Python newb here looking for some assistance...
For a variable number of dicts in a python list like:
list_dicts = [
{'id':'001', 'name':'jim', 'item':'pencil', 'price':'0.99'},
{'id':'002', 'name':'mary', 'item':'book', 'price':'15.49'},
{'id':'002', 'name':'mary', 'item':'tape', 'price':'7.99'},
{'id':'003', 'name':'john', 'item':'pen', 'price':'3.49'},
{'id':'003', 'name':'john', 'item':'stapler', 'price':'9.49'},
{'id':'003', 'name':'john', 'item':'scissors', 'price':'12.99'},
]
I'm trying to find the best way to group dicts where the value of key "id" is equal, then add/merge any unique key:value and create a new list of dicts like:
list_dicts2 = [
{'id':'001', 'name':'jim', 'item1':'pencil', 'price1':'0.99'},
{'id':'002', 'name':'mary', 'item1':'book', 'price1':'15.49', 'item2':'tape', 'price2':'7.99'},
{'id':'003', 'name':'john', 'item1':'pen', 'price1':'3.49', 'item2':'stapler', 'price2':'9.49', 'item3':'scissors', 'price3':'12.99'},
]
So far, I've figured out how to group the dicts in the list with:
myList = itertools.groupby(list_dicts, operator.itemgetter('id'))
But I'm struggling with how to build the new list of dicts to:
1) Add the extra keys and values to the first dict instance that has the same "id"
2) Set the new name for "item" and "price" keys (e.g. "item1", "item2", "item3"). This seems clunky to me, is there a better way?
3) Loop over each "id" match to build up a string for later output
I've chosen to return a new list of dicts only because of the convenience of passing a dict to a templating function where setting variables by a descriptive key is helpful (there are many vars). If there is a cleaner more concise way to accomplish this, I'd be curious to learn. Again, I'm pretty new to Python and in working with data structures like this.
Try to avoid complex nested data structures. I believe people tend to grok them only while they are intensively using the data structure. After the program is finished, or is set aside for a while, the data structure quickly becomes mystifying.
Objects can be used to retain or even add richness to the data structure in a saner, more organized way. For instance, it appears the item
and price
always go together. So the two pieces of data might as well be paired in an object:
class Item(object):
def __init__(self,name,price):
self.name=name
self.price=price
Similarly, a person seems to have an id
and name
and a set of possessions:
class Person(object):
def __init__(self,id,name,*items):
self.id=id
self.name=name
self.items=set(items)
If you buy into the idea of using classes like these, then your list_dicts
could become
list_people = [
Person('001','jim',Item('pencil',0.99)),
Person('002','mary',Item('book',15.49)),
Person('002','mary',Item('tape',7.99)),
Person('003','john',Item('pen',3.49)),
Person('003','john',Item('stapler',9.49)),
Person('003','john',Item('scissors',12.99)),
]
Then, to merge the people based on id
, you could use Python's reduce
function,
along with take_items
, which takes (merges) the items from one person and gives them to another:
def take_items(person,other):
'''
person takes other's items.
Note however, that although person may be altered, other remains the same --
other does not lose its items.
'''
person.items.update(other.items)
return person
Putting it all together:
import itertools
import operator
class Item(object):
def __init__(self,name,price):
self.name=name
self.price=price
def __str__(self):
return '{0} {1}'.format(self.name,self.price)
class Person(object):
def __init__(self,id,name,*items):
self.id=id
self.name=name
self.items=set(items)
def __str__(self):
return '{0} {1}: {2}'.format(self.id,self.name,map(str,self.items))
list_people = [
Person('001','jim',Item('pencil',0.99)),
Person('002','mary',Item('book',15.49)),
Person('002','mary',Item('tape',7.99)),
Person('003','john',Item('pen',3.49)),
Person('003','john',Item('stapler',9.49)),
Person('003','john',Item('scissors',12.99)),
]
def take_items(person,other):
'''
person takes other's items.
Note however, that although person may be altered, other remains the same --
other does not lose its items.
'''
person.items.update(other.items)
return person
list_people2 = [reduce(take_items,g)
for k,g in itertools.groupby(list_people, lambda person: person.id)]
for person in list_people2:
print(person)
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