How convert list to nested dictionary?

[macm]

Hi Folks

How convert list to nested dictionary?

l ['k1', 'k2', 'k3', 'k4', 'k5']
result

{'k1': {'k2': {'k3': {'k4': {'k5': {}}}}}}

Regards

macm

 

[Chris Rebert]

Hi Folks

How convert list to nested dictionary?

l ['k1', 'k2', 'k3', 'k4', 'k5']
result

{'k1': {'k2': {'k3': {'k4': {'k5': {}}}}}}

We don't do homework.
Hint: Iterate through the list in reverse order, building up your
result. Using reduce() is one option.

Cheers,
Chris

 

[macm]

Hi Chris

Thanks for your hint.

I am reading this

http://www.amk.ca/python/writing/functional

Do you have good links or books to me learn "Functional Programming"?

but I am not asking "...because is easy but because is hard."

Show me, please! if you can.

Thanks is advance.

Best regards

macm

Hi Folks

How convert list to nested dictionary?

l ['k1', 'k2', 'k3', 'k4', 'k5']
result

{'k1': {'k2': {'k3': {'k4': {'k5': {}}}}}}

We don't do homework.
Hint: Iterate through the list in reverse order, building up your
result. Using reduce() is one option.

Cheers,
Chris

 

[Chris Rebert]

Hi Folks

How convert list to nested dictionary?

l
['k1', 'k2', 'k3', 'k4', 'k5']
result
{'k1': {'k2': {'k3': {'k4': {'k5': {}}}}}}

We don't do homework.
Hint: Iterate through the list in reverse order, building up your
result. Using reduce() is one option.

Thanks for your hint.

Do you have good links or books to me learn "Functional Programming"?

Relevant to the particular problem you posed:
http://en.wikipedia.org/wiki/Fold_(higher-order_function)

Show me, please! if you can.

I will give you this further hint:

def reducer(accumulator, elem):
# if accumulator = {'k5': {} }
# and elem = 'k4'
# then we want to return {'k4': {'k5': {} } }
now_implement_me()

l = ['k1', 'k2', 'k3', 'k4', 'k5']
result = reduce(reducer, reversed(l), {})

Note that:
reduce(reducer, reversed(l), {})
Is basically equivalent to:
reducer( reducer( reducer( reducer( reducer({}, 'k5'), 'k4'), 'k3'),
'k2'), 'k1')

Cheers,
Chris

 

[Arnaud Delobelle]

Hi Folks

How convert list to nested dictionary?

l ['k1', 'k2', 'k3', 'k4', 'k5']
result

{'k1': {'k2': {'k3': {'k4': {'k5': {}}}}}}

Regards

macm

reduce(lambda x,y: {y:x}, reversed(l), {})

 

[Boris Borcic]

Hi Folks

How convert list to nested dictionary?

l ['k1', 'k2', 'k3', 'k4', 'k5']
result

{'k1': {'k2': {'k3': {'k4': {'k5': {}}}}}}

Regards

macm

reduce(lambda x,y: {y:x}, reversed(l), {})

d={}
while L : d={L.pop():d}

 

[Peter Otten]

Hi Folks

How convert list to nested dictionary?

l
['k1', 'k2', 'k3', 'k4', 'k5']
result
{'k1': {'k2': {'k3': {'k4': {'k5': {}}}}}}

Regards

macm

reduce(lambda x,y: {y:x}, reversed(l), {})

d={}
while L : d={L.pop():d}

Iterating over the keys in normal order:
.... outer[key] = inner = {}
.... outer = inner
....{'a': {'b': {'c': {'d': {'e': {}}}}}}

The "functional" variant:
{'a': {'b': {'c': {'d': {'e': {}}}}}}

In a single expression if you are willing to pay the price:
"abcde", ({},)*2)[0]
{'a': {'b': {'c': {'d': {'e': {}}}}}}

Peter

 

[macm]

Hi Folks

thanks a lot all. All solutions work fine.

while I am doing my home work.
Reading "Learning Python" and much more.

Let me ask again to close my doubts:

l = ['k1', 'k2', 'k3', 'k4', 'k5']
d = reduce(lambda x,y: {y:x}, reversed(l), {'/':[1,2,3]})
d {'k1': {'k2': {'k3': {'k4': {'k5': {'/': [1, 2, 3]}}}}}}
d['k1']['k2']['k3']['k4']['k5'] {'/': [1, 2, 3]}
d['k1']['k2']['k3']['k4']['k5']['/'] [1, 2, 3]

now I want generate the "index" to access the element.

==> d['k1']['k2']['k3']['k4']['k5']['/'] from l

So again I have only.

l = ['k1', 'k2', 'k3', 'k4', 'k5']

z = ?magicCode?

z = d['k1']['k2']['k3']['k4']['k5']['/']

z [1, 2, 3]
z.append('4')
z

[1, 2, 3, 4]

Best Regards

macm

Hi Folks
How convert list to nested dictionary?
l
['k1', 'k2', 'k3', 'k4', 'k5']
result
{'k1': {'k2': {'k3': {'k4': {'k5': {}}}}}}
Regards
macm
reduce(lambda x,y: {y:x}, reversed(l), {})

d={}
while L : d={L.pop():d}

Iterating over the keys in normal order:

...     outer[key] = inner = {}
...     outer = inner
...>>> d

{'a': {'b': {'c': {'d': {'e': {}}}}}}

The "functional" variant:

{'a': {'b': {'c': {'d': {'e': {}}}}}}

In a single expression if you are willing to pay the price:

"abcde", ({},)*2)[0]
{'a': {'b': {'c': {'d': {'e': {}}}}}}

Peter

 

[Peter Otten]

thanks a lot all. All solutions work fine.

while I am doing my home work.
Reading "Learning Python" and much more.

Let me ask again to close my doubts:

l = ['k1', 'k2', 'k3', 'k4', 'k5']
d = reduce(lambda x,y: {y:x}, reversed(l), {'/':[1,2,3]})
d {'k1': {'k2': {'k3': {'k4': {'k5': {'/': [1, 2, 3]}}}}}}
d['k1']['k2']['k3']['k4']['k5'] {'/': [1, 2, 3]}
d['k1']['k2']['k3']['k4']['k5']['/'] [1, 2, 3]

now I want generate the "index" to access the element.

==> d['k1']['k2']['k3']['k4']['k5']['/'] from l

So again I have only.

l = ['k1', 'k2', 'k3', 'k4', 'k5']

z = ?magicCode?

z = d['k1']['k2']['k3']['k4']['k5']['/']

You'll eventually have to start and write your first line of code. Why not
doing it right now? It is sure more rewarding than copying other people's
canned solutions and it can even be fun.

Peter

 

[macm]

Hi Peter

Thanks a lot for your tips and codes,

Cake Recipes are good to learn! So I post just basic issues.

Hopping a good soul like you can help me!

But I am still learning... : )

Best Regards

macm

thanks a lot all. All solutions work fine.

while I am doing my home work.
Reading "Learning Python" and much more.

Let me ask again to close my doubts:

l = ['k1', 'k2', 'k3', 'k4', 'k5']
d = reduce(lambda x,y: {y:x}, reversed(l), {'/':[1,2,3]})
d

{'k1': {'k2': {'k3': {'k4': {'k5': {'/': [1, 2, 3]}}}}}}

d['k1']['k2']['k3']['k4']['k5'] {'/': [1, 2, 3]}
d['k1']['k2']['k3']['k4']['k5']['/']

[1, 2, 3]

now I want generate the "index" to access the element.

==> d['k1']['k2']['k3']['k4']['k5']['/'] from l

So again I have only.

l = ['k1', 'k2', 'k3', 'k4', 'k5']

z = ?magicCode?

z = d['k1']['k2']['k3']['k4']['k5']['/']

You'll eventually have to start and write your first line of code. Why not
doing it right now? It is sure more rewarding than copying other people's
canned solutions and it can even be fun.

Peter

 

[macm]

Ok. Done

.... nest2 = nest
.... for key in path[:-1]:
.... nest2 = nest2[key]
.... nest2[path[-1]].append(val)
.... return nest
....

l = ['k1','k2','k3','k4','k5']
ndict = reduce(lambda x,y: {y:x}, reversed(l), {'/':[1,2,3]})
x = l + list('/')
print appendNested(ndict, x, 5) {'k1': {'k2': {'k3': {'k4': {'k5': {'/': [1, 2, 3, 5]}}}}}}

I need make lambda but now is friday night here and I guess I will go
drink a beer or three!

Cheers!

macm

Hi Peter

Thanks a lot for your tips and codes,

Cake Recipes are good to learn! So I post just basic issues.

Hopping a good soul like you can help me!

But I am still learning... : )

Best Regards

macm

thanks a lot all. All solutions work fine.
while I am doing my home work.
Reading "Learning Python" and much more.
Let me ask again to close my doubts:
l = ['k1', 'k2', 'k3', 'k4', 'k5']
d = reduce(lambda x,y: {y:x}, reversed(l), {'/':[1,2,3]})
d
{'k1': {'k2': {'k3': {'k4': {'k5': {'/': [1, 2, 3]}}}}}}
d['k1']['k2']['k3']['k4']['k5']
{'/': [1, 2, 3]}
d['k1']['k2']['k3']['k4']['k5']['/']
[1, 2, 3]
now I want generate the "index" to access the element.
==> d['k1']['k2']['k3']['k4']['k5']['/'] from l
So again I have only.
l = ['k1', 'k2', 'k3', 'k4', 'k5']
z = ?magicCode?
z = d['k1']['k2']['k3']['k4']['k5']['/']

You'll eventually have to start and write your first line of code. Why not
doing it right now? It is sure more rewarding than copying other people's
canned solutions and it can even be fun.

Peter

 

[Arnaud Delobelle]

Hi Folks

How convert list to nested dictionary?

l
['k1', 'k2', 'k3', 'k4', 'k5']
result
{'k1': {'k2': {'k3': {'k4': {'k5': {}}}}}}

Regards

macm

reduce(lambda x,y: {y:x}, reversed(l), {})

d={}
while L : d={L.pop():d}

Iterating over the keys in normal order:
... outer[key] = inner = {}
... outer = inner
...{'a': {'b': {'c': {'d': {'e': {}}}}}}

The "functional" variant:
{'a': {'b': {'c': {'d': {'e': {}}}}}}

In a single expression if you are willing to pay the price:
"abcde", ({},)*2)[0]
{'a': {'b': {'c': {'d': {'e': {}}}}}}

Peter

The most basic functional version:

f = lambda l: {l[0]: f(l[1:])} if l else {}
f('abcde')

{'a': {'b': {'c': {'d': {'e': {}}}}}}

 

[Alexander Gattin]

Hello,

How convert list to nested dictionary?

l
['k1', 'k2', 'k3', 'k4', 'k5']
result
{'k1': {'k2': {'k3': {'k4': {'k5': {}}}}}}

http://www.amk.ca/python/writing/functional

so, why didn't you try python's reduce?

IMO this is trivial:

l ['k1', 'k2', 'k3', 'k4', 'k5']
reduce(lambda x,y: {y:x}, l.__reversed__()) {'k1': {'k2': {'k3': {'k4': 'k5'}}}}
reduce(lambda x,y: {y:x}, l.__reversed__(), dict()) {'k1': {'k2': {'k3': {'k4': {'k5': {}}}}}}

second reduce uses empty dict for seed.