Comparing keys of a dictionary in Python and extracting corresponding values

Question

I have two dictionaries:-

dict_1 = {'a': ["['b','c','d','e']"],'b': ["['a','c','d','e']"],'p': ["['q','r','s']"]}   
dict_2 = {'a': ["['x','y','z','n']"],'b': ["['u','v','w','x','y','z','n']"]}   

I want to extract value of all the key for which exit in both dictionaries, such that at time I can get only one value of corresponding key at time from a dictionary.

Expected output

kindly help me to get the output.

Answer 1

I don't know what's a dataframe. Here is the code and sample output:

dict_one = {1: [1, 2, 3, 4]}
dict_two = {1: [7, 8, 9, 10]}
all_keys = set(dict_one).union(set(dict_two))
rows = []
for key in all_keys:
    dict_one_values = []
    if key in dict_one:
        dict_one_values = dict_one[key]
    dict_two_values = []
    if key in dict_two:
        dict_two_values = dict_two[key]
    for i in range(min(len(dict_one_values), len(dict_two_values))):
        rows.append([key, dict_one_values[i], dict_two_values[i]])
for row in rows:
    print(*row)

output:

1 1 7
1 2 8
1 3 9
1 4 10

Answer 2

I will assume this input:

dict_1 = {'a': ['b','c','d','e'],'b': ['a','c','d','e'],'p': ['q','r','s']}

dict_2 = {'a': ['x','y','z','n'],'b': ['u','v','w','x','y','z','n']}

if not you can easily obtain it with dict_i = {k: literal_eval(v) for k,v in dict_i.items()}

You can use a comprehension to select the expected data, and use it to build your dataframe:

data = [(k, v1, v2) for k in dict_1 if k in dict_2
        for v1, v2 in zip(dict_1[k], dict_2[k])]

df = pd.DataFrame(data)

it gives:

   0  1  2
0  a  b  x
1  a  c  y
2  a  d  z
3  a  e  n
4  b  a  u
5  b  c  v
6  b  d  w
7  b  e  x