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I have a dictionary defined as:
letters = {'a': 2, 'b': 1, 'c': 5}
I want to add values to this dictionary based on two lists: one which contains the keys and another which contains the values.
key_list = [a, c]
value_list = [2, 5]
This should give the output:
{a: 4, b: 1, c: 10}
Any ideas on how I can accomplish this? I am new to working with the dictionary structure so I apologise if this is extremely simple.
Thanks.
You can zip the two lists and then add to the dictionary as so;
letters = {'a': 2, 'b': 1, 'c': 5}
key_list = ['a', 'c']
value_list = [2, 5]
for k,v in zip(key_list, value_list):
letters[k] = letters.get(k, 0) + v
Using the dictionary's get() method as above allows you to add letters that aren't already in the dictionary.
for i in range(len(key_list)):
letters[key_list[i]] += value_list[i]
You can simply add or modify values from a dictionary using the key
For example:
letters = {'a': 2, 'b':1 , 'c': 5}
letters['a'] += 2
letters['c'] += 5
print(letters)
output = {'a': 4, 'b': 1, 'c': 10}
Given
listOfDict = [{'ref': 1, 'a': 1, 'b': 2, 'c': 3},
{'ref': 2, 'a': 4, 'b': 5, 'c': 6},
{'ref': 3, 'a': 7, 'b': 8, 'c': 9}]
Lets' consider a list of permutable integer
[7,8,9]=[7,9,8]=[8,7,9]=[8,9,7]=[9,7,8]=[9,8,7] # (3!)
Each of this list has a unique mapping ref, so how given for (8,7,9) can I get ref=3 ?
Also in real case I might until 10 (a,b,c,d,e,f,g,h,i,j)...
You can generate a dictionary that maps the values as frozenset to the value of ref:
listOfDict = [{'ref': 1, 'a': 1, 'b': 2, 'c': 3},
{'ref': 2, 'a': 4, 'b': 5, 'c': 6},
{'ref': 3, 'a': 7, 'b': 8, 'c': 9}]
keys = ['a', 'b', 'c']
out = {frozenset(d[k] for k in keys): d['ref'] for d in listOfDict}
# {frozenset({1, 2, 3}): 1,
# frozenset({4, 5, 6}): 2,
# frozenset({7, 8, 9}): 3}
example:
check = frozenset((8,7,9))
out[check]
# 3
but I don't know in advance the name of the other keys!
Then use this approach:
out = {}
for d in listOfDict:
d2 = d.copy() # this is to avoid modifying the original object
out[frozenset(d2.values())] = d2.pop('ref')
out
or as a comprehension:
out = dict(((d2:=d.copy()).pop('ref'), frozenset(d2.values()))[::-1]
for d in listOfDict)
Here is a commented solution to your problem. The idea is to compare the sorted list of the values in a, b, c etc with the sorted values in list_of_ints. The sorted values will be the same for all permutations of a given set of numbers.
def get_ref(list_of_ints):
# Loop through dictionaries in listOfDict.
for dictionary in listOfDict:
# Get list of values in each dictionary.
vals = [dictionary[key] for key in dictionary if key != "ref"]
if sorted(vals) == sorted(list_of_ints):
# If sorted values are equal to sorted list of ints, return ref.
return dictionary["ref"])
By the way, I believe it would be cleaner to structure this data as a dict of dicts in the following way:
dicts = {
1: {'a': 1, 'b': 2, 'c': 3},
2: {'a': 4, 'b': 5, 'c': 6},
3: {'a': 7, 'b': 8, 'c': 9}
}
The code would then be:
def get_ref(list_of_ints):
for ref, dictionary in dicts.items():
if sorted(dictionary.values()) == sorted(list_of_ints):
return ref
Assuming that all integers in the permutations are unique, the code can be simplified further using sets instead of sorted lists.
Since its a list of dict I can call each dict as it self by using for loop
and record the first number on ref
for i in listOfDict:
ref_num=i["ref"]
and to turn dictunary to list we simply use:
z=list(i.values())
then the last step is to find if its the same input list if so we print/return the ref number
if z[1:]==InputList:
return ref_num
and the code should be like this:
listOfDict = [
{"ref": 1,
"a": 1,
"b": 2,
"c": 3},
{"ref": 2,
"a": 4,
"b": 5,
"c": 6},
{"ref": 3,
"a": 7,
"b": 8,
"c": 9},]
def find_ref_Num(InputList):
for i in listOfDict:
ref_num=i["ref"]
z=list(i.values())
if z[1:]==InputList:
return ref_num
print ("your ref number is: "+str(find_ref_Num([7,8,9])))
I have a dictionary with product names and prices:
products = {'a': 2, 'b': 3, 'c': 4, 'd': 5, 'e': 6, 'f': 7, 'g': 8}
And a list with amounts of each product:
amounts = [3, 0, 5, 1, 3, 2, 0]
I want to get an output shown there total price of that order.
Not using functions I seem to get it right:
products = {'a': 2, 'b': 3, 'c': 4, 'd': 5, 'e': 6, 'f': 7, 'g': 8}
amounts = [3, 0, 5, 1, 3, 2, 0]
res_list = []
order = []
for value in products.values():
res_list.append(value)
for i in range(0, len(res_list)):
order.append(amounts[i] * res_list[i])
total = sum(order)
print(res_list)
print(order) #this line and the one above are not really necessary
print(total)
Output : 63
But when I try using this code within a function I am having some problems. this is what I have tried:
products = {'a': 2, 'b': 3, 'c': 4, 'd': 5, 'e': 6, 'f': 7, 'g': 8}
amounts = [3, 0, 5, 1, 3, 2, 0]
#order = []
def order(prod):
res_list = []
for value in prod.values():
res_list.append(value)
return res_list
prices = order(products)
print(prices)
def order1(prices):
order =[]
for i in range(0, len(prices)):
order.append(amounts[i] * prices[i])
total = sum(order)
return total
print(order1(prices))
Not working the way it is intended.
Thanks for all the help I am learning.
The immediate problem is that your lines:
total = sum(order)
return total
are indented too much, so that they are inside the for loop. Outside of a function, the bug does not matter too much, because all that happens is that the total is recalculated on every iteration but the final value is the one that is used. But inside the function, what will happen is that it will return on the first iteration.
Reducing the indentation so that it is outside the for loop will fix this.
def order1(prices):
order =[]
for i in range(0, len(prices)):
order.append(amounts[i] * prices[i])
total = sum(order)
return total
However, separate from that, you are relying on the order within the dictionary, which is only guaranteed for Python 3.7 and more recent. If you want to allow the code to be run reliably on earlier versions of Python, you can use an OrderedDict.
from collections import OrderedDict
products = OrderedDict([('a', 2), ('b', 3), ('c', 4), ('d', 5),
('e', 6), ('f', 7), ('g', 8)])
Incidentally, your order function is unnecessary. If you want to convert products.values() (a dictionary values iterator) to a list, just use:
prices = list(products.values())
Also, in order1 it is unnecessary to build up an order list and sum it - you could use:
total = 0
for i in range(0, len(prices)):
total += amounts[i] * prices[i]
That is probably enough to be getting on with for now, but if you wish to make a further refinement, then look up about how zip is used, and think how it could be used with your loop over amounts and prices.
Just zip products.values() and amounts, find the product of each pair, and then finally sum the result
>>> products = {'a': 2, 'b': 3, 'c': 4, 'd': 5, 'e': 6, 'f': 7, 'g': 8}
>>> amounts = [3, 0, 5, 1, 3, 2, 0]
>>>
>>> sum(i*j for i,j in zip(products.values(), amounts))
63
You can do this.
products = {'a': 2, 'b': 3, 'c': 4, 'd': 5, 'e': 6, 'f': 7, 'g': 8}
amounts = [3, 0, 5, 1, 3, 2, 0]
def order(products, amounts):
res_list = []
order = []
for value in products.values():
res_list.append(value)
for i in range(0, len(res_list)):
order.append(amounts[i] * res_list[i])
total = sum(order)
print(res_list)
print(order) #this line and the one above are not really necessary
print(total)
return(total)
order(products, amounts)
You don't really need to iterate twice assuming that the amount of items in products and in amounts is the same.
products = {'a': 2, 'b': 3, 'c': 4, 'd': 5, 'e': 6, 'f': 7, 'g': 8}
amounts = [3, 0, 5, 1, 3, 2, 0]
def order(products: dict, amounts: list):
total = 0
for idx, (_key, val) in enumerate(products.items()):
total = total + amounts[idx] * val
return total
print(order(products, amounts))
Note: The order of the items in the dictionary is not guaranteed, you might want to look into different data structures that link together products and amounts in a better way, i.e.:
products = {'a': 2, 'b': 3, 'c': 4, 'd': 5, 'e': 6, 'f': 7, 'g': 8}
amounts = {'a': 3, 'b': 0, 'c': 5, 'd': 1, 'e': 3, 'f': 2, 'g': 0}
In this way you could do this:
def order(products: dict, amounts: dict):
total = 0
for key, val in products.items():
total = total + val * amounts[key]
return total
print(order(products, amounts))
Once we're at it, let's get fancy with numpy, since in the end, you just want the dot product prices x amounts:
import numpy as np
total = np.dot(list(products.values()), amounts)
63
But seriously, I'd strictly use either lists or dicts for both datasets, not mix them up, since that can seriously cause problems with order syncronisation between them, even if you are on Python 3.7 with the changes made there as mentioned.
Im trying to multiple some values from dictionary
example
price_list = {'a': 3, 'b': 2, 'c': 5, 'd': 10}
when i type
total=sum(price_list.values())
print("Total sum is ",total)
it result 20
But now i want to multiple a with 3, b with 5, c with 2 and d with 3 and my desired output to be 59. What is easiest way to do that?
Assuming your numbers are stored in the list, iterate through the values, and multiply with your required number like so
price_dict = {'a': 3, 'b': 2, 'c': 5, 'd': 10}
numbers_dict = {'a': 3, 'b': 5, 'c': 2, 'd': 3}
result = 0
for key, value in price_dict.items():
result += numbers_dict[key] * value
print(result)
#59
You can just perform operations on the dictionary item like you would any other variable:
# multiply 'a' by 3
price_list['a'] *= 3
Try this:
price_list = {'a': 3, 'b': 2, 'c': 5, 'd': 10}
numbers = [3,5,2,3]
for k,n in list(zip(price_list, numbers)):
price_list[k] *= n
then the price list will change, you can use sum as you did to calculate the result.
How can I iterate over only X number of dictionary items? I can do it using the following bulky way, but I am sure Python allows a more elegant way.
d = {'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5}
x = 0
for key in d:
if x == 3:
break
print key
x += 1
If you want a random sample of X values from a dictionary you can use random.sample on the dictionary's keys:
from random import sample
d = {'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5}
X = 3
for key in sample(d, X):
print key, d[key]
And get output for example:
e 5
c 3
b 2