find extra key in dict by value-CodePudding

I have to dicts like this

new_ids = {10722774: 5537170, 10722775: 5537171, 10722776: 5537172, 10722777: 5537173, 10722778: 5537174, 10722779: 5537175, 10722780: 5537176, 10722781: 5537177, 10722782: 5537178}

old_ids = {10722773: 5537179, 10722764: 5537170, 10722765: 5537171, 10722766: 5537172, 10722767: 5537173, 10722768: 5537174, 10722769: 5537175, 10722770: 5537176, 10722771: 5537177, 10722772: 5537178}

values in them are the same, just new_ids have 9 of them and old_ids have 10. how is to find this extra key that exists in new_ids?

value of 5537179 in old_ids does not exist in new_ids, so i need to get a key 10722773

CodePudding user response：

Given the clarification:

new_ids = {
    10722774: 5537170,
    10722775: 5537171,
    10722776: 5537172,
    10722777: 5537173,
    10722778: 5537174,
    10722779: 5537175,
    10722780: 5537176,
    10722781: 5537177,
    10722782: 5537178,
}
old_ids = {
    10722773: 5537179,
    10722764: 5537170,
    10722765: 5537171,
    10722766: 5537172,
    10722767: 5537173,
    10722768: 5537174,
    10722769: 5537175,
    10722770: 5537176,
    10722771: 5537177,
    10722772: 5537178,
}

# Flip key-value to value-key
flipped_old = {v: k for k, v in old_ids.items()}
flipped_new = {v: k for k, v in new_ids.items()}
flipped_combined = flipped_old | flipped_new
# Find keys that aren't shared between the flipped dicts
uncommon_keys = set(flipped_old) ^ set(flipped_new)
# Generate a dict that unflips the key-values containing only the uncommon keys
uncommon = {flipped_combined[v]: v for v in uncommon_keys}
print(uncommon)

prints

{10722773: 5537179}

CodePudding user response：

First way

_1. only key


new_ids = {10722774: 5537170, 10722775: 5537171, 10722776: 5537172, 10722777: 5537173, 10722778: 5537174, 10722779: 5537175, 10722780: 5537176, 10722781: 5537177, 10722782: 5537178}

old_ids = {10722773: 5537179, 10722764: 5537170, 10722765: 5537171, 10722766: 5537172, 10722767: 5537173, 10722768: 5537174, 10722769: 5537175, 10722770: 5537176, 10722771: 5537177, 10722772: 5537178}

notLst = [k for k,v in old_ids.items() if v not in new_ids.values()]

print(notLst)

OUTPUT

[10722773]

_2. both key and value


new_ids = {10722774: 5537170, 10722775: 5537171, 10722776: 5537172, 10722777: 5537173, 10722778: 5537174, 10722779: 5537175, 10722780: 5537176, 10722781: 5537177, 10722782: 5537178}

old_ids = {10722773: 5537179, 10722764: 5537170, 10722765: 5537171, 10722766: 5537172, 10722767: 5537173, 10722768: 5537174, 10722769: 5537175, 10722770: 5537176, 10722771: 5537177, 10722772: 5537178}

notLst = {k:v for k,v in old_ids.items() if v not in new_ids.values()}

print(notLst)

Output

{10722773: 5537179}

There are more ways too. Fastest way also

Second Way

You can use a set way too to solve the problem

new_ids = {10722774: 5537170, 10722775: 5537171, 10722776: 5537172, 10722777: 5537173, 10722778: 5537174, 10722779: 5537175, 10722780: 5537176, 10722781: 5537177, 10722782: 5537178}

old_ids = {10722773: 5537179, 10722764: 5537170, 10722765: 5537171, 10722766: 5537172, 10722767: 5537173, 10722768: 5537174, 10722769: 5537175, 10722770: 5537176, 10722771: 5537177, 10722772: 5537178}

# n = set.symmetric_difference(set(new_ids.values()),set(old_ids.values()))
n = set.symmetric_difference(*map(set,(new_ids.values(),old_ids.values())))

notLst = {k:v for k,v in old_ids.items() if v in n} # both key Value
# notLst = [k for k,v in old_ids.items() if v in n] # Only key

print(notLst)

Output

{10722773: 5537179} # for both key value
[10722773] # for only key

Time difference between my second way and the @AKK way

from timeit import timeit

new_ids = {10722774: 5537170, 10722775: 5537171, 10722776: 5537172, 10722777: 5537173, 10722778: 5537174, 10722779: 5537175, 10722780: 5537176, 10722781: 5537177, 10722782: 5537178}

old_ids = {10722773: 5537179, 10722764: 5537170, 10722765: 5537171, 10722766: 5537172, 10722767: 5537173, 10722768: 5537174, 10722769: 5537175, 10722770: 5537176, 10722771: 5537177, 10722772: 5537178}

def myway():
    # n = set.symmetric_difference(set(new_ids.values()),set(old_ids.values()))
    n = set.symmetric_difference(*map(set,(new_ids.values(),old_ids.values())))

    notLst = {k:v for k,v in old_ids.items() if v in n} # both key Value
    # notLst = [k for k,v in old_ids.items() if v in n] # Only key

    return notLst


def AKKway():

    flipped_old = {v: k for k, v in old_ids.items()}
    flipped_new = {v: k for k, v in new_ids.items()}
    flipped_combined = flipped_old | flipped_new
    # Find keys that aren't shared between the flipped dicts
    uncommon_keys = set(flipped_old) ^ set(flipped_new)
    # Generate a dict that unflips the key-values containing only the uncommon keys
    uncommon = {flipped_combined[v]: v for v in uncommon_keys}
    return (uncommon)


print("My way",timeit(myway,number=10000))
print("Akk way",timeit(AKKway,number=10000))

OUTPUT

My way 0.010995300021022558
Akk way 0.028010500012896955

CodePudding user response：

I suggest making a condition to find the extra key within the new_ids.

Note: I am just trying to comment since I do not have a 50 reputation