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.
Related
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])))
Suppose that the range of parameters of interest are given a dictionary that contains the range for each parameter of interest:
G = {'a': [1,2], 'b': [3], 'c': [1, 2.5] }
The goal is to extract every parameter configuration on this grid. In the example above, there are 4 such, corresponding to 2 values of a, and two values of b:
G1 = {'a': 1, 'b': 3, 'c': 1 }
G2 = {'a': 2, 'b': 3, 'c': 1 }
G3 = {'a': 1, 'b': 3, 'c': 2.5 }
G4 = {'a': 2, 'b': 3, 'c': 2.5 }
It's straightforward to write two nested for loops to produce all such configurations, it becomes less trivial how to do it for a general case, when there are a variable number of lists in G.
The only solution that comes to my mind is to create a multi-index vector vec=[0,0] which is as long as the number of parameters, and increment to iterate over all possible configurations: [0,0] -> [1,0] -> [0,1] -> [1,1]:
G = {'a': [1,2], 'b': [3], 'c': [1, 2.5] }
def get_configs(G):
keys = list(G.keys())
lists = list(G.values())
sizes = [len(l) for l in lists]
num_confs = np.prod(sizes)
index = [0]*(len(G)+1)
configs = []
while len(configs)<num_confs:
configs.append( {keys[i]: lists[i][index[i]] for i in range(len(G))})
index[0] += 1
cur = 0
while len(configs)<num_confs and index[cur]>=sizes[cur]:
index[cur]=0
cur += 1
index[cur] += 1
return configs
configs = get_configs(G)
print(configs)
However, the solution seems a bit over-complicated and ugly. Is there a clean solution using python?
Here is a generalizable implementation using itertools.product:
from itertools import product
def dict_configs(d):
for vcomb in product(*d.values()):
yield dict(zip(d.keys(), vcomb))
Usage:
>>> G = {'a': [1,2], 'b': [3], 'c': [1, 2.5] }
>>> for config in dict_configs(G):
... print(config)
...
{'a': 1, 'b': 3, 'c': 1}
{'a': 1, 'b': 3, 'c': 2.5}
{'a': 2, 'b': 3, 'c': 1}
{'a': 2, 'b': 3, 'c': 2.5}
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
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.