update dockerfile
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4cc4e991ba
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@ -53,6 +53,8 @@
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### python_wechat.py: 玩点好玩的--自己写一个微信小助手
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### python_csv.py: Python中CSV文件的简单读写
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### python_numpy.py: 使用numpy进行矩阵操作
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===================================================================================================
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### 您可以fork该项目, 并在修改后提交Pull request
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@ -369,7 +369,7 @@
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A = 1 if X else (2 if Y else 3)
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# 也可以使用and-or语句(一条语句实现多个if-else)
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a = 6
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result = (a > 20 and "big than 20" or a > 10 and "big than 10" or a > 5 and "big than 5") # 返回"big than 20"
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result = (a > 20 and "big than 20" or a > 10 and "big than 10" or a > 5 and "big than 5") # 返回"big than 5"
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#-- Python的while语句或者for语句可以带else语句 当然也可以带continue/break/pass语句
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while a > 1:
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@ -0,0 +1,318 @@
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# _*_coding:utf-8-*_
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import numpy as np
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# 定义矩阵变量并输出变量的一些属性
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# 用np.array()生成矩阵
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arr=np.array([[1,2,3],
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[4,5,6]])
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print(arr)
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print('number of arr dimensions: ',arr.ndim)
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print('~ ~ ~ shape: ',arr.shape)
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print('~ ~ ~ size: ', arr.size)
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# 输出结果:
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[[1 2 3]
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[4 5 6]]
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number of arr dimensions: 2
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~ ~ ~ shape: (2, 3)
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~ ~ ~ size: 6
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# 定义一些特殊矩阵
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# 指定矩阵数据类型
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arr=np.array([[1,2,3],
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[4,5,6]],
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dtype=np.float64) # 我的电脑np.int是int32,还可以使用np.int32/np.int64/np.float32/np.float64
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print(arr.dtype)
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# 用np.zeros()生成全零矩阵
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arr_zeros=np.zeros( (2,3) )
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print(arr_zeros)
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# 用np.ones()生成全一矩阵
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arr_ones=np.ones( (2,3) )
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print(arr_ones)
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# 生成随机矩阵np.random.random()
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arr_random=np.random.random((2,3))
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print(arr_random)
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# 用np.arange()生成数列
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arr=np.arange(6,12)
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print(arr)
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# 用np.arange().reshape()将数列转成矩阵
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arr=np.arange(6,12).reshape( (2,3) )
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print(arr)
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# 用np.linspace(开始,结束,多少点划分线段),同样也可以用reshape()
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arr=np.linspace(1,5,3)
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print(arr)
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# 矩阵运算
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arr1=np.array([1,2,3,6])
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arr2=np.arange(4)
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# 矩阵减法,加法同理
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arr_sub=arr1-arr2
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print(arr1)
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print(arr2)
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print(arr_sub)
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# 矩阵乘法
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arr_multi=arr1**3 # 求每个元素的立方,在python中幂运算用**来表示
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print(arr_multi)
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arr_multi=arr1*arr2 # 元素逐个相乘
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print(arr_multi)
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arr_multi=np.dot(arr1, arr2.reshape((4,1))) # 维度1*4和4*1矩阵相乘
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print(arr_multi)
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arr_multi=np.dot(arr1.reshape((4,1)), arr2.reshape((1,4))) # 维度4*1和1*4矩阵相乘
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print(arr_multi)
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arr_multi=arr1.dot(arr2.reshape((4,1))) # 也可以使用矩阵名.doc(矩阵名)
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print(arr_multi)
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# 三角运算:np.sin()/np.cos()/np.tan()
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arr_sin=np.sin(arr1)
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print(arr_sin)
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# 逻辑运算
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print(arr1<3) # 查看arr1矩阵中哪些元素小于3,返回[ True True False False]
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# 矩阵求和,求矩阵最大最小值
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arr1=np.array([[1,2,3],
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[4,5,6]])
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print(arr1)
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print(np.sum(arr1)) # 矩阵求和
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print(np.sum(arr1,axis=0)) # 矩阵每列求和
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print(np.sum(arr1,axis=1).reshape(2,1)) # 矩阵每行求和
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print(np.min(arr1)) # 求矩阵最小值
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print(np.min(arr1,axis=0))
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print(np.min(arr1,axis=1))
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print(np.max(arr1)) # 求矩阵最大值
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print(np.mean(arr1)) # 输出矩阵平均值,也可以用arr1.mean()
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print(np.median(arr1)) # 输出矩阵中位数
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# 输出矩阵某些值的位置
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arr1=np.arange(2,14).reshape((3,4))
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print(arr1)
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print(np.argmin(arr1)) # 输出矩阵最小值的位置,0
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print(np.argmax(arr1)) # 输出矩阵最大值的位置,11
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print(np.cumsum(arr1)) # 输出前一个数的和,前两个数的和,等等
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print(np.diff(arr1)) # 输出相邻两个数的差值
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arr_zeros=np.zeros((3,4))
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print(np.nonzero(arr_zeros)) #输出矩阵非零元素位置,返回多个行向量,第i个行向量表示第i个维度
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print(np.nonzero(arr1))
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print(np.sort(arr1)) # 矩阵逐行排序
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print(np.transpose(arr1)) # 矩阵转置,也可以用arr1.T
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print(np.clip(arr1,5,9)) #将矩阵中小于5的数置5,大于9的数置9
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# numpy索引
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arr1=np.array([1,2,3,6])
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arr2=np.arange(2,8).reshape(2,3)
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print(arr1)
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print(arr1[0]) # 索引从0开始计数
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print(arr2)
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print(arr2[0][2]) # arr[行][列],也可以用arr[行,列]
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print(arr2[0,:]) # 用:来代表所有元素的意思
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print(arr2[0,0:3]) # 表示输出第0行,从第0列到第2列所有元素
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# 注意python索引一般是左闭右开
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# 通过for循环每次输出矩阵的一行
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for row in arr2:
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print(row)
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# 如果要每次输出矩阵的一列,就先将矩阵转置
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arr2_T=arr2.T
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print(arr2_T)
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for row in arr2_T:
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print(row)
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# 将矩阵压成一行逐个输出元素
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arr2_flat=arr2.flatten()
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print(arr2_flat)
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for i in arr2.flat: # 也可以用arr2.flatten()
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print(i)
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# 矩阵合并与分割
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# 矩阵合并
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arr1=np.array([1,2,3,6])
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arr2=np.arange(4)
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arr3=np.arange(2,16+1,2).reshape(2,4)
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print(arr1)
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print(arr2)
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print(arr3)
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arr_hor=np.hstack((arr1,arr2)) # 水平合并,horizontal
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arr_ver=np.vstack((arr1,arr3)) # 垂直合并,vertical
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print(arr_hor)
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print(arr_ver)
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# 矩阵分割
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print('arr3: ',arr3)
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print(np.split(arr3,4,axis=1)) # 将矩阵按列均分成4块
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print(np.split(arr3,2,axis=0)) # 将矩阵按行均分成2块
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print(np.hsplit(arr3,4)) # 将矩阵按列均分成4块
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print(np.vsplit(arr3,2)) # 将矩阵按行均分成2块
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print(np.array_split(arr3,3,axis=1)) # 将矩阵进行不均等划分
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# numpy复制:浅复制,深复制
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# 浅复制
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arr1=np.array([3,1,2,3])
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print(arr1)
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a1=arr1
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b1=a1
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# 通过上述赋值运算,arr1,a1,b1都指向了同一个地址(浅复制)
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print(a1 is arr1)
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print(b1 is arr1)
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print(id(a1))
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print(id(b1))
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print(id(arr1))
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# 会发现通过b1[0]改变内容,arr1,a1,b1的内容都改变了
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b1[0]=6
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print(b1)
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print(a1)
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print(arr1)
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# 深复制
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arr2=np.array([3,1,2,3])
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print('\n')
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print(arr2)
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b2=arr2.copy() # 深复制,此时b2拥有不同于arr2的空间
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a2=b2.copy()
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# 通过上述赋值运算,arr1,a1,b1都指向了不同的地址(深复制)
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print(id(arr2))
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print(id(a2))
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print(id(b2))
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# 此时改变b2,a2的值,互不影响
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b2[0]=1
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a2[0]=2
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print(b2)
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print(a2)
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print(arr2)
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# 线性代数模块(linalg)
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# 求范数
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a=np.array([5,12])
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print(a)
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b=np.linalg.norm(a) # norm表示范数,默认求2范数,ord=1求1范数,ord=np.inf求无穷范数
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print(b)
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# 求矩阵的迹、行列式、秩、特征值、特征向量
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b = np.array([
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[1, 2, 3],
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[4, 5, 6],
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[7, 8, 9]
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])
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print(np.trace(b)) # 15,求矩阵的迹(主对角线上各个元素的总和)
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c=np.linalg.det(b)
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print(c) # 输出一个很小的值6.66133814775e-16,求矩阵的行列式值
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# 如果希望输出为0,使用round(c, 2),四舍五入保留小数点后两位
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# 不过对精度要求高可以使用decimal模块
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c=np.linalg.matrix_rank(b)
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print(c) # 2,求矩阵的秩
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u,v=np.linalg.eig(b) # u为特征值
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print(u)
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print(v)
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# 矩阵分解
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# Cholesky分解并重建
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d = np.array([
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[2, 1],
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[1, 2]
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])
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l = np.linalg.cholesky(d)
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print(l) # 得到下三角矩阵
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e=np.dot(l, l.T)
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print(e) # 重建得到矩阵d
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# 对不正定矩阵,进行SVD分解并重建
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U, s, V = np.linalg.svd(d)
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S = np.array([
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[s[0], 0],
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[0, s[1]]
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])
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print(np.dot(U, np.dot(S, V))) # 重建得到矩阵d
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# 矩阵乘法
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# https://docs.scipy.org/doc/numpy/reference/generated/numpy.dot.html#numpy.dot
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print(np.dot(3, 4)) # 12,0-D矩阵相乘(也就是标量相乘)
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print(np.dot([2j, 3j], [2j, 3j])) # (-13+0j),1-D矩阵相乘(实际上是向量做点积)
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a=[[1, 0], [0, 1]]
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b=[[4, 1, 0], [2, 2, 0]]
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print(np.dot(a, b))
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'''
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array([[4, 1],
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[2, 2]])
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2-D矩阵相乘
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这里是2*2矩阵和2*3矩阵相乘,结果为2*3矩阵
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'''
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a=[[1, 0], [1, 2]]
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b=[2,2]
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c=np.dot(a,b)
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print(c)
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'''
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[2 6]
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注意这里b是向量
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numpy处理时并不是按照矩阵乘法规则计算
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而是向量点积
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也就是np.dot([1, 0],[1, 2])和np.dot([1, 2],[2,2])
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'''
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# 再做个实验来区别向量乘法和矩阵乘法
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b=np.array([
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[1, 2, 3],
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[4, 5, 6],
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[7, 8, 9]
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])
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# 这里插播一下,np.array([1,0,1])是3维向量,而不是1*3的矩阵
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c1=np.array([[1,0,2]])
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print(c1.shape) # (1, 3),这是一个1*3的矩阵
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c2=np.array([1,0,2])
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print(c2.shape) # (3,),这是一个3维向量
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# print(np.dot(b,c1)) # 报错,不符合矩阵乘法规则
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print(np.dot(b,c2)) # [ 7 16 25],点积运算
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print(np.dot(c1,b)) # [[15 18 21]],矩阵乘法运算规则
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print(np.dot(c2,b)) # [15 18 21],点积运算
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# 还要补充一下,如果是用python自带的*运算符计算则是广播机制
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print(b*c1) # print(b*c2)结果一样
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'''
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[[ 1 0 6]
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[ 4 0 12]
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[ 7 0 18]]
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'''
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print(b+c1) # print(b*c2)结果一样
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'''
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[[ 2 2 5]
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[ 5 5 8]
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[ 8 8 11]]
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'''
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Loading…
Reference in New Issue