How to get the first value in a Python dictionary

Question

I have a dictionary like this:

myDict = { 'BigMeadow2_U4': (1609.32, 22076.38, 3.98), 'MooseRun': (57813.48, 750187.72, 231.25), 'Hwy14_2': (991.31, 21536.80, 6.47) }

How can I get the first value of each item in my dicitionary?

I want in the end a list:

myList = [1609.32,57813.48,991.31]

A dict is unordered. This means that there is no ordering of the elements in the dictionary - you need to access the values by the keys that they are associated with. So please explain why the elements in myList are ordered the way they are (first element of the values associated with keys in sorted order, perhaps?) — inspectorG4dget
– inspectorG4dget, Commented Feb 21, 2014 at 9:44
Dict is unordered. There is an Ordered Dict you can use but I forget if it's built in so you might need to import it. — TimothyH
– TimothyH, Commented Dec 20, 2023 at 20:57
Just looked it up. Dicts are now guaranteed to be ordered so you only need OrderedDict if you need the extra features. Because of this just iterate the dict. — TimothyH
– TimothyH, Commented Dec 20, 2023 at 20:59

laur · Accepted Answer · 2024-11-08 17:44:06Z

82

Modern Python 3.x's can use iterator that has less overhead than constructing a list.

first_value = next(iter(my_dict.values()))

Note that if the dictionary is empty you will get StopIteration exception and not None. To produce None in case of an empty dict, provide a default value to next():

first_value = next(iter(my_dict.values()), None)

Since Python 3.7+ this is guaranteed to give you the first item.

edited Nov 8, 2024 at 17:44

laur

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answered Mar 17, 2022 at 21:36

Mikko Ohtamaa

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6 Comments

DavidW Over a year ago

As of Python 3.7 dicts are now guaranteed to be ordered by insertion order

Mikko Ohtamaa Over a year ago

But this is not guaranteed to be the case in the future Python versions or non-CPython interpreters. It applies only for a certain CPython version.

DavidW Over a year ago

No. It's a documented feature of Python, so other implementations should do it too if they claim to implement Python>=3.7. So PyPy does implement it (at least for PyPy 3.7+). So it's guaranteed into the future as much as anything in Python is.

Emanuel P Over a year ago

"Guaranteed into the future" with Python still means that in a few months it'll probably require a new unintuitive joke of a syntax. In order to get the first value, get a list, construct an iterator, and call the function "next". Yeah, sure, whatever Python. I'll just spend half my workday googling again and looking forward to working with a real language again.

Mikko Ohtamaa Over a year ago

@EmanuelP if you have a bad day and want to rant, I suggest posting on Reddit. People here try to be helpful. We cannot fix all the issues in the world, like war, hunger and legacy quirks of programming languages. But we can make the day of some people better.

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Andrii Rusanov · Accepted Answer · 2014-02-21 09:49:49Z

13

Try this way:

my_list = [elem[0] for elem in your_dict.values()]

Offtop: I think you shouldn't use camelcase, it isn't python way

UPD: inspectorG4dget notes, that result won't be same. It's right. You should use collections.OrderedDict to implement this correctly.

from collections import OrderedDict my_dict = OrderedDict({'BigMeadow2_U4': (1609.32, 22076.38, 3.98), 'MooseRun': (57813.48, 750187.72, 231.25), 'Hwy14_2': (991.31, 21536.80, 6.47) })

edited Feb 21, 2014 at 9:49

answered Feb 21, 2014 at 9:40

Andrii Rusanov

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5 Comments

inspectorG4dget Over a year ago

Result won't be the same, always, given that a dictionary is an unordered data structure. Needs a stronger justification for the ordering of the elements of my_list

zhangxaochen Over a year ago

I can't find a linking saying that CamcelCase is not python way. legacy.python.org/dev/peps/pep-0008/#descriptive-naming-styles

zhangxaochen Over a year ago

they are just different naming styles, and many people just like the CamcelCase more ;P

Andrii Rusanov Over a year ago

Ok, as I now CamelCase isn't good for function names and variables. I googled a little and find this: stackoverflow.com/questions/159720/… Anyway, it isn't strict rule and I think you shouldn't be worried about that. It was just my notice

PPC Over a year ago

Isn't this one-liner going to iterate the whole dictionary when you only want to peek at one item?

WeCanDoItGuys · Accepted Answer · 2024-02-29 22:02:16Z

A dictionary is not indexed, but it is in some way, ordered. The following would give you the first existing value:

list(myDict.values())[0]

Houcheng · Accepted Answer · 2014-02-21 09:47:46Z

one lines...

myList = [myDict [i][0] for i in sorted(myDict.keys()) ]

the result:

>>> print myList [1609.32, 991.31, 57813.48]

linpingta · Accepted Answer · 2014-02-21 09:49:30Z

0

myList = [] for k,v in myDict.items() myList.append(v[0])

answered Feb 21, 2014 at 9:49

linpingta

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Comments

DominiCane · Accepted Answer · 2016-04-16 09:21:55Z

0

If you want ordered dictionary, use:

from collections import OrderedDict ordered = OrderedDict( ('BigMeadow2_U4', (1609.32, 22076.38, 3.98)), ('MooseRun', (57813.48, 750187.72, 231.25)), ('Hwy14_2', (991.31, 21536.80, 6.47)) ) first_values = [v[0] for v in ordered.values()]

The output order will be exactly as your input order.

edited Apr 16, 2016 at 9:21

answered Feb 21, 2014 at 9:55

DominiCane

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4 Comments

Silmathoron Over a year ago

Careful with that: since the input dictionary is unordered, they is no way of knowing how the items will be ordered when the OrderedDict is created from a regular dict. (see penultimate line in the class's doc)

DominiCane Over a year ago

@Silmathoron , you are wrong. Read in the docs: Ordered dictionaries are just like regular dictionaries but they remember the order that items were inserted

Silmathoron Over a year ago

Read my comment again ;) I'm saying you cannot tell if 'BigMeadow2_U4' will be the first value if you create the OrderedDict from a plain dict instead of adding them in the right order.

DominiCane Over a year ago

oh, yes, in this case you're right, changed it to tuples

Community · Accepted Answer · 2020-06-20 09:12:55Z

Try this.Type the dictionary and the position you want to print in the function

d = {'Apple': 1, 'Banana': 9, 'Carrot': 6, 'Baboon': 3, 'Duck': 8, 'Baby': 2} print(d) def getDictKeyandValue(dict,n): c=0 mylist=[] for i,j in d.items(): c+=1 if c==n: mylist=[i,j] break return mylist print(getDictKeyandValue(d,2))

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