机器学习入门------python基础

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环境:Python 3.6.4 |Anaconda, Inc.

Python常用容器类型

1.list

 1 l = [1, a, 2, b]
 2 print(type(l))
 3 print(修改前:, l)
 4 
 5 # 修改list的内容
 6 l[0] = 3
 7 print(修改后:, l)
 8 
 9 # 末尾添加元素
10 l.append(4)
11 print(添加后:, l)
12 
13 # 遍历list
14 print(遍历list(for循环):)
15 for item in l:
16     print(item)
17     
18 # 通过索引遍历list
19 print(遍历list(while循环):)
20 i = 0
21 while i != len(l):
22     print(l[i])
23     i += 1
24     
25 # 列表合并
26 print(列表合并(+):, [1, 2] + [3, 4])
27 
28 # 列表重复
29 print(列表重复(*):, [1, 2] * 5)
30 
31 # 判断元素是否在列表中
32 print(判断元素存在(in):, 1 in [1, 2])
<class ‘list‘>
修改前: [1, ‘a‘, 2, ‘b‘]
修改后: [3, ‘a‘, 2, ‘b‘]
添加后: [3, ‘a‘, 2, ‘b‘, 4]
遍历list(for循环):
3
a
2
b
4
遍历list(while循环):
3
a
2
b
4
列表合并(+): [1, 2, 3, 4]
列表重复(*): [1, 2, 1, 2, 1, 2, 1, 2, 1, 2]
判断元素存在(in): True

2.tuple

 1 t = (1, a, 2, b)
 2 print(type(t))
 3 
 4 #元组的内容不能修改,否则会报错
 5 # t[0] = 3 
 6 
 7 # 遍历tuple
 8 print(遍历list(for循环):)
 9 for item in t:
10     print(item)
11     
12 # 通过索引遍历tuple
13 print(遍历tuple(while循环):)
14 i = 0
15 while i != len(t):
16     print(t[i])
17     i += 1
18     
19 # 解包 unpack
20 a, b, _, _ = t
21 print(unpack: , c)
22 
23 # 确保unpack接收的变量个数和tuple的长度相同,否则报错
24 # 经常出现在函数返回值的赋值时
25 # a, b, c = t
<class ‘tuple‘>
遍历list(for循环):
1
a
2
b
遍历tuple(while循环):
1
a
2
b
 
---------------------------------------------------------------------------
NameError                                 Traceback (most recent call last)
<ipython-input-3-88506d8b1a51> in <module>()
     19 # 解包 unpack
     20 a, b, _, _ = t
---> 21 print(‘unpack: ‘, c)
     22 
     23 # 确保unpack接收的变量个数和tuple的长度相同,否则报错

NameError: name ‘c‘ is not defined

3.dictionary

 1 d = {小象学院: http://www.chinahadoop.cn/,
 2     百度: https://www.baidu.com/,
 3     阿里巴巴: https://www.alibaba.com/,
 4     腾讯: https://www.tencent.com/}
 5 
 6 print(通过key获取value: , d[小象学院])
 7 
 8 # 遍历key
 9 print(遍历key: )
10 for key in d.keys():
11     print(key)
12     
13 # 遍历value
14 print(遍历value: )
15 for value in d.values():
16     print(value)
17     
18 # 遍历item
19 print(遍历item: )
20 for key, value in d.items():
21     print(key + :  + value)
22 
23 # format输出格式
24 print(format输出格式:)
25 for key, value in d.items():
26     print({}的网址是{}.format(key, value))
通过key获取value:  http://www.chinahadoop.cn/
遍历key: 
小象学院
百度
阿里巴巴
腾讯
遍历value: 
http://www.chinahadoop.cn/
https://www.baidu.com/
https://www.alibaba.com/
https://www.tencent.com/
遍历item: 
小象学院: http://www.chinahadoop.cn/
百度: https://www.baidu.com/
阿里巴巴: https://www.alibaba.com/
腾讯: https://www.tencent.com/
format输出格式:
小象学院的网址是http://www.chinahadoop.cn/
百度的网址是https://www.baidu.com/
阿里巴巴的网址是https://www.alibaba.com/
腾讯的网址是https://www.tencent.com/

4.set

 1 print(创建set:)
 2 my_set = {1, 2, 3}
 3 print(my_set)
 4 my_set = set([1, 2, 3, 2])
 5 print(my_set)
 6 
 7 print(添加单个元素:)
 8 my_set.add(3)
 9 print(添加3, my_set)
10 
11 my_set.add(4)
12 print(添加4, my_set)
13 
14 print(添加多个元素:)
15 my_set.update([4, 5, 6])
16 print(my_set)
创建set:
{1, 2, 3}
{1, 2, 3}
添加单个元素:
添加3 {1, 2, 3}
添加4 {1, 2, 3, 4}
添加多个元素:
{1, 2, 3, 4, 5, 6}

5.Counter

  • 初始化
1 import collections
2 
3 c1 = collections.Counter([a, b, c, a, b, b])
4 c2 = collections.Counter({a:2, b:3, c:1})
5 c3 = collections.Counter(a=2, b=3, c=1)
6 
7 print(c1)
8 print(c2)
9 print(c3)
Counter({‘b‘: 3, ‘a‘: 2, ‘c‘: 1})
Counter({‘b‘: 3, ‘a‘: 2, ‘c‘: 1})
Counter({‘b‘: 3, ‘a‘: 2, ‘c‘: 1})
  • 更新内容
1 # 注意这里是做“加法”,不是“替换”
2 c1.update({a: 4, c: -2, d: 4})
3 print(c1)
Counter({‘a‘: 6, ‘d‘: 4, ‘b‘: 3, ‘c‘: -1})
  • 访问内容
1 print(a=, c1[a])
2 print(b=, c1[b])
3 # 对比和dict的区别
4 print(e=, c1[e])
a= 6
b= 3
e= 0
  • element()方法
1 for element in c1.elements():
2     print(element)
d
d
d
d
b
b
b
a
a
a
a
a
a
  • most_common()方法
1 c1.most_common(3)
2 [(a, 6), (d, 4), (b, 3)]

6.defaultdict

1 # 统计每个字母出现的次数
2 s = chinadoop
3 
4 # 使用Counter
5 print(collections.Counter(s))
Counter({‘o‘: 2, ‘d‘: 1, ‘c‘: 1, ‘p‘: 1, ‘a‘: 1, ‘n‘: 1, ‘h‘: 1, ‘i‘: 1})
1 # 使用dict
2 counter = {}
3 for c in s:
4     if c not in counter:
5         counter[c] = 1
6     else:
7         counter[c] += 1
8         
9 print(counter.items())
dict_items([(‘d‘, 1), (‘c‘, 1), (‘p‘, 1), (‘a‘, 1), (‘o‘, 2), (‘n‘, 1), (‘h‘, 1), (‘i‘, 1)])
1 # 使用defaultdict
2 counter2 = collections.defaultdict(int)
3 for c in s:
4     counter2[c] += 1
5 print(counter2.items())
dict_items([(‘d‘, 1), (‘c‘, 1), (‘p‘, 1), (‘a‘, 1), (‘o‘, 2), (‘n‘, 1), (‘h‘, 1), (‘i‘, 1)])
1 # 记录相同元素的列表
2 colors = [(yellow, 1), (blue, 2), (yellow, 3), (blue, 4), (red, 1)]
3 d = collections.defaultdict(list)
4 for k, v in colors:
5     d[k].append(v)
6 
7 print(d.items())
dict_items([(‘blue‘, [2, 4]), (‘yellow‘, [1, 3]), (‘red‘, [1])])

7.map函数

 1 import math
 2 
 3 print(示例1,获取两个列表对应位置上的最小值:)
 4 l1 = [1, 3, 5, 7, 9]
 5 l2 = [2, 4, 6, 6, 9]
 6 mins = map(min, l1, l2)
 7 print(mins)
 8 
 9 # map()函数操作时,直到访问数据时才会执行
10 for item in mins:
11     print(item)
12 
13 print(示例2,对列表中的元素进行平方根操作:)
14 squared = map(math.sqrt, l2)
15 print(squared)
16 print(list(squared))
示例1,获取两个列表对应位置上的最小值:
<map object at 0x0000019AF8B0CDD8>
1
3
5
6
9
示例2,对列表中的元素进行平方根操作:
<map object at 0x0000019AF8A79DD8>
[1.4142135623730951, 2.0, 2.449489742783178, 2.449489742783178, 3.0]

8.匿名函数lambda

 1 # my_func = lambda a, b, c: a * b
 2 # print(my_func)
 3 # print(my_func(1, 2, 3))
 4 
 5 # 结合map
 6 print(lambda结合map)
 7 l1 = [1, 3, 5, 7, 9]
 8 l2 = [2, 4, 6, 8, 10]
 9 result = map(lambda x, y: x * 2 + y, l1, l2)
10 print(list(result))
lambda结合map
[4, 10, 16, 22, 28]

9.python操作csv数据文件

1 import csv
2 
3 with open(grades.csv) as csvfile:
4     grades_data = list(csv.DictReader(csvfile))
5     
6 print(记录个数:, len(grades_data))
7 print(前2条记录:, grades_data[:2])
8 print(列名:, list(grades_data[0].keys()))
记录个数: 2315
前2条记录: [OrderedDict([(‘student_id‘, ‘B73F2C11-70F0-E37D-8B10-1D20AFED50B1‘), (‘assignment1_grade‘, ‘92.73394640624123‘), (‘assignment1_submission‘, ‘2015-11-02 06:55:34.282000000‘), (‘assignment2_grade‘, ‘83.03055176561709‘), (‘assignment2_submission‘, ‘2015-11-09 02:22:58.938000000‘), (‘assignment3_grade‘, ‘67.16444141249367‘), (‘assignment3_submission‘, ‘2015-11-12 08:58:33.998000000‘), (‘assignment4_grade‘, ‘53.01155312999494‘), (‘assignment4_submission‘, ‘2015-11-16 01:21:24.663000000‘), (‘assignment5_grade‘, ‘47.710397816995446‘), (‘assignment5_submission‘, ‘2015-11-20 13:24:59.692000000‘), (‘assignment6_grade‘, ‘38.16831825359636‘), (‘assignment6_submission‘, ‘2015-11-22 18:31:15.934000000‘)]), OrderedDict([(‘student_id‘, ‘98A0FAE0-A19A-13D2-4BB5-CFBFD94031D1‘), (‘assignment1_grade‘, ‘86.79082085792986‘), (‘assignment1_submission‘, ‘2015-11-29 14:57:44.429000000‘), (‘assignment2_grade‘, ‘86.29082085792986‘), (‘assignment2_submission‘, ‘2015-12-06 17:41:18.449000000‘), (‘assignment3_grade‘, ‘69.7726566863439‘), (‘assignment3_submission‘, ‘2015-12-10 08:54:55.904000000‘), (‘assignment4_grade‘, ‘55.0981253490751‘), (‘assignment4_submission‘, ‘2015-12-13 17:32:30.941000000‘), (‘assignment5_grade‘, ‘49.5883128141676‘), (‘assignment5_submission‘, ‘2015-12-19 23:26:39.285000000‘), (‘assignment6_grade‘, ‘44.62948153275085‘), (‘assignment6_submission‘, ‘2015-12-21 17:07:24.275000000‘)])]
列名: [‘student_id‘, ‘assignment1_grade‘, ‘assignment1_submission‘, ‘assignment2_grade‘, ‘assignment2_submission‘, ‘assignment3_grade‘, ‘assignment3_submission‘, ‘assignment4_grade‘, ‘assignment4_submission‘, ‘assignment5_grade‘, ‘assignment5_submission‘, ‘assignment6_grade‘, ‘assignment6_submission‘]
1 avg_assign1 = sum([float(row[assignment1_grade]) for row in grades_data]) / len(grades_data) 
2 print(assignment1平均分数:, avg_assign1)
assignment1平均分数: 74.5357320747794
1 assign1_sub_month = set(row[assignment1_submission][:7] for row in grades_data)
2 print(assign1_sub_month)
{‘2016-02‘, ‘2015-09‘, ‘2016-01‘, ‘2016-04‘, ‘2016-03‘, ‘2016-06‘, ‘2016-08‘, ‘2015-10‘, ‘2016-05‘, ‘2016-07‘, ‘2015-12‘, ‘2015-11‘}

科学计算库NumPy

1 import numpy as np

1. 创建Array

1 my_list = [1, 2, 3]
2 x = np.array(my_list)
3 
4 print(列表:, my_list)
5 print(Array: , x)
列表: [1, 2, 3]
Array:  [1 2 3]
1 np.array([1, 2, 3]) - np.array([4, 5, 6])
array([-3, -3, -3])
1 m = np.array([[1, 2, 3], [4, 5, 6]])
2 print(m)
3 print(shape: , m.shape)
[[1 2 3]
 [4 5 6]]
shape:  (2, 3)
1 n = np.arange(0, 30, 2)
2 print(n)
[ 0  2  4  6  8 10 12 14 16 18 20 22 24 26 28]
1 n = n.reshape(3, 5)
2 print(reshape后: )
3 print(n)
reshape后: 
[[ 0  2  4  6  8]
 [10 12 14 16 18]
 [20 22 24 26 28]]
1 print(ones:\n, np.ones((3, 2)))
2 print(zeros:\n, np.zeros((3, 2)))
3 print(eye:\n, np.eye(3))
4 print(diag:\n, np.diag(my_list))
ones:
 [[1. 1.]
 [1. 1.]
 [1. 1.]]
zeros:
 [[0. 0.]
 [0. 0.]
 [0. 0.]]
eye:
 [[1. 0. 0.]
 [0. 1. 0.]
 [0. 0. 1.]]
diag:
 [[1 0 0]
 [0 2 0]
 [0 0 3]]
1 print(*操作:\n, np.array([1, 2, 3] * 3))
2 print(repeat:\n, np.repeat([1, 2, 3], 3))
*操作:
 [1 2 3 1 2 3 1 2 3]
repeat:
 [1 1 1 2 2 2 3 3 3]
1 p1 = np.ones((3, 3))
2 p2 = np.arange(9).reshape(3, 3)
3 print(纵向叠加: \n, np.vstack((p1, p2)))
4 print(横向叠加: \n, np.hstack((p1, p2)))
纵向叠加: 
 [[ 1.  1.  1.]
 [ 1.  1.  1.]
 [ 1.  1.  1.]
 [ 0.  1.  2.]
 [ 3.  4.  5.]
 [ 6.  7.  8.]]
横向叠加: 
 [[ 1.  1.  1.  0.  1.  2.]
 [ 1.  1.  1.  3.  4.  5.]
 [ 1.  1.  1.  6.  7.  8.]]

2. Array操作

1 p1 = np.array([[1, 1, 1], [1, 1, 1],[1,1,1]])
2 p2 = np.arange(9).reshape(3, 3)3 print(p1: \n, p1)
4 print(p2: \n, p2)
5 
6 print(p1 + p2 = \n, p1 + p2)
7 print(p1 * p2 = \n, p1 * p2)
8 print(p2^2 = \n, p2 ** 2)
9 print(p1.p2 = \n, p1.dot(p2))
p1: 
 [[1 1 1]
 [1 1 1]
 [1 1 1]]
p2: 
 [[0 1 2]
 [3 4 5]
 [6 7 8]]
p1 + p2 = 
 [[1 2 3]
 [4 5 6]
 [7 8 9]]
p1 * p2 = 
 [[0 1 2]
 [3 4 5]
 [6 7 8]]
p2^2 = 
 [[ 0  1  4]
 [ 9 16 25]
 [36 49 64]]
p1.p2 = 
 [[ 9 12 15]
 [ 9 12 15]
 [ 9 12 15]]
1 p3 = np.arange(6).reshape(2, 3)
2 print(p3形状: , p3.shape)
3 print(p3)
4 p4 = p3.T
5 print(转置后p3形状: , p4.shape)
6 print(p4)
p3形状:  (2, 3)
[[0 1 2]
 [3 4 5]]
转置后p3形状:  (3, 2)
[[0 3]
 [1 4]
 [2 5]]
1 p3 = np.arange(6).reshape(2, 3)
2 print(p3数据类型:, p3.dtype)
3 print(p3)
4 
5 p5 = p3.astype(float)
6 print(p5数据类型:, p5.dtype)
7 print(p5)
p3数据类型: int32
[[0 1 2]
 [3 4 5]]
p5数据类型: float64
[[0. 1. 2.]
 [3. 4. 5.]]
a = np.array([-4, -2, 1, 3, 5])
print(sum: , a.sum())
print(min: , a.min())
print(max: , a.max())
print(mean: , a.mean())
print(std: , a.std()) //标准差
print(argmax: , a.argmax()) //argmax(f(x))是使得 f(x)取得最大值所对应的变量x
print(argmin: , a.argmin()) //argmax(f(x))是使得 f(x)取得最小值所对应的变量x
sum:  3
min:  -4
max:  5
mean:  0.6
std:  3.2619012860600183
argmax:  4
argmin:  0

3. 索引与切片

1 # 一维array
2 s = np.arange(13) ** 2
3 print(s: , s)
4 print(s[0]: , s[0])
5 print(s[4]: , s[4])
6 print(s[0:3]: , s[0:3])
7 print(s[[0, 2, 4]]: , s[[0, 2, 4]])
s:  [  0   1   4   9  16  25  36  49  64  81 100 121 144]
s[0]:  0
s[4]:  16
s[0:3]:  [0 1 4]
s[[0, 2, 4]]:  [ 0  4 16]
1 # 二维array
2 r = np.arange(36).reshape((6, 6))
3 print(r: \n, r)
4 print(r[2, 2]: \n, r[2, 2]) //对应矩阵第三行第三列
5 print(r[3, 3:6]: \n, r[3, 3:6]) //对应第四行第四列到第7列的数(只表示该行的数)
r: 
 [[ 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]]
r[2, 2]: 
 14
r[3, 3:6]: 
 [21 22 23]
1 r = np.arange(36).reshape((6, 6))
2 r > 30
array([[False, False, False, False, False, False],
       [False, False, False, False, False, False],
       [False, False, False, False, False, False],
       [False, False, False, False, False, False],
       [False, False, False, False, False, False],
       [False,  True,  True,  True,  True,  True]])
1 # 过滤
2 print(r[r > 30])
3 
4 # 将大于30的数赋值为30
5 r[r > 30] = 30
6 print(r)
[31 32 33 34 35]
[[ 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 30 30 30 30 30]]
1 # copy()操作
2 r2 = r[:3, :3]
3 print(r2)
[[ 0  1  2]
 [ 6  7  8]
 [12 13 14]]
1 # 将r2内容设置为0
2 r2[:] = 0
3 
4 # 查看r的内容
5 print(r)
[[ 0  0  0  3  4  5]
 [ 0  0  0  9 10 11]
 [ 0  0  0 15 16 17]
 [18 19 20 21 22 23]
 [24 25 26 27 28 29]
 [30 30 30 30 30 30]]
1 r3 = r.copy()
2 r3[:] = 0
3 print(r)
[[ 0  0  0  3  4  5]
 [ 0  0  0  9 10 11]
 [ 0  0  0 15 16 17]
 [18 19 20 21 22 23]
 [24 25 26 27 28 29]
 [30 30 30 30 30 30]]

4. 遍历 Array

1 import numpy as np
2 t = np.random.randint(0, 10, (4, 3))
3 print(t)
[[3 2 7]
 [4 9 1]
 [1 3 0]
 [0 9 1]]
1 for row in t:
2     print(row)
[3 2 7]
[4 9 1]
[1 3 0]
[0 9 1]
1 # 使用enumerate()
2 for i, row in enumerate(t):
3     print(row {} is {}.format(i, row))
row 0 is [3 2 7]
row 1 is [4 9 1]
row 2 is [1 3 0]
row 3 is [0 9 1]
1 t2 = t ** 2
2 print(t2)
[[ 9  4 49]
 [16 81  1]
 [ 1  9  0]
 [ 0 81  1]]
1 # 使用zip对两个array进行遍历计算
2 for i, j in zip(t, t2):
3     print({} + {} = {}.format(i, j, i + j))
[3 2 7] + [ 9  4 49] = [12  6 56]
[4 9 1] + [16 81  1] = [20 90  2]
[1 3 0] + [1 9 0] = [ 2 12  0]
[0 9 1] + [ 0 81  1] = [ 0 90  2]




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