Pythonic way to create empty map of vector of vector of vector

I have the following C++ code

std::map<std::string, std::vector<std::vector<std::vector<double> > > > details
details["string"][index][index].push_b开发者_开发技巧ack(123.5);

May I know what is the Pythonic to declare an empty map of vector of vector of vector? :p

I try to have

self.details = {}
self.details["string"][index][index].add(value)

I am getting

KeyError: 'string'

Probably the best way would be to use a dict for the outside container with strings for the keys mapping to an inner dictionary with tuples (the vector indices) mapping to doubles:

 d = {'abc': {(0,0,0): 1.2, (0,0,1): 1.3}}

It's probably less efficient (less time-efficient at least, it's actually more space-efficient I would imagine) than actually nesting the lists, but IMHO cleaner to access:

>>> d['abc'][0,0,1]
1.3

Edit

Adding keys as you went:

d = {} #start with empty dictionary
d['abc'] = {} #insert a new string key into outer dict
d['abc'][0,3,3] = 1.3 #insert new value into inner dict
d['abc'][5,3,3] = 2.4 #insert another value into inner dict
d['def'] = {} #insert another string key into outer dict
d['def'][1,1,1] = 4.4
#...
>>> d
{'abc': {(0, 3, 3): 1.3, (5, 3, 3): 2.4}, 'def': {(1, 1, 1): 4.4}}

Or if using Python >= 2.5, an even more elegant solution would be to use defaultdict: it works just like a normal dictionary, but can create values for keys that don't exist.

import collections
d = collections.defaultdict(dict)   #The first parameter is the constructor of values for keys that don't exist
d['abc'][0,3,3] = 1.3
d['abc'][5,3,3] = 2.4
d['def'][1,1,1] = 4.4
#...
>>> d
defaultdict(<type 'dict'>, {'abc': {(0, 3, 3): 1.3, (5, 3, 3): 2.4}, 'def': {(1, 1, 1): 4.4}})

Python is a dynamic (latent-typed) language, so there is no such thing as a "map of vector of vector of vector" (or "dict of list of list of list" in Python-speak). Dicts are just dicts, and can contain values of any type. And an empty dict is simply: {}

create dict that contains a nested list which inturn contains a nested list

dict1={'a':[[2,4,5],[3,2,1]]}

dict1['a'][0][1]
4

Using collections.defaultdict, you can try the following lambda trick below. Note that you'll encounter problems pickling these objects.

from collections import defaultdict

# Regular dict with default float value, 1D
dict1D = defaultdict(float)
val1 = dict1D["1"] # string key type; val1 == 0.0 by default

# 2D
dict2D = defaultdict(lambda: defaultdict(float))
val2 = dict2D["1"][2] # string and integer key types; val2 == 0.0 by default

# 3D
dict3D = defaultdict(lambda: defaultdict(lambda: defaultdict(float)))
val3 = dict3D[1][2][3] # val3 == 0.0 by default

# N-D, arbitrary nested defaultdicts
dict4D = defaultdict(lambda: defaultdict(lambda: defaultdict(lambda: defaultdict(str))))
val4 = dict4D["abc"][10][9][90] # val4 == '' by default

You can basically nest as many of these defaultdict collection types. Also, note that they behave like regular python dictionaries that can take the usual key types (non-mutable and hashable). Best of luck!