1.列表排序
- def que6():
- # 6.输入三个整数x, y, z,形成一个列表,请把这n个数由小到大输出。
- # 程序分析:列表有sort方法,所以把他们组成列表即可。
- li = np.random.randint(-100, 100, size=10)
- # 就地转化
- li = li.tolist()
- # 用sort()结果
- li_sort = sorted(li, reverse = False)
- print('用sort方法,重新排列结果:{}'.format(li_sort))
-
- # 不用sort方法,自己写排序方法做,
-
- # 冒泡排序
- def bubbleSort(m):
- m = m.copy()
- for time in range(1, len(m)):
- for index in range(len(m) - time):
- if m[index] > m[index+1]:
- m[index], m[index+1] = m[index+1] , m[index]
- return m
-
- # 选择排序
- def selectSort(m):
- m = m.copy()
- for seat_L in range(len(m)-1):
- for seat_R in range(seat_L+1, len(m)):
- if m[seat_L] > m[seat_R]:
- m[seat_L], m[seat_R] = m[seat_R], m[seat_L]
- return m
-
- # 插入排序1(内部写成函数):
- def insertSort_1(m):
- result = []
-
- # 单个元素k插入列表li
- def to_insert(li, k):
- # 标识符
- tab = False
-
- # 寻找插入位置
- # 循环次数应该至少大于列表长度+1,None也占一位(空列表),即认为扑克牌最末尾还有一张‘空牌’
- for i in range(len(li) + 1):
- # 修改标识符,标志‘遍历完后的下一个循环’,即在和‘空牌’比较
- if i == (len(li)):
- tab = True
-
- # 如果在对li[-1]比较完成(包含)之前,且寻找到位置,即把扑克从左往右比较一遍
- if not tab and k < li[i]:
- li.insert(i, k)
- break
- # 如果遍历完成,多循环一次,即和‘空牌’不需要比较,直接把牌替换掉‘空牌’
- if tab:
- li.append(k)
- return li
- # 遍历列表
- # result = result[:1]
- for length in range(len(m)):
- result = to_insert(result, m[length])
-
- # print(result,m[length])
- return result
-
- # 插入排序2(直接嵌套循环):
- def insertSort2(m):
- m = m.copy()
- result = m[:1]
- for index_choose in range(1, len(m)):
- # 手上已经有index_choose张牌,比较第index_choose+1张牌则append
- # 逐个比对手上的牌,如果都对比了一遍,则加到最后
- for index_insert in range(len(result) + 1):
- print(result, index_insert,'\n',m, index_choose,'\n\n')
- if index_insert != index_choose and m[index_choose] < result[index_insert] :
- result.insert(index_insert, m[index_choose])
- break
- if index_insert == index_choose:
- result.append(m[index_choose])
- # print(result, m[index_choose])
- return result
-
-
-
- # print(li)
- print('插入排序:',insertSort3(li))
- print('选择排序:',selectSort(li))
- print('冒泡排序:',bubbleSort(li))
- que6()
2.调换字python教程典键值
- # 1. 调换元素.\
- def que1():
- d={1:"one",2:"two"}
-
- # 方法1 --- 动态赋值
- def method1(d):
- d = d.copy()
- result = {}
- for k,v in d.items():
- result[v] = k
- return result
-
- # 方法2 --- 生成器
- def method2(d):
- d = d.copy()
- result = {v:k for k,v in d.items()}
- return result
-
- # 方法3 --- 由值寻找键
- def method3(d):
- d = d.copy()
- # 由键寻找值
- def match(dic, b):
- return [k for k,v in dic.items() if v == b]
- # 先生成key-None,再赋值
- result = {}
- result = result.fromkeys(d.values())
- for k in result.keys():
- result[k] = match(d, k)[0]
- return result
-
- # 方法4 --- 列表转字典 < 直接转换/动态赋值 >
- def method4(d):
- d = d.copy()
- key = d.keys()
- val = d.values()
- data = list(zip(key, val))
-
- # 方法4-1
- result1 = {}
- for i in range(len(data)):
- result1[data[i][1]] = data[i][0]
-
- # 方法4-2
- result2 = dict(zip(val, key))
-
- return result1, result2
-
- print('新列表动态赋值方法:{}'.format(method1(d)))
- print('生成器方法:{}'.format(method2(d)))
- print('由键寻值方法:{}'.format(method3(d)))
- print('动态赋值列表转字典方法:{}'.format(method4(d)[0]))
- print('直接列表转字典方法:{}'.format(method4(d)[1]))
- # que1()
3.删除列表中的重复元素
- #遇到问题没人解答?小编创建了一个Python学习交流群:531509025
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-
- # 3. 删除重复元素list =[1,2,5,4,1,5,6,8,0,2,5]
- a = np.random.randint(-100, 100, size=10)
- a = a.tolist()
-
- def method1(a):
- a = a.copy()
- a = set(a)
- return a
- def method2(a):
- b = a.copy()
- c = 0
- for i in range(len(a)-1):
- if b[i+c] in b[:i+c]+b[i+c+1:]:
- b.pop(i+c)
- c -= 1
- return b
- print('集合法:',method1(a))
- print('遍历法:',method2(a))
4.输出质数
- def prime(end):
-
- prime_list = []
- if end <= 1:
- print('必须大于1')
- else:
- # prime_list.append(2)
- for i in range(2, end+1, 1):
- count = 0
- if i == 2:
- if i%2 != 0:
- prime_list.append(2)
- else:
- for m in range(2, i):
- # 能够整除,则跳出循环
- if (i % m) == 0:
- # print(i, m)
- break
- # 否则计数+1
- else:
- count += 1
- # 判断是否整除完成(0/n)
- if count == i - 2:
- prime_list.append(i)
-
- print(count, i, m)
-
- return (prime_list)
-
- num = int(input('想输出2到多少?'))
- print(prime(num))
5.判断是一年中第几天
- def que3():
- # 3.输入某年某月某日,判断这一天是这一年的第几天?:
- # 闰年判断函数
- def judge_leap(num):
- date = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
- # (四年一闰 and not百年) or 四百年闰
- if (num % 4 == 0 and num % 100 != 0) or num % 400 ==0:
- date[1] =29
- return date
-
- # 格式转换
- date = (input('请输入一个日期,格式如:“2018.02.12”:'))
- date_list = (list(map(int, (date.split('.')))))
- # 遍历计算天数
- day = date_list[2]
- for i in range(date_list[1]):
- day += judge_leap(date_list[0])[i]
- print('{}月{}日是{}年的第{}天\n'.format(date_list[1], date_list[2], date_list[0], day))
- # que3()
6.猜数字
- # 重新猜数字
- import random
-
- def judge_num(num, num_random):
- if num > num_random:
- print('It\'s too big')
- return 1
- elif num < num_random:
- print('It\'s too small')
- return 1
- else:
- print("Congratulation!! That\' right!")
- return 0
-
- # 产生随机数
- num_start = int(input('Digital lower limit of guess number:\n'))
- num_end = int(input('Digital upper limit of guess number:\n'))
- num_random = random.randint(num_start, num_end)
-
- # 参数初始化
- result = 1 # 判断结果
- i = 0 # 循环次数
- frequency = 3 # 循环限制次数
-
- # 提示总猜测次数、剩余次数
- print('WARNING: You have【{}】 chances you guess '.format(frequency), end = '--&&>>--')
- print('【{}】 chances left now:\n'.format(frequency - i +1))
-
- while result and i != frequency:
- # 猜数字
- num = int(input('Please guess a int_number:\n'))
- result = judge_num(num, num_random)
- i += 1
7.进制转换
- # 任意进制转十进制
- def other_to_decimal(hex, num):
- # 整型转化为列表,
- num_str = str(num)
- # map()将List对象中的元素(list类型)转化为集合(set)类型
- num_list = list(map(int, num_str))
- # 列表反序
- num_list = num_list[::-1]
- print(list(map(int, num_str)))
-
- # 获取数字位数
- digit = len(num_list)
- num_decimal = 0
-
- # 累加
- for i in range(digit):
- numi = num_list[i]
- # print(numi, hex**i)
- num_decimal += numi*(hex**i) # 对每一位数的幂指数累加
-
- return num_decimal
-
- # 十进制转任意进制
- def decimal_to_other(hex, num):
- # 获取数字位数
- digit = len(str(num))
-
- num_hex = []
- quotient = 1
-
- # 相除,余数计入列表num_hex
- while quotient:
- # 取余和取商
- quotient = num // hex
- remainder = num % hex
- # print(quotient, remainder)
- # 余数计入列表
- num_hex.append(remainder)
- # 商做下一次循环
- num = quotient
-
- # 列表反序,通过切片和sort()函数可以实现
- num_hex = num_hex[::-1]
- # num_hex.sort(reverse=True)
-
- # 如果超过十进制,用ASCII码转化为字母
- for i in range(len(num_hex)):
- if num_hex[i] > 9:
- num_hex[i] = chr(int(num_hex[i])+87)
- # print(num_hex)
-
- # 列表转化为字符串
- result = (''.join('%s' %m for m in num_hex))
- return result
-
-
- Type = bool(input("十进制转任意进制请输入1,任意进制转十进制请输入0\n"))
- if Type:
- hex = int(input("需要把十进制转换为多少进制?请输入正整数\n"))
- num = int(input("需要转换的数字是:"))
- print("换算结果是:", decimal_to_other(hex, num))
- else:
- hex = int(input("需要把多少进制转换为十进制?请输入正整数\n年"))
- num = int(input("需要转换的数字是:"))
- print("换算结果是:", other_to_decimal(hex, num))