CODEWITHSUNDEEP
pythonpython-3.xdictionary
SujitS
SujitS

Reputation: 11423

pandas dataframe to dictionary value

I have three column datagram intended to convert in dictionary in the format given:

datagram:

user_id item_id ratings
3         2       3
3         3       4
1         3       1
2         1       4

No of user = 3

NO of item = 3

ratings = 0 to 5

dictionary=

{user_id1:[rating_for_item1, rating_for_item2, rating_for_item3],
 user_id2:[.same as previous.],
 user_id3:[..same as prev..]}

eg,

{1:[0,0,1], 2:[4,0,0], 3:[0,3,4]}

SO, far I could do is to output like:

{1:{3:1}, 2:{1:4}, 3:{2:3, 3:4}} #{user_id:{item_id:rating}.....}

The code for above output is like:

import pandas as pd
data = {}
cols = ['user_id', 'item_id', 'ratings']
pf = pd.read_csv('filename', sep='\t', names= cols)
for user, item, rate in pf.values: data.setdefault(user,{})[item] = rate
print data

What is missing in my code, or am I completely in wrong path. Please help.

Upvotes: 3

Views: 5044

Answers (2)

DSM
DSM

Reputation: 353059

I would pivot and then build the dict. For example:

pdf = df.pivot("user_id", "item_id").fillna(0)
d = {k: v.tolist() for k,v in pdf.iterrows()}

produces

>>> d
{1: [0.0, 0.0, 1.0], 2: [4.0, 0.0, 0.0], 3: [0.0, 3.0, 4.0]}

First, the frame:

>>> df
   user_id  item_id  ratings
0        3        2        3
1        3        3        4
2        1        3        1
3        2        1        4

Pivot:

>>> pdf = df.pivot("user_id", "item_id")
>>> pdf
         ratings        
item_id        1   2   3
user_id                 
1            NaN NaN   1
2              4 NaN NaN
3            NaN   3   4

Replace the NaNs by 0:

>>> pdf = df.pivot("user_id", "item_id").fillna(0)
>>> pdf
         ratings      
item_id        1  2  3
user_id               
1              0  0  1
2              4  0  0
3              0  3  4

And build a row-wise dictionary using a dictionary comprehension:

>>> d = {k: v.tolist() for k,v in pdf.iterrows()}
>>> d
{1: [0.0, 0.0, 1.0], 2: [4.0, 0.0, 0.0], 3: [0.0, 3.0, 4.0]}

There are lots of ways to do this last step, including dict(zip(pdf.index, pdf.values.tolist())), but many of them don't generalize as easily when you want to tweak it a little.

Upvotes: 2

DJG
DJG

Reputation: 6543

How about processing what you have into what you want like so:

from collections import defaultdict

processed_data = defaultdict(list)
for k,v in data.items():
    for idx in range(1, 4): # Make sure we check each item
                            # from (1 to 3 inclusive) for each iteration
                            # of the dictionary
        val = v.get(idx, 0)
        processed_data[k].append(val)

processed_data yields:

defaultdict(<type 'list'>, {1: [0, 0, 1], 2: [4, 0, 0], 3: [0, 3, 4]})

If you would like to convert this back to a regular dictionary (from a defaultdict,) then do the following:

dict(processed_data)

which yields

{1: [0, 0, 1], 2: [4, 0, 0], 3: [0, 3, 4]}

Upvotes: 1

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