一、課後習題2.4

1、求均值、中位數和標準差

age = [23, 23, 27, 27, 39, 41, 47, 49, 50, 52, 54, 54, 56, 57, 58, 58, 60, 61]
fat = [9.5, 26.5, 7.8, 17.8, 31.4, 25.9, 27.4, 27.2, 31.2, 34.6, 42.5, 28.8, 33.4, 30.2, 34.1, 32.9, 41.2, 35.7]

def mean(List):
    """計算列表List的均值並返回"""
    return sum(List) / len(List)

def medium(List):
    """計算列表List的中位數"""
    List = sorted(List)
    n = len(List)
    mid = int(n / 2)
    if(n % 2 == 0):  # 列表長度爲偶數
        return (List[mid - 1] + List[mid]) / 2
    return List[mid]

def std(List):
    """計算列表的標準差"""
    n = len(List)
    men = mean(List)
    sum = 0
    for i in range(n):
        sum += (List[i] - men)**2
    ret = (sum / (n - 1))**0.5
    return ret

if __name__ == '__main__':
    print("age的均值爲: {}, 中位數爲: {}, 標準差爲: {}".format(mean(age), medium(age), std(age)))
    print("fat的均值爲: {}, 中位數爲: {}, 標準差爲: {}".format(mean(fat), medium(fat), std(fat)))

$\quad$ 程序運行結果如下：

2、繪製盒圖

$\quad$ 盒圖的形式如下：

$\quad$ 在python的pandas庫中，我們可以直接得到數據的統計信息，也可以直接畫出盒圖，程序如下：

import pandas as pd
import matplotlib.pyplot as plt

age = [23, 23, 27, 27, 39, 41, 47, 49, 50, 52, 54, 54, 56, 57, 58, 58, 60, 61]
fat = [9.5, 26.5, 7.8, 17.8, 31.4, 25.9, 27.4, 27.2, 31.2, 34.6, 42.5, 28.8, 33.4, 30.2, 34.1, 32.9, 41.2, 35.7]

def boxPlot(List, title):
    """繪製列表List的盒圖"""
    df = pd.DataFrame(List)
    df.plot.box(title=title)  # 繪製盒圖名稱
    plt.grid(linestyle="--", alpha=0.3)
    plt.show()

if __name__ == '__main__':
    print(pd.DataFrame(age).describe())  # 對數據的各個屬性進行描述
    boxPlot(age, title="age")
    print(pd.DataFrame(fat).describe())  # 對數據的各個屬性進行描述
    boxPlot(fat, title="fat")

age的描述爲：

count  18.000000  數據數目
mean   46.444444  均值
std    13.218624  方差
min    23.000000  最小值
25%    39.500000  下四分位數Q1
50%    51.000000  中位數
75%    56.750000  上四分位數Q3
max    61.000000  最大值

fat描述爲

count  18.000000
mean   28.783333
std     9.254395
min     7.800000
25%    26.675000
50%    30.700000
75%    33.925000
max    42.500000

age的盒圖爲：
fat的盒圖爲：

3、繪製這兩個變量的散點圖和q-q圖

$\quad$ python上繪製q-q圖不是很方便，可在matlab中完成

age = [23 23 27 27 39 41 47 49 50 52 54 54 56 57 58 58 60 61];
fat = [9.5 26.5 7.8 17.8 31.4 25.9 27.4 27.2 31.2 34.6 42.5 28.8 33.4 30.2 34.1 32.9 41.2 35.7];
qqplot(age, fat);
title('q-q圖');

二、課後習題2.8

target = [1.4, 1.6]  # 查詢點
# data存放x1到x5五個點的信息，x1 = [1.5, 1.7, 0]表示x1座標爲(1.5, 1.7)，距離查詢點距離爲0
data = [[1.5, 1.7, 0], [2, 1.9, 0], [1.6, 1.8, 0], [1.2, 1.5, 0], [1.5, 1.0, 0]]

def Euclid(A, B):
    """計算A和B的歐式距離"""
    n = min(len(A), len(B))
    ret = 0
    for i in range(n):
        ret += (A[i] - B[i])**2
    return ret**0.5

def Manhattan(A, B):
    """計算A和B的曼哈頓距離"""
    n = min(len(A), len(B))
    ret = 0
    for i in range(n):
        ret += abs(A[i] - B[i])
    return ret

def Supremum(A, B):
    """計算A和B的上確界距離"""
    n = min(len(A), len(B))
    ret = 0
    for i in range(n):
        ret = max(ret, abs(A[i] - B[i]))
    return ret

def Cosine(A, B):
    """計算A和B的餘弦距離"""
    n = min(len(A), len(B))
    up = 0
    for i in range(n):
        up += A[i] * B[i]
    normA, normB = 0, 0
    for i in range(n):
        normA += A[i]**2
        normB += B[i]**2
    normA, normB = normA**0.5, normB**0.5
    return up / (normA * normB)

def solve(method):
    n = len(data)
    for i in range(n):
        data[i][2] = method(data[i][:2], target)  # 用傳入的method的方法計算距離
    ret = sorted(data, key=lambda item: item[2])  # 按照距離從小到大排序
    print(method.__name__, ret)
    
if __name__ == '__main__':
    methods = [Euclid, Manhattan, Supremum, Cosine]
    for method in methods:
        solve(method)

$\quad$ 輸出結果如下，每一行表示在該距離衡量方法下查詢點到 $x_1,\cdots,x_5$ 這5個點距離從小到大的排序值。例如Euclid方法下表示查詢點距離點(1.5,1.7)距離最小，距離是1.414。

Euclid [[1.5, 1.7, 0.14142135623730948], [1.2, 1.5, 0.22360679774997896], [1.6, 1.8, 0.28284271247461906], [1.5, 1.0, 0.608276253029822], [2, 1.9, 0.6708203932499369]]
Manhattan [[1.5, 1.7, 0.19999999999999996], [1.2, 1.5, 0.30000000000000004], [1.6, 1.8, 0.40000000000000013], [1.5, 1.0, 0.7000000000000002], [2, 1.9, 0.8999999999999999]]
Supremum [[1.5, 1.7, 0.10000000000000009], [1.2, 1.5, 0.19999999999999996], [1.6, 1.8, 0.20000000000000018], [2, 1.9, 0.6000000000000001], [1.5, 1.0, 0.6000000000000001]]
Cosine [[1.5, 1.0, 0.9653633930282662], [2, 1.9, 0.9957522612528874], [1.2, 1.5, 0.9990282349375618], [1.6, 1.8, 0.9999694838187877], [1.5, 1.7, 0.999991391443956]]

$\quad$ 對於第二問，將數據規範化後再計算查詢點與這五個點的歐氏距離，程序如下：

target = [1.4, 1.6]  # 查詢點
# data存放x1到x5五個點的信息，x1 = [1.5, 1.7, 0]表示x1座標爲(1.5, 1.7)，距離查詢點距離爲0
data = [[1.5, 1.7, 0], [2, 1.9, 0], [1.6, 1.8, 0], [1.2, 1.5, 0], [1.5, 1.0, 0]]
def normlize(List):
    """規範化數據，使得List的範數爲1"""
    l2 = (List[0]**2 + List[1]**2)**0.5
    List[0] /= l2
    List[1] /= l2
    return List
def Euclid(A, B):
    """計算A和B的歐式距離"""
    n = min(len(A), len(B))
    ret = 0
    for i in range(n):
        ret += (A[i] - B[i])**2
    return ret**0.5
def solve():
    """將數據規範化後計算距離並排序"""
    n = len(data)
    for i in range(n):
        data[i][:2] = normlize(data[i][:2])
        data[i][2] = Euclid(data[i][:2], normlize(target))  # 用傳入的method的方法計算距離
    ret = sorted(data, key=lambda item: item[2])  # 按照距離從小到大排序
    print(method.__name__, ret)
    
if __name__ == '__main__':
    solve()

$\quad$ 結果如下：

Euclid [[0.6616216370868464, 0.7498378553650925, 0.004149350803200864], [0.6643638388299198, 0.7474093186836597, 0.007812321193114019], [0.6246950475544242, 0.7808688094430303, 0.044085486555962686], [0.7249994335944139, 0.6887494619146931, 0.09217091457843411], [0.8320502943378437, 0.5547001962252291, 0.2631980507972417]]

課後習題3.3

$\quad$ 這裏給出第一問平滑的程序：

def smooth(data, k):
    """k爲光滑箱深度，data爲需平滑的數據，這裏我們假設data長度能整除k"""
    ret = []
    n = len(data) // k
    for i in range(n):
        sum = 0
        for j in range(i*k, (i+1)*k):
            sum += data[j]
        avg = sum / k
        for j in range(k):
            ret.append(avg)
    return ret

if __name__ == '__main__':
    age = [13, 15, 16, 16, 19, 20, 20, 21, 22, 22,
           25, 25, 25, 25, 30, 33, 33, 35, 35, 35,
           35, 36, 40, 45, 46, 52, 70]
    age = smooth(age, k=3)
    print(age)

$\quad$ 運行結果：

[14.666666666666666, 14.666666666666666, 14.666666666666666, 18.333333333333332, 18.333333333333332, 18.333333333333332, 21.0, 21.0, 21.0, 24.0, 24.0, 24.0, 26.666666666666668, 26.666666666666668, 26.666666666666668, 33.666666666666664, 33.666666666666664, 33.666666666666664, 35.0, 35.0, 35.0, 40.333333333333336, 40.333333333333336, 40.333333333333336, 56.0, 56.0, 56.0]

課後習題3.7

$\quad$ 給出程序如下：

import numpy as np

def minmaxScaler(data, target):
    """將數據規範化到[0, 1]區間"""
    Min, Max = data[0], data[0]
    n = len(data)
    for i in range(n):
        Min = min(Min, data[i])
        Max = max(Max, data[i])
    target = (target - Min) / (Max - Min)
    return target

def zScaler(data, target):
    """將data進行z-score規範化"""
    Mean, std = np.mean(data), np.std(data)
    target = (target - Mean) / std
    return target

def pivotScaler(data, target):
    """將data使用小數定標規範化"""
    Max = np.max(data)
    j = len(str(Max))
    return target / (10**j)

if __name__ == '__main__':
    age = [13, 15, 16, 16, 19, 20, 20, 21, 22, 22,
           25, 25, 25, 25, 30, 33, 33, 35, 35, 35,
           35, 36, 40, 45, 46, 52, 70]

    print(minmaxScaler.__name__, minmaxScaler(age, 35))
    print(zScaler.__name__, zScaler(age, 35))
    print(pivotScaler.__name__, pivotScaler(age, 35))

$\quad$ 結果如下：

minmaxScaler 0.38596491228070173
zScaler 0.3966110348537352
pivotScaler 0.35

數據挖掘課程作業代碼實現

一、課後習題2.4

1、求均值、中位數和標準差

2、繪製盒圖

3、繪製這兩個變量的散點圖和q-q圖

二、課後習題2.8

課後習題3.3

課後習題3.7

課後習題3.12

關於遊戲付費的一點想法

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快速計算數組中前n個數的均值和方差

python3實現無限次翻譯

Leetcode 20. 有效的括號 0ms通過

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