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pythondictionarynested
Chinmaya Barik
Chinmaya Barik

Reputation: 1

Python dictionary conversion

I have a Python dictionary:

adict = {
    'col1': [
        {'id': 1, 'tag': '#one#two'},
        {'id': 2, 'tag': '#two#'},
        {'id': 1, 'tag': '#one#three#'}
    ]
}

I want the result as follows:

Id tag
1  one,two,three
2  two

Could someone please tell me how to do this?

Upvotes: 0

Views: 89

Answers (5)

shalini M
shalini M

Reputation: 1

dic=[{'col1':[{'id':1,'tag':'#one#two'},{'id':2,'tag':'#two#'},{'id':1,'tag':'#one#three#'}]}]

row=[]
for key in dic:
    data=key['col1']
    for rows in data:
        row.append(rows)
df=pd.DataFrame(row)
print(df)

o

Upvotes: 0

Chinmaya Barik
Chinmaya Barik

Reputation: 1

I tried as below

import pandas as pd
a = {'col1':[{'id':1, 'tag':'#one#two'},{'id':2, 'tag':'#two#'},{'id':1, 'tag':'#one#three#'}]}

df = pd.DataFrame(a)
df[["col1", "col2"]] = pd.DataFrame(df.col1.values.tolist(), index = df.index)
df['col1'] = df.col1.str.replace('#', ',')
df = df.groupby(["col2"])["col1"].apply(lambda x : ",".join(set(''.join(list(x)).strip(",").split(","))))

O/P:
col2
1    one,two,three
2    two

Upvotes: 0

jezrael
jezrael

Reputation: 862601

If order of tags is important first remove trailing # and split by #, then per groups remove duplicates and join:

df = pd.DataFrame(d['col1'])
df['tag'] = df['tag'].str.strip('#').str.split('#')
f = lambda x: ','.join(dict.fromkeys([z for y in x for z in y]).keys())
df = df.groupby('id')['tag'].apply(f).reset_index()
print (df)
   id            tag
0   1  one,two,three
1   2            two

If order of tags is not important for remove duplicates use sets:

df = pd.DataFrame(d['col1'])
df['tag'] = df['tag'].str.strip('#').str.split('#')
f = lambda x: ','.join(set([z for y in x for z in y]))
df = df.groupby('id')['tag'].apply(f).reset_index()
print (df)
   id            tag
0   1  three,one,two
1   2            two

Upvotes: 0

Bharath_Raja
Bharath_Raja

Reputation: 642

Here is a simple code 

import pandas as pd

d = {'col1':[{'id':1,'tag':'#one#two'},{'id':2,'tag':'#two#'},{'id':1,'tag':'#one#three#'}]}


df = pd.DataFrame(d)

df['Id'] = df.col1.apply(lambda x: x['id'])

df['tag'] = df.col1.apply(lambda x: ''.join(list(','.join(x['tag'].split('#')))[1:]))

df.drop(columns = 'col1', inplace = True)

Output:
Id Tag
1  one, two
2  two
1  one, three

Upvotes: 1

Rajith Thennakoon
Rajith Thennakoon

Reputation: 4130

Try this

import pandas as pd
d={'col1':[{'id':1,'tag':'#one#two'},{'id':2,'tag':'#two#'},{'id':1,'tag':'#one#three#'}]}

df = pd.DataFrame()
for i in d:
    for k in d[i]:
        t = pd.DataFrame.from_dict(k, orient='index').T
        t["tag"] = t["tag"].str.replace("#",",")
        df = pd.concat([df,t])

tf = df.groupby(["id"])["tag"].apply(lambda x : ",".join(set(''.join(list(x)).strip(",").split(","))))

Upvotes: 2

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