, 890])]
import numpy as np
# The predictions are in three separate numpy arrays. Concatenate them into one.
# We concatenate them on axis 0, as they only have one axis.
predictions = np.concatenate((predictions[0],predictions[1],predictions[2]), axis=0)
test_idices = np.concatenate((test_idices[0],test_idices[1],test_idices[2]),axis=0)
# Map predictions to outcomes (only possible outcomes are 1 and 0)
predictions[predictions > .5] = 1 #代表预测
predictions[predictions <=.5] = 0
# accuracy = len(predictions[predictions == titanic["Survived"].iloc[test_idices].values]) / len(predictions)
#应为predict是针对索引0,1，2.依次比较的结果，所以有等价写法
accuracy = len(predictions[predictions == titanic["Survived"].values]) / len(predictions)
print(accuracy)
0.7833894500561167
2.2 使用逻辑回归模型（Logistics Regress其实是分类器模型，名字有点混淆）构建分类器
from sklearn.model_selection import cross_val_score
from sklearn.linear_model import LogisticRegression
# Initialize our algorithm
alg = LogisticRegression(random_state=1,solver='liblinear')
# Compute the accuracy score for all the cross validation folds. (much simpler than what we did before!)
scores = cross_val_score(alg, titanic[predictors], titanic["Survived"], cv=3)
# Take the mean of the scores (because we have one for each fold)
print(scores.mean())
0.7878787878787877
2.2.1 重新导入数据训练
titanic_test = pandas.read_csv("test.csv")
titanic_test["Age"] = titanic_test["Age"].fillna(titanic["Age"].median())
titanic_test["Fare"] = titanic_test["Fare"].fillna(titanic_test["Fare"].median())
titanic_test.loc[titanic_test["Sex"] == "male", "Sex"] = 0
titanic_test.loc[titanic_test["Sex"] == "female", "Sex"] = 1
titanic_test["Embarked"] = titanic_test["Embarked"].fillna("S")
titanic_test.loc[titanic_test["Embarked"] == "S", "Embarked"] = 0
titanic_test.loc[titanic_test["Embarked"] == "C", "Embarked"] = 1
titanic_test.loc[titanic_test["Embarked"] == "Q", "Embarked"] = 2
from sklearn.model_selection import KFold,cross_val_score
from sklearn.ensemble import RandomForestClassifier
predictors = ["Pclass", "Sex", "Age", "SibSp", "Parch", "Fare", "Embarked"]
# Initialize our algorithm with the default paramters
# n_estimators is the number of trees we want to make
# min_samples_split is the minimum number of rows we need to make a split
# min_samples_leaf is the minimum number of samples we can have at the place where a tree branch ends (the bottom points of the tree)
alg = RandomForestClassifier(random_state=1, n_estimators=10, min_samples_split=2, min_samples_leaf=1)
# Compute the accuracy score for all the cross validation folds. (much simpler than what we did before!)
#kf = model_selection.KFold(titanic.shape[0], n_folds=3, random_state=1)
kf = KFold(n_splits=5,random_state=2)
scores = cross_val_score(alg, titanic[predictors], titanic["Survived"], cv=kf)
# Take the mean of the scores (because we have one for each fold)
print(scores.mean())
0.8013935095097608
alg = RandomForestClassifier(random_state=1, n_estimators=100, min_samples_split=4, min_samples_leaf=2)
# Compute the accuracy score for all the cross validation folds. (much simpler than what we did before!)
kf = KFold( 3, random_state=1)
scores = cross_val_score(alg, titanic[predictors], titanic["Survived"], cv=kf)
# Take the mean of the scores (because we have one for each fold)
print(scores.mean())
0.8148148148148148
# Generating a familysize column
titanic["FamilySize"] = titanic["SibSp"] + titanic["Parch"]
# The .apply method generates a new series
# lambda arg1,arg2,.....argn:expression
titanic["NameLength"] = titanic["Name"].apply(lambda x: len(x))
Note: 关键字lambda表示匿名函数，lambda arg1,arg2,…argn:expression
- 冒号:之前的a,b,c表示它们是这个函数的参数
- 匿名函数不需要return来返回值，表达式本身结果就是返回值
# 正则表达式处理模块RE
import re
# A function to get the title from a name.
def get_title(name):
# Use a regular expression to search for a title. Titles always consist of capital and lowercase letters, and end with a period.
title_search = re.search(' ([A-Za-z]+)\.', name)
# If the title exists, extract and return it.
if title_search:
return title_search.group(1)
return ""
# Get all the titles and print how often each one occurs.
titles = titanic["Name"].apply(get_title)
print(pandas.value_counts(titles))
# Map each title to an integer. Some titles are very rare, and are compressed into the same codes as other titles.
title_mapping = {"Mr": 1, "Miss": 2, "Mrs": 3, "Master": 4, "Dr": 5, "Rev": 6, "Major": 7, "Col": 7, "Mlle": 8, "Mme": 8, "Don": 9, "Lady": 10, "Countess": 10, "Jonkheer": 10, "Sir": 9, "Capt": 7, "Ms": 2}
for k,v in title_mapping.items():
titles[titles == k] = v
# Verify that we converted everything.
print(pandas.value_counts(titles))
# Add in the title column.
titanic["Title"] = titles
Mr 517
Miss 182
Mrs 125
Master 40
Dr 7
Rev 6
Mlle 2
Col 2
Major 2
Jonkheer 1
Lady 1
Sir 1
Capt 1
Don 1
Countess 1
Ms 1
Mme 1
Name: Name, dtype: int64
1 517
2 183
3 125
4 40
5 7
6 6
7 5
10 3
8 3
9 2
Name: Name, dtype: int64
3.1.1 在决策树模型中选择最优的特征个数，并绘制各个特征对分类影响的重要程度
import numpy as np
from sklearn.feature_selection import SelectKBest, f_classif
from sklearn.model_selection import cross_val_score
import matplotlib.pyplot as plt
predictors = ["Pclass", "Sex", "Age", "SibSp", "Parch", "Fare", "Embarked", "FamilySize", "Title", "NameLength"]
# Perform feature selection
selector = SelectKBest(f_classif, k=5)
selector.fit(titanic[predictors], titanic["Survived"])
# Get the raw p-values for each feature, and transform from p-values into scores
scores = -np.log10(selector.pvalues_)
# Plot the scores. See how "Pclass", "Sex", "Title", and "Fare" are the best?
plt.bar(range(len(predictors)), scores)
plt.xticks(range(len(predictors)), predictors, rotation='vertical')
plt.title("importance of features")
plt.xlabel("features")
plt.ylabel("importance")
plt.show()
# Pick only the four best features.
predictors = ["Pclass", "Sex", "Fare", "Title"]
alg = RandomForestClassifier(random_state=1, n_estimators=50, min_samples_split=8, min_samples_leaf=4)
score = cross_val_score(alg,titanic[predictors],titanic["Survived"],cv=5)
print("预测的得分值: " ,score.mean())
预测的得分值: 0.8193563556396282
3. 使用集成算法求解
from sklearn.ensemble import GradientBoostingClassifier
import numpy as np
# The algorithms we want to ensemble.
# We're using the more linear predictors for the logistic regression, and everything with the gradient boosting classifier.
algorithms = [
[GradientBoostingClassifier(random_state=1, n_estimators=25, max_depth=3), ["Pclass", "Sex", "Age", "Fare", "Embarked", "FamilySize", "Title",]],
[LogisticRegression(random_state=1,solver='liblinear'), ["Pclass", "Sex", "Fare", "FamilySize", "Title", "Age", "Embarked"]]
]
# Initialize the cross validation folds
kf = KFold(n_splits=3, random_state=1)
predictions = []
for train, test in kf.split(titanic):
train_target = titanic["Survived"].iloc[train]
full_test_predictions = []
# Make predictions for each algorithm on each fold
for alg, predictors in algorithms:
# Fit the algorithm on the training data.
alg.fit(titanic[predictors].iloc[train,:], train_target)
# Select and predict on the test fold.
# The .astype(float) is necessary to convert the dataframe to all floats and avoid an sklearn error.
test_predictions = alg.predict_proba(titanic[predictors].iloc[test,:].astype(float))[:,1]
full_test_predictions.append(test_predictions)
# Use a simple ensembling scheme -- just average the predictions to get the final classification.
test_predictions = (full_test_predictions[0] + full_test_predictions[1]) / 2
# Any value over .5 is assumed to be a 1 prediction, and below .5 is a 0 prediction.
test_predictions[test_predictions <= .5] = 0
test_predictions[test_predictions > .5] = 1
predictions.append(test_predictions)
# Put all the predictions together into one array.
predictions = np.concatenate(predictions, axis=0)
# Compute accuracy by comparing to the training data.
accuracy = len(predictions[predictions == titanic["Survived"]]) / len(predictions)
print('模型精确度：',accuracy)
模型精确度： 0.8215488215488216
titles = titanic_test["Name"].apply(get_title)
# We're adding the Dona title to the mapping, because it's in the test set, but not the training set
title_mapping = {"Mr": 1, "Miss": 2, "Mrs": 3, "Master": 4, "Dr": 5, "Rev": 6, "Major": 7, "Col": 7, "Mlle": 8, "Mme": 8, "Don": 9, "Lady": 10, "Countess": 10, "Jonkheer": 10, "Sir": 9, "Capt": 7, "Ms": 2, "Dona": 10}
for k,v in title_mapping.items():
titles[titles == k] = v
titanic_test["Title"] = titles
# Check the counts of each unique title.
print(pandas.value_counts(titanic_test["Title"].values))
# Now, we add the family size column.
titanic_test["FamilySize"] = titanic_test["SibSp"] + titanic_test["Parch"]
1 240
2 79
3 72
4 21
7 2
6 2
10 1
5 1
dtype: int64
3.2构建集成算法，对一个未知结果的数据集进行预测
predictors = ["Pclass", "Sex", "Age", "Fare", "Embarked", "FamilySize", "Title"]
algorithms = [
[GradientBoostingClassifier(random_state=1, n_estimators=25, max_depth=3), predictors],
[LogisticRegression(random_state=1,solver='liblinear'), ["Pclass", "Sex", "Fare", "FamilySize", "Title", "Age", "Embarked"]]
]
full_predictions = []
for alg, predictors in algorithms:
# Fit the algorithm using the full training data.
alg.fit(titanic[predictors], titanic["Survived"])
# Predict using the test dataset. We have to convert all the columns to floats to avoid an error.
predictions = alg.predict_proba(titanic_test[predictors].astype(float))[:,1]
full_predictions.append(predictions)
# The gradient boosting classifier generates better predictions, so we weight it higher.
predictions = (full_predictions[0] * 3 + full_predictions[1]) / 4
predictions[predictions > .5] = 1
predictions[predictions <= .5] = 0
predictions
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