How to split a dictionary?

Question

I want to split a dictionary:

Dict1 = {1 : "dallas", 2 : "texas", 3 : "California" , 4 : "New Orleans"}

How do I split it into a half so that the result is:

Dict1 = {1: "dallas" , 2: "texas"}
Dict2 = {3 : "California" , 4 : "New Orleans"}

Answer 1

well in short you can't, but to do this you'll have to use a list to separate dict which means that the splitting will be the order of the original dict, if you have like

data = {2: "foo", 1: "bar"}

the splitted dict will be in that order.
So back to the splitting, dict have a method called "items()" which returns an iterator that contains (key, value) format on each elements so we want to take that in a list like

items = list(Dict1.items()) # we use list() because items() returns an iterator instead of a list

now we have the dict converted to a list and we can go ahead and split the list.

items = list(Dict1.items())
center = len(items)//2
Dict1 = dict(items[:center])
Dict2 = dict(items[center:])

and now we have the dict splitted yay!
also this code only works if the dict has even elements like 10 or 2 elements

Answer 2

also a suggestion: don't use PascalCase for normal variable instead use snake_case so instead of Dict1 you would use dict1. NEVER USE camelCase IN PYTHON

Answer 3

and extra information: __setitem__ doesn't call update and update doesn't call __setitem__

Answer 4

Rei,
I tried your logic in a code and it works really well 👍

It also works for odd number of items, even without the check for odd/even number of items.

When dictionary has odd number of items, the second split will get more items than the first. By incrementing <center> the second split will get more.

Can't wait for the OP to try it out!

Answer 5

Well provided that the length is divisible by 2, try this.

def dict_split(dictionary):
dict1 = dict()
dict 2 = dict ()
for key, value in dictionary.items():
if len(dict1) < len(dictionary)/2:
dict1.update({key: value})
else:
dict2.update({key: value})
return dict 1, dict2
def main():
dictionary = {1:"dallas", 2:"texas", 3:"California, 4 : " Orleans"}
dict1, dict2 = dict_split(dictionary)
print (dict1)
print (dict2)
if __name__ == "__main__":
main()

Answer 6

example:

if (len(items) % 2) != 0:
center += 1

Answer 7

Rei why not camelCase?

Answer 8

Or something like this:

d1 = {"e":0, "f":0}

print(d1.items()[0:(len(d1)/2)])
print(d1.items()[(len(d1)/2)::])

if items %2!=0 parts of same size not possible, as far I get it.
I tryed with 3 elements in dict, it split it in groups of 1 and 2 elements.

Answer 9

Nwachi Faith btw its kinda not useful using dict.update() at this case, since we're just setting a key might as well just dict1[key] = value

Answer 10

Nwachi Faith but update is supposed to be similar to list.extend, for adding a single element to a dict by using dict.update is overkill, you're better off with dict[key] = value, unless you already have a dict you want to use or multiple key to map

Answer 11

What should it be like when the original dictionary contains odd number of items?

Answer 12

well it'll be the same except you swap the assignment, so Dict1 is Dict2 and Dict2 is Dict1

Answer 13

oh wait nevermind that's bad

Answer 14

okay so basically you can check if the length of the items is an odd number so then we can add 1 to the center variable

Answer 15

oh my bad

Answer 16

Oma Falk because python naming convention, for classes its mandatory to use PascalCase while normal variable or functions are snake_case camelCase are never used in python except for old modules, and using camelCase is generally discouraged in python

Answer 17

Rei It does the same thing. update () method also adds key: value pairs to the dictionary as much dict [key] = value does. So win win.