python pandas dataframe to dictionary

Question

See the docs for to_dict. You can use it like this:

df.set_index('id').to_dict()

And if you have only one column, to avoid the column name is also a level in the dict (actually, in this case you use the Series.to_dict()):

df.set_index('id')['value'].to_dict()

mydict = dict(zip(df.id, df.value))

If you want a simple way to preserve duplicates, you could use groupby:

>>> ptest = pd.DataFrame([['a',1],['a',2],['b',3]], columns=['id', 'value']) 
>>> ptest
  id  value
0  a      1
1  a      2
2  b      3
>>> {k: g["value"].tolist() for k,g in ptest.groupby("id")}
{'a': [1, 2], 'b': [3]}

The answers by joris in this thread and by punchagan in the duplicated thread are very elegant, however they will not give correct results if the column used for the keys contains any duplicated value.

For example:

>>> ptest = p.DataFrame([['a',1],['a',2],['b',3]], columns=['id', 'value']) 
>>> ptest
  id  value
0  a      1
1  a      2
2  b      3

# note that in both cases the association a->1 is lost:
>>> ptest.set_index('id')['value'].to_dict()
{'a': 2, 'b': 3}
>>> dict(zip(ptest.id, ptest.value))
{'a': 2, 'b': 3}

If you have duplicated entries and do not want to lose them, you can use this ugly but working code:

>>> mydict = {}
>>> for x in range(len(ptest)):
...     currentid = ptest.iloc[x,0]
...     currentvalue = ptest.iloc[x,1]
...     mydict.setdefault(currentid, [])
...     mydict[currentid].append(currentvalue)
>>> mydict
{'a': [1, 2], 'b': [3]}

Simplest solution:

df.set_index('id').T.to_dict('records')

Example:

df= pd.DataFrame([['a',1],['a',2],['b',3]], columns=['id','value'])
df.set_index('id').T.to_dict('records')

If you have multiple values, like val1, val2, val3,etc and u want them as lists, then use the below code:

df.set_index('id').T.to_dict('list')

You can use 'dict comprehension'

my_dict = {row[0]: row[1] for row in df.values}