Question

我试图创建一个将浏览数组的代码，当最终用户希望继续前进时，它会进入数组的开头。当在数组的开头并且用户希望向后移动时，它将进入数组的最后。虽然我能够从一个方面看，但我似乎无法连续走另一条路？当我进入P时，外观完美无缺，并会不断询问。虽然当我输入F时，循环在一次按下后停止。帮助我让F继续像p !!

#declaring array names.
longitude=[]; latitude=[]; messagetext=[];encryptions=[];
input_file = open('messages.txt', 'r')

#read file
lines_in_file_array = input_file.read().splitlines()


#appending the lines in a file to select records.
for line in lines_in_file_array:
     record_array = line.split(',')
     longitude.append(record_array[0])
     latitude.append(record_array[1])
     messagetext.append(record_array[2])

#Stop reading from file
input_file.close()

#This encrypts the message by turning each character into their individual
#ascii values, adding 2, then converting those ascii values back to that
#values character.
def encrypt():
    temporary_array=[]
    for index in range(len(messagetext)):
        x=messagetext[index]
        x=([ord(character)+2 for character in x])
        codedx=''.join([chr(character) for character in x])
        temporary_array.append(codedx)
    global temporary_array


def navigation():
    # Index position
    i = 0;

    # Loop forever
    while True:


     # Get the user's input, and store the response in answer
        answer = input("See Entry? P/F)?")

        # If the user entered lower case or upper case Y
        if answer.lower() == "f":

            # print the message
            print(messagetext[i % len(messagetext)])
            print(temporary_array[i % len(temporary_array)])
            print("")

            # and add to the index counter
            i = i + 1

        if answer.lower() == "p":

            # print the message
            print(messagetext[i % len(messagetext)])
            print(temporary_array[i % len(temporary_array)])
            print("")

            # and take away from the index counter
            i = i - 1


        # Otherwise leave the loop
        else:
            break


encrypt()
navigation()

Answer 1

你说“如果f，这个;如果p，这个;否则就会破坏;” “else”语句仅适用于p，而不是f。

我所说的是，您检查if answer.lower == 'p'的部分不应该说if，应该说elif：

if answer.lower() == "f":
    i = i + 1

elif answer.lower() == "p":
    i = i - 1

else:
    break

Answer 2

使用itertools.cycle：

a = ["spam", "bacon", "baked beans"]

# This will never end.
for food in itertools.cycle(a):
    print(a)

Answer 3

查看以下内容是否有效，请说ls是列表。

from itertools import cycle
rls = reversed(ls)
z = zip(cycle(ls), cycle(rls))
while True:
    choice = input("n / p: ")
    n, p = next(z)
    result = n if choice == "n" else p
    print(result)

看它是否符合您的要求。如果它确实很好，因为这里没有索引操作。如果没有，请发表评论。

Answer 4

一些变化：

其他：中断仅适用于p测试。通过添加elif
而不是在i开始0作为i启动索引位置None，以允许在第一个周期进行独特测试

在打印元素之前的第一个增量索引之外的循环中。这会将代码配置为在用户在p和n之间切换时不重复元素，反之亦然。

from sys import version_info
messagetext = ['one', 'two', 'three', 'four']

def navigation(messagetext):
    # Index position
    i = None
    py3 = version_info[0] > 2 #creates boolean value for test that Python major version > 2

    # Loop forever
    while True:
       # Get the user's input, and store the response in answer
        if py3:
            answer = input('See Entry? P/F?')
        else:
            answer = raw_input('See Entry? P/F?')

        # If the user entered lower case or upper case Y
        if answer.lower() == 'f':
            i = 0 if i == None else i + 1
            print(messagetext[i % len(messagetext)])
            print("")

        elif answer.lower() == 'p':
            i = -1 if i == None else i - 1
            # print the message
            print(messagetext[i % len(messagetext)])
            print("")

        # Otherwise leave the loop
        else:
            break

navigation(messagetext)

如何在Python中浏览数组？

4 个答案: