How to access the first and the last elements in a dictionary?
Question:
Before posting, I have already gone through Access an arbitrary element in a dictionary in Python, but I’m uncertain about this.
I have a long dictionary and I’ve to get the values of its first and last keys. I can use dict[dict.keys()[0]] and dict[dict.keys()[-1]] to get the first and last elements, but since the key:value pairs are outputted in a random form(as in the positioning of the key:value pairs is random), will the solution provided in this link always work?
Answers:
Python dictionaries are unordered, so “first” and “last” isn’t defined. Instead, you can sort your keys, and then access the element associated with the first and last key in your sorted set.
EDIT:
The OP clarified that by “first” and “last” he meant the order in which keys were added to the dictionary. collections.OrderedDict should work for this case.
There is no such thing as “first” or “last” key in dictionary, which does not guarantee any particular ordering. So there is no possibility to get “first” or “last” element. You can only create your own wrapper around python dict, which will store the information about “first” and “last” object
Something like
class MyDict:
def __init__(self):
self.first=None
self.last=None
self.dict={}
def add( key, value ):
if self.first==None: self.first=key
self.last=key
self.dict[key]=value
def get( key ):
return self.dict[key]
def first():
return self.dict[ self.first ]
def last():
return self.dict[ self.last ]
Although as it was pointed out in the comment there is already a class OrderedDict: http://docs.python.org/2/library/collections.html#collections.OrderedDict
Ordered dictionaries are just like regular dictionaries but they remember
the order that items were inserted. When iterating over an ordered
dictionary, the items are returned in the order their keys were first
added.
Use an OrderedDict, because a normal dictionary doesn’t preserve the insertion order of its elements when traversing it. Here’s how:
# import the right class
from collections import OrderedDict
# create and fill the dictionary
d = OrderedDict()
d['first'] = 1
d['second'] = 2
d['third'] = 3
# retrieve key/value pairs
els = list(d.items()) # explicitly convert to a list, in case it's Python 3.x
# get first inserted element
els[0]
=> ('first', 1)
# get last inserted element
els[-1]
=> ('third', 3)
def dictionarySortingExample(yourDictionary):
#get all the keys and store them to a list
allKeys = yourDictionary.keys()
#sort the list of keys
allKeysSorted = sorted(allKeys)
#retrieve the first and last keys in the list
firstKey = allKeysSorted[0]
lastKey = allKeysSorted[-1]
#retrive the values from the dictionary
firstValue = yourDictionary[firstKey]
lastValue = yourDictionary[lastKey]
print "---Sorted Dictionary---"
print "original dictionary: " + str(yourDictionary)
print "list of all keys: " + str(allKeys)
print "ordered list of all keys: " + str(allKeysSorted)
print "first item in sorted dictionary: " + str(firstKey) + ":" + str(firstValue)
print "last item in sorted dictionary: " + str(lastKey) + ":" + str(lastValue)
example dictionary sorting
sampleDictionary = {4:”four”, “Cranberry”:2, 3:”three”, 2:”two”, “Apple”:3, 1:”one”, “Bananna”:1}
dictionarySortingExample(sampleDictionary)
If working with Python 3.6+ you can do a one liner:
First:
list({'fist': 1, 'second': 2, 'last': 3}.items())[0]
=> ('first', 1)
Last:
list({'fist': 1, 'second': 2, 'third': 3}.items())[-1]
=> ('third', 1)
This is the case because Python 3.6+ default dictionary preserves insertion order. This is also mentioned in the documentation:
Dictionaries preserve insertion order. Note that updating a key does not affect the order. Keys added after deletion are inserted at the end.
and
Changed in version 3.7: Dictionary order is guaranteed to be insertion order. This behavior was an implementation detail of CPython from 3.6.
You can do it by using list().
dir = dict()
dir['Key-3'] = 'Value-3' # Added First Item
dir['Key-2'] = 'Value-2' # Added Second Item
dir['Key-4'] = 'Value-4' # Added Third Item
dir['Key-1'] = 'Value-1' # Added Fourth Item
lst = list(dir.items()) # For key & value
# lst = list(dir.keys()) # For keys
# lst = list(dir.values()) # For values
print('First Element:- ', lst[0])
print('Last Element:- ', lst[-1])
Output:-
First Element:- (‘Key-3’, ‘Value-3’)
Last Element:- (‘Key-1’, ‘Value-1’)
With OrderedDict you can use iterators
d = OrderedDict(a=1, b=2, c=3)
next(iter(d)) # returns 'a'
next(reversed(d) # returns 'c'
Obviously it’s too late to answer, yet I’d like to add the following to the awesome answers above
dct = {"first": 1, "second": 2, "last": 3}
first, *_, last = dct.items()
print("*** Items ***")
print("First Item:", first)
print("Last Item:", last)
first, *_, last = dct.keys()
print("*** Keys ***")
print("First Key:", first)
print("Last Key:", last)
first, *_, last = dct.values()
print("*** Values ***")
print("First Value:", first)
print("Last Value:", last)
*** Items ***
First Item: ('first', 1)
Last Item: ('last', 3)
*** Keys ***
First Key: first
Last Key: last
*** Values ***
First Value: 1
Last Value: 3
Before posting, I have already gone through Access an arbitrary element in a dictionary in Python, but I’m uncertain about this.
I have a long dictionary and I’ve to get the values of its first and last keys. I can use dict[dict.keys()[0]] and dict[dict.keys()[-1]] to get the first and last elements, but since the key:value pairs are outputted in a random form(as in the positioning of the key:value pairs is random), will the solution provided in this link always work?
Python dictionaries are unordered, so “first” and “last” isn’t defined. Instead, you can sort your keys, and then access the element associated with the first and last key in your sorted set.
EDIT:
The OP clarified that by “first” and “last” he meant the order in which keys were added to the dictionary. collections.OrderedDict should work for this case.
There is no such thing as “first” or “last” key in dictionary, which does not guarantee any particular ordering. So there is no possibility to get “first” or “last” element. You can only create your own wrapper around python dict, which will store the information about “first” and “last” object
Something like
class MyDict:
def __init__(self):
self.first=None
self.last=None
self.dict={}
def add( key, value ):
if self.first==None: self.first=key
self.last=key
self.dict[key]=value
def get( key ):
return self.dict[key]
def first():
return self.dict[ self.first ]
def last():
return self.dict[ self.last ]
Although as it was pointed out in the comment there is already a class OrderedDict: http://docs.python.org/2/library/collections.html#collections.OrderedDict
Ordered dictionaries are just like regular dictionaries but they remember
the order that items were inserted. When iterating over an ordered
dictionary, the items are returned in the order their keys were first
added.
Use an OrderedDict, because a normal dictionary doesn’t preserve the insertion order of its elements when traversing it. Here’s how:
# import the right class
from collections import OrderedDict
# create and fill the dictionary
d = OrderedDict()
d['first'] = 1
d['second'] = 2
d['third'] = 3
# retrieve key/value pairs
els = list(d.items()) # explicitly convert to a list, in case it's Python 3.x
# get first inserted element
els[0]
=> ('first', 1)
# get last inserted element
els[-1]
=> ('third', 3)
def dictionarySortingExample(yourDictionary):
#get all the keys and store them to a list
allKeys = yourDictionary.keys()
#sort the list of keys
allKeysSorted = sorted(allKeys)
#retrieve the first and last keys in the list
firstKey = allKeysSorted[0]
lastKey = allKeysSorted[-1]
#retrive the values from the dictionary
firstValue = yourDictionary[firstKey]
lastValue = yourDictionary[lastKey]
print "---Sorted Dictionary---"
print "original dictionary: " + str(yourDictionary)
print "list of all keys: " + str(allKeys)
print "ordered list of all keys: " + str(allKeysSorted)
print "first item in sorted dictionary: " + str(firstKey) + ":" + str(firstValue)
print "last item in sorted dictionary: " + str(lastKey) + ":" + str(lastValue)
example dictionary sorting
sampleDictionary = {4:”four”, “Cranberry”:2, 3:”three”, 2:”two”, “Apple”:3, 1:”one”, “Bananna”:1}
dictionarySortingExample(sampleDictionary)
If working with Python 3.6+ you can do a one liner:
First:
list({'fist': 1, 'second': 2, 'last': 3}.items())[0]
=> ('first', 1)
Last:
list({'fist': 1, 'second': 2, 'third': 3}.items())[-1]
=> ('third', 1)
This is the case because Python 3.6+ default dictionary preserves insertion order. This is also mentioned in the documentation:
Dictionaries preserve insertion order. Note that updating a key does not affect the order. Keys added after deletion are inserted at the end.
and
Changed in version 3.7: Dictionary order is guaranteed to be insertion order. This behavior was an implementation detail of CPython from 3.6.
You can do it by using list().
dir = dict()
dir['Key-3'] = 'Value-3' # Added First Item
dir['Key-2'] = 'Value-2' # Added Second Item
dir['Key-4'] = 'Value-4' # Added Third Item
dir['Key-1'] = 'Value-1' # Added Fourth Item
lst = list(dir.items()) # For key & value
# lst = list(dir.keys()) # For keys
# lst = list(dir.values()) # For values
print('First Element:- ', lst[0])
print('Last Element:- ', lst[-1])
Output:-
First Element:- (‘Key-3’, ‘Value-3’)
Last Element:- (‘Key-1’, ‘Value-1’)
With OrderedDict you can use iterators
d = OrderedDict(a=1, b=2, c=3)
next(iter(d)) # returns 'a'
next(reversed(d) # returns 'c'
Obviously it’s too late to answer, yet I’d like to add the following to the awesome answers above
dct = {"first": 1, "second": 2, "last": 3}
first, *_, last = dct.items()
print("*** Items ***")
print("First Item:", first)
print("Last Item:", last)
first, *_, last = dct.keys()
print("*** Keys ***")
print("First Key:", first)
print("Last Key:", last)
first, *_, last = dct.values()
print("*** Values ***")
print("First Value:", first)
print("Last Value:", last)
*** Items ***
First Item: ('first', 1)
Last Item: ('last', 3)
*** Keys ***
First Key: first
Last Key: last
*** Values ***
First Value: 1
Last Value: 3