How to add values in keys in a dictionary inside a loop?

Question

I have the following list:

x=['a','3','4','b','1','2','c','4','5']

How can i make the following dictionary:

b = {'a':[3,4],'b':[1,2],'c':[4,5]}

I tried the following:

Category = defaultdict(int)
for i in a:
    if Varius.is_number(i)==False:
        Category[i]=[]
        keys.append(i)
    else:
        Category[keys(i)] = i

The keys are created but after i have problem to insert the values.(is_number is a function which check if the value of the list is number or string).First day away of MATLAB.First day in Python..

Answer 1

Here an example that actually uses the feature that defaultdict provides over the normal dict:

from collections import defaultdict

x=['a','3','4','b','1','2','c','4','5']

key='<unknown>' # needed if the first value of x is a number
category = defaultdict(list)  # defaultdict with list
for i in x:
    if i.isalpha():
        key = i;
    else:
        category[key].append(i) # no need to initialize with an empty list

print category

Also: you should use lower case names for class instances. Uppercase names are usually reserved for classes. Read pep8 for a style guide.

Answer 2

If the key/value entries in your list x are evenly spaced, here's an alternative way to build the dictionary. It uses a few of Python's built in features and functions:

>>> keys = x[::3]
>>> values = [map(int, pair) for pair in zip(x[1::3], x[2::3])]
>>> dict(zip(keys, values))
{'a': [3, 4], 'b': [1, 2], 'c': [4, 5]}

To explain what's being used here:

• list slicing to create new lists from x: x[start:stop:step]
• here zip takes two lists and makes a list of tuples containing the n-th elements of each list
• map(int, pair) turns a tuple of digit strings into a list of integers
• values is constructed with list comprehension - the map function is applied to each pair
• dict turns a list of pairs into dictionary keys/values

Answer 3

You can use itertools.groupby :

>>> from itertools import groupby
>>> l=[list(g) for k,g in groupby(x,lambda x : x.isalpha())]
>>> p=[l[i:i+2] for i in range(0,len(l),2)]
>>> {i[0]:map(int,j) for i,j in p}
{'a': [3, 4], 'c': [4, 5], 'b': [1, 2]}

Answer 4

Assuming you have a key on the first positions of your list (x) and that after a letter, there will be a number:

from collections import defaultdict

x=['a','3','4','b','1','2','c','4','5']

key = x[0]
Category = defaultdict(int)
for i in x:
    if i.isalpha():
        key = i;
    else:
        Category[key].append(i)

print Category

Answer 5

Using itertools.groupby and some iterator stuff:

>>> from itertools import groupby  
>>> it = (next(g) if k else map(int, g) for k, g in groupby(x, str.isalpha))
>>> {k: next(it) for k in it}
{'a': [3, 4], 'c': [4, 5], 'b': [1, 2]}

Here the first iterator will yield something like:

>>> [next(g) if k else map(int, g) for k, g in groupby(x, str.isalpha)]
['a', [3, 4], 'b', [1, 2], 'c', [4, 5]]

Now as this iterator is going to yield key-value alternately, we can loop over this iterator and get the next item from it(i.e the value) using next()

>>> it = (next(g) if k else map(int, g) for k, g in groupby(x, str.isalpha))
>>> for k in it:                                                            
    print k,'-->' ,next(it)
...     
a --> [3, 4]
b --> [1, 2]
c --> [4, 5]

---

There's another way to consume this iterator that is using zip, but it's a little hard to understand IMO:

>>> it = (next(g) if k else map(int, g) for k, g in groupby(x, str.isalpha))
>>> dict(zip(*[it]*2))
{'a': [3, 4], 'c': [4, 5], 'b': [1, 2]}

Answer 6

Simple solution using just dict and list:

>>> category = {}
>>> for i in x:
...     if i.isalpha():
...         items = category[i] = []
...     elif category:
...         items.append(i)
... 
>>> print(category)
{'c': ['4', '5'], 'a': ['3', '4'], 'b': ['1', '2']}