python小题

Estinien

请问下这个题怎么解

请描述 x 的哪些值使这个布尔值成为真
(x.count('p') == 2 and 'o' in x and x[-1] == 'n') or (len(x) == 5 and x[3] == x[4])

灰晨

x是什么

3236654291

from random import *
x = ['p','o','n']

lst = []

num = 10000


while num > 0:
    if (x.count('p') == 2 and 'o' in x and x[-1] == 'n') or \
       (len(x) == 5 and x[3] == x[4]):
        lst.append(x)
        x = ['p','o','n']

    else:
        x.insert(randint(0,len(x)),choice(x))

    num -= 1


if len(lst):
    for el in range(len(lst)-1, -1, -1):
        if lst.count(lst[el]) > 1:
            lst.pop(el)

for i in lst:
    print(i,'\n')


print('over')

随机生产结果，行不？

阿奇_o

>>> x = 'ppon'
>>> (x.count('p') == 2 and 'o' in x and x[-1] == 'n')
True
>>> x2 = 'abc22'
>>> (len(x2) == 5 and x2[3] == x2[4])
True

由以上试验，再根据or的特性，可以推出 x 可以是 'ppon', 'popn', 'oppn' 
或者是 五个字符长度的 结尾是任意两个相同的字符，即可，如 'abcxx', 'xyz..', '123**', .... 

白two

俩括号中间用的 or 连接，所以 or 左右两边有一个为真即可

先看左边：
x.count('p') == 2 and 'o' in x and x[-1] == 'n'
三者间用的 and 连接，所以必须同时满足三个条件，即：
x 里面要有两个 p 字符，至少得有一个 o 字符，且最后一个字符为 n ；

再看右边：
len(x) == 5 and x[3] == x[4]
两者用 and 连接，也必须同时满足两个条件，即：
x 的长度等于 5 ， x 第 4 个值等于第 5 个值；

例如：

>>> x = ["p","p","o",1,2,3,4,"n"]#满足右边
>>> print((x.count('p') == 2 and 'o' in x and x[-1] == 'n') or (len(x) == 5 and x[3] == x[4]))
True
>>> x = ('p','p','1','o','2','n')#满足右边
>>> print((x.count('p') == 2 and 'o' in x and x[-1] == 'n') or (len(x) == 5 and x[3] == x[4]))
True
>>> x = "oppn" #满足右边
>>> print((x.count('p') == 2 and 'o' in x and x[-1] == 'n') or (len(x) == 5 and x[3] == x[4]))
True
>>> x = "xyyyy"#满足左边
>>> print((x.count('p') == 2 and 'o' in x and x[-1] == 'n') or (len(x) == 5 and x[3] == x[4]))
True
>>> x = [1,2,3,4,4]#满足左边
>>> print((x.count('p') == 2 and 'o' in x and x[-1] == 'n') or (len(x) == 5 and x[3] == x[4]))
True

YaoShi

先看最外层    是or   那只要两边有一边为真结果就是真了
左边为真的情况 ： x满足  'p'字符有且仅有两个  'o'字符个数大于等于一个   以字符'n'结尾
右边为真的情况 ： x长度为五    最后两个字符一样
贴几个例子

>>> x='ppon'
>>> (x.count('p') == 2 and 'o' in x and x[-1] == 'n') or (len(x) == 5 and x[3] == x[4])
True
>>> x='asdwqdpasdpozn'
>>> (x.count('p') == 2 and 'o' in x and x[-1] == 'n') or (len(x) == 5 and x[3] == x[4])
True
>>> x='asdxx'
>>> (x.count('p') == 2 and 'o' in x and x[-1] == 'n') or (len(x) == 5 and x[3] == x[4])
True

lufeixiong

or两边有一边为真结果就是真了
左边为真 ： x满足p字符有且仅有两个o字符个数大于等于一个并以字符n结尾
右边为真 ： x长度为五，最后两个字符一样

星行123

#三种情况
#1、字符串形式：
#1.1、最后一个字符是n，含有o，两个p的任意字符串：
>>> x = 'ppon'
>>> (x.count('p') == 2 and 'o' in x and x[-1] == 'n') or (len(x) == 5 and x[3] == x[4])
True
#1.2、字符长度为5，最后两个字符相同
>>> x = 'abcdd'
>>> (x.count('p') == 2 and 'o' in x and x[-1] == 'n') or (len(x) == 5 and x[3] == x[4])
True
#2、列表形式
#2.1、最后一个元素为‘n’，含有‘o’和两个‘p’
>>> x = ['p','p','o','n']
>>> (x.count('p') == 2 and 'o' in x and x[-1] == 'n') or (len(x) == 5 and x[3] == x[4])
True