CODEWITHSUNDEEP
pythonlistdictionarytuples
CosimoCD
CosimoCD

Reputation: 3770

Python how to convert a list of dict to a list of tuples

I have a list of dict that looks like this:

list=[{u'hello':['001', 3], u'word':['003', 1], u'boy':['002', 2]}, 
     {u'dad':['007', 3], u'mom':['005', 3], u'honey':['002', 2]} ]

What I need is to iterate on my list in order to create list of tuples like this:

new_list=[('hello','001', 3), ('word','003',1), ('boy','002', 2)
           ('dad','007',3), ('mom', '005', 3), ('honey','002',2)]

NOTE! the numbers with the zeros ('001',003'... and so on) must be considerated as a string.

Is there anybody whom can help me?

Upvotes: 4

Views: 10280

Answers (5)

sathish
sathish

Reputation: 149

list=[{u'hello':['001', 3], u'word':['003', 1], u'boy':['002', 2]}, 
     {u'dad':['007', 3], u'mom':['005', 3], u'honey':['002', 2]} ] 

 >>> [map(lambda (k,v): (k,)+tuple(v), dictionary.iteritems()) for dictionary in list]

[[(u'boy', '002', 2), (u'word', '003', 1), (u'hello', '001', 3)], [(u'dad', '007', 3), (u'honey', '002', 2), (
u'mom', '005', 3)]]
>>>

Upvotes: 2

Jitsusama
Jitsusama

Reputation: 755

Assuming that you don't care about having the output re-coded as ASCII as your example shows, that you don't care about preserving order and that you want to merge items in case they appear more than once across the union of the listed dictionaries, this code example should work:

>>> input_list = [
...    {u'hello': ['001', 3], u'word': ['003', 1], u'boy': ['002', 2]},
...    {u'dad': ['007', 3], u'mom': ['005', 3], u'honey': ['002', 2]}] 
...
>>> temporary_dict = {}
>>> output_list = []
>>> for dictionary in input_list:
...     for key in dictionary.keys():
...         if key in temporary_dict:
...             temporary_dict[key] += dictionary[key]
...         else:
...             temporary_dict[key] = dictionary[key]
...
>>> for key in temporary_dict.keys():
...     output_list.append(tuple([key] + temporary_dict[key]))
...
>>> print(output_list)
[(u'dad', '007', 3), (u'boy', '002', 2), (u'word', '003', 1),
 (u'honey', '002', 2), (u'mom', '005', 3), (u'hello', '001', 3)]

Note how I first concatenate the keys from all of the listed dictionaries into a temporary_dict and then I iterate over that dictionary, making a concatenated list of the dictionary key as well as that key's values (temporary_dict[key]), and append them to a new output list.

Please let me know if any of my assumptions are incorrect.

Upvotes: 1

Ma0
Ma0

Reputation: 15204

List-comprehension one-liner for Python 3.5+:

my_list=[{u'hello':['001', 3], u'word':['003', 1], u'boy':['002', 2]},
         {u'dad':['007', 3], u'mom':['005', 3], u'honey':['002', 2]}]

result = [(k, *v) for f in my_list for k,v in f.items()]

# [('word', '003', 1), ('hello', '001', 3), ('boy', '002', 2), ('dad', '007', 3), ('honey', '002', 2), ('mom', '005', 3)]

PS: Do not use the variable name list since it is a python built-in.

For older versions of Python see answer by @WillemVanOnsem which does not make use of the starred expressions (*v).

Upvotes: 2

Zaccharie Ramzi
Zaccharie Ramzi

Reputation: 2316

If your values in the dictionary are always lists of length 2, you could do:

new_list = [ (key, value[0], value[1]) for dict in list for key, value in dict.iteritems() ]

Upvotes: 1

willeM_ Van Onsem
willeM_ Van Onsem

Reputation: 476669

You can use list comprehension for that:

new_list = [(key,)+tuple(val) for dic in list for key,val in dic.items()]

Here we iterate over all dictonaries in list. For every dictionary we iterate over its .items() and extract the key and value and then we construct a tuple for that with (key,)+val.

Whether the values are strings or not is irrelevant: the list comprehension simply copies the reference so if the original elements were Foos, they remain Foos.

Finally note that the dictionaries are unordered, so the order is undetermined. However if a dictionary d1 occurs before a dictionary d2, all the elements of the first will be placed in the list before the tuples of the latter. But the order of tuples for each individual dictionary is not determined.

Upvotes: 6

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