Python dictionary - Copy and flip values-CodePudding

How to check if a key of multiple dictionary is the same and if it is copy the value and paste it to a new value but flipped.

dicts = [
{'A':1, 'B':2, 'C':10, 'D':100, 'E':None},
{'A':2, 'B':3, 'C':10, 'D':200, 'E':None},
{'A':3, 'B':4, 'C':20, 'D':300, 'E':None},
{'A':4, 'B':5, 'C':20, 'D':400, 'E':None},
{'A':5, 'B':6, 'C':30, 'D':500, 'E':None},
{'A':6, 'B':7, 'C':30, 'D':600, 'E':None}]

for x in dicts:
  for y in dicts:
    if y['C'] == x['C']:
       # do something and flip the A value as below and copy it to E
  print(dicts)
[{'A':1, 'B':2, 'C':10, 'D':100, 'E':2},
{'A':2, 'B':3, 'C':10, 'D':200, 'E':1},
{'A':3, 'B':4, 'C':20, 'D':300, 'E':4},
{'A':4, 'B':5, 'C':20, 'D':400, 'E':3},
{'A':5, 'B':6, 'C':30, 'D':500, 'E':7},
{'A':6, 'B':7, 'C':30, 'D':600, 'E':6}]

If there's just one match, just copy the value from A to E.

CodePudding user response：

You can solve this in linear time (i.e. not using a double loop) by keeping references to the found matches.

I am assuming here that there is a maximum of 2 matches per C:

dicts = [
{'A':1, 'B':2, 'C':10, 'D':100, 'E':None},
{'A':2, 'B':3, 'C':10, 'D':200, 'E':None},
{'A':3, 'B':4, 'C':20, 'D':300, 'E':None},
{'A':4, 'B':5, 'C':20, 'D':400, 'E':None},
{'A':5, 'B':6, 'C':30, 'D':500, 'E':None},
{'A':6, 'B':7, 'C':30, 'D':600, 'E':None}]

refs = {}

for i, d in enumerate(dicts):
    if d['C'] in refs:    # we already found the first match
        j = refs[d['C']]
        # swap the values
        dicts[j]['E'], dicts[i]['E'] = dicts[i]['A'], dicts[j]['A']
    else:                 # this is the first match
        refs[d['C']] = i  # keep a reference of the index
        # copy A to E in case of a single match
        d['E'] = d['A']

dicts

output:

[{'A': 1, 'B': 2, 'C': 10, 'D': 100, 'E': 2},
 {'A': 2, 'B': 3, 'C': 10, 'D': 200, 'E': 1},
 {'A': 3, 'B': 4, 'C': 20, 'D': 300, 'E': 4},
 {'A': 4, 'B': 5, 'C': 20, 'D': 400, 'E': 3},
 {'A': 5, 'B': 6, 'C': 30, 'D': 500, 'E': 6},
 {'A': 6, 'B': 7, 'C': 30, 'D': 600, 'E': 5}]

CodePudding user response：

if there are always 2 possile duplicate of 'C' you can just make sure that x and y is different and then assign 'E':

dicts = [
{'A':1, 'B':2, 'C':10, 'D':100, 'E':None},
{'A':2, 'B':3, 'C':10, 'D':200, 'E':None},
{'A':3, 'B':4, 'C':20, 'D':300, 'E':None},
{'A':4, 'B':5, 'C':20, 'D':400, 'E':None},
{'A':5, 'B':6, 'C':30, 'D':500, 'E':None},
{'A':6, 'B':7, 'C':30, 'D':600, 'E':None}]
for x in dicts:
  for y in dicts:
    if y['C'] == x['C'] and y is not x:
        x['E'] = y['A']
        y['E'] = x['A']
print(dicts)

Output:

[{'A': 1, 'B': 2, 'C': 10, 'D': 100, 'E': 2},
 {'A': 2, 'B': 3, 'C': 10, 'D': 200, 'E': 1},
 {'A': 3, 'B': 4, 'C': 20, 'D': 300, 'E': 4},
 {'A': 4, 'B': 5, 'C': 20, 'D': 400, 'E': 3},
 {'A': 5, 'B': 6, 'C': 30, 'D': 500, 'E': 6},
 {'A': 6, 'B': 7, 'C': 30, 'D': 600, 'E': 5}]

CodePudding user response：

Use itertools.combinations to loop over your dicts:

import itertools

dicts = [
{'A':1, 'B':2, 'C':10, 'D':100, 'E':None},
{'A':2, 'B':3, 'C':10, 'D':200, 'E':None},
{'A':3, 'B':4, 'C':20, 'D':300, 'E':None},
{'A':4, 'B':5, 'C':20, 'D':400, 'E':None},
{'A':5, 'B':6, 'C':30, 'D':500, 'E':None},
{'A':6, 'B':7, 'C':30, 'D':600, 'E':None}]

for x, y in itertools.combinations(dicts, 2):
    if x['E'] is None: # skip if already set
        if x['C'] == y['C']:
            x['E'] = y['A']
            y['E'] = x['A']

This works because there is only ever one matching pair in the example dicts. In case there is not, you should specify how to handle that case.

CodePudding user response：

Beware of the BigO : your 2 nested for loops equal to O(n^2).
You can reduce it with an hashtable.

dicts = [
    {"A": 1, "B": 2, "C": 10, "D": 100, "E": None},
    {"A": 2, "B": 3, "C": 10, "D": 200, "E": None},
    {"A": 3, "B": 4, "C": 20, "D": 300, "E": None},
    {"A": 4, "B": 5, "C": 20, "D": 400, "E": None},
    {"A": 5, "B": 6, "C": 30, "D": 500, "E": None},
    {"A": 6, "B": 7, "C": 30, "D": 600, "E": None},
]

c = {}

for i in range(len(dicts)):
    if not c.get(dicts[i]["C"]):
        c.update({dicts[i]["C"]: {"upper_row": i}})
    else:
        c[dicts[i]["C"]].update({"lower_row": i})

print(dicts)
print(c)

for value, duplicates in c.items():
    lower = duplicates.get("lower_row")
    if duplicates.get("lower_row"):
        upper = duplicates.get("upper_row")
        dicts[upper]["E"] = dicts[lower]["A"]
        dicts[lower]["E"] = dicts[upper]["A"]

print(dicts)

Output :

"""
Original state of dicts :

[
{'A': 1, 'B': 2, 'C': 10, 'D': 100, 'E': None},
{'A': 2, 'B': 3, 'C': 10, 'D': 200, 'E': None},
{'A': 3, 'B': 4, 'C': 20, 'D': 300, 'E': None},
{'A': 4, 'B': 5, 'C': 20, 'D': 400, 'E': None},
{'A': 5, 'B': 6, 'C': 30, 'D': 500, 'E': None},
{'A': 6, 'B': 7, 'C': 30, 'D': 600, 'E': None}]

c hashtable :

{10: {'upper_row': 0, 'lower_row': 1}, 20: {'upper_row': 2, 'lower_row': 3}, 30: {'upper_row': 4, 'lower_row': 5}}

Final state of dicts: 

[
{'A': 1, 'B': 2, 'C': 10, 'D': 100, 'E': 2}, 
{'A': 2, 'B': 3, 'C': 10, 'D': 200, 'E': 1},
{'A': 3, 'B': 4, 'C': 20, 'D': 300, 'E': 4},
{'A': 4, 'B': 5, 'C': 20, 'D': 400, 'E': 3},
{'A': 5, 'B': 6, 'C': 30, 'D': 500, 'E': 6},
{'A': 6, 'B': 7, 'C': 30, 'D': 600, 'E': 5}]
"""