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.
Related
My goal is to find a more efficient way to get all combinations of 1 to r mixed elements, where each family of element potentially has a different count and r is a parameter. The elements can be any (hashable) type. The result is a list of Counter-like dictionaries.
Here is an example data:
example = {1e-8: 3, "k": 2}
r = 5 # sum(example.values()) == 5 therefore all possible combinations for this example
The expected result is the following:
[{1e-08: 1},
{'k': 1},
{1e-08: 2},
{1e-08: 1, 'k': 1},
{'k': 2},
{1e-08: 3},
{1e-08: 2, 'k': 1},
{1e-08: 1, 'k': 2},
{1e-08: 3, 'k': 1},
{1e-08: 2, 'k': 2},
{1e-08: 3, 'k': 2}]
... correspondong to every possible combinations of 1, 2, 3, 4 and 5 elements.
The order preservation of the list is preferable (since Python 3.7+ preserves the order of keys inside dictionaries) but not mandatory.
Here is the solution I currently use:
from more_itertools import distinct_combinations
from collections import Counter
def all_combis(elements, r=None):
if r is None:
r = sum(elements.values())
# "Flattening" by repeating the elements according to their count
flatt = []
for k, v in elements.items():
flatt.extend([k] * v)
for r in range(1, r+1):
for comb in distinct_combinations(flatt, r):
yield dict(Counter(comb))
list(all_combis(example))
# > The expected result
A real-life example has 300 elements distributed among 15 families. It is processed in ~13 seconds with a value of r=10 for about 2 million combinations, and ~31 seconds with r=11 for 4.5 million combinations.
I'm guessing there are better ways which avoid "flattening" the elements and/or counting the combinations, but I struggle to find any when each element has a different count.
Can you design a more time-efficient solution ?
The keys are a bit of a distraction. They can be added in later. Mathematically, what you have is a vector of bounds, together with a global bound, and want to generate all tuples where each element is bounded by its respective bound, and the total is bounded by the global bound. This leads to a simple recursive approach based on the idea that if
(a_1, a_2, ..., a_n) <= (b_1, b_2, ..., b_n) with a_1 + ... a_n <= k
then
(a_2, ..., a_n) <= (b_2, ..., b_n) with a_2 + ... a_n <= k - a_1
This leads to something like:
def bounded_tuples(r,bounds):
n = len(bounds)
if r == 0:
return [(0,)*n]
elif n == 0:
return [()]
else:
tuples = []
for i in range(1+min(r,bounds[0])):
tuples.extend((i,)+t for t in bounded_tuples(r-i,bounds[1:]))
return tuples
Note that this includes the solution with all 0's -- which you exclude, but that can be filtered out and the keys reintroduced:
def all_combis(elements, r=None):
if r is None:
r = sum(elements.values())
for t in bounded_tuples(r,list(elements.values())):
if max(t) > 0:
yield dict(zip(elements.keys(),t))
For example:
example = {1e-8: 3, "k": 2}
for d in all_combis(example):
print(d)
Output:
{1e-08: 0, 'k': 1}
{1e-08: 0, 'k': 2}
{1e-08: 1, 'k': 0}
{1e-08: 1, 'k': 1}
{1e-08: 1, 'k': 2}
{1e-08: 2, 'k': 0}
{1e-08: 2, 'k': 1}
{1e-08: 2, 'k': 2}
{1e-08: 3, 'k': 0}
{1e-08: 3, 'k': 1}
{1e-08: 3, 'k': 2}
Which is essentially what you have. The code could obviously be tweaked to eliminate dictionary entries with the value 0.
Timing with larger examples seems to suggest that my approach isn't any quicker than yours, though it still might give you some ideas.
As #John Coleman said without the keys you may be able to speed things up.
This recursive approach starts at the end of the list and iterates until either the max sum is reached, or the max value of that element.
It returns a list, but as #John Coleman also showed, it is easy to add the keys later.
From my tests it appears to run in about half the time as your current implementation.
def all_combis(elements, r=None):
if r is None:
r = sum(elements)
if r == 0:
yield [0] * len(elements)
return
if not elements:
yield []
return
elements = list(elements)
element = elements.pop(0)
for i in range(min(element + 1, r + 1)):
for combi in all_combis(elements, r - i):
yield [i] + combi
example = {1e-8: 3, "k": 2}
list(all_combis([val for val in example.values()]))
Output:
[[0, 0], [0, 1], [0, 2], [1, 0], [1, 1], [1, 2], [2, 0], [2, 1], [2, 2], [3, 0], [3, 1], [3, 2]]
Suppose I have the following dictionary:
{'a': 0, 'b': 1, 'c': 2, 'c.1': 3, 'd': 4, 'd.1': 5, 'd.1.2': 6}
I wish to write an algorithm which outputs the following:
{
"a": 0,
"b": 1,
"c": {
"c": 2,
"c.1": 3
},
"d":{
"d": 4,
"d.1": {
"d.1": 5,
"d.1.2": 6
}
}
}
Note how the names are repeated inside the dictionary. And some have variable level of nesting (eg. "d").
I was wondering how you would go about doing this, or if there is a python library for this? I know you'd have to use recursion for something like this, but my recursion skills are quite poor. Any thoughts would be highly appreciated.
You can use a recursive function for this or just a loop. The tricky part is wrapping existing values into dictionaries if further child nodes have to be added below them.
def nested(d):
res = {}
for key, val in d.items():
t = res
# descend deeper into the nested dict
for x in [key[:i] for i, c in enumerate(key) if c == "."]:
if x in t and not isinstance(t[x], dict):
# wrap leaf value into another dict
t[x] = {x: t[x]}
t = t.setdefault(x, {})
# add actual key to nested dict
if key in t:
# already exists, go one level deeper
t[key][key] = val
else:
t[key] = val
return res
Your example:
d = {'a': 0, 'b': 1, 'c': 2, 'c.1': 3, 'd': 4, 'd.1': 5, 'd.1.2': 6}
print(nested(d))
# {'a': 0,
# 'b': 1,
# 'c': {'c': 2, 'c.1': 3},
# 'd': {'d': 4, 'd.1': {'d.1': 5, 'd.1.2': 6}}}
Nesting dictionary algorithm ...
how you would go about doing this,
sort the dictionary items
group the result by index 0 of the keys (first item in the tuples)
iterate over the groups
if there are is than one item in a group make a key for the group and add the group items as the values.
Slightly shorter recursion approach with collections.defaultdict:
from collections import defaultdict
data = {'a': 0, 'b': 1, 'c': 2, 'c.1': 3, 'd': 4, 'd.1': 5, 'd.1.2': 6}
def group(d, p = []):
_d, r = defaultdict(list), {}
for n, [a, *b], c in d:
_d[a].append((n, b, c))
for a, b in _d.items():
if (k:=[i for i in b if i[1]]):
r['.'.join(p+[a])] = {**{i[0]:i[-1] for i in b if not i[1]}, **group(k, p+[a])}
else:
r[b[0][0]] = b[0][-1]
return r
print(group([(a, a.split('.'), b) for a, b in data.items()]))
Output:
{'a': 0, 'b': 1, 'c': {'c': 2, 'c.1': 3}, 'd': {'d': 4, 'd.1': {'d.1': 5, 'd.1.2': 6}}}
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.
I have the following:
d = {"a":3,"b":2,"c":3,"d":2,"e":2,"f":3,"g":4, "h":6}
m = {v: i+1 for i,v in enumerate(sorted(set(d.values()),reverse=True))}
r = {k:m[d[k]] for k in d}
where r is:
{'a': 3, 'd': 4, 'b': 4, 'c': 3, 'e': 4, 'f': 3, 'g': 2, 'h': 1}
So "h" has the highest value, 6, in d so it is remapped to 1 in r. Then 'g' is ranked 2 since it has the next highest value, 4 in d.
My solution works fine but I was wondering if there is a more elegant solution.
Python dicts don't keep order. If you want that you need an OrderedDict.
Use Counter to get the ranks. Then turn that into a list of tuples or into an OrderedDict.
from collections import Counter, OrderedDict
d = {"a":3,"b":2,"c":3,"d":2,"e":2,"f":3,"g":4, "h":6}
c = Counter(d)
# if you want a list of tuples
ranked_list = [(pair[0],rank+1) for rank,pair in enumerate(c.most_common())]
# [('h', 1),('g', 2),('f', 3),('a', 4),('c', 5),('b', 6),('d', 7), ('e', 8)]
# if you want a dict:
ranked_dict = OrderedDict(ranked_list)
# OrderedDict([('h', 1),('g', 2),('f', 3),('a', 4),('c', 5),('b', 6),('d', 7), ('e', 8)])
You can use this:
d = {"a":3,"b":2,"c":3,"d":2,"e":2,"f":3,"g":4, "h":6}
# sort the dictionary items by -value, throw away old value and use the
# enumerate position starting at 1 instead - no backreferencing in the old
# dict needed here
k = {k:idx for idx,(k,_) in enumerate(sorted(d.items(), key = lambda x:-x[1]),1)}
print(k)
Output:
{'h': 1, 'g': 2, 'a': 3, 'c': 4, 'f': 5, 'b': 6, 'd': 7, 'e': 8}
def ranker(d):
ranks = sorted(set(d.values()),reverse=True)
ranks = {r:i+1 for i,r in enumerate(ranks)}
return {k: ranks[v] for k,v in d.items()}
I have a dictionary that looks like this :
{1224:{'A': 6, 'B': 4, 'C': 5}, 1225: {'A': 6, 'B': 6, 'C': 5}}
I want to store the total of A in each key and get a result like this :
{1224:{'A': 6, 'B': 4, 'C': 5, 'Total_A' : 6}, 1225: {'A': 6, 'B': 6, 'C': 5, 'Total_A' : 12}}
Total_A being the A value in first key (1224) + A value in next key (1225).
I tried this :
for d in celldict.values():
sum = 0
sum += d.get('A',0)
d['TOTAL_A'] = sum
But it doesn't sum anything and it only returns the A value for each key every time.
i think you should know what happened in the loop.
correct answer is follow:
sum = 0
for d in celldict.values():
sum += d.get('A',0)
d['TOTAL_A'] = sum
The problem is, that you reset sum on each iteration. This is why sum never accumulates the previous values.
Move sum = 0 outside the loop.