Question

我有一个单词列表，我的输入是列表前半部分的单词之一（在这种情况下，此列表是德语歌曲）。现在，我将这个单词的长度跳到列表中这个单词的长度，f.e。列表中的第一个单词为“ Es”，长度为2。现在我们从“ Es”开始计数2次，然后按单词“ zwei”着陆。我应该浏览此列表，并检查单词是否再次出现在列表的后半部分。

程序可以运行，但是问题出在输入上。它花了一个字，看看该理论是否有效。这是歌曲列表：“ hin”是上半部分的最后一个单词

song = [
    "Es", "gingen", "zwei", "Parallelen",
    "ins", "Endlose", "hinaus",
    "zwei", "kerzengerade", "Seelen",
    "und", "aus", "solidem", "Haus", 

    "Sie", "wollten", "sich", "nicht", "schneiden", 
    "bis", "an", "ihr", "seliges", "Grab", 
    "Das", "war", "nun", "einmal", "der", "beiden", 
    "geheimer", "Stolz", "und", "Stab", 

    "Doch", "als", "sie", "zehn", "Lichtjahre", 
    "gewandert", "neben", "sich", "hin", #End of the first half of the song
    "da", "wards", "dem", "einsamen", "Paare", 
    "nicht", "irdisch", "mehr", "zu", "Sinn", 

    "Warn", "sie", "noch", "Parallelen",
    "Sie", "wusstens", "selber", "nicht", 
    "sie", "flossen", "nur", "wie", "zwei", "Seelen",
    "zusammen", "durch", "ewiges", "Licht", 

    "Das", "ewige", "Licht", "durchdrang", "sie",
    "da", "wurden", "sie" "eins", "in", "ihm", 
    "die", "Ewigkeit", "verschlang", "sie",
    "als", "wie", "zwei", "Seraphim"]

我希望输入的内容是列表前半部分的所有单词（在这种情况下为歌曲），而不仅仅是一个单词。因此，它只是打印出每一行中每个单词的结果（在这种情况下，结果是列表）。

我希望它能在理论的前半部分立即打印出每个单词的每个结果。这将比输出：

理论有效/无效

结果1

理论有效/无效

Result2

理论有效/无效

Result3

依此类推...

代码如下：

with open('C:/Users/xy/Desktop/BWINF/parallelen.txt', 'r') as f:

song = f.read()

noneed = "–?,.;:"
for char in noneed:
    song = song.replace(char, "")
song = song.split()

def Parallelen(listSong):

originalWord = input("Enter a word: ")
originalWordSaved = originalWord
theorie_list = [] # The list for found words
index = song.index(originalWord) # Get the index of the first instance of "word"
indexOriginal = song.index(originalWordSaved)
wordCount = song.count(originalWord)

while True:
    if indexOriginal > 42:
        print("Word is in the second half")
        break
    if wordCount <= 1:
        print("Word appears only 1 time and therefore can't appear one more time")
        print("Theorie doesn't work")
        break  
    try:
        theorie_list.append(listSong[index])
        theorie_list.append(len(listSong[index]))
        index += len(listSong[index]) 
        if listSong[index] == originalWordSaved:
            theorie_list.append(listSong[index])
            theorie_list.append(len(listSong[index]))
            print("Theorie works")
            break
    except:
        print("Theorie doesn't work") 
        break

return theorie_list

print(Parallelen(song))

Answer 1

这是处理列表第一部分中每个单词的函数的版本。该问题指出该程序可以正常运行，因此主要逻辑未更改。

这些是已进行的更改：

删除input语句
在列表的第一部分添加for循环
打印正在处理的单词（出于调试和可读性考虑，可能会删除）
删除多余的originalWordSaved和originalIndex变量
将过滤逻辑（该单词在列表的后半部分，或出现少于两次）移出while循环，因为每个单词只需要运行一次
将代码移至无法导致IndexError脱离try / except块的位置
专门捕获IndexError，因为这是代码可能引发的唯一错误：应避免使用裸露的except语句
打印列表而不是返回列表

def Parallelen(listSong):

    # Magic number: length of list is 89
    halfway = 42

    for originalWord in listSong[: halfway + 1]:
        print("\nProcessing {!r}".format(originalWord))
        theorie_list = []  # The list for found words
        index = song.index(originalWord)
        # Get the index of the first instance of "word"
        index = song.index(originalWord)
        if index > halfway:
            print("Word is in the second half")
            continue
        wordCount = song.count(originalWord)
        # The word must appear at least once
        if wordCount == 1:
            print("Word appears only 1 time and therefore can't appear one more time")
            print("Theorie doesn't work")
            continue

        while True:
            theorie_list.append(listSong[index])
            theorie_list.append(len(listSong[index]))
            index += len(listSong[index])
            try:
                if listSong[index] == originalWord:
                    theorie_list.append(listSong[index])
                    theorie_list.append(len(listSong[index]))
                    print("Theorie works")
                    break
            except IndexError:
                print("Theorie doesn't work")
                break
        print(theorie_list)

    return


Parallelen(listSong)

列表中特定数量的输入元素

1 个答案: