Question

作为Python的初学者，我让老师完成了这些任务，而我被其中一项困住了。这是关于使用for循环在单词中找到辅音，然后使用这些辅音创建一个字符串。

我的代码是：

consonants = ["qwrtpsdfghjklzxcvbnm"]
summer_word = "icecream"

new_word = ""

for consonants in summer_word:
    new_word += consonants

ANSWER = new_word

我得到了for循环，但这是我没有真正得到的串联。如果我使用new_word = []，它将成为一个列表，所以我应该使用""？如果您将多个字符串或字符连接起来，它应该成为一个字符串，对吗？如果您有一个int，则还必须使用str(int)来进行连接。但是，如何创建这串辅音呢？我认为我的代码不错，但无法播放。

致谢

Answer 1

您的循环当前仅循环遍历summer_word的字符。您在“用于辅音...”中使用的“辅音”名称只是一个虚拟变量，它实际上并未引用您定义的辅音。尝试这样的事情：

consonants = "qwrtpsdfghjklzxcvbnm" # This is fine don't need a list of a string.
summer_word = "icecream"

new_word = ""

for character in summer_word: # loop through each character in summer_word
    if character in consonants: # check whether the character is in the consonants list
        new_word += character
    else:
        continue # Not really necessary by adds structure. Just says do nothing if it isn't a consonant.

ANSWER = new_word

Answer 2

Python中的字符串已经是字符列表，可以这样处理：

In [3]: consonants = "qwrtpsdfghjklzxcvbnm"

In [4]: summer_word = "icecream"

In [5]: new_word = ""

In [6]: for i in summer_word:
   ...:     if i in consonants:
   ...:         new_word += i
   ...:

In [7]: new_word
Out[7]: 'ccrm'

Answer 3

是的，如果字符是数字，则必须使用str（int）将其转换为字符串类型。

let material = SCNMaterial()

material.diffuse.contents = UIImage(named: "Albedo.png")
material.ambientOcclusion.contents = UIImage(named: "AO.png")
material.normal.contents = UIImage(named: "Normals.png")

在for循环中，您正在评估'辅音'不是元音也不是int。希望对您有帮助。

Answer 4

这里的问题是，您已将变量辅音创建为列表，其中包含字符串。所以去掉方括号就可以了

Answer 5

consonants = "qwrtpsdfghjklzxcvbnm"
summer_word = "icecream"

new_word = ""


for letter in summer_word:
    if letter in consonants:
      new_word += letter

print(new_word)

一个较短的将是

consonants = "qwrtpsdfghjklzxcvbnm"
summer_word = "icecream"

new_word = ""

new_word = [l for l in summer_word if l in consonants]
print("".join(new_word))

Python-使用for循环创建字符串

5 个答案: