CODEWITHSUNDEEP
pythonlistdictionary
Thaiman Thatte
Thaiman Thatte

Reputation: 35

create a dictionary for a certain range in Python

I have a following dictionary:

d1 = {}

I need to assign it the following values, but only the first n values from the following list:

all_examples=   ['A,1,1', 'B,2,1', 'C,4,4', 'D,4,5']

Final output: d1={"A":[1,1],"B":[2,1]}

1st version of the Question:

I have a following dictionary:

d1 = {'UID': 'A12B4', 'name': 'John', 'email': '[email protected]'}

How do I create a dictionary d2 such that d2 has the first n number of keys and values from d1?

2nd version of the Question

I have a following dictionary: d1 = {}

I need to assign it the following values, but only the first n values from the following list:

all_examples=   ['A,1,1', 'B,2,1', 'C,4,4', 'D,4,5']

Upvotes: 0

Views: 194

Answers (1)

zmo
zmo

Reputation: 24802

How do I create a dictionary d2 such that d2 has the first n number of keys and values from d1?

I'm sorry to tell you, but you shouldn't do that. At least, not with the stock dict. The reason is that no order is guaranteed in a dict, and whenever you'll add or remove a value, the order of the dict CAN change.

But if you really insist you could do:

>>> d1 = {1:'a', 3:'b', 2:'c', 4:'d'}
>>> d1
{1: 'a', 2: 'c', 3: 'b', 4: 'd'}
>>> d.keys()[:2]
[1, 2]
>>> {k : d[k] for k in d.keys()[:2]}
{1: 'a', 2: 'c'}

but my suggestion to you is to use an OrderedDict object, that will guarantee the order of the elements within the dict:

>>> od = OrderedDict()
>>> od[1] = 'a'
>>> od[3] = 'b'
>>> od[2] = 'c'
>>> od[4] = 'd'
>>> od
OrderedDict([(1, 'a'), (3, 'b'), (2, 'c'), (4, 'd')])
>>> od.keys()[:2]
[1, 3]
>>> OrderedDict([(k, od[k]) for k in od.keys()[:2]])
OrderedDict([(1, 'a'), (3, 'b')])

Ok, looks like you just changed radically your question, so here's the new answer:

>>> all_examples=   ['A,1,1', 'B,2,1', 'C,4,4', 'D,4,5']
>>> n = 2
>>> all_examples[:n]
['A,1,1', 'B,2,1']

here's how you select up to n values of a list.

once again, you're changing your question…

Final output: d1={"A":[1,1],"B":[2,1]}

well you can just do:

>>> {elt[0] : (elt[1], elt[2]) for elt in [elt.split(',') for elt in all_examples[:n]]}
{'A': ('1', '1'), 'B': ('2', '1')}

HTH

Upvotes: 2

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