python：不可散列的类型错误
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python: unhashable type error
提问by l--''''''---------''''''''''''
Traceback (most recent call last):
File "<pyshell#80>", line 1, in <module>
do_work()
File "C:\pythonwork\readthefile080410.py", line 14, in do_work
populate_frequency5(e,data)
File "C:\pythonwork\readthefile080410.py", line 157, in populate_frequency5
data=medications_minimum3(data,[drug.upper()],1)
File "C:\pythonwork\readthefile080410.py", line 120, in medications_minimum3
counter[row[11]]+=1
TypeError: unhashable type: 'list'
I am getting the above error on this line:
我在这一行收到上述错误：
data=medications_minimum3(data,[drug.upper()],1)
(I have also tried drug.upper() without brackets)
（我也试过没有括号的 drug.upper()）
Here is a preview of this function:
下面是这个函数的预览：
def medications_minimum3(c,drug_input,sample_cutoff): #return sample cut off for # medications/physician
d=[]
counter=collections.defaultdict(int)
for row in c:
counter[row[11]]+=1
for row in c:
if counter[row[11]]>=sample_cutoff:
d.append(row)
write_file(d,'/pythonwork/medications_minimum3.csv')
return d
Does anyone know what I am doing wrong here?
有谁知道我在这里做错了什么？
I know that what must be wrong is the way I am calling this function, because I call this function from a different location and it works fine:
我知道一定是我调用这个函数的方式有问题，因为我从不同的位置调用这个函数并且它工作正常：
d=medications_minimum3(c,drug_input,50)
Thank you very much for your help!
非常感谢您的帮助！
采纳答案by John La Rooy
I don't think converting to a tuple is the right answer. You need go and look at where you are calling the function and make sure that cis a list of list of strings, or whatever you designed this function to work with
我不认为转换为元组是正确的答案。您需要查看调用函数的位置，并确保这c是一个字符串列表，或者您设计此函数使用的任何内容
For example you might get this error if you passed [c]to the function instead of c
例如，如果您传递[c]给函数而不是c
回答by John Kugelman
counter[row[11]]+=1
You don't show what datais, but apparently when you loop through its rows, row[11]is turning out to be a list. Lists are mutable objects which means they cannot be used as dictionary keys. Trying to use row[11]as a key causes the defaultdictto complain that it is a mutable, i.e. unhashable, object.
你没有显示是什么data，但显然当你遍历它的行时，结果row[11]是一个list. 列表是可变对象，这意味着它们不能用作字典键。尝试row[11]用作键会导致defaultdict抱怨它是可变的，即不可散列的对象。
The easiest fix is to change row[11]from a listto a tuple. Either by doing
最简单的解决方法是row[11]从 a更改list为 a tuple。要么通过做
counter[tuple(row[11])] += 1
or by fixing it in the caller before datais passed to medications_minimum3. A tuple simply an immutable list, so it behaves exactly like a list does except you cannot change it once it is created.
或者在data传递给medications_minimum3. 元组只是一个不可变的列表，因此它的行为与列表完全一样，只是一旦创建就无法更改。
回答by S.Lott
File "C:\pythonwork\readthefile080410.py", line 120, in medications_minimum3
counter[row[11]]+=1
TypeError: unhashable type: 'list'
row[11]is unhashable. It's a list. That is precisely (and only) what the error message means. You might not like it, but that is the error message.
row[11]是不可哈希的。这是一个清单。这正是（并且仅）错误消息的含义。您可能不喜欢它，但这是错误消息。
Do this
做这个
counter[tuple(row[11])]+=1
Also, simplify.
另外，简化。
d= [ row for row in c if counter[tuple(row[11])]>=sample_cutoff ]
回答by chryss
As Jim Garrison said in the comment, no obvious reason why you'd make a one-element list out of drug.upper()(which implies drug is a string).
正如吉姆加里森在评论中所说的那样，没有明显的理由让你制作一个单元素列表drug.upper()（这意味着药物是一个字符串）。
But that's not your error, as your function medications_minimum3()doesn't even use the second argument (something you should fix).
但这不是您的错误，因为您的函数medications_minimum3()甚至不使用第二个参数（您应该解决的问题）。
TypeError: unhashable type: 'list'usually means that you are trying to use a list as a hash argument (like for accessing a dictionary). I'd look for the error in counter[row[11]]+=1-- are you sure that row[11]is of the right type? Sounds to me it might be a list.
TypeError: unhashable type: 'list'通常意味着您正在尝试使用列表作为哈希参数（例如访问字典）。我会寻找错误counter[row[11]]+=1- 您确定这row[11]是正确的类型吗？在我看来，这可能是一个列表。
