How to join dictionaries created with for loop?

Question

I'm new in Python.

I created a code that should allow me to find the percentage of items that follows a given item in a list.

Given a list:

list1=["a", "b", "a", "c", "a", "b", "c", "d", "e", "a", "b", "d", "e", "a", "c"]

I would like to find, for each, say, "a", in which percentage every item is following. The code returns:

[(33, 'a'), (25, 'b'), (16, 'e'), (16, 'd'), (16, 'c')]
[(30, 'a'), (20, 'e'), (20, 'd'), (20, 'c'), (20, 'b')]
[(25, 'e'), (25, 'd'), (25, 'b'), (25, 'a'), (12, 'c')]
[(33, 'e'), (33, 'd'), (33, 'b'), (33, 'a')]
[]  

The output is right, and it's what i wanted.
But I would also like to sum every key of the different dictionaries, so I can have something like:

[(121, 'a'), (103, 'b'), (94, 'e'), (94, 'd'), (48, 'c')]

I didn't find a way to do that. I know that there are some ways to sum values of every key in different dictionaries, but the problem here is that dictionaries are created inside a for loop, because i need as dictionaries as much target items are (in this case, "a").

I tried to iterate in every dict with

   for key, value in dictio.items():
        dictio[key]=value + dictio.get(key, 0)
        print (dictio)

But the result is a mess, and it's not even far from what I would like to have.

I would like to know from you if it is possible to join multiple dictionaries, without knowing their number (because they are created in a for loop).

And, as I would like to understand better Python logic, I would like not to use external libraries, if it's possibile.

Thank you in advance!

Niccolò

Answer 1

Just a lazy way using Counter

from collections import Counter
d = Counter()


mylist = [[(33, 'a'), (25, 'b'), (16, 'e'), (16, 'd'), (16, 'c')],
            [(30, 'a'), (20, 'e'), (20, 'd'), (20, 'c'), (20, 'b')],
            [(25, 'e'), (25, 'd'), (25, 'b'), (25, 'a'), (12, 'c')],
            [(33, 'e'), (33, 'd'), (33, 'b'), (33, 'a')],
            []]

for i in mylist:
    d.update(dict([(m,n) for n,m in i]))
>>>[(j,i) for i,j in d.items()]
[(121, 'a'), (48, 'c'), (103, 'b'), (94, 'e'), (94, 'd')]

To sort

>>>sorted([(j,i) for i,j in d.items()], key=lambda x:x[1])
[(121, 'a'), (103, 'b'), (48, 'c'), (94, 'd'), (94, 'e')]

To get percent(assuming)

>>>[(j*100/sum(d.values()),i) for i,j in d.items()] # caution==> sum(d.values()) save in a variable, otherwise it will execute in every iteration
[(26, 'a'), (10, 'c'), (22, 'b'), (20, 'e'), (20, 'd')]

Answer 2

The following will sum your keys and calculate the percentages:

import collections, itertools

d = collections.Counter()

mylist = [[(33, 'a'), (25, 'b'), (16, 'e'), (16, 'd'), (16, 'c')],
            [(30, 'a'), (20, 'e'), (20, 'd'), (20, 'c'), (20, 'b')],
            [(25, 'e'), (25, 'd'), (25, 'b'), (25, 'a'), (12, 'c')],
            [(33, 'e'), (33, 'd'), (33, 'b'), (33, 'a')],
            []]

for count, item in itertools.chain.from_iterable(mylist):
    d.update(itertools.repeat(item, count))

print "Usage order:", d.most_common()
lsorted = sorted(d.items())
print "Key order:", lsorted

total = sum(d.values())
print "Percentages:", [(key, (value * 100.0)/total) for key,value in lsorted]

Giving:

Usage order: [('a', 121), ('b', 103), ('e', 94), ('d', 94), ('c', 48)]
Key order: [('a', 121), ('b', 103), ('c', 48), ('d', 94), ('e', 94)]
Percentages: [('a', 26.304347826086957), ('b', 22.391304347826086), ('c', 10.434782608695652), ('d', 20.434782608695652), ('e', 20.434782608695652)]

Answer 3

If you need unique followers for each item in a list you can consider taking only first occurrence of each item then counting items after it, in this case item "e" will have no new items after it (0%). But if the question is number of occurrence of an item after a given element I would proceed as follows:

list1=["a", "b", "a", "c", "a", "b", "c", "d", "e", "a", "b", "d", "e", "a", "c"]
indexlist=[list1.index(item) for item in list(set(list1))]
newlist=[list1[j] for j in sorted(indexlist)]

for item in newlist:
    print '\n',item,'Followers:'
    a=list1[list1.index(item)+1:]
    for follower in a:
        if item!=follower:
            fol=(follower,Counter.get(Counter(a),follower)*100.0/Counter.get(Counter(list1),follower))
    print fol,'round'