不管哪个领域，都可以在上升期做科研，在平稳期做业务，在饱和期做教育，显然Andrew Ng是个明白人。
No matter what field you are in, you can do research in the growing period, dedicate into the industry in the steady period, and develop education in the saturation period. Obviously, Andrew Ng is a sensible person.
按理说，我们二十五六岁，买车买房，工作体面，不能再抱怨了。但是房，车，体面，这些都是大路商品。我有，其他人可能有我的一百倍，一千倍还多。但是十年的青春时光，每个人都只有一次。
我希望看到这里的人，可以记得未来很重要，但是二十多岁一定要快意，可以努力奋斗也可以玩，可以富有也可以贫穷，但不能扭曲，要快意。
当你26岁决定自己要不要成为30岁的博士时，要记得这件事。
To be honest, we have a decent job, house, car at the age of 25, should not complain more. However, cars/houses/good jobs, all of them, are general commodities, others may have them of ten times or even of hundred times than of what I have. But the ten years of youth, everyone has only one time.
So, please remember this thing, when you are 26 years old and decide whether you want to be a 30-year-old Doctor.
其实我们都已经意识到了，机器学习正在朝着超大规模参数的趋势发展。以后决定机器学习准确率的是GPU的数量和数据的使用权，而不再是研究机器学习的人。
Actually, we have been aware of the trending of massive scale parameters in machine learning. The number of CPUs and the access of data determines the final performance, but not the person who researches machine learning algorithms.
A method of computer multiplication and division is proposed which uses binary logarithms. The logarithm of a binary number may be determined approximately from the number itself by simple shifting and counting. A simple add or subtract and shift operation is all that is required to multiply or divide.
#include<stdio.h>
int main() {
float a = 12.3f;
float b = 4.56f;
int c = *(int*)&a + *(int*)&b - 0x3f800000;
printf("Approximate result：%f\n", *(float*)&c);
printf("Accurate result：%f\n", a * b);
return 0;
}
Step 1. Suspend your web server (nginx for example)sudo service nginx stop
Step 2. Renew your certificatesudo letsencrypt certonly --standalone --email {email address} -d elfsong.cn -d {web address}
Step 3. Restart your web serversudo service nginx start
Step 4. Check the statussudo service nginx status
有机械者必有机事，有机事者必有机心。机心存于胸中，则纯白不备；纯白不备，则神生不定；神生不定者，道之所不载也。吾非不知，羞而不为也。
import numpy as np
import networkx as nx
from functools import reduce
import matplotlib.pyplot as plt
connect_graph = np.array([[0, 1, 0, 0, 0],
[0, 0, 0, 1, 0],
[0, 0, 0, 1, 0],
[0, 0, 0, 0, 1],
[0, 0, 1, 0, 0]])
def ring_add(a, b):
return a or b
def ring_multi(a, b):
return a and b
def dot_product(i, j):
row = connect_graph[i]
column = connect_graph[:,j]
return reduce(ring_add, [ring_multi(a, b) for a, b in zip(row, column)])
def next_generation(connect_graph):
candidate_number = connect_graph.shape[0]
new_connect_graph = np.zeros((candidate_number, candidate_number))
for i in range(candidate_number):
for j in range(candidate_number):
new_connect_graph[i][j] = dot_product(i,j)
return new_connect_graph
new_connect_graph = next_generation(connect_graph)
def draw_graph(connect_graph):
G = nx.DiGraph()
candidate_number = connect_graph.shape[0]
node_name = list(range(candidate_number))
G.add_nodes_from(node_name)
for i in range(candidate_number):
for j in range(candidate_number):
if connect_graph[i][j]:
G.add_edge(i, j)
nx.draw(G, with_labels=True)
plt.show()
draw_graph(new_connect_graph)
他说：”好可怕呀！”,
然后转身投入了战斗。
class Test():
def heap_sort(self, nums):
i, l = 0, len(nums)
self.nums = nums
# 构造大顶堆，从非叶子节点开始倒序遍历，因此是l//2 -1 就是最后一个非叶子节点
for i in range(l//2-1, -1, -1):
self.build_heap(i, l-1)
print(nums)
# 上面的循环完成了大顶堆的构造，那么就开始把根节点跟末尾节点交换，然后重新调整大顶堆
for j in range(l-1, -1, -1):
nums[0], nums[j] = nums[j], nums[0]
self.build_heap(0, j-1)
return nums
def build_heap(self, i, l):
"""构建大顶堆"""
nums = self.nums
left, right = 2*i+1, 2*i+2 ## 左右子节点的下标
large_index = i
if left <= l and nums[i] < nums[left]:
large_index = left
if right <= l and nums[left] < nums[right]:
large_index = right
# 通过上面跟左右节点比较后，得出三个元素之间较大的下标，如果较大下表不是父节点的下标，说明交换后需要重新调整大顶堆
if large_index != i:
nums[i], nums[large_index] = nums[large_index], nums[i]
self.build_heap(large_index, l)