Question

现在我的所有代码都是在我的所有单词中找到所有a。我想要它做的就是找到以我输入的字母开头的单词来找到单词。例如。 A会找到activity，again和ago字样。我试着寻找每一个地方，找不到我需要的答案。

dictionary=["activity","again","ago","begin","behaviour","beyond","camp","cannon","cell","discussion","doctor","display","else","estimate","establish","fudge","flight","fight","gear","great","grunt","how","hoe","house","impact","image","implication","just","job","judge","keep","key","kai"]
newlist=[]
choice = ''
while choice != 'q':

    choice = input("?")
    if choice == 'a':
    for a in dictionary:
        newlist.append(choice.lower())
        print(newlist)

Answer 1

dictionary=["activity","again","ago","begin","behaviour","beyond","camp","cannon","cell","discussion","doctor","display","else","estimate","establish","fudge","flight","fight","gear","great","grunt","how","hoe","house","impact","image","implication","just","job","judge","keep","key","kai"]
newlist=[]
choice = ''
while choice != 'q':

choice = input("?")
if choice == 'a':
    for item in dictionary:
        if(item[0] == choice):
            newlist.append(item)
    print(newlist)

您的上述代码未检查字典中每个项目的第一个字符。它只是遍历字典中的每个项目，并将“选择”（在本例中为“a”）附加到“newList”，然后在“newList”中打印当前项目

要解决此问题，您需要根据您的选择检查字典中每个项目的第一个字母，然后只将这些特定项目附加到“newList”

Answer 2

Python字符串实际上有一个startswith方法:)在您的情况下，它等同于string[0]。

dictionary=["activity","again","ago","begin","behaviour","beyond","camp","cannon","cell","discussion","doctor","display","else","estimate","establish","fudge","flight","fight","gear","great","grunt","how","hoe","house","impact","image","implication","just","job","judge","keep","key","kai"]
newlist=[]
choice = ''
while choice != 'q':
    choice = input("?")
    for a in dictionary:
        if a.startswith(choice.lower()):
            newlist.append(a)
    print(newlist)

Answer 3

任何人都需要代码并运行代码打印一段时间。这应该有助于：

count = 0
while count < len(newlist):
    count = count + 1
    print(newlist)
    length = len(newlist)

我需要一个程序的帮助，该程序可以找到以我输入的字母开头的单词

3 个答案: