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Efficiently convert Dict to NamedTuple in Julia


By : Ermen
Date : September 06 2020, 04:00 PM
With these it helps I would like to have an interface that accepts a Dict or a NamedTuple as input, but then always converts the input to a NamedTuple. , What about:
code :
unzip(d::Dict) = (;(p.first => unzip(p.second) for p in d)...)
unzip(d) = d
julia> unzip(dd)
(a = 1, b = (y = 4, x = 3))


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How to convert a nested namedtuple to a dict?


By : Mike gauthier
Date : March 29 2020, 07:55 AM
This might help you You can unnest namedtuples by recursively unpacking ._asdict().items(), checking the instance of the value type and unpacking the nested values accordingly. The instance checking of a namedtuple takes a little extra effort to make sure the nested namedtuples aren't seen by the unpacker as pure tuples.
code :
def isnamedtupleinstance(x):
    _type = type(x)
    bases = _type.__bases__
    if len(bases) != 1 or bases[0] != tuple:
        return False
    fields = getattr(_type, '_fields', None)
    if not isinstance(fields, tuple):
        return False
    return all(type(i)==str for i in fields)

def unpack(obj):
    if isinstance(obj, dict):
        return {key: unpack(value) for key, value in obj.items()}
    elif isinstance(obj, list):
        return [unpack(value) for value in obj]
    elif isnamedtupleinstance(obj):
        return {key: unpack(value) for key, value in obj._asdict().items()}
    elif isinstance(obj, tuple):
        return tuple(unpack(value) for value in obj)
    else:
        return obj

# data = TypeToolObjectSetting(version=1, xx=0.0, ..
unpacked_data = unpack(data)

Convert Dict to DataFrame in Julia


By : Vincent Milotti
Date : March 29 2020, 07:55 AM
this will help Suppose I have a Dict defined as follows: , Another method could be
code :
DataFrame(Any[values(x)...],Symbol[map(symbol,keys(x))...])
DataFrame(;[Symbol(k)=>v for (k,v) in x]...)
x = Dict{Symbol,Array{Integer,1}}(:A => [1,2,3], :B => [4,5,6])
df = DataFrame(;x...)

Convert Dict to Array{Tuple} in Julia


By : TOMOT
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further You can also use collect(d), which again gives an array of Pairs of the form k=>v.
These can be indexed just as if it were a tuple:
code :
julia> p = 3=>4
3=>4

julia> p[1]
3

julia> p[2]
4

julia> a, b = p
3=>4

julia> a
3

julia> b
4

Pythonic Way to Convert Dictionary to namedtuple, or Another Hashable dict-like?


By : sram
Date : March 29 2020, 07:55 AM
around this issue To create the subclass, you may just pass the keys of a dict directly:
code :
MyTuple = namedtuple('MyTuple', sorted(d))
my_tuple = MyTuple(**d)
>>> from types import SimpleNamespace
>>> SimpleNamespace(**d)
namespace(a=1, b=2, c=3, d=4)
>>> from box import Box
>>> b = Box(d, frozen_box=True)
>>> hash(b)
7686694140185755210
>>> b.a
1
>>> b['a']
1

Python: Convert string form with unicode of namedtuple to namedtuple


By : Chawn
Date : March 29 2020, 07:55 AM
This might help you You need to use eval() please follow the official documentation https://docs.python.org/3/library/functions.html#eval or https://docs.python.org/3/library/functions.html#exec
code :
x = 1
eval('x+1')
eval("Job(name=u'MyJob', type=u'datamart', status=u'complete', ended=152717815282, id=u'2001840', baseMetaPath=u'/datamarts/2001546', updated=1527178152000, archiveUnarchive=0")
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