有没有办法在具有唯一标准 Python 的列表中查找元素？

Question

由于某种原因 .index 和 .find 在我的程序中不起作用。

基本上，我希望程序根据设置的条件找到列表的索引号。

例如这个列表：

['Synonyms: Pocket Monsters, Indigo League, Adventures on the Orange Islands, The Johto Journeys, Johto League Champions, Master Quest', 'Japanese: ポケットモンスター', 'Type: TV', 'Episodes: 276', 'Status: Finished Airing', 'Aired: Apr 1, 1997 to Nov 14, 2002', 'Premiered: Spring 1997', 'Broadcast: Thursdays at 19:00 (JST)', 'Producers: TV Tokyo, TV Tokyo Music, Studio Jack', 'Licensors: VIZ Media, 4Kids Entertainment', 'Studios: OLM', 'Source: Game', 'Genres: Action, Adventure, Comedy, Kids, Fantasy', 'Duration: 24 min. per ep.', 'Rating: PG - Children', 'Score: 7.341 (scored by 291,570 users)', 'Ranked: #21572', 'Popularity: #287', 'Members: 504,076', 'Favorites: 4,076', '']

我想找到“流派”位置的索引号。我尝试这样做 .index("Genres:") 但没有找到索引号并返回错误。 我需要这个来查找索引号，因为对于本网站上的其他页面，“流派”处于不同的位置

这是我尝试过的，只是返回了一个错误

GIndex = Information.index("Genres")
print (Information[GIndex])
Genre = (Information[GIndex])

Answer 1

您也可以使用列表推导：

word = 'Genres'
results = [i for i, l in enumerate(lst) if word in l]

Answer 2

要使索引正常工作，它需要完全匹配。这将检查“流派”是否包含在列表中的任何字符串中并打印其索引

word = 'Genres'
for i, item in enumerate(lst):
    if word in item:
        print(i, item)

你可以很容易地把它变成一个函数来返回 i 是索引

Answer 3

你可以使用enumerate()

l=['Synonyms: Pocket Monsters, Indigo League, Adventures on the Orange Islands, The Johto Journeys, Johto League Champions, Master Quest', 'Japanese: ポケットモンスター', 'Type: TV', 'Episodes: 276', 'Status: Finished Airing', 'Aired: Apr 1, 1997 to Nov 14, 2002', 'Premiered: Spring 1997', 'Broadcast: Thursdays at 19:00 (JST)', 'Producers: TV Tokyo, TV Tokyo Music, Studio Jack', 'Licensors: VIZ Media, 4Kids Entertainment', 'Studios: OLM', 'Source: Game', 'Genres: Action, Adventure, Comedy, Kids, Fantasy', 'Duration: 24 min. per ep.', 'Rating: PG - Children', 'Score: 7.341 (scored by 291,570 users)', 'Ranked: #21572', 'Popularity: #287', 'Members: 504,076', 'Favorites: 4,076', '']
for i, s in enumerate(l):
        if "Genres" in s:
              print(i)
>>>12

Answer 4

在您的列表中没有值为"Genres" 的元素。 "Genres" 和 "Genres: Action, Adventure, Comedy, Kids, Fantasy" 不相等。如果你想找到以"Genres"开头的元素，你可以写一个非常简单的for循环，比如

for index, item in enumerate(information_list):
   if item.startswith('Genres'):
      print(index, item)
      break

@9769953 建议的更好方法是使用字典。在 dict 中查找的效率几乎是恒定的时间，您的数据将整齐地组织，键对应于属性名称，值对应于属性值。

字典应该是这样的

information_dict = {
  "Synonyms": "Pocket Monsters, Indigo League, Adventures on the Orange Islands, The Johto Journeys, Johto League Champions, Master Quest",
  "Japanese": "ポケットモンスター",
  "Type": "TV",
  "Episodes": "276",
  "Status": "Finished Airing",
  "Aired": "Apr 1, 1997 to Nov 14, 2002",
  "Premiered": "Spring 1997",
  "Broadcast": "Thursdays at 19:00 (JST)",
  "Producers": "TV Tokyo, TV Tokyo Music, Studio Jack",
  "Licensors": "VIZ Media, 4Kids Entertainment",
  "Studios": "OLM",
  "Source": "Game",
  "Genres": "Action, Adventure, Comedy, Kids, Fantasy",
  "Duration": "24 min. per ep.",
  "Rating": "PG - Children",
  "Score": "7.341 (scored by 291,570 users)",
  "Ranked": "#21572",
  "Popularity": "#287",
  "Members": "504,076",
  "Favorites": "4,076"
}

通过information_dict["Genres"]，你可以获得"Action, Adventure, Comedy, Kids, Fantasy"的值。