2、切片
nd
array([[7, 9, 2, 3], [0, 2, 7, 3], [1, 9, 0, 1], [4, 1, 2, 8]])
nd[0:100] # 左闭右开的区间,右边可以无限大
array([[7, 9, 2, 3], [0, 2, 7, 3], [1, 9, 0, 1], [4, 1, 2, 8]])
lp[0:100]
[[1, 2, 3], [4, 5, 6], [7, 8]]
nd[:2]
array([[7, 9, 2, 3], [0, 2, 7, 3]])
nd[1:] • 1
array([[0, 2, 7, 3], [1, 9, 0, 1], [4, 1, 2, 8]])
nd[3:0:-1] # 如果步长为负数，代表从后往前数，要求区间也是倒着的 • 1 • 2
array([[4, 1, 2, 8], [1, 9, 0, 1], [0, 2, 7, 3]])
nd • 1
array([[7, 9, 2, 3], [0, 2, 7, 3], [1, 9, 0, 1], [4, 1, 2, 8]])
nd[:,0::2] • 1
array([[7, 2], [0, 7], [1, 0], [4, 2]])
nd[1:3,0:2] # 即切行又切列 • 1
array([[0, 2], [1, 9]])
把girl倒过来
girl
array([[[225, 231, 231], [229, 235, 235], [222, 228, 228], ..., [206, 213, 162], [211, 213, 166], [217, 220, 173]], [[224, 230, 230], [229, 235, 235], [223, 229, 229], ..., [206, 213, 162], [211, 213, 166], [217, 220, 173]], [[224, 230, 230], [229, 235, 235], [223, 229, 229], ..., [206, 213, 162], [211, 213, 166], [219, 221, 174]], ..., [[175, 187, 213], [180, 192, 218], [175, 187, 213], ..., [155, 162, 180], [153, 160, 178], [156, 163, 181]], [[175, 187, 213], [180, 192, 218], [174, 186, 212], ..., [155, 162, 180], [153, 160, 178], [155, 162, 180]], [[177, 189, 215], [181, 193, 219], [174, 186, 212], ..., [155, 162, 180], [153, 160, 178], [156, 163, 181]]], dtype=uint8)
plt.imshow(girl[::-2,::-2]) plt.show()
拼图小游戏：把女孩放在老虎背上
t = tigger.copy() # • 1
plt.imshow(tigger) plt.show()
girl2 = plt.imread("./source/girl2.jpg") plt.imshow(girl2) plt.show()
# 给老虎挖坑 tigger[150:450,300:600] = girl2
plt.imshow(tigger) plt.show()
3、变形
reshape()
resize()
tigger.shape • 1
(786, 1200, 3) • 1
nd = np.random.randint(0,10,size=12) nd
array([4, 0, 1, 1, 8, 7, 7, 5, 3, 0, 7, 3]) • 1
nd.shape • 1
(12,) • 1
nd.reshape((3,2,2,1)) # 参数为一个元组，代表的就是要把nd变成一个什么形状
array([[[[4], [0]], [[1], [1]]], [[[8], [7]], [[7], [5]]], [[[3], [0]], [[7], [3]]]])
nd • 1
array([4, 0, 1, 1, 8, 7, 7, 5, 3, 0, 7, 3])
nd.reshape((3,2))#cannot reshape array of size 12 into shape (3,8) # 变形的时候size要保持一致
--------------------------------------------------------------------------- ValueError Traceback (most recent call last) <ipython-input-94-dda3397392b8> in <module>() ----> 1 nd.reshape((3,2))#cannot reshape array of size 12 into shape (3,8) ValueError: cannot reshape array of size 12 into shape (3,2)
nd.resize((2,6)) • 1
nd • 1
array([[4, 0, 1, 1, 8, 7], [7, 5, 3, 0, 7, 3]])
【注意】
1）形变之前和形变之后的数组的size要保持一致，否则无法形变 2）reshape()函数是把原数组拷贝副本以后对副本进行形变，并且把形变的结果返回 3）resize()函数在原来的数组上进行形变，不需要返回结果
4、级联
级联：就是按照指定的维度把两个数组连在一起
nd1 = np.random.randint(0,10,size=(4,4)) nd2 = np.random.randint(20,40,size=(3,4)) • 1 • 2
print(nd1) print(nd2)
[[2 5 6 1] [4 8 0 5] [9 4 7 8] [4 3 0 8]] [[38 22 25 38] [22 38 30 21] [23 34 28 26]]
# 将两个数组进行级联 np.concatenate([nd1,nd2],axis=0) # 参数1，是一个列表（或者元组），列表中是参与级联的那些数组 # 参数axis默认为0代表在行上（第0个维度）进行级联，1代表在列上（第1个维度）进行级联
array([[ 2, 5, 6, 1], [ 4, 8, 0, 5], [ 9, 4, 7, 8], [ 4, 3, 0, 8], [38, 22, 25, 38], [22, 38, 30, 21], [23, 34, 28, 26]])
np.concatenate([nd1,nd2],axis=1) # 列级联需要行数一致
--------------------------------------------------------------------------- ValueError Traceback (most recent call last) <ipython-input-102-0a76346b819d> in <module>() ----> 1 np.concatenate([nd1,nd2],axis=1) ValueError: all the input array dimensions except for the concatenation axis must match exactly
nd3 = np.random.randint(0,10,size=(4,3)) nd3
array([[1, 3, 7], [9, 5, 3], [9, 0, 2], [0, 7, 4]])
nd1
array([[2, 5, 6, 1], [4, 8, 0, 5], [9, 4, 7, 8], [4, 3, 0, 8]])
np.concatenate([nd1,nd3]) # 列数不一致，不能进行行级联
--------------------------------------------------------------------------- ValueError Traceback (most recent call last) <ipython-input-106-871caaeeb895> in <module>() ----> 1 np.concatenate([nd1,nd3]) ValueError: all the input array dimensions except for the concatenation axis must match exactly
np.concatenate([nd1,nd3],axis=1)
array([[2, 5, 6, 1, 1, 3, 7], [4, 8, 0, 5, 9, 5, 3], [9, 4, 7, 8, 9, 0, 2], [4, 3, 0, 8, 0, 7, 4]])
推广
1)形状一致才可以级联
nd4 = np.random.randint(0,10,size=(1,2,3)) nd5 = np.random.randint(0,10,size=(1,4,3)) print(nd4) print(nd5)
[[[2 9 8] [9 5 6]]] [[[9 9 6] [8 3 4] [8 7 7] [0 6 6]]]
np.concatenate([nd4,nd5],axis=1)
array([[[2, 9, 8], [9, 5, 6], [9, 9, 6], [8, 3, 4], [8, 7, 7], [0, 6, 6]]])
nd6 = np.random.randint(0,10,size=4) nd6
array([3, 5, 3, 6]) • 1
2)维度不一致不能级联
np.concatenate([nd1,nd6])
--------------------------------------------------------------------------- ValueError Traceback (most recent call last) <ipython-input-124-6dd6213f71bc> in <module>() ----> 1 np.concatenate([nd1,nd6]) ValueError: all the input arrays must have same number of dimensions
级联需要注意的问题：
1）维度必须一样 2）形状必须相符（axis等于哪个维度，我们去掉这个维度以后，剩余的形状必须一致） 3）级联方向可以有axis来指定，默认是0
针对于二维数组还有hstack和vstack
nd = np.random.randint(0,10,size=(10,1)) nd
array([[1], [7], [6], [9], [0], [4], [6], [2], [0], [8]])
np.hstack(nd)
array([1, 7, 6, 9, 0, 4, 6, 2, 0, 8])
nd1 = np.random.randint(0,10,size=(10,2)) nd1
array([[4, 4], [3, 1], [3, 3], [9, 6], [5, 1], [4, 7], [3, 3], [4, 3], [7, 9], [6, 5]])
np.hstack(nd1)
array([4, 4, 3, 1, 3, 3, 9, 6, 5, 1, 4, 7, 3, 3, 4, 3, 7, 9, 6, 5])
np.vstack(nd1)
array([[4, 4], [3, 1], [3, 3], [9, 6], [5, 1], [4, 7], [3, 3], [4, 3], [7, 9], [6, 5]])
nd2 = np.random.randint(0,10,size=10) nd2
array([1, 7, 4, 3, 9, 0, 3, 3, 2, 5])
np.vstack(nd2)
array([[1], [7], [4], [3], [9], [0], [3], [3], [2], [5]])
np.hstack(nd2)
array([1, 7, 4, 3, 9, 0, 3, 3, 2, 5])
hstack()把列数组改成行数组，把二维数组改成一维
vstack()把行数组改成列数组，把一维数组改成二维（把一维数组中的每一个元素作为一行）
5、切分
切分就是把一个数组切成多个
vsplit()
hsplit()
split()
nd = np.random.randint(0,100,size=(5,6)) nd
array([[17, 47, 83, 33, 69, 24], [60, 4, 34, 29, 75, 60], [33, 55, 67, 1, 76, 82], [31, 92, 1, 14, 83, 95], [59, 88, 81, 49, 70, 11]])
# 水平方向上切分 np.hsplit(nd,[1,4,5,8,9]) # 参数1，代表被切分的数组，参数2，是一个列表，代表了切分点的位置
[array([[17], [60], [33], [31], [59]]), array([[47, 83, 33], [ 4, 34, 29], [55, 67, 1], [92, 1, 14], [88, 81, 49]]), array([[69], [75], [76], [83], [70]]), array([[24], [60], [82], [95], [11]]), array([], shape=(5, 0), dtype=int32), array([], shape=(5, 0), dtype=int32)]
# 竖直方向上切分 np.vsplit(nd,[1,3,5])
[array([[17, 47, 83, 33, 69, 24]]), array([[60, 4, 34, 29, 75, 60], [33, 55, 67, 1, 76, 82]]), array([[31, 92, 1, 14, 83, 95], [59, 88, 81, 49, 70, 11]]), array([], shape=(0, 6), dtype=int32)]
split()函数
nd
array([[17, 47, 83, 33, 69, 24], [60, 4, 34, 29, 75, 60], [33, 55, 67, 1, 76, 82], [31, 92, 1, 14, 83, 95], [59, 88, 81, 49, 70, 11]])
np.split(nd,[1,2],axis=0) # axis默认为0代表在第0个维度上进行切分，1代表切的是第1个维度
[array([[17, 47, 83, 33, 69, 24]]), array([[60, 4, 34, 29, 75, 60]]), array([[33, 55, 67, 1, 76, 82], [31, 92, 1, 14, 83, 95], [59, 88, 81, 49, 70, 11]])]
推广
nd1 = np.random.randint(0,10,size=(3,4,5)) nd1
array([[[5, 7, 8, 7, 9], [3, 6, 1, 9, 0], [6, 0, 2, 6, 9], [4, 5, 5, 3, 9]], [[6, 7, 6, 2, 3], [3, 0, 0, 5, 3], [9, 9, 0, 6, 2], [5, 4, 5, 4, 4]], [[8, 7, 4, 8, 9], [2, 2, 1, 7, 3], [2, 2, 9, 4, 7], [7, 3, 9, 4, 1]]])
np.split(nd1,[2],axis=2)
[array([[[5, 7], [3, 6], [6, 0], [4, 5]], [[6, 7], [3, 0], [9, 9], [5, 4]], [[8, 7], [2, 2], [2, 2], [7, 3]]]), array([[[8, 7, 9], [1, 9, 0], [2, 6, 9], [5, 3, 9]], [[6, 2, 3], [0, 5, 3], [0, 6, 2], [5, 4, 4]], [[4, 8, 9], [1, 7, 3], [9, 4, 7], [9, 4, 1]]])]
6、副本
nd = np.random.randint(0,100,size=6) nd
array([34, 69, 14, 2, 48, 74])
nd1 = nd # 数组之间的赋值只是对地址一次拷贝，数组对象本身并没有被拷贝 • 1 • 2
nd1 • 1
array([34, 69, 14, 2, 48, 74])
nd1[0] = 100 • 1
nd1 • 1
array([100, 69, 14, 2, 48, 74]) • 1
nd
array([100, 69, 14, 2, 48, 74]) • 1
nd2 = nd.copy() # copy函数是把nd引用的那个数组也拷贝一份副本，并且把这个副本的地址存入了nd2 • 1 • 2
nd2[0] = 200000 • 1
nd
array([100, 69, 14, 2, 48, 74]) • 1
nd1 • 1
array([100, 69, 14, 2, 48, 74])
nd2 • 1
array([200000, 69, 14, 2, 48, 74]) • 1
讨论：由列表创建数组的过程有木有副本的创建
l = [1,2,3] l • 1 • 2
[1, 2, 3] • 1
nd = np.array(l) nd
array([1, 2, 3]) • 1
nd[0] = 1000 • 1
l • 1
[1, 2, 3]
说明：由列表创建数组的过程就是把列表拷贝出一个副本，然后把这个副本中的元素类型做一个统一化，然后放入数组对象中
四、ndarray的聚合操作
聚合操作指的就是对数组内部的数据进行某些特性的求解
1、求和
nd = np.random.randint(0,10,size=(3,4)) nd • 1 • 2
array([[5, 9, 6, 8], [3, 7, 1, 9], [5, 7, 6, 3]])
nd.sum() # 完全聚合 • 1
69 • 1
nd.sum(axis=0) # 对行进行聚合（即对第0个维度进行聚合）
array([13, 23, 13, 20]) • 1
nd.sum(axis=1) # 对列进行聚合（即对第1个维度进行聚合） • 1
array([28, 20, 21])
推广
nd = np.random.randint(0,10,size=(2,3,4)) nd
array([[[1, 0, 0, 3], [9, 6, 1, 8], [4, 9, 3, 9]], [[8, 0, 4, 3], [3, 0, 1, 8], [8, 0, 7, 4]]])
nd.sum() • 1
99 • 1
nd.sum(axis=0)
array([[ 9, 0, 4, 6], [12, 6, 2, 16], [12, 9, 10, 13]])
nd.sum(axis=2) • 1
array([[ 4, 24, 25], [15, 12, 19]]) • 1 • 2
聚合操作的规律：通过axis来改变聚合轴，axis=x的时候，第x的维度就会消失，把这个维度上对应的元素进行聚合
练习：给定一个4维矩阵，如何得到最后两维的和？
nd1 = np.random.randint(0,10,size=(2,3,4,5)) nd1
array([[[[3, 2, 9, 4, 0], [1, 0, 2, 3, 7], [4, 8, 6, 6, 5], [2, 3, 4, 1, 5]], [[3, 2, 0, 1, 3], [7, 3, 3, 4, 1], [0, 4, 0, 6, 9], [3, 8, 6, 0, 5]], [[5, 1, 3, 5, 0], [1, 4, 1, 8, 0], [9, 1, 9, 6, 5], [6, 1, 8, 5, 1]]], [[[7, 5, 3, 4, 5], [7, 8, 6, 7, 2], [9, 9, 5, 3, 4], [9, 2, 9, 7, 2]], [[3, 2, 9, 7, 7], [0, 8, 1, 3, 0], [1, 5, 5, 6, 5], [4, 8, 7, 2, 9]], [[1, 3, 5, 0, 6], [6, 0, 3, 5, 6], [2, 4, 6, 9, 0], [8, 7, 4, 0, 6]]]])
写法一
nd1.sum(axis=2).sum(axis=2) • 1
array([[ 75, 68, 79], [113, 92, 81]]) • 1 • 2
写法二
nd1.sum(axis=-1).sum(axis=-1) • 1
array([[ 75, 68, 79], [113, 92, 81]]) • 1 • 2
写法三
nd1.sum(axis=(-1,-2)) • 1
array([[ 75, 68, 79], [113, 92, 81]]) • 1 • 2
2、最值
nd
array([[[1, 0, 0, 3], [9, 6, 1, 8], [4, 9, 3, 9]], [[8, 0, 4, 3], [3, 0, 1, 8], [8, 0, 7, 4]]])
nd.sum(axis=-1) • 1
array([[ 4, 24, 25], [15, 12, 19]]) • 1 • 2
nd.max() • 1
9 • 1
nd.max(axis=-1)
array([[3, 9, 9], [8, 8, 8]]) • 1 • 2
nd.max(axis=1)
array([[9, 9, 3, 9], [8, 0, 7, 8]]) • 1 • 2
nd.min(axis=0)
array([[1, 0, 0, 3], [3, 0, 1, 8], [4, 0, 3, 4]])
