Populating a pandas dataframe from an odd dictionary - python-3.x

Question

I have a dictionary as follows:
{'header_1': ['body_1', 'body_3', 'body_2'],
'header_2': ['body_6', 'body_4', 'body_5'],
'header_4': ['body_7', 'body_8'],
'header_3': ['body_9'],
'header_9': ['body_10'],
'header_10': []}
I would like to come up with a dataframe like this:
+----+----------+--------+
| ID | header | body |
+----+----------+--------+
| 1 | header_1 | body_1 |
+----+----------+--------+
| 2 | header_1 | body_3 |
+----+----------+--------+
| 3 | header_1 | body_2 |
+----+----------+--------+
| 4 | header_2 | body_6 |
+----+----------+--------+
| 5 | header_2 | body_4 |
+----+----------+--------+
| 6 | header_2 | body_5 |
+----+----------+--------+
| 7 | header_4 | body_7 |
+----+----------+--------+
Where blank items (such as for the key header_10 in the dict above) would receive a value of None. I have tried a number of varieties for df.loc such as:
for header_name, body_list in all_unique.items():
for body_name in body_list:
metadata.loc[metadata.index[-1]] = [header_name, body_name]
To no avail. Surely there must be a quick way in Pandas to append rows and autoincrement the index? Something similar to the SQL INSERT INTO statement only using pythonic code?

Answer 1

Use dict comprehension for add Nones for empty lists and then flatten for list of tuples:
d = {'header_1': ['body_1', 'body_3', 'body_2'],
'header_2': ['body_6', 'body_4', 'body_5'],
'header_4': ['body_7', 'body_8'],
'header_3': ['body_9'],
'header_9': ['body_10'],
'header_10': []}
d = {k: v if bool(v) else [None] for k, v in d.items()}
data = [(k, y) for k, v in d.items() for y in v]
df = pd.DataFrame(data, columns= ['a','b'])
print (df)
a b
0 header_1 body_1
1 header_1 body_3
2 header_1 body_2
3 header_2 body_6
4 header_2 body_4
5 header_2 body_5
6 header_4 body_7
7 header_4 body_8
8 header_3 body_9
9 header_9 body_10
10 header_10 None
Another solution:
data = []
for k, v in d.items():
if bool(v):
for y in v:
data.append((k, y))
else:
data.append((k, None))
df = pd.DataFrame(data, columns= ['a','b'])
print (df)
a b
0 header_1 body_1
1 header_1 body_3
2 header_1 body_2
3 header_2 body_6
4 header_2 body_4
5 header_2 body_5
6 header_4 body_7
7 header_4 body_8
8 header_3 body_9
9 header_9 body_10
10 header_10 None

Answer 2

If the dataset is too big, this solution would be slow, but it should still work.
for key in data.keys():
vals= data[key]
# Create temp df with data from a single key
t_df = pd.DataFrame({'header':[key]*len(vals),'body':vals})
# Append it to your full dataframe.
df = df.append(t_df)

Answer 3

This is another unnesting problem again
Borrow Jez's setting up for your d
d = {k: v if bool(v) else [None] for k, v in d.items()}
1st convert your dict into dataframe
df=pd.Series(d).reset_index()
df.columns
Out[204]: Index(['index', 0], dtype='object')
Then using this function in here
yourdf=unnesting(df,[0])
yourdf
Out[208]:
0 index
0 body_1 header_1
0 body_3 header_1
0 body_2 header_1
1 body_6 header_2
1 body_4 header_2
1 body_5 header_2
2 body_7 header_4
2 body_8 header_4
3 body_9 header_3
4 body_10 header_9
5 None header_10
def unnesting(df, explode):
idx=df.index.repeat(df[explode[0]].str.len())
df1=pd.concat([pd.DataFrame({x:np.concatenate(df[x].values)} )for x in explode],axis=1)
df1.index=idx
return df1.join(df.drop(explode,1),how='left')