I have two lists:
header = ["Name", "Age"]
detail = ["Joe", 22, "Dave", 43, "Herb", 32]
And would like to create a list of dictonaries like this:
[{"Name": "Joe", "Age": 22}, {"Name": "Dave", "Age": 32}, {"Name": "Herb", "Age": 32}]
This method zip gets me partially there, but only adds the first set of values to the dictionary:
>>> dict(zip(header, detail))
{'Age': 22, 'Name': 'Joe'}
How can I output as one dictionary for all values in the detail
list? I found this answer, but this depends on detail
containing nested lists.
>>> detail = ["Joe", 22, "Dave", 43, "Herb", 32]
>>> d = dict(zip(detail[::2], detail[1::2]))
>>> d
{'Herb': 32, 'Dave': 43, 'Joe': 22}
For your new/edited question:
>>> d = [dict(zip(header, items)) for items in zip(detail[::2],detail[1::2])]
>>> d
[{'Age': 22, 'Name': 'Joe'}, {'Age': 43, 'Name': 'Dave'}, {'Age': 32, 'Name': 'H
erb'}]
For such tasks I prefer functional approach.
Here is a recipe for grouper:
def grouper(n, iterable, fillvalue=None):
"Collect data into fixed-length chunks or blocks"
# grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx
args = [iter(iterable)] * n
return izip_longest(fillvalue=fillvalue, *args)
By using it, we may advance trough detail
by groups of 2
:
>>> groups = grouper(len(header),detail)
>>> list(groups)
[('Joe', 22), ('Dave', 43), ('Herb', 32)]
And then we can use this iterator to create dictionaries as you need:
>>> [dict(zip(header,group)) for group in groups]
[{'Age': 22, 'Name': 'Joe'}, {'Age': 43, 'Name': 'Dave'}, {'Age': 32, 'Name': 'Herb'}]
To clarify, zip(header,group)
gives this:
>>> zip(["Name", "Age"],('Joe', 22))
[('Name', 'Joe'), ('Age', 22)]
And summoning dict
constructor gives this:
>>> dict([('Name', 'Joe'), ('Age', 22)])
{'Age': 22, 'Name': 'Joe'}
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