CODEWITHSUNDEEP
pythonpython-3.x
user7800892
user7800892

Reputation:

Return the first key in Dictionary - Python 3

I'm trying to print the key and value using the below code : 

data = {"Key1" : "Value1", "Key2" : "Value2"}
print(data.keys()[1])   
print(data.values()[1])

I get the error 

'dict_keys' object does not support indexing

What is wrong with the mentioned code?

Accepted Output :

Key2
Value2

Upvotes: 11

Views: 47957

Answers (7)

dtar
dtar

Reputation: 1669

For Python 3 you can simply do 

first_key = next(iter(data.keys()))
print(first_key)   # 'Key1'

Upvotes: 1

cssyphus
cssyphus

Reputation: 40038

Shortest:

mydict = {"Key1" : "Value1", "Key2" : "Value2"}
print( next(iter(mydict)) ) # 'Key1'

For both the key and value:

print( next(iter( mydict.items() )) ) # ('Key1', 'Value1')

Upvotes: 16

BoarGules
BoarGules

Reputation: 16952

It does rather depend on what you mean by first. In Python 3.6, entries in a dictionary are ordered by the key, but probably not quite in the way you expect.

To take your example:

>>> data = {"Key1" : "Value1", "Key2" : "Value2"}

Now add the key 0:

>>> data[0] = "Value0"
>>> data
{'Key1': 'Value1', 'Key2': 'Value2', 0: 'Value0'}

So the zero comes at the end. But if you construct the dict from scratch, like this:

>>> data = {0: "Value0", "Key1" : "Value1", "Key2" : "Value2"}

you get this result instead

>>> data
{0: 'Value0', 'Key1': 'Value1', 'Key2': 'Value2'}

This illustrates the principle that you should not depend on the ordering, which is defined only by the dict implementation, which, in CPython 3.6 and later, is order of entry. To illustrate that point in a different way:

>>> data = {0: "Value0", "Key1" : "Value1", "Key2" : "Value2"}
>>> sorted(data.keys())
Traceback (most recent call last):
  File "<pyshell#42>", line 1, in <module>
    sorted(data.keys())
TypeError: '<' not supported between instances of 'str' and 'int'

Guido has this to say on the subject:

I'd like to handwave on the ordering of ... dicts. Yes, in CPython 3.6 and in PyPy they are all ordered, but it's an implementation detail. I don't want to force all other implementations to follow suit. I also don't want too many people start depending on this, since their code will break in 3.5. (Code that needs to depend on the ordering of keyword args or class attributes should be relatively uncommon; but people will start to depend on the ordering of all dicts all too easily. I want to remind them that they are taking a risk, and their code won't be backwards compatible.)

The full post is here.

Upvotes: 1

Mohamed Ali JAMAOUI
Mohamed Ali JAMAOUI

Reputation: 14689

in python3 data.keys() returns a dict_keys object, so, in general, apply list on it to be able to index/slice it: 

data = {"Key1" : "Value1", "Key2" : "Value2"}
print(data.keys()) 
# output >>> dict_keys(['Key1', 'Key2'])
print(list(data.keys())[1])   
# output >>> Key2
print(list(data.values())[1])
# output >>> Value2

For your specific case, you need to convert the dictionary to an ordered one to conserve the order and get the first element as follows: 

from collections import OrderedDict 
data = {"Key1" : "Value1", "Key2" : "Value2"}
data = OrderedDict(data)
print(data)
# output >>> OrderedDict([('Key1', 'Value1'), ('Key2', 'Value2')])
print(list(data.keys())[0])
# output >>> Key1

Edit:

Based on comments from @Mseifert (thanks), preserving the order after conversion from the unordered dictionary to the ordered one is only an implementation detail that works in python3.6 and we cannot rely on, here's the discussion Mseifert shared:

So the correct way to do what you want is to explicitly define the order 

from collections import OrderedDict
data = OrderedDict([('Key1', 'Value1'), ('Key2', 'Value2')])
print(list(data.keys())[0])

Upvotes: 14

techkuz
techkuz

Reputation: 3956

Dicts represent an unordered datatype, therefore there is no indexing. There is an option in collections. Try OrderedDict.

import collections

d = collections.OrderedDict()
d['Key1'] = 'Value1'
d['Key2'] = 'Value2' # that's how to add values to OrderedDict

items = list(d.items())
print(items)
#[('Key1', 'Value1'), ('Key2', 'Value2')]
print(items[0])
#('Key1', 'Value1')
print(items[1])
#('Key2', 'Value2')

Upvotes: 0

MSeifert
MSeifert

Reputation: 152657

Dictionaries are unordered and in the newer Python versions the hashes of strings are randomized (per session). So you have to accept that what you get as "n"-th key (or value) of a dictionary isn't predictable (at least when the keys are strings).

But if you just want the element that happens to be "first" (or "second") just use list to convert the dict_keys to an sequence that can be indexed:

print(list(data.keys())[1])   
print(list(data.values())[1])

However, my suggestion would be to use an OrderedDict instead of a normal dict to make the result deterministic:

from collections import OrderedDict
data = OrderedDict([("Key1", "Value1"), ("Key2", "Value2")])
print(list(data.keys())[1])     # Key2
print(list(data.values())[1])   # Value2

Upvotes: 3

Richard Neumann
Richard Neumann

Reputation: 3361

In Python, dicts are unordered data structures. Therefore there is no first key. If you want to have a dict-like data structure with internal ordering of keys, have a look at OrderedDict.

Upvotes: 1

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