How do I compare two lists of dict? The result should be the odd ones out from the list of dict B.

Example:

ldA = [{'user':"nameA", 'a':7.6, 'b':100.0, 'c':45.5, 'd':48.9},
       {'user':"nameB", 'a':46.7, 'b':67.3, 'c':0.0, 'd':5.5}]


ldB =[{'user':"nameA", 'a':7.6, 'b':99.9, 'c':45.5, 'd':43.7},
      {'user':"nameB", 'a':67.7, 'b':67.3, 'c':1.1, 'd':5.5},
      {'user':"nameC", 'a':89.9, 'b':77.3, 'c':2.2, 'd':6.5}]

Here I want to compare ldA with ldB. It should print the below output.

ldB -> {user:"nameA",  b:99.9, d:43.7}
ldB -> {user:"nameB",  a:67.7, c:1.1 }
ldb -> {user:"nameC", a:89.9, b:77.3, c:2.2, d:6.5}

I have gone through the below link, but there it return onlys the name, but I want name and value like above.

List of Dicts comparision to match between lists and detect value changes in Python

For a general solution, consider the following. It will properly diff, even if the users are out of order in the lists.

def dict_diff ( merge, lhs, rhs ):
    """Generic dictionary difference."""
    diff = {}
    for key in lhs.keys():
          # auto-merge for missing key on right-hand-side.
        if (key not in rhs):
            diff[key] = lhs[key]
          # on collision, invoke custom merge function.
        elif (lhs[key] != rhs[key]):
            diff[key] = merge(lhs[key], rhs[key])
    for key in rhs.keys():
          # auto-merge for missing key on left-hand-side.
        if (key not not lhs):
            diff[key] = rhs[key]
    return diff

def user_diff ( lhs, rhs ):
    """Merge dictionaries using value from right-hand-side on conflict."""
    merge = lambda l,r: r
    return dict_diff(merge, lhs, rhs)

import copy

def push ( x, k, v ):
    """Returns copy of dict `x` with key `k` set to `v`."""
    x = copy.copy(x); x[k] = v; return x

def pop ( x, k ):
    """Returns copy of dict `x` without key `k`."""
    x = copy.copy(x); del x[k]; return x

def special_diff ( lhs, rhs, k ):
      # transform list of dicts into 2 levels of dicts, 1st level index by k.
    lhs = dict([(D[k],pop(D,k)) for D in lhs])
    rhs = dict([(D[k],pop(D,k)) for D in rhs])
      # diff at the 1st level.
    c = dict_diff(user_diff, lhs, rhs)
      # transform to back to initial format.
    return [push(D,k,K) for (K,D) in c.items()]

Then, you can check the solution:

ldA = [{'user':"nameA", 'a':7.6, 'b':100.0, 'c':45.5, 'd':48.9},
       {'user':"nameB", 'a':46.7, 'b':67.3, 'c':0.0, 'd':5.5}]
ldB =[{'user':"nameA", 'a':7.6, 'b':99.9, 'c':45.5, 'd':43.7},
      {'user':"nameB", 'a':67.7, 'b':67.3, 'c':1.1, 'd':5.5},
      {'user':"nameC", 'a':89.9, 'b':77.3, 'c':2.2, 'd':6.5}]
import pprint
if __name__ == '__main__':
    pprint.pprint(special_diff(ldA, ldB, 'user'))