# -*- coding: utf-8 -*-
"""算法逻辑训练""""
from collections import Counter, defaultdict
# one : 以下是用户id对应的技能爱好，比如(0, "Hadoop")表示0号用户爱好是Hadoop
# 根据以下数据集，给一个用户id，算出与该用户技能爱好最多的其他用户（可能有多个）
interests = [
(0, "Hadoop"), (0, "Big Data"), (0, "HBase"), (0, "Java"),
(0, "Spark"), (0, "Storm"), (0, "Cassandra"),
(1, "NoSQL"), (1, "MongoDB"), (1, "Cassandra"), (1, "HBase"),
(1, "Postgres"), (2, "Python"), (2, "scikit-learn"), (2, "scipy"),
(2, "numpy"), (2, "statsmodels"), (2, "pandas"), (3, "R"), (3, "Python"),
(3, "statistics"), (3, "regression"), (3, "probability"),
(4, "machine learning"), (4, "regression"), (4, "decision trees"),
(4, "libsvm"), (5, "Python"), (5, "R"), (5, "Java"), (5, "C++"),
(5, "Haskell"), (5, "programming languages"), (6, "statistics"),
(6, "probability"), (6, "mathematics"), (6, "theory"),
(7, "machine learning"), (7, "scikit-learn"), (7, "Mahout"),
(7, "neural networks"), (8, "neural networks"), (8, "deep learning"),
(8, "Big Data"), (8, "artificial intelligence"), (9, "Hadoop"),
(9, "Java"), (9, "MapReduce"), (9, "Big Data")
]
def get_set():
"""每个用户对应的所有技能爱好"""
a = set()
for i, j in interests:
a.add(i)
r_list = list(a)
r_dic = {num: set() for num in r_list}
for a in r_dic:
for i, j in interests:
if i == a:
r_dic[a].add(j)
return r_dic
def get_person(id):
if id not in get_set():
print('用户无效')
return None
r_dic = get_set()
obj = r_dic[id]
same_num = [(0, 0)]
for i in r_dic:
s_list = [i[0] for i in same_num]
if i == id or i in s_list:
continue
num = len(r_dic[i] & obj)
if num > same_num[0][1]:
same_num=[]
same_num.append((i, num))
continue
if num == same_num[0][1]:
same_num.append((i, num))
return same_num
print(get_person(0))
# two : 以下是一个元素是序号和值组成的元祖的列表，相同序号的元祖的值相加，得到一个没有重复序号的新列表
# a = [(105, 1), (101, 2), (104, 3), (105, 4)] --> a = [(105, 5), (101, 2), (104, 3)] 顺序可以改变
# 方法一：
a = [(105, 1), (101, 2), (104, 3), (105, 4)]
r_set = {i[0] for i in a}
b = [[i, 0] for i in list(r_set)]
for i in b:
for j in a:
if i[0] == j[0]:
i[1] += j[1]
c = [tuple(i) for i in b]
print(c)
# 方法二：
d = defaultdict(list) # 设置字典d的value默认是列表
for k, v in a:
d[k].append(v)
r_list = [(i, sum(d[i])) for i in d]
print(r_list)
# three : 实现字符串形式的二进制加法运算：
# '1100' + '10101' = '100001'
# 方法一：
def my_bites(a, b):
"""将两个字符串二进制数变成列表，并将元素少的用0补充与元素多的等长，再相加"""
middle = 0 # 进位数（只可能是1或者0）
result = ''
a_list = [int(i) for i in a][::-1]
b_list = [int(i) for i in b][::-1]
max_s, min_s = (a_list, b_list) if len(a) > len(b) else (b_list, a_list)
for _ in range(len(max_s)-len(min_s)):
min_s.append(0)
for i in range(len(max_s)):
if max_s[i]+min_s[i]+middle >= 2:
result += str(max_s[i]+min_s[i]+middle-2)
middle = 1
else:
result += str(max_s[i] + min_s[i] + middle)
middle = 0
if middle:
result += '1'
return result[::-1]
print(my_bites('1100','10101'))
# 方法二：
def add_bin(a, b):
a_b = int(a, 2)
b_b = int(b, 2)
return bin(a_b+b_b)[2:]
print(add_bin('1100', '10101'))
# four: 实现机械学习KNN算法
def distance(a, b):
"""计算两个向量距离"""
sum_d = 0
for i in range(len(a)):
sum_d += (a[i] - b[i]) ** 2
return sum_d**0.5
def my_knn(test_data, cla, data, k):
d_list = [(cla[i], distance(data, test_data[i])) for i in range(len(test_data))]
d_list.sort(key=lambda x: x[1])
r_list = [i[0] for i in d_list][:k]
r_c = Counter(r_list).most_common()[0][0]
return r_c
# 测试KNN算法
test_data = [(1, 150), (4, 123), (130, 2), (148, 8)]
cla = ['love', 'love', 'action', 'action']
data = (45, 46)
k = 3
print(my_knn(test_data, cla, data, k))
# five : 有一千盏灯，全都是灭的，第一个人从第一盏灯开始全都按一下，第二个人
# 从第二盏灯开始往后的等都按一下，第三个人从第三盏灯开始往后的等都按
# 一下，以此类推，有一千个人，问最后有多少灯是亮着的，多少灯是灭的
# 0 代表灯是灭的状态 , 1 代表灯是亮的状态
a = [0 for _ in range(1000)]
for i in range(1000):
i += 1
for j in range(1, 1001):
if i//j == 0:
a[i-1] = 0 if a[i-1] else 1
print(a)
print(f'灭的灯共:{a.count(0)}盏')
print(f'亮的灯共:{a.count(1)}盏')
# six : 替换某个字符串中的一个或者多个子串
# 方法一
def str_replace(ostr, old_str, new_str):
list_s = ostr.split(old_str)
str_result = ''
for strs in list_s[:-1]:
str_result += strs+new_str
return str_result+list_s[-1]
a = 'heluollo luoshuxiaoluoll'
print(str_replace(a, 'luo', '##'))
# 方法二
def str_replace1(ostr, old_str, new_str):
return ostr.replace(old_str, new_str)
print(str_replace1(a, 'luo', '##'))
# seven : 冒泡排序算法
def my_sort(list1):
for i in range(len(list1)):
for j in range(i+1, len(list1)):
if list1[i] > list1[j]:
list1[i], list1[j] = list1[j], list1[i]
return list1
# eight : 100课糖随机分配给10个小朋友，求糖果获得第n多的小朋友获得的糖果（不能用语言自带的排序）
def my_sorted(ch_list, k):
for i in range(len(ch_list)):
for j in range(len(ch_list)-1):
if ch_list[i] > ch_list[j]:
ch_list[i], ch_list[j] = ch_list[j], ch_list[i]
num = ch_list[0]
n = 1
for i in ch_list[1:]:
if i == num:
continue
else:
n += 1
num = i
if n == k:
return i
else:
print('输入错误')
ch_list = [10, 30, 5, 7, 12, 20, 3, 3, 10]
k = 3
print(my_sorted(ch_list, 3))
# night ： 归并排序算法
def split_data(data):
if len(data) == 1:
return data
mid = len(data)//2
left_data = data[:mid]
right_data = data[mid:]
left = split_data(left_data)
right = split_data(right_data)
return merge_sort(left,right)
def merge_sort(left_list,right_list):
result = []
while len(left_list)>0 and len(right_list)>0:
if left_list[0] >= right_list[0]:
result.append(left_list.pop(0))
else:
result.append(right_list.pop(0))
result.extend(left_list)
result.extend(right_list)
return result
test_list = [1, 2, 1, 5, 6, 7, 10, 43, 24, 45, 23]
print(split_data(test_list))
# ten : 快速排序
def quick_sort(lis):
if len(lis) > 1:
left, right = [], []
mid = lis[len(lis)//2]
lis.remove(mid)
for i in lis:
if i >= mid:
right.append(i)
else:
left.append(i)
return quick_sort(left)+[mid]+quick_sort(right)
return lis
# eleven : 二分查找
def bin_search(link, val):
min = 0
max = len(link)-1
while min <= max:
mid = (min+max)//2
if val == link[mid]:
return mid
elif val > link[mid]:
min = mid+1
elif val < link[mid]:
max = mid-1
return None
# twelve : 斐波那契数列
def feibolaqi(num):
a, b, c = 0, 1, []
for i in range(num-1):
c.append(b)
a, b = b, a+b
c.append(b)
return b,c
其中：b代表第num个斐波那契数，c代表前num个斐波拉契数
# thirteen : 写一个长链接转短连接算法
# 方案：用一个很大的数作为key,长链接作为值，存起来，将这个key转为62进制数对应保存为短连接：
def convert(num):
global all_chars
all_chars = '0123456789ABCDEFGHIJKLMNOPQRSTUVWSYZabcdefghijklmnopqrstuvwxyz'
result = []
while num > 0:
result.insert(0, all_chars[num % 62])
num //= 62
return ''.join(result)
# fourteen : 判断一个数是不是快乐数
# 快乐数：对于正整数，每一次将该数替换为它每个
# 位置上的数字的平方和，然后重复这个过程直到这个数
# 变为1，也可能是无限循环始终变不到1，如果可以变为1，
# 那么这个数就是快乐数
# fifteen : 1,2,3,4 四个数中选三个数组成三位数，输出所有三位数（三位数每个位置上的数不重复）
a = [1, 2, 3, 4]
for i in a:
b = a[:]
b.remove(i)
for j in b:
c = b[:]
c.remove(j)
for k in c:
print(f'{i}{j}{k}')
Coding
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