newbe - 1 year ago 148

Python Question

How do I compare two lists of

`dict`

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.

Answer Source

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 (not rhs.has_key(key)):
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 (not lhs.has_key(key)):
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'))
```