Python: 每日一题 36

willLin

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solomonxian

如果知道pin=1357，马上想到懒人方法，不过没有通用性

1. for i in [1,2,4]:
   
2.     for j in [2,3,6]:
   
3.         for k in [2,4,5,6,8]:
   
4.             for m in [4,7,8]:
   
5.                 print(i,j,k,m)

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那还是用个字典存起来，发现itertools模块有计算笛卡尔积的工具

1. from itertools import product
   
2. 
   
3. pins = {1:[1,2,4], 2:[1,2,3,5], 3:[2,3,6], 4:[1,4,5,7], 5:[2,4,5,6,8], \
   
4.         6:[3,5,6,9], 7:[4,7,8], 8:[5,7,8,9], 9:[6,8,9], 0:[8,0]}
   
5. 
   
6. pin = '1357'
   
7. # pin = input("请输入密码：")
   
8. try:
   
9.     # 将输入生成二维列表，再解包，用product工具将笛卡尔积打印出来
   
10.     for i in product(*[pins[int(j)] for j in pin]):
    
11.         print(i)
    
12. except ValueError:
    
13.     print("请输入0-9正整数集合！")

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然后再试下不用工具，怎么求笛卡尔积，结果只能想到用神奇的eval()函数凑出来

1. def cartesian_product(times,sets):
   
2.     """参数times是列表个数，sets是包含各个列表的列表，返回笛卡尔积"""
   
3.     string_1 = "("
   
4.     string_2 = ""
   
5.     for i in range(times):
   
6.         string_1 += "a"*(i+1) + ',' # 最后会多一个逗号，切片去掉
   
7.         string_2 += " for "+ "a"*(i+1)+" in "+str(sets[i]) # 确保关键字左右有空格
   
8.     # 把字符串组合成列表推导式
   
9.     string = "[" + string_1[:-1] + ")" + string_2 + "]"
   
10.     output = eval(string)
    
11.     print(*output)
    
12. 
    
13. a = []
    
14. pin = input("请输入数字，我没有写防错，你输入了其他就直接报警哦：")
    
15. for j in pin:
    
16.     a.append(pins[int(j)])
    
17. cartesian_product(len(a),a)

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yyf900524

答案

大头目

大佬你的题目都好绕啊。。。基本看一遍是看不明白的

小庆子

1. import itertools as t
   
2. 
   
3. def getPINs(pin):
   
4.     pin = list(pin)
   
5.     for i in range(len(pin)):
   
6.         if int(pin[i]) == 1:
   
7.             pin[i] = [2,4]
   
8.         elif int(pin[i]) == 2:
   
9.             pin[i] = [1,3,5]
   
10.         elif int(pin[i]) == 3:
    
11.             pin[i] = [2,6]
    
12.         elif int(pin[i]) == 4:
    
13.             pin[i] = [1,5,7]
    
14.         elif int(pin[i]) == 5:
    
15.             pin[i] = [2,4,6,8]
    
16.         elif int(pin[i]) == 6:
    
17.             pin[i] = [3,5,9]
    
18.         elif int(pin[i]) == 7:
    
19.             pin[i] = [4,8]
    
20.         elif int(pin[i]) == 8:
    
21.             pin[i] = [5,7,9,0]
    
22.         elif int(pin[i]) == 9:
    
23.             pin[i] = [6,8]
    
24.         else:
    
25.             pin[i] = [8]
    
26.     pins = []
    
27.     for i in t.product(pin[0],pin[1],pin[2],pin[3]):
    
28.         pins.append(i)
    
29.     print(pins)
    
30. 
    
31. 
    
32. '''pin = input('间谍看到的数字：')'''
    
33. pin = '1357'
    
34. getPINs(pin)

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不知道怎么把product里面加入未知数量的iter

萧丹夜

有点意思，成功是成功了，但是不好

1. ## 代码
   
2. import test
   
3. import itertools
   
4. 
   
5. def get_pins(x):
   
6.     form = [[0,8],[1,2,4],[2,1,3,5],[3,2,6],[4,1,5,7],[5,2,4,6,8],[6,3,5,9],[7,4,8],[8,5,7,9,0],[9,6,8]]
   
7.     n = len(x)
   
8.     y = []
   
9.     result = []
   
10.     for each in x:
    
11.         y.append(int(each))
    
12.     if n == 1:
    
13.         comb = itertools.product(form[y[0]])
    
14.     elif n == 2:
    
15.         comb = itertools.product(form[y[0]],form[y[1]])
    
16.     elif n == 3:
    
17.         comb = itertools.product(form[y[0]],form[y[1]],form[y[2]])
    
18.     elif n == 4:
    
19.         comb = itertools.product(form[y[0]],form[y[1]],form[y[2]],form[y[3]])
    
20.     elif n == 5:
    
21.         comb = itertools.product(form[y[0]],form[y[1]],form[y[2]],form[y[3]],form[y[4]])
    
22.     else:
    
23.         return '排列过大'
    
24.     for i in comb:
    
25.         z = ''
    
26.         for j in i:
    
27.             z += str(j)
    
28.         result.append(z)
    
29.     return result
    
30.    
    
31. 
    
32. 
    
33. 
    
34. 
    
35. ## 测试
    
36. expectations = [('8', ['5','7','8','9','0']),
    
37.                 ('11',["11", "22", "44", "12", "21", "14", "41", "24", "42"]),
    
38.                 ('369', ["339","366","399","658","636","258","268","669","668","266","369","398","256","296","259","368","638","396","238","356","659","639","666","359","336","299","338","696","269","358","656","698","699","298","236","239"])]
    
39. 
    
40. for tup in expectations:
    
41.   test.assert_equals(sorted(get_pins(tup[0])), sorted(tup[1]), 'PIN: ' + tup[0])
    
42. 
    
43. ## 结果
    
44. Success！
    
45. Success！
    
46. Success！

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小强工作室

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lyslyslys

学习学习

double-lee

虽然有基本思路,但总会卡在某个地方想挺久的,看来还只是刚碰到python的一点点皮毛...

3333332

谢

捞鬼

捞

panheng

交作业。请问不用模块怎么计算不同长度和可变数量的笛卡尔积。想了一下午没头绪

1. import itertools
   
2. def getPins(PIN=1357):
   
3.     dict1 = {'1': '124', '2': '2135', '3': '326', '4': '4157', '5': '52468', '6': '6359', '7': '748', '8': '80579',
   
4.              '9': '968', '0': '08'}
   
5.     temp = []
   
6.     for each in str(PIN):
   
7.         temp.append(dict1.get(each))  # 每个数字可能的字符串
   
8.     list1 = list(itertools.product(*temp))  # 计算笛卡尔乘积，返回密码的元组
   
9.     pinlist = [''.join(each) for each in list1]  # 提取元组字符串拼接
   
10.     print(pinlist)
    
11.     return pinlist

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Jung

def Fun36():
    for i in [1,2,4]:
        for j in [3,2,6]:
            for k in [5,2,4,6,8]:
                for g in [7,4,8]
                    print(i,j,k,g)

仅针对该题，的有效组合

Geoffreylee

1. from functools import reduce
   
2. password_dict = {'0': [0, 8], '1': [1, 2, 4], '2': [1, 2, 3, 5], '3': [2, 3, 6],
   
3.                  '4': [1, 4, 5, 7], '5': [2, 4, 5, 6, 8], '6': [3, 5, 6, 9],
   
4.                  '7': [4, 7, 8], '8': [0, 5, 7, 8, 9], '9': [6, 8, 9]}
   
5. 
   
6. 
   
7. def combine_list(list1, list2):
   
8.     return [str(i)+str(j) for i in list1 for j in list2]
   
9. 
   
10. 
    
11. def f_36(string: str) -> list:
    
12.     iter_list = []
    
13.     for i in string:
    
14.         iter_list.append(password_dict[i])
    
15. 
    
16.     return reduce(combine_list, iter_list)
    
17. 
    
18. 
    
19. print(f_36('1357'))
    

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19971023

1

nononoyes

666