Using variable name as name of dictionary

Question

I want to accomplish the following:

vname = [ 'name1', 'name2']

for i in len(vname):
  vname[i] = {} 
  vname[i]['key1'] = 4*i

I expect to create two dictionaries name1 and name2 such that

name1['key1'] = 0
name2['key1'] = 4

your list contains strings not dictionaries, when you do vname[i] = {} you replace the string with a dictionary so the variables name1 and name2 never exist, also I assume you mean to use 4*i instead of 4*1 since the latter is just 4. — Tadhg McDonald-Jensen
– Tadhg McDonald-Jensen, Commented Jun 6, 2016 at 15:05

alecxe · Accepted Answer · 2016-06-06 15:05:44Z

1

I'd use a nested dictionary instead, generating values with the help of enumerate():

>>> {key: {'key1': index * 4} for index, key in enumerate(vname)}
{'name2': {'key1': 4}, 'name1': {'key1': 0}}

answered Jun 6, 2016 at 15:05

alecxe

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user2390182 · Accepted Answer · 2016-06-06 15:14:15Z

0

In order to dynamically create variables of a certain name, you could add them to the global namespace:

>>> vname = [ 'name1', 'name2']
>>> for i in range(len(vname)):
...     globals()[vname[i]] = {}
...     globals()[vname[i]]['key1'] = 4*i
... 
>>> name1
{'key1': 0}
>>> name2
{'key1': 4}

However, I have never encountered a situation where this was useful or necessary. If you are trying to sth. like this, you are very likely better off with a different (and better behaved) data structure like the nested dict suggested by @alexce.

answered Jun 6, 2016 at 15:14

user2390182

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