Obtaining length of list as a value in dictionary in Python 2.7

Question

I have two lists and dictionary as follows:

>>> var1=[1,2,3,4]
>>> var2=[5,6,7]
>>> dict={1:var1,2:var2}

I want to find the size of the mutable element from my dictionary i.e. the length of the value for a key. After looking up the help('dict'), I could only find the function to return number of keys i.e. dict.__len__(). I tried the Java method(hoping that it could work) i.e. len(dict.items()[0]) but it evaluated to 2.

I intend to find this:
Length of value for first key: 4
Length of value for second key: 3

when the lists are a part of the dictionary and not as individual lists in case their length is len(list).

Any suggestions will be of great help.

Answer 1

dict.items() is a list containing all key/value-tuples of the dictionary, e.g.:

[(1, [1,2,3,4]), (2, [5,6,7])]

So if you write len(dict.items()[0]), then you ask for the length of the first tuple of that items-list. Since the tuples of dictionaries are always 2-tuples (pairs), you get the length 2. If you want the length of a value for a given key, then write:

len(dict[key])

Aso: Try not to use the names of standard types (like str, dict, set etc.) as variable names. Python does not complain, but it hides the type names and may result in unexpected behaviour.

Answer 2

You can do this using a dict comprehension, for example:

>>> var1 = [1,2,3,4]
>>> var2 = [5,6,7]
>>> d = {1:var1, 2:var2}
>>> lengths = {key:len(value) for key,value in d.iteritems()}
>>> lengths
{1: 4, 2: 3}

Your "Java" method would also nearly have worked, by the way (but is rather unpythonic). You just used the wrong index:

>>> d.items()
[(1, [1, 2, 3, 4]), (2, [5, 6, 7])]
>>> d.items()[0]
(1, [1, 2, 3, 4])
>>> len(d.items()[0][1])
4