是否可以将熊猫系列添加到列表中
是否可以将熊猫系列添加到列表中
提问于 2019-06-26 03:19:14
我最近一直在做一个项目，它预测了梦幻超级联赛中最好的球队。在成功地分析了不同的特征和参数之后，我遇到了以下问题："TypeError：'Series‘对象是可变的，因此它们不能被散列“。
我已经完成了代码的第一部分，但收到一个错误。我在网上找了找，但找不到解决办法。其中一个解决方案说，您不能将序列追加到列表中。这是真的吗？对于同样的问题，可能的解决方案是什么。我已经走得太远了，我真的想把这件事做好。
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 teammy_team()
在运行代码之后，我得到了这个错误：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回答 1
Stack Overflow用户
发布于 2019-06-26 03:28:12
熊猫框架是可变的。因此，它们不能用作dict的键，也不能用作set的元素。
看看第一个堆栈跟踪中的第11行。我已经将其重新格式化，使其具有可读性。
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):我们这里有一个player not in injured子句。前者被定义为
player = Top_points.loc[ind] 我认为它的类型是Series。
现在我们有了处理in操作符的__contains__方法的第二个堆栈跟踪。在其中，我假设self是injured，key是player。事实上，它不能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中提取名称或其他ID，并在injured中搜索它；也许我会在此过程中将injured转换为一组名称。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/56760712
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