numpy 进阶

1 内存模块



In [11]:

    
import numpy as np
x = np.array([1,2,3],dtype=np.int32)
x.data









    Out[11]:





<read-write buffer for 0x109974990, size 12, offset 0 at 0x10997fab0>



In [12]:

    
str(x.data)









    Out[12]:





'\x01\x00\x00\x00\x02\x00\x00\x00\x03\x00\x00\x00'

数值类型为np.int32,表明一个数字由四个字节表示，一个字节有8bit,而0x的十六进制中，一个数字代表了4个bit。

查看数据在内存中的位置



In [13]:

    
x.__array_interface__['data'][0]









    Out[13]:





4358785056

查看数据详细信息



In [16]:

    
x.__array_interface__









    Out[16]:





{'data': (4358785056, False),
 'descr': [('', '<i4')],
 'shape': (3,),
 'strides': None,
 'typestr': '<i4',
 'version': 3}

2 Data types

type scalar type of the data,int8, int16
itemsize
data size of data block
byteorder
byte order big-endian > litte-endian < not applicabel
fields
sub-dtypes
shape
shape of the array



In [35]:

    
np.dtype(int).type









    Out[35]:





numpy.int64



In [36]:

    
np.dtype(int).byteorder









    Out[36]:





'='



In [37]:

    
np.dtype(int).shape









    Out[37]:





()

3 Strides

多少字节从一个元素跳到下一个元素



In [46]:

    
x = np.array([[1,2,3],
             [4,5,6],
             [7,8,9]],dtype=np.int8)
str(x.data)
x.strides









    Out[46]:





(3, 1)

表明：跳转到下一行需要3个字节，跳转到下一列需要1个字节。



In [48]:

    
byte_offset = 3*1 + 1*2
x.flat[byte_offset]









    Out[48]:





6



In [49]:

    
x[1,2]









    Out[49]:





6

4 改变数据类型



In [41]:

    
x = np.array([1,2,3,4],dtype=np.uint8)
str(x.data)









    Out[41]:





'\x01\x02\x03\x04'



In [42]:

    
x.dtype='<i2'
x









    Out[42]:





array([ 513, 1027], dtype=int16)



In [44]:

    
0x0201,0x0403









    Out[44]:





(513, 1027)

little-endian: least significant byte is on the left in memory

5 CPU 缓冲机制



In [57]:

    
x = np.zeros((10000,))
y = np.zeros((10000*67,))[::67]
%timeit x.sum()









    



The slowest run took 5.01 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 6.19 µs per loop



In [58]:

    
%timeit y.sum()









    



The slowest run took 116.58 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 25.2 µs per loop



In [59]:

    
x.strides,y.strides









    Out[59]:





((8,), (536,))



In [ ]: