Python如何将dict列表转换为元组列表

Question

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?

Answer 1

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 dic tonaries in list . For every dic tionary we iterate over its .items() and extract the key and val ue 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 Foo s, they remain Foo s.

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.

Answer 2

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 ).

Answer 3

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)]]
>>>

Answer 4

如果字典中的值始终是长度为2的列表，则可以执行以下操作：

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

Answer 5

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.