Question

所以我正在尝试制作一个添加，删除，搜索，保存和加载列表的程序

除了在列表中搜索第一个字母之外，我还设法使一切正常工作

程序将询问用户要搜索的字母然后它将在列表中搜索用户输入的第一个字母

我所要做的就是：

mylist = ["hello","world","how","you","doing"]
for word in mylist:
  print (word[0])

确实打印出每个单词的前字母

h
w
h
y
d
>>>

但是我想做什么

mylist = ["hello","world","how","you","doing"]
letter = input("input a letter you would like to search ").lower()

将在整个列表中搜索用户输入的字母

找到所有以字母开头的单词后，我希望它打印出找到的单词数，然后打印出与用户要求的字母一起找到的单词

Answer 1

在startswith中使用列表理解

[i for i in mylist if i.startswith(letter)]

这与您正在执行的操作类似，但是您并未检查以下情况

mylist = ["hello","world","how","you","doing"]
for word in mylist:
      if word[0]==letter:
            print(word)

为了安全起见，请对列表中出现的字符串也.lower()使用

mylist = ["Hello","world","how","you","doing"]
letter = 'h'
[i for i in mylist if i.lower().startswith(letter)] #["Hello","how"]

Answer 2

for word in list:
    if input() in word:
        print(word)

那条线上的东西。它从列表中获取单词，然后针对该单词测试用户输入。

Answer 3

您可以使用列表推导：

mylist = ["hello","world","how","you","doing"]
letter = input("input a letter you would like to search ").lower()
result = [word for word in mylist if word.lower().startswith(letter)]
print(len(result))
print(result)

您可以详细了解List Comprehensions in the docs

的功能

编辑（受其他答案启发）：使用word.startswith()代替索引word[0]可以避免空字符串的问题。

Answer 4

mylist = ["hello","world","how","you","doing"]
letter = input("input a letter you would like to search ").lower()
    input a letter you would like to search 'h'

list_starting_with_letter = [i for i in mylist if i[0].lower()==letter.lower()]
list_starting_with_letter
    ['hello', 'how']

print('Number of such words: '+ str(len(list_starting_with_letter)))
    Number of such words: 2

Answer 5

https://ideone.com/U6ZM2l

mylist = ["hello","world","how","you","doing"]
letter = input("input a letter you would like to search ").lower()

words = list(filter(lambda s: s.startswith(letter), mylist))

print( len(words) )

for w in words: print ("\n"+w)

Answer 6

希望这对您有帮助

FanOut

如何在以用户要求的特定字母开头的列表中查找单词

6 个答案: