[Django]-Error: "dictionary update sequence element #0 has length 1; 2 is required" on Django 1.4

Just ran into this problem. I don’t know if it’s the same thing that hit your code, but for me the root cause was because I forgot to put name= on the last argument of the url (or path in Django 2.0+) function call.

For instance, the following functions throw the error from the question:

url(r'^foo/(?P<bar>[A-Za-z]+)/$', views.FooBar.as_view(), 'foo')
path('foo/{slug:bar}/', views.FooBar, 'foo')

But these actually work:

url(r'^foo/(?P<bar>[A-Za-z]+)/$', views.FooBar.as_view(), name='foo')
path('foo/{slug:bar}/', views.FooBar, name='foo')

The reason why the traceback is unhelpful is because internally, Django wants to parse the given positional argument as the keyword argument kwargs, and since a string is an iterable, an atypical code path begins to unfold. Always use name= on your urls!

I got this error when I was messing around with string and dictionary.

dict1 = {'taras': 'vaskiv', 'iruna': 'vaskiv'}
str1 = str(dict1)
dict(str1)
*** ValueError: dictionary update sequence element #0 has length 1; 2 is required

So what you actually got to do to get dict from string is:

dic2 = eval(str1)
dic2
{'taras': 'vaskiv', 'iruna': 'vaskiv'}

Or in matter of security we can use literal_eval

from ast import literal_eval

Error in your question is raised when you try something like following:

>>> a_dictionary = {}
>>> a_dictionary.update([[1]])
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: dictionary update sequence element #0 has length 1; 2 is required

It’s hard to tell where is the cause in your code unless you show your code, full traceback.

I faced the above mentioned problem when I forgot to pass a keyword argument name to url() function.

Code with error

 url(r"^testing/$", views.testing, "testing")

Code without error

url(r"^testing/$", views.testing, name="testing")

So finally I removed the above error in this way. It might be something different in your case. So check your url patterns in urls.py.

Here is how I encountered this error in Django and fixed it:

Code with error

urlpatterns = [path('home/', views.home, 'home'),]

Correction

urlpatterns = [path('home/', views.home, name='home'),]

Here is the reproduced error.

>>> d = {}
>>> d.update([(1,)])
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: dictionary update sequence element #0 has length 1; 2 is required
>>> 
>>> d
{}
>>> 
>>> d.update([(1, 2)])
>>> d
{1: 2}
>>> 
>>> d.update('hello_some_string')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>  
ValueError: dictionary update sequence element #0 has length 1; 2 is required
>>> 

If you give the sequence and any element length is 1 and required two then we will get this kind of error.
See the above code. First time I gave the sequence with tuple and it’s length 1, then we got the error and dictionary is not updated. second time I gave inside tuple with with two elements, dictionary got updated.

Solution»

Pass a keyword argument name with value as your view name e.g home or home-view etc. to url() function.

Throws Error»

url(r'^home$', 'common.views.view1', 'home'),

Correct»

url(r'^home$', 'common.views.view1', name='home'),

I got the same issue and found that it was due to wrong parameters.
In views.py, I used:

return render(request, 'demo.html',{'items', items})    

But I found the issue: {'items', items}. Changing to {'items': items} resolved the issue.

Another scenario that causes this error:

dict('{"a":1}')  # gives the error

One way to achieve what you want is to use eval

eval('{"a":1}')  # gives {"a":1}

urlpatterns = [ 
    path('',HomePageView.as_view(),name='home'),
]

I had the same problem and it turned out that missing ‘name’ in the urls.py was the cause of problem.

In my case, my get_context_data in one of my views was returning return render(self.request, 'es_connection_error.html', {'error':error}); in a try/catch block instead of returning context

The error should be with the params. Please verify that the params is a dictionary object. If it is just a list/tuple of arguments use only one * (*params) instead of two * (**params). This will explode the list/tuple into the proper amount of arguments.

Or, if the params is coming from some other part of code as a JSON file, please do json.loads(params), because the JSON objects sometimes behave as string and so you need to make it as a JSON using load from string (loads).

super(HStoreDictionary, self).__init__(value, **params)

Hope this helps!

check in your dictionary whether if you have single or double Quotation mark in your key or value!

dict1 = {'hello': 'world', 'programmer's': 'have girlfriend'}

to resolve it you can simply convert your dictionary in to string and then use replace method!

I faced this problem when I was trying to convert json string to dict.

Input: '{\r\n "resource_id": "id",\r\n "resource_type": "resource"\r\n}'

Using dict() to convert the string to dict gave me this error. The correct way would be to use json module.

Example: json.loads(input_str)

I encountered this issue when trying to invoke the update method with a parameter of a wrong type.
The expected dict was:

{'foo': True}

The one that was passed was:

{'foo': "True"}

make sure you check all the parameters you pass are of the expected type.

I too had a similar type of problem . The solution is simple . just dont try to enter NULL or None value in values or u might have to use Something like this
dic.update([(key,value)])

I hit this error calling:

dict(my_data)

I fixed this with:

import json

json.loads(my_data)

You are sending one parameter incorrectly; it should be a dictionary object:

• Wrong: func(a=r)

• Correct: func(a={'x':y})

This behavior will also occurs when using get_or_create method in the following example below one will obviously get that error:

state,_ = Status.objects.get_or_create('Pending')

To resolve this you will add the the respective key and value.

i.e like this state,_ = Status.objects.get_or_create(name='Pending')

Please check your URL path I fixed this issue by changing the URL.

path(‘reset_password_email/’, requestpasswordresetemail, "request-rest-email")

Instead of use

path(‘reset_password_email/’, requestpasswordresetemail, name="request-rest-email")

I was getting this issue so this is how I solved it ( by doing an eval and strip )

aggr = {}
with open("output.txt", "r") as f:
    for line in f.readlines():
        for k, v in dict(eval(line.strip())).items():
            aggr.setdefault(k, 0)
            aggr[k] = aggr[k] + v
    
print(aggr)

print(aggr)

The data I was reading was a text file with a dictionary objects

{'unknown': 298, 'a4a815d631c805ccd10dd2f1548baa57': 9724, '22b5a0ff959ce0b4036716cc0c2df68b': 1341, 'd186fde596dffaab46260765c7fcba61': 2052}
{'unknown': 323, '49e5357782510659cf083356f7d2a1ab': 9826, 'a4a815d631c805ccd10dd2f1548baa57': 9812, 'b615930608b8dcd217de7904d4463efb': 2409}

from ast import literal_eval

c = "{'taras': 'vaskiv', 'iruna': 'vaskiv'}"
c= dict(literal_eval(c))

I got this error when I was messing around with string and dictionary.

dict1 = {'taras': 'vaskiv', 'iruna': 'vaskiv'}
str1 = str(dict1)
dict(str1)

or to be specific

query_result = req_db_session.query(app_config_model.APPCONFIG).with_entities(app_config_model.APPCONFIG.value).filter(app_config_model.APPCONFIG.name == 'test_me').first()
print(query_result, type(query_result))
print('--------------------------')
print(type(query_result))
# prints <class 'sqlalchemy.util._collections.result'>
print(dict(query_result))
# throws error

*** ValueError: dictionary update sequence element #0 has length 1; 2 is required
So what you actually worked was:

query_result = req_db_session.query(app_config_model.APPCONFIG).with_entities(app_config_model.APPCONFIG.value).filter(app_config_model.APPCONFIG.name == 'test_me').first()
print(query_result, type(query_result))
print('--------------------------')
print(type(query_result.value))
# prints <type 'unicode'>
print(json.loads(query_result.value))
# prints parsed dictionary