[Homework 4] 文句處理

[joyshiang]

提交連結

https://judge.ccclub.io/status/fb9ebc040faeda02efa59f621ef2881b

程式碼

keyword_lst = input().split()
def length(s):
    return(len(s))
keyword_lst1 = sorted(keyword_lst, reverse=True, key=length)
keyword_lst2 = ['「{}」'.format(i) for i in keyword_lst1]#將關鍵字加上引號
keyword_lst3 = []#將關鍵字中的關鍵字也加上括號
for key2 in keyword_lst2:
    for key1 in keyword_lst1:
        if key2.find(key1) != -1: # key2內有找到key1
            if key2 != '「' + key1 + '」':
                a = '「' + key1 + '」'
                b = key2.replace(key1, a)
                keyword_lst3.append(b)
                break
        else:
            keyword_lst3.append(key2)
            break

sentence_lst = []
while True: #不定行處理end
    sentence = input()
    if sentence == "end":
        break
    sentence_lst.append(sentence)

for s in sentence_lst:# 替換關鍵字
    for key1, key2 in zip(keyword_lst1, keyword_lst2):
        s = s.replace(key1, key2)
        for key2, key3 in zip(keyword_lst2, keyword_lst3):
            s = s.replace(key3, key2)
    print(s)

錯誤訊息

問題描述

嘗試檢查每個list，發現keyword_lst3跑測資三的時候是['「郭「婞淳」」', '「婞淳」', '「金牌」']，結果也是對的。但跑測資一時keyword_lst3卻是這樣:['「apple」', '「cute」', '「app」']，最後結果是:
I have an 「「app」le」
you are so 「cute」

想不太到位甚麼結果會變成這樣，還是說中文英文的邏輯不一樣?

[SleeperShark]

你邏輯是對的，但條件設得不好，到底要怎麼設計lst3的elem的迴圈結構可以再想一下。

以下是我寫的方法，你可以參考一下，想一下為甚麼這樣寫得出來

for key2 in keyword_lst2:
    temp = key2[:]

    for key1 in keyword_lst1:

        if temp.find(key1) != -1 and temp != f'「{key1}」':
            idx = temp.find(key1)
            temp = f'{temp[:idx]}「{key1}」{temp[idx+len(key1):]}'

    keyword_lst3.append(temp)

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