Flattening a list with a nested dictionary - list

Question

I have a list that includes a nested dictionary:
l =[('M', 1, 2, {'t': (2, 1.0)}), ('L', 2, 4, {'b': (4, 0.25), 'fi': (4, 0.75)}),
('J', 4, 5, {'a': (5, 0.2), 'w': (5, 0.2), 'wh': (5, 0.4), 'en': (5, 0.2)}),
('T', 4, 6, {'sl': (6, 0.5), 'f': (6, 0.1), 'pz': (6, 0.17), 'al': (6, 0.1)}),
('P', 5, 5, {'tr': (5, 0.2), 'in': (5, 0.2), 'fa': (5, 0.2), 'if': (5, 0.2)})]
I would like to flat out this list, in order to have a plain list like this:
[('M', 1, 2, 't', 2, 1.0, 'L', 2, 4, 'b', 4, 0.25, 'fi', 4, 0.7),
('J', 4, 5, 'a', 5, 0.2, 'w', 5, 0.2, 'wh', 5, 0.4, 'en', 5, 0.2)]
I tried some flatten functions, but I got confused with how to flatten the dictionary within the list. I am new to python, as you can tell. Could anyone help me with getting around this.

Answer 1

I could only think of below brute force method...
your given iterator:
d =[('M', 1, 2, {'t': (2, 1.0)}), ('L', 2, 4, {'b': (4, 0.25), 'fi': (4, 0.75)}),
('J', 4, 5, {'a': (5, 0.2), 'w': (5, 0.2), 'wh': (5, 0.4), 'en': (5, 0.2)}),
('T', 4, 6, {'sl': (6, 0.5), 'f': (6, 0.1), 'pz': (6, 0.17), 'al': (6, 0.1)}),
('P', 5, 5, {'tr': (5, 0.2), 'in': (5, 0.2), 'fa': (5, 0.2), 'if': (5, 0.2)})]
My solution:
l = [ ]
for item in d:
for i in item:
if type(i) is dict:
for j in i.items():
for p in j:
l.append(p) if not isinstance(p,tuple) else [l.extend(k for k in p)]
else:
l.append(i)
print l
output:
['M', 1, 2, 't', 2, 1.0, 'L', 2, 4, 'fi', 4, 0.75, 'b', 4, 0.25, 'J', 4, 5, 'a', 5, 0.20000000000000001, 'en', 5, 0.20000000000000001, 'w', 5, 0.20000000000000001, 'wh', 5, 0.40000000000000002, 'T', 4, 6, 'pz', 6, 0.17000000000000001, 'f', 6, 0.10000000000000001, 'al', 6, 0.10000000000000001, 'sl', 6, 0.5, 'P', 5, 5, 'fa', 5, 0.20000000000000001, 'if', 5, 0.20000000000000001, 'tr', 5, 0.20000000000000001, 'in', 5, 0.20000000000000001]
Hope this helps :)