pandas dataframe: df.shape (86, 245) However, when I do this: df[0, :] I get the error: *** TypeError: unhashable type How do I fix this? This means that when you try to hash an unhashable I have the foll. Conversely, you can also do. TypeError: unhashable type: 'list' You can resolve this issue by casting list to tuple . Copy link Quote reply jtscs commented Jun 23, 2018. This was before the holidays. unhashable type list python. I know I have errors all through the code. There are no duplicates allowed. Apple disclaims any and all liability for the acts, omissions and conduct of any third parties in connection with or related to your use of the site. The benefits of a set are: very fast membership testing along with being able to use powerful set operations, like union, difference, and intersection. Benjamin Schmitt. What you need is to get just the first item in list, written like so k = list[0]. eg: [[1,2,3,4],[4,5,6,7]] Als ik de ingebouwde set-functie gebruik om duplicaten uit deze lijst te verwijderen, krijg ik de foutmelding. If you want to unpack the sets, maybe import itertools and then you can print(max(list(itertools.chain.from_iterable(dict1.values())),key=dict2.get)) b) list and set both neemt Iterable als parameter. I want to get the count of words based on those users which are named as gold_users. Let us first understand what is hashable and unhasable. Typeerror: unhashable type: 'list' set. What you need is to get just the first item in list, written like so k = list[0].The same for v = list[j + 1:] which should just be v = list[2] for the third element of the list returned from the call to readline.split(" "). The reason you’re getting the unhashable type: 'list' exception is because k = list[0:j] sets k to be a “slice” of the list, which is another, usually shorter, list. Enter. Out of types predefined by Python only the immutable ones, such as strings, numbers, and tuples, are Lists have an unmutable equivalent, called a 'tuple'. The reason you’re getting the unhashable type: 'list' exception is because k = list[0:j] sets k to be a “slice” of the list, which is logically another, often shorter, list. Tuple and List. TypeError: unhashable type: 'set' So, I either need to resolve why I am receiving this and fix it or try another approach, of course. I have a dict (IDMapping) that I'm looking up to in order to get a value, which I am then using to lookup to another dict ... You can create a set holding the different IDs and then compare the size of that set to the total number of quests. unhashable type list set. Sets are a datatype that allows you to store other immutable types in an unsorted way. I have segregated a list of users based on certain conditions. set cheat sheet type set use Used for storing immutable data types uniquely. my_set = set(my_list) or, for Python 3, my_set = {*my_list} to create a set from a list. The python error TypeError: unhashable type: ‘list’ occurs when you add a list to a Python Set or as a dictionary key. Unhashable Type List Set. But I am receiving this error: Runtime 2. python提示：TypeError: unhashable type: 'list' Enter. my_list = list(my_set) or, for Python 3, my_list = [*my_set] to create a list from a set. TypeError: unhashable type: 'dict' The problem is that a list/dict can't be used as the key in a dict, since dict keys need to be immutable and unique. Unhashable Type List Python. The list is an unhashable object. Lists are used to store multiple items in a single variable. Het is geen hash-type. Now, we write a for loop that goes through our list of cakes and finds the ones that have been sold more than five times. Tuples can be used as keys if they contain only strings, numbers, or tuples; if a tuple contains any mutable object either directly or indirectly, it cannot be used as a key. TypeError: unhashable type: 'list' De code die ik gebruik is. TypeError: unhashable type: 'list' usually means that you are trying to use a list as an hash argument.The only types of values not acceptable as keys are values containing lists or dictionaries or other mutable types that are compared by value rather than by object identity, the reason being that the efficient implementation of dictionaries requires a key’s hash value to remain constant. But that's not your error, as your function medications_minimum3() doesn't even use the second argument (something you should fix). Lists are created using square brackets: So your program is trying to find the value of the set itself, and that's not possible. Go back. I have the following dataframe comments. Dictionaries cannot be sliced like a list. 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