Unhashable type : 'list' error in Python

Question

I have already checked some question-answers related to Unhashable type : 'list' in stackoverflow, but none of them helped me. Here is my partial code:

keyvalue = {};
input_list_new = input_list;

for file in os.listdir('Directory/'):
    pathname = os.path.join('Directory/', file)
    dict_table = []; # creates blank 2d array
    with open(pathname) as dict_file:
        for line in dict_file:
            dict_table.append(line.strip().split("\t"))
    dict_list = [];
    for i in list(range(0, len(dict_table))):
        dict_list.append(dict_table[i][0])
    matched = list(set(dict_list) & set(input_list)) 
    for i in list(range(0, len(matched))):
        temp = [dict_table[0][0]] + dict_table[(dict_list.index(matched[i]))]

        input_list_new.insert((input_list_new.index(matched[i])), temp)

        dict = {matched[i] : temp}
        keyvalue.update(dict)

where dict_table is a list of lists, dict_list is just a list & keyvalue is a python dictionary. The entire code runs well when ever I comment the line input_list_new.insert((input_list_new.index(matched[i])), temp), but without that line being commented, it shows Unhashable type : 'list' error.

can you show the error exactly ? in which line is the error coming ?> — Kaustav Datta
– Kaustav Datta, Commented Jan 18, 2016 at 6:41
Please paste the full Traceback of the error into your question; put it in a code block to preserve formatting. BTW, it's a bad idea to use dict as a variable name as that shadows the built-in dict type, which can lead to mysterious bugs if you attempt to create a dictionary using dict(). That's not a problem for the partial code you've posted, but it may affect subsequent code in that function / module. — PM 2Ring
– PM 2Ring, Commented Jan 18, 2016 at 6:54
Use pieces from the problem line to see which of those is causing the problem. That is, put matched[i] on its on line, put input...() on its own, etc — hpaulj
– hpaulj, Commented Jan 18, 2016 at 7:00
Some place or other you are using a list as a key to a dictionary or set, e.g. ` set([[1,2],3])` produces this error. So does {}[[1,2]]. — hpaulj
– hpaulj, Commented Jan 18, 2016 at 7:02

Martin Konecny · Accepted Answer · 2016-01-18 06:41:31Z

1

The error message does not correspond to the line

input_list_new.insert((input_list_new.index(matched[i])), temp)

being commented out. If anything, the culprit is the line

dict = {matched[i] : temp}

The problem should be that you are trying to use a list as your dictionary key. Here is the problem in a simple reproducible way:

{[]: 5}  # throws TypeError: unhashable type: 'list'

The reason it fails, is because a dictionary is also be called a hashmap (in other languages). The key must be hashable, and in Python trying to use a list as the key throws your error becauses lists are not hashable (see "hashable" entry here).

Avoid using a list as your dictionary key will fix the problem.

edited Jan 18, 2016 at 6:41

answered Jan 18, 2016 at 6:37

Martin Konecny

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3 Comments

Koushik Khan Over a year ago

Hey Martin, thanks for your suggestion. But I'm not using a list as a key of the python dictionary, matched[i] is just a string.

Martin Konecny Over a year ago

is it possible that a list is somehow inserted into matched? Can you do print(matched[i]) each time before dict = {matched[i] : temp} and see the last value before it crashes? Also it would be very useful to see what line number the error happens on.

mike3996 Over a year ago

A dictionary is not necessarily a hashmap so the terms aren't synonymous.

Kaustav Datta · Accepted Answer · 2016-01-18 07:46:21Z

1

Here's my guess .... You mentioned that the error is with the line matched = list(set(dict_list) & set(input_list)) .. That is probably because in either input_list or dict_list, you have a list within a list .... the items in a set need to hashable, and hence immutable .. for example you cannot do set([1,5,6,[1,3]]) ... that will give the unhashable type list error ... but you can do set([1,5,6,(1,3)]) because a tuple is immutable and hashable

answered Jan 18, 2016 at 7:46

Kaustav Datta

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