Question

我的代码是关于使用每个单词/数字的第一个字符/数字作为句子/短语中的字符来创建密码，并将其打印出来。

示例：停止并闻到350＆＃34;玫瑰＆＃34;。 - ＆GT; Sast3r。（忽略引用使用r代替）

使用列表会很容易，但是在我的代码的分配中你不能使用它们。所以，到目前为止，我还不知道该怎么做

功能：

def create_password(phrase):
q = "'"  # quotations
dq = '"'  # double quotes

password = phrase[0]

for i in phrase:
    x = phrase.find(" ")
    if i.isalnum:
        password += phrase[x + 1]
    elif x == q or x == dq:
        password += phrase[x + 2]

return password

主：

# Imports
from credentials import create_password

# Inputs
phrase = str(input("Enter a sentence or phrase: "))

# Outputs
password = create_password(phrase)
print(password)

Answer 1

我认为通过整个短语而不用担心分裂空间更为直接。而是跟踪你是否刚刚看到一个空间。您只想在看到空格后添加角色。

def create_password(phrase):
    q = "'"  # quotations
    dq = '"'  # double quotes

    #Initialize the password to be an empty string
    password = ""

    #We are at the start of a new word (want to add first index to password)
    new_word = True

    #Walk through every character in the phrase
    for char in phrase:

        #We only want to add char to password if the following is all true:
        #(1) It's a letter or number
        #(2) It's at the start of a new word
        #(3) It's not a single quote
        #(4) It's not a double quote
        if char.isalnum and new_word:
            if char != q and char != dq:
                password += char
                new_word = False #<-- After adding char, we are not at a new word

        #If we see a space then we are going to be at a new word
        elif char == " ":
            new_word = True

    return password

p = create_password('Stop and smell the 350 "roses"')
print(p)

输出：

Sast3r

Answer 2

你肯定是在正确的轨道上！使用str.find()方法绝对是可行的方法！

但是，您需要了解str.find()方法的作用。签名：

str.find(sub [,start [,end) -> int

    # sub -> character to find
    # start -> where the function should start looking in the string
    # end -> where the function should stop looking

    # Returns a number, which is the place it found the character. 
    # If it didn't find anything, then return -1.

在不告诉函数从哪里开始的情况下，它总是会在字符串中找到第一个出现的字符。它不会知道你正在浏览字符串的每个字符。

所以让我们稍微改变一下：

for char_index in xrange(len(phrase)):
    # Tell the method to look after char_index: we've already looked before this!
    x = phrase.find(' ', char_index) index

    if phrase[x+1].isalnum(): # It's a function, notice the brackets?
        password += phrase[x + 1]
    elif phrase[x+2] == q or phrase[x+2] == dq:
        password += phrase[x + 2]

希望这可以获得您想要的密码。

Answer 3

优先使用内置函数，例如，每次找到空间的位置时，为什么不直接按照spilled函数的空间，使字符串直接到字符列表，每个element是一个单词，然后删除列表中的每个元素。

def create_password(phrase):
    password = ''
    phrase_list = phrase.split(' ')
    print (phrase_list)

    for i in phrase_list:
        print (i[0])
        password += i[0]
    return password

if __name__ == '__main__':
    # Inputs
    phrase = str(input("Enter a sentence or phrase: "))

    # Outputs
    password = create_password(phrase)
    print(password)

Answer 4

尝试获取字符串中的第一个字符，然后是空格后面的每个字符。看起来你有正确的想法。

Python：取第一个没有列表的角色

4 个答案: