numpy array using data processing

meshgrid function

understanding:

Two-dimensional coordinate system, X-axis can take three values 1,2,3, Y-axis can take three values 7,8, ask how many points the coordinates can be obtained?
Apparent is 6:
(1,7) ( 2,7) (3,7)
(1,8) (2,8) (3,8)

>>> import numpy as np#导入numpy
>>> a=np.array([1,2,3])#创建一维数组
>>> b=np.array([7,8])
>>> res=np.meshgrid(a,b)#获取所有点的横坐标和纵坐标
>>> res#返回list,有两个元素,第一个元素是X轴的取值,第二个元素是Y轴的取值
[array([[1, 2, 3],
       [1, 2, 3]]), array([[7, 7, 7],
       [8, 8, 8]])]

Generating a rendered image coordinates:

>>> import numpy as np
>>> points=np.arange(-5,5,0.01)#从-5到5步长为0.01
>>> xs,ys=np.meshgrid(points,points)
>>> ys
array([[-5.  , -5.  , -5.  , ..., -5.  , -5.  , -5.  ],
       [-4.99, -4.99, -4.99, ..., -4.99, -4.99, -4.99],
       [-4.98, -4.98, -4.98, ..., -4.98, -4.98, -4.98],
       ...,
       [ 4.97,  4.97,  4.97, ...,  4.97,  4.97,  4.97],
       [ 4.98,  4.98,  4.98, ...,  4.98,  4.98,  4.98],
       [ 4.99,  4.99,  4.99, ...,  4.99,  4.99,  4.99]])
>>> z=np.sqrt(xs**2+ys**2)#函数的求值运算
>>> z
array([[7.07106781, 7.06400028, 7.05693985, ..., 7.04988652, 7.05693985,
        7.06400028],
       [7.06400028, 7.05692568, 7.04985815, ..., 7.04279774, 7.04985815,
        7.05692568],
       [7.05693985, 7.04985815, 7.04278354, ..., 7.03571603, 7.04278354,
        7.04985815],
       ...,
       [7.04988652, 7.04279774, 7.03571603, ..., 7.0286414 , 7.03571603,
        7.04279774],
       [7.05693985, 7.04985815, 7.04278354, ..., 7.03571603, 7.04278354,
        7.04985815],
       [7.06400028, 7.05692568, 7.04985815, ..., 7.04279774, 7.04985815,
        7.05692568]])
>>> import matplotlib.pyplot as plt
>>> plt.imshow(z, cmap=plt.cm.gray); plt.colorbar()
<matplotlib.image.AxesImage object at 0x00000000087D6EB8>
<matplotlib.colorbar.Colorbar object at 0x000000000D8F4FD0>
>>> plt.title("Image plot of $\sqrt{x^2 + y^2}$ for a grid of values")
Text(0.5, 1.0, 'Image plot of $\\sqrt{x^2 + y^2}$ for a grid of values')
>>>plt.show()

The conditional logic array operation expressed as

According to the selection of the array of values cond, cond if the value is True, then select the corresponding value x, otherwise selecting a corresponding value of y, such cond [0] = True, then the result [0] It should be placed at yarr [0] value

>>> xarr=np.array([1.1,1.2,1.3,1.4,1.5])
>>> yarr=np.array([2.1,2.2,2.3,2.4,2.5])
>>> cond=np.array([True,False,True,True,False])
>>> result=[(x if c else y) for x,y,c in zip(xarr,yarr,cond)]
>>> result
[1.1, 2.2, 1.3, 1.4, 2.5]

where method

Use vector

where the above methods may be substituted wording

>>> result=np.where(cond,xarr,yarr)
>>> result
array([1.1, 2.2, 1.3, 1.4, 2.5])

Scalar

np.where second and third parameters need not be an array, which can be a scalar value. In the data analysis, where typically for generating a new array in accordance with another array. Suppose there is a matrix composed of random data, you wish to replace all positive values 2, -2 replace all negative values. If the use of np.where, will be very simple:

>>> arr=np.random.randn(4,4)
>>> arr
array([[-0.02461196, -0.11867552,  0.49004256, -0.25236766],
       [ 1.66484493, -0.41180743, -0.34837293, -2.61422661],
       [ 0.51476568, -0.41451909,  1.28553372,  0.13866799],
       [-0.94795594, -1.49302099,  1.18480636,  0.20310613]])
>>> arr>0
array([[False, False,  True, False],
       [ True, False, False, False],
       [ True, False,  True,  True],
       [False, False,  True,  True]])
>>> np.where(arr>0,2,-2)
array([[-2, -2,  2, -2],
       [ 2, -2, -2, -2],
       [ 2, -2,  2,  2],
       [-2, -2,  2,  2]])

Scalar Vector bound

If the corresponding location is True, the new array corresponds to position 2, otherwise it is the original value

>>> np.where(arr>0,2,arr)
array([[-0.02461196, -0.11867552,  2.        , -0.25236766],
       [ 2.        , -0.41180743, -0.34837293, -2.61422661],
       [ 2.        , -0.41451909,  2.        ,  2.        ],
       [-0.94795594, -1.49302099,  2.        ,  2.        ]])

Statistical Methods

mean function

Averaging all the elements

>>> import numpy as np
>>> arr=np.array([[1,2,3],[4,5,6],[7,8,9]])
>>> arr.mean()
5.0

>>> np.mean(arr)
5.0

sum function

Sum of all the elements

>>> arr.sum()
45

Alternatively axis evaluation

>>> arr.mean(axis=1)#列求平均值
array([2., 5., 8.])
>>> arr.mean(axis=0)#行求平均值
array([4., 5., 6.])
>>> arr.sum(axis=0)
array([12, 15, 18])#行求和
>>> arr.sum(axis=1)
array([ 6, 15, 24])#列求和

cumsum function

1. The one-dimensional input a (List may be, may be Array, assuming a = [1, 2, 3, 4, 5, 6, 7], and is added to the current column prior to the current row, as follows:

>>>import numpy as np
>>> a=[1,2,3,4,5,6,7]
>>> np.cumsum(a)
array([  1,   3,   6,  10,  15,  21,  28,  36,  45,  55,  75, 105])

2. For the two-dimensional input a, axis = 0 (line 1 does not move, the accumulated row to another row 1); axis = 1 (into the innermost layer, is converted into the column processing does not move the first column, the first. an additive to other columns), as follows:

>>>import numpy as np
>>> c=[[1,2,3],[4,5,6],[7,8,9]]
>>> np.cumsum(c,axis=0)
array([[ 1,  2,  3],
       [ 5,  7,  9],
       [12, 15, 18]])
>>> np.cumsum(c,axis=1)
array([[ 1,  3,  6],
       [ 4,  9, 15],
       [ 7, 15, 24]])

cumprod function

Similarly multiplicative method

>>> arr.cumprod(axis=1)
array([[  1,   2,   6],
       [  4,  20, 120],
       [  7,  56, 504]], dtype=int32)