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Python Find permutable list in a dict list

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])))

Combinations with max length and per element max repetition values

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]]

Nesting dictionary algorithm

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}}}

Merge 2 dictionary and Remove duplicate and subtract value of keys of duplicate element

I have a 2 text file in which I m storing dictionary values and reading it.
so I m reading file1.txt and storing the value in the counter variable and for file2.txt I m storing the value in the counter1 variable.
when i m printing counter and counter1 it gives following output.
counter= {'laptop': 5, 'chair':6 , 'tv': 3}
counter1= {'laptop': 2, 'chair': 4, 'tv': 4,'plant':2}
p=len(counter)
for z in range (0 ,p):
if (list(counter.values())[v]!=list(counter1.values())[v]):
print("Objects value on main list %s and Second List %s "%(list(counter.items())[v],list(counter1.items())[v]))
else:
pass
v=v+1
i'd like to return this:
[{'laptop': 3}, {'chair': 2}, {'tv': 1}, {'plant': 2}]
subtract the value of counter and counter1 and print the result.

use this
counter= {'laptop': 5, 'chair':6 , 'tv': 3}
counter1= {'laptop': 2, 'chair': 4, 'tv': 4,'plant':2}
out = []
for key, val in counter.items():
if key in counter1:
out.append({key:abs(val-counter1[key])})
else:
out.append({key:val})
for key, val in counter1.items():
if key not in counter:
out.append({key:val})
print(out)

counter= {'laptop': 5, 'chair':6 , 'tv': 3}
counter1= {'laptop': 2, 'chair': 4, 'tv': 4,'plant':2}
out = [{key : abs(counter1[key] - counter[key])} if ((key in counter) and (key in counter1)) else {key : counter[key]} if (key in counter) else {key:counter1[key]} for key in set(counter.keys() + counter1.keys())]
print out
It seems like it can be done in 1 line, this should work.

Python set an item value based other items in a dictionary

A = {0:{a:1, b:7}, 1:{a:5,b:5}, 2:{a:4,b:6}}
I want to attach an item guess to each sub dictionary based on the value b accounting of all b's in each sub dictionary.
Saying, in Dictionary A:
0-b-7 percentage of b: 7/(7+5+6)
1-b-5 percentage of b: 5/(7+5+6)
2-b-6 percentage of b: 1 - 7/(7+5+6) - 5/(7+5+6)
The desired Dictionary should be like
A = {0:{a:1, b:7, 'guess': 7/(7+5+6)},
1:{a:5,b:5, 'guess': 5/(7+5+6)},
2:{a:4,b:6, 'guess': 1 - 7/(7+5+6) - 5/(7+5+6)}}
I don't know how to incorporate the other two b's for a specific subdictionary.

One approach is to precompute the sum of all bs and then use it to add a new key-value pair to your dictionary.
b_total = float(sum(A[k]['b'] for k in A))
for k in A:
A[k]['guess'] = A[k]['b'] / b_total
#{0: {'a': 1, 'b': 7, 'guess': 0.3888888888888889},
# 1: {'a': 5, 'b': 5, 'guess': 0.2777777777777778},
# 2: {'a': 4, 'b': 6, 'guess': 0.3333333333333333}}

A = {0:{"a":1, "b":7}, 1:{"a":5,"b":5}, 2:{"a":4,"b":6}}
char = "b"
denominator = 0
# =========================
# First Calculate the sum
# =========================
for key in A:
inner_map = A[key]
denominator += inner_map[char]
# ========================================
# Now insert the new key to the inner_map
# ========================================
for key in A:
inner_map = A[key]
inner_map["guess"] = inner_map[char]/denominator
print(A)
Output:
{0: {'a': 1, 'b': 7, 'guess': 0.3888888888888889}, 1: {'a': 5, 'b': 5, 'guess': 0.2777777777777778}, 2: {'a': 4, 'b': 6, 'guess': 0.3333333333333333}}

Try this:
def add_calc(my_dict):
total_guesses = sum(map(lambda x: my_dict.get(x).get('b'), my_dict))
for item in my_dict.itervalues():
item.update({'guess': 1.0 * item.get('b') / total_guesses})
return my_dict
d = add_calc(A)
{0: {'a': 1, 'b': 7, 'guess': 0.3888888888888889},
1: {'a': 5, 'b': 5, 'guess': 0.2777777777777778},
2: {'a': 4, 'b': 6, 'guess': 0.3333333333333333}}
I'm on Python 2 btw, you didn't specify version

You can use dictionary unpacking:
A = {0:{'a':1, 'b':7}, 1:{'a':5, 'b':5}, 2:{'a':4, 'b':6}}
results = {a:{**b, **{'guess':b['b']/float(sum(c['b'] for _, c in A.items()))}} for a, b in A.items()}
Output:
{0: {'guess': 0.3888888888888889, 'b': 7, 'a': 1}, 1: {'guess': 0.2777777777777778, 'b': 5, 'a': 5}, 2: {'guess': 0.3333333333333333, 'b': 6, 'a': 4}}