CODEWITHSUNDEEP
pythonarraysdictionary
Greg Schlaepfer
Greg Schlaepfer

Reputation: 25

Two-dimensional array/dictionary combination

Is there a way to define a two-dimensional array/dictionary combination, wherein the first value is enumerative and the second associative? The end result would ideally look like this, where the first is a simple index, and the second a key->value pair.

data[0]["Name"] = ...

Thanks in advance!

Upvotes: 1

Views: 7242

Answers (3)

dawg
dawg

Reputation: 103874

Sure -- a list of dictionaries:

>>> LoD=[{c:i for i,c in enumerate(li)} for li in ('abc','def','ghi')]
>>> LoD
[{'c': 2, 'b': 1, 'a': 0}, {'f': 2, 'e': 1, 'd': 0}, {'g': 0, 'i': 2, 'h': 1}]
>>> LoD[2]['g']
0
>>> LoD[2]['h']
1

Be sure that you use dict methods on the dicts and list methods on the lists:

OK:

>>> LoD[2]['new']='new value'
>>> LoD
[{'c': 2, 'b': 1, 'a': 0}, {'f': 2, 'e': 1, 'd': 0}, {'g': 0, 'new': 'new value', 'i': 2, 'h': 1}]

>>> LoD.append({'new key':'new value'})
>>> LoD
[{'c': 2, 'b': 1, 'a': 0}, {'f': 2, 'e': 1, 'd': 0}, {'g': 0, 'new': 'new value', 'i': 2, 'h': 1}, {'new key': 'new value'}]

No go:

>>> LoD['new']='test'
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: list indices must be integers, not str

>>> LoD[2].append('something')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'dict' object has no attribute 'append'

Upvotes: 0

Stephen Rodriguez
Stephen Rodriguez

Reputation: 1037

dicts = [  {"name": "Tom", "age": 10 },
           {"name": "Tom", "age": 10 },
           {"name": "Tom", "age": 10 }  ]

print dicts[0]['name']

Essentially, you would be creating a list of dictionaries. Props to mhlester for having the right answer earlier as a comment.

Upvotes: 0

mhlester
mhlester

Reputation: 23231

Expanding my comment, a list of dicts:

>>> list_of_dicts = [{'first_name':'greg', 'last_name':'schlaepfer'},
...                  {'first_name':'michael', 'last_name':'lester'}]
>>>
>>> list_of_dicts[0]['first_name']
'greg'

Upvotes: 5

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