dimension of X and then the next 2 matrices as the width and height prediction result of box. The last dimension of the length 7 is clearly divided. Here score = nd.sigmoid (score_pred) and XY = Nd.sigmoid (xy_pred) are normalized because the score range is between 0 and 1, because the relative coordinates of the grid cell are used, So need 0 to 1 range (can see the original Figure3 bx and by calculation, where the model predicted that the XY corresponds to Figure3 in the TX and Ty). Transform_
numpy Array Base Operation
1. Array index Access
#!/usr/bin/env python
# encoding:utf-8
import numpy as np
B = Np.array ([[1,2,3],[4,5,6],[7,8,9],[10,11,12 ]],dtype=int)
C = b[0,1] #1行 Second cell element
# output: 2
d = b[:,1] #所有行 Second cell element
# output: [2 5 8 11]
2. Array of combinations (functions)
'''# combination function#创建两个测试数组# Arange creates a one-dimensional one with 9 elements# reshape method, you can create a
matrix is a standardized covariance matrix.
#获取相关系数矩阵
cm = Np.corrcoef (data[cols].values. T)
#设置字的比例
sns.set (font_scale=1.5)
#绘制相关系数图
HM = Sns.heatmap (cm,cbar=true,annot=true,square=true , fmt= ". 2f",
annot_kws={"size": 15},yticklabels=cols,xticklabels=cols)
plt.show ()
Through the correlation coefficient matrix, we can find that the correlation between Lstat and Medv is the largest (-0.74), and the second is that RM and Medv are the most
Original URL:
Http://www.cnblogs.com/denny402/p/5072746.html
This article explains some of the other common layers, including: Softmax_loss layer, Inner product layer, accuracy layer, reshape layer and dropout layer, and their parameter configuration.
1, Softmax-loss
The Softmax-loss layer and the Softmax layer are calculated roughly the same. Softmax is a classifier that calculates the probability of a class (likelihood) and is a generalization of
-*
Save_net.ckpt.index
Save_net.ckpt.meta
The following is the read code.
import tensorflow as tfimport osimport numpy as npos.environ['TF_CPP_MIN_LOG_LEVEL'] = '2'w = tf.Variable(np.arange(9).reshape((3,3)),dtype=tf.float32)b = tf.Variable(np.arange(3).reshape((1,3)),dtype=tf.float32)a = tf.Variable(np.arange(4).reshape((2,2)),dtype=tf.float32)saver =tf.t
multiplication of the numbers in shape
Ndim: Number of dimensions of the array
C= Np.array ([[[[[[
1,4,7],[2,5,8]]
,
[
[3,6,9],[6,6,6]]]
)
print (c)
output:
[[[1 4 7]
[2 5 8]]
[[3 6 9]
[6 6]]]
print (C.ndim) # The latitude of the array is: 6
print (c.dtype) # array element data type is: Int32
Print (c.shape) # Array of dimensions of each latitude: (2, 2, 3) The number of elements in the array (c.size) #:
d= c.astype (float)
print (D.dtype) #数组元素数据类型为: float64
, "border=" 0 "alt=" clip_ image002 "src=" http://s3.51cto.com/wyfs02/M00/75/04/wKiom1YwhyWgPrBgAACJBv2Tmk0515.jpg "height=" 243 "/>R1 (config) #priority-list 1 protocol IP? \ \ can define four priority levelsHighMediumNormalLowR1 (config) #priority-list 1 Protocol IP high TCP 23 \ \ We put telenet traffic at the highest priorityR1 (config) #priority-list 1 interface f0/0 medium \ \ Put the traffic received on an interface in a medium priorityR1 (config) #priority-list 1 Default LowR1 (config) #
Question link:
Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4527
Problem description James recently liked a game named ten drops of water.
The game is played in a 6*6 square, with a drop of water or no water on each grid. Water Drops are classified into four levels: 1 ~ 4. Initially, you have ten drops of water. By adding water to the Water Drop in the gri
])
Ndarray also supports multi-dimensional array slicing, which can be generated by modifying the shape attribute of a one-dimensional array or calling its reshape method:
In[68]: a = arange(0, 24).reshape(2, 3, 4)In[69]: aOut[69]: array([[[ 0, 1, 2, 3], [ 4, 5, 6, 7], [ 8, 9, 10, 11]], [[12, 13, 14, 15], [16, 17, 18, 19], [20, 21, 22, 23]]])
The index of a multi-dimensional array is actually
123456789 How is arithmetic equal to 1? -Abccsss's answer
Assume that each number can occur only once.
Reply content:Mathematica Code
More Concise
Det/@N@Range@9~Permutations~{9}~ArrayReshape~{9!,3,3}//Max
The above uses MATLAB brute force to crack (enumeration kind of case), not the output determinant of the same other situation, seemingly basic seconds out.
Max_det = 0;Init_perm = Reshape(1:9, [3, 3]);all_perms = perms(1:9); for I = 1:size(all_p
instance of the Ndarray class to invoke these methods.
>>> a= Np.random.random ((2,3)) >>> a array ([[0.65806048, 0.58216761, 0.59986935], [0.6004008, 0.41965453, 0.71487337]]) >>> a.sum () 3.5750261436902333 >>> a.min () 0.41965453489104032 > >> A.max () 0.71487337095581649
These operations treat arrays as a one-dimensional linear list. However, you can perform the corresponding operation on the specified axis by specifying the axis parameter (that is, the row of the array):
>>>
omitted slice. The syntax of multi-dimensional slicing is sequence [start1: end1, start2: end2], or the ellipsis, sequence [..., Start1: end1]. The slice object can also be implemented by the built-in function slice ().
Selection of two-dimensional arrays:The syntax of multi-dimensional array slicing is sequence [start1: end1, start2: end2 ,..., Startn: endn] we use a 3x3 two-dimensional array to demonstrate the selection problem:
>>> b = np.arange(9).resh
123456789 how is the calculation equal to 1? -The abccsss answer assumes that each number can only appear once. 123456789 how is the calculation equal to 1? -Abccsss answer
It is assumed that each number can only appear once. Reply: Mathematica code
Relatively simple
Det/@N@Range@9~Permutations~{9}~ArrayReshape~{9!,3,3}//Max
The above uses Matlab brute-force cracking (enumeration cases). No other cases with the same determinant have been output yet, which seems to be generated in seconds.
operations, such as calculating the sum of all elements of an array, are implemented as methods of The ndarray class. these methods need to be called by instances of The ndarray class during use.
>>> a= np.random.random((2,3)) >>> a array([[ 0.65806048, 0.58216761, 0.59986935], [ 0.6004008, 0.41965453, 0.71487337]]) >>> a.sum() 3.5750261436902333 >>> a.min() 0.41965453489104032 >>> a.max() 0.71487337095581649
These operations regard arrays as a one-dimensional linear list. Howev
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