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Identify changed values between two list of dictionaries

I have two lists of dictionaries as below:
old = [{'a': 'aa', 'b': 'bb', 'c': 'cc'}, {'a': 'aaa', 'b': 'bbb', 'c': 'ccc'}]
new = [{'a': 'aa', 'b': 'boy', 'c': 'cc'}, {'a': 'aaa', 'b': 'bbb', 'c': 'cat'}]
In every dictionary in new and old list, the keys are 'a', 'b', 'c'. I need to identify the difference if any in the values. In the example above ideally I want the output to be 'b' and 'c' as the values for those keys have changed. Just identify keys whose values have been modified.
How should I implement this? Thanks!

Try this snippet to see it helps you:
Please ask if there is any questions.
d1 = [{'a': 'aa', 'b': 'bb', 'c': 'cc'},
{'a': 'aaa', 'b': 'bbb', 'c': 'ccc'}]
d2 = [{'a': 'aa', 'b': 'boy', 'c': 'cc'},
{'a': 'aaa', 'b': 'bbb', 'c': 'cat'}]
for d1, d2 in zip(d1, d2):
#print(d1, d2) # get each dict from d1 and d2
for k in d1: # start to check the key from old (d1)
if k in d2 and d1[k] != d2[k]:
print(k) # b then c

To compare two dictionaries, use something like this:
having dictionaries d1 and d2
differences={key: ( d1[key], d2[key]) for key in d1 if key not in d2 or d1[key]!=d2[key]}

This will compare values from the first dictionary and check if its value is different from the second dictionary. If so, it will yield it.
However, this function will NOT yield keys that exist in one dictionary but not the other.
This is a generator function, so remember to iterate along it or evaluate it with list(...).
old = [{'a': 'aa', 'b': 'bb', 'c': 'cc'}, {'a': 'aaa', 'b': 'bbb', 'c': 'ccc'}]
new = [{'a': 'aa', 'b': 'boy', 'c': 'cc'}, {'a': 'aaa', 'b': 'bbb', 'c': 'cat'}]
def difference(a, b):
for first, second in zip(a, b):
for key, value in first.items():
if key in second:
if value != second[key]:
yield key
print(list(difference(old, new)))
> python .\diff.py
['b', 'c']

Separating nested list and dictionary in separate columns

I created an function to gather the following sample list below:
full_list = ['Group1', [{'a':'1', 'b':'2'},{'c':'3', 'x':'1'}]
'Group2', [{'d':'7', 'e':'18'}],
'Group3', [{'m':'21'}, {'n':'44','p':'13'}]]
As you can see some of the elements inside the lists are made up of key-value pair dictionaries.
And these dictionaries are of different sizes (number of kv pairs).
Can anyone suggest what to use in python to display this list in separate columns?
Group1 Group2 Group3
{'a':'1', 'b':'2'} {'d':'7', 'e':'18'} {'m':'21'}
{'c':'3', 'x':'1'} {'n':'44','p':'13'}
I am not after a solution but rather a point in the right direction for a novice like me.
I have briefly looked at itertools and pandas dataframes
Thanks in advance

Here is one way:
First extract the columns and the data:
import pandas as pd
columns = full_list[::2]
#['Group1', 'Group2', 'Group3']
data = full_list[1::2]
#[[{'a': '1', 'b': '2'}, {'c': '3', 'x': '1'}],
# [{'d': '7', 'e': '18'}],
# [{'m': '21'}, {'n': '44', 'p': '13'}]]
Here the [::2] means iterate from begin to end but only every 2 items and so does [1::2] but it starts iterating from index 1 (second position)
Then create a pd.DataFrame:
df = pd.DataFrame(data)
#0 {'a': '1', 'b': '2'} {'c': '3', 'x': '1'}
#1 {'d': '7', 'e': '18'} None
#2 {'m': '21'} {'n': '44', 'p': '13'}
Ooops but the columns and rows are transposed so we need to convert it:
df = df.T
Then add the columns:
df.columns = columns
And there we have it:
Group1 Group2 Group3
0 {'a': '1', 'b': '2'} {'d': '7', 'e': '18'} {'m': '21'}
1 {'c': '3', 'x': '1'} None {'n': '44', 'p': '13'}

Swap keys of nested dictionaries

I have a dictionary as follows:
Each key has a dictionary associated with it.
dict_sample = {'a': {'d0': '1', 'd1': '2', 'd2': '3'}, 'b': {'d0': '1'}, 'c': {'d1': '1'}}
I need the output as follows:
output_dict = {'d0': {'a': 1, 'b': 1}, 'd1': {'a': 2, 'c': 1}, 'd2': {'a': 3}}
I'd appreciate any help on the pythonic way to achieve this. Thank You !

I believe this produces the desired output
>>> from collections import defaultdict
>>> d = defaultdict(dict)
>>>
>>> dict_sample = {'a': {'d0': '1', 'd1': '2', 'd2': '3'}, 'b': {'d0': '1'}, 'c': {'d1': '1'}}
>>>
>>> for key, value in dict_sample.items():
... for k, v in value.items():
... d[k][key] = v
...
>>> d
defaultdict(<class 'dict'>, {'d0': {'a': '1', 'b': '1'}, 'd1': {'a': '2', 'c': '1'}, 'd2': {'a': '3'}})

You can use dict.setdefault on a new dict with a nested loop:
d = {}
# for each key and sub-dict in the main dict
for k1, s in dict_sample.items():
# for each key and value in the sub-dict
for k2, v in s.items():
# this is equivalent to d[k2][k1] = int(v), except that when k2 is not yet in d,
# setdefault will initialize d[k2] with {} (a new dict)
d.setdefault(k2, {})[k1] = int(v)
d would become:
{'d0': {'a': 1, 'b': 1}, 'd1': {'a': 2, 'c': 1}, 'd2': {'a': 3}}

Get dictionary contains in list if key and value exists

How to get complete dictionary data inside lists. but first I need to check them if key and value is exists and paired.
test = [{'a': 'hello' , 'b': 'world', 'c': 1},
{'a': 'crawler', 'b': 'space', 'c': 5},
{'a': 'jhon' , 'b': 'doe' , 'c': 8}]
when I try to make it conditional like this
if any((d['c'] is 8) for d in test):
the value is True or False, But I want the result be an dictionary like
{'a': 'jhon', 'b': 'doe', 'c': 8}
same as if I do
if any((d['a'] is 'crawler') for d in test):
the results is:
{'a': 'crawler', 'b': 'space', 'c': 5}
Thanks in advance

is tests for identity, not for equality which means it compares the memory address not the values those variables are storing. So it is very likely it might return False for same values. You should use == instead to check for equality.
As for your question, you can use filter or list comprehensions over any:
>>> [dct for dct in data if dct["a"] == "crawler"]
>>> filter(lambda dct: dct["a"] == "crawler", data)
The result is a list containing the matched dictionaries. You can get the [0]th element if you think it contains only one item.

Use comprehension:
data = [{'a': 'hello' , 'b': 'world', 'c': 1},
{'a': 'crawler', 'b': 'space', 'c': 5},
{'a': 'jhon' , 'b': 'doe' , 'c': 8}]
print([d for d in data if d["c"] == 8])
# [{'c': 8, 'a': 'jhon', 'b': 'doe'}]

While Dynamically creating dictionaries with a reference dictionary, why is the reference dictionaries getting modified? (Python)

I was writing a Python Code that creates dictionaries dynamically,initializes it to a reference dictionary, and modifying a particular value in the dictionary. But,I found that not only I am getting unexpected results,but the reference dictionary is also getting modified.
My Code:
tdict={'a':'1','b':'2','c':'3'}
newdict={}
for i in range(5):
newdict['name'+str(i)]=tdict
newdict['name'+str(i)]['a']='value'+str(i)
print 'tdict: ',tdict
print 'newdict: ',newdict
And the result:
tdict: {'a': 'value0', 'c': '3', 'b': '2'}
tdict: {'a': 'value1', 'c': '3', 'b': '2'}
tdict: {'a': 'value2', 'c': '3', 'b': '2'}
tdict: {'a': 'value3', 'c': '3', 'b': '2'}
tdict: {'a': 'value4', 'c': '3', 'b': '2'}
newdict: {'name4': {'a': 'value4', 'c': '3', 'b': '2'}, 'name2': {'a': 'value4', 'c': '3', 'b': '2'}, 'name3': {'a': 'value4', 'c': '3', 'b': '2'}, 'name0': {'a': 'value4', 'c': '3', 'b': '2'}, 'name1': {'a': 'value4', 'c': '3', 'b': '2'}}
whereas I expected my 'newdict' to be like:
newdict: {'name4': {'a': 'value4', 'c': '3', 'b': '2'}, 'name2': {'a': 'value2', 'c': '3', 'b': '2'}, 'name3': {'a': 'value3', 'c': '3', 'b': '2'}, 'name0': {'a': 'value0', 'c': '3', 'b': '2'}, 'name1': {'a': 'value1', 'c': '3', 'b': '2'}}
Can anyone please help me figuring out why this is happening? Also, why is the reference dictionary 'tdict' getting changed when I am not assigning any any value to it?
Thanks in advance

You are storing a reference to tdict in every value of your newdict dictionary:
newdict['name'+str(i)]=tdict
You are then modifying the key 'a' of tdict by doing
# newdict['name'+str(i)] is a reference to tdict
newdict['name'+str(i)]['a']='value'+str(i)
# this is equivalent to doing
tdict['a']='value'+str(i)
What you maybe want is storing a copy of tdict in your newdict dictionary:
newdict['name'+str(i)]=dict(tdict)
Creating a new dictionary by using an existing dictionary as constructor argument creates a shallow copy where you can assign new values to existing keys. What you cannot (or what you don't want) is modifying mutable values in this dictionary. Example:
>>> a={'a': 1, 'b': 2, 'c': [1,2,3]}
>>> b=dict(a)
>>> b['a']=9
>>> a
{'a': 1, 'c': [1, 2, 3], 'b': 2}
>>> b
{'a': 9, 'c': [1, 2, 3], 'b': 2}
>>> b['c'].append(99)
>>> a
{'a': 1, 'c': [1, 2, 3, 99], 'b': 2}
>>> b
{'a': 9, 'c': [1, 2, 3, 99], 'b': 2}
If you want to modify mutable values in a dictionary you need to create a deep copy:
>>> import copy
>>> a={'a': 1, 'b': 2, 'c': [1,2,3]}
>>> b=copy.deepcopy(a)
>>> b['a']=9
>>> b['c'].append(99)
>>> a
{'a': 1, 'c': [1, 2, 3], 'b': 2}
>>> b
{'a': 9, 'c': [1, 2, 3, 99], 'b': 2}

Is just cause you are making a reference to tdict and not a copy. In order to copy you can either use
newdict['name'+str(i)] = tdict.copy()
or
newdict['name'+str(i)] = dict(tdict)
Hope it helps