是否可以将熊猫系列附加到列表中

Posted 2023-03-12

【中文标题】是否可以将熊猫系列附加到列表中

【英文标题】：Is it possible to append a pandas series to a list

【发布时间】：2019-11-07 16:27:19

【问题描述】：

我最近一直在做一个项目，预测梦幻超级联赛中最优秀的球队。在成功分析不同的特征和参数后，由于以下“TypeError：'Series'对象是可变的，因此它们不能被散列”，我被卡住了

我已经完成了第一部分代码的编写，但收到了一个错误。我在网上搜索但找不到解决方案。一种解决方案说您不能将系列附加到列表中。真的吗？以及相同的可能解决方案是什么。我已经走得太远了，我真的很想把这件事做好。

def my_team (budget = 100, star_player_limit = 3, gk = 2, df = 5, mid = 5, fwd = 3 ): # Pass constraints to function
    team = [ ]                          # List of team to be returned
    star_position = [ ]                 # list containing position of starplayer
    star_player_limit = star_player_limit
    budget = budget
    injured = dataset2.loc[(dataset2.loc[:,"Status"] == 'injured'),:] # Keeping a check of injury status
    positions = "GKP":gk,"DEF":df,"MID":mid,"FWD":fwd       # Dict accounting for no. of postions left to fill
    for ind in Top_points.index:       # Looping through the dataframe of players
        player = Top_points.loc[ind]   # Row of Dataframe one at a time
        star_position.append(player.Position)    # Checking position of star player
        if len(team) < star_player_limit and player not in injured and budget > player.Cost and positions[player.Position] > 0 and player.Position not in star_position:
            team.append(player)
            budget -= player.Cost
            positions[player.Position] -= 1

    return team

我的团队（）

运行代码后出现此错误：TypeError: 'Series' objects are mutable, thus they cannot be hashed。

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-150-a7d781e901c6> in <module>()
----> 1 my_team()

<ipython-input-149-ec17dbd9b9ba> in my_team(budget, star_player_limit, gk, df, mid, fwd)
      9         player = Top_points.loc[ind]
     10         star_position.append(player.Position)
---> 11         if len(team) < star_player_limit and player not in injured and budget > player.Cost and positions[player.Position] > 0 and player.Position not in star_position:
     12             team.append(player)
     13             budget -= player.Cost

~\Anaconda3\lib\site-packages\pandas\core\generic.py in __contains__(self, key)
   1517     def __contains__(self, key):
   1518         """True if the key is in the info axis"""
-> 1519         return key in self._info_axis
   1520 
   1521     @property

~\Anaconda3\lib\site-packages\pandas\core\indexes\base.py in __contains__(self, key)
   2018     @Appender(_index_shared_docs['__contains__'] % _index_doc_kwargs)
   2019     def __contains__(self, key):
-> 2020         hash(key)
   2021         try:
   2022             return key in self._engine

~\Anaconda3\lib\site-packages\pandas\core\generic.py in __hash__(self)
   1487     def __hash__(self):
   1488         raise TypeError('0!r objects are mutable, thus they cannot be'
-> 1489                         ' hashed'.format(self.__class__.__name__))
   1490 
   1491     def __iter__(self):

TypeError: 'Series' objects are mutable, thus they cannot be hashed

【参考方案1】：

Pandas 框架是可变的。因此，它们不能用作字典的键或集合的元素。

查看第一个堆栈跟踪中的第 11 行。我已将其重新格式化为可读。

if (len(team) < star_player_limit and
    player not in injured and
    budget > player.Cost and
    positions[player.Position] > 0 and
    player.Position not in star_position):

我们在这里有一个player not in injured 子句。前者定义为

player = Top_points.loc[ind] 

我想它的类型是Series。

现在我们有了处理in 运算符的__contains__ 方法的第二个堆栈跟踪。其中，我想self 是injured，key 是player。 确实不能hash(player)。

（它不能是第二个in 子句，因为start_position 是一个普通的Python 列表，而__contains__ 堆栈跟踪来自pandas。）

我会从player 中提取姓名或其他ID，并在injured 中搜索；也许我会把injured 变成一组名字。

【讨论】：

感谢您的回复，但我确实研究了“ind”的类型，结果发现它是一个“int”，正如预期的那样测试代码为ind在 Top_points.index 中： print(type(ind)) row = Top_points.loc[ind] print(row.Position) print(type(row)) print('\n') 输出： MID

嘿，我错了，因为我读错了堆栈跟踪！我正在重写我的答案。