Three lists zipped into list of dicts

{k : {key_1 : v1, key_2 : v2} for k,v1,v2 in zip(idx, l_1, l_2)}

Solution 1: You may use zip twice (actually thrice) with dictionary comprehension to achieve this as:

idx = ['a', 'b', 'c', 'd']l_1 = [1, 2, 3, 4]l_2 = [5, 6, 7, 8]keys = ['mkt_o', 'mkt_c']   # yours keys in another listnew_dict = {k: dict(zip(keys, v)) for k, v in zip(idx, zip(l_1, l_2))}

Solution 2: You may also use zip with nested list comprehension as:

new_dict = dict(zip(idx, [{key_1: i, key_2: j} for i, j in zip(l_1, l_2)]))

Solution 3: using dictionary comprehension on top of zip as shared in DYZ's answer:

new_dict = {k : {key_1 : v1, key_2 : v2} for k,v1,v2 in zip(idx, l_1, l_2)}

All the above solutions will return new_dict as:

{     'a': {'mkt_o': 1, 'mkt_c': 5},      'b': {'mkt_o': 2, 'mkt_c': 6},      'c': {'mkt_o': 3, 'mkt_c': 7},     'd': {'mkt_o': 4, 'mkt_c': 8} }

You're working with dicts, lists, indices, keys and would like to transpose the data. It might make sense to work with pandas (DataFrame, .T and .to_dict):

>>> import pandas as pd>>> idx = ['a', 'b', 'c', 'd']>>> l_1 = [1, 2, 3, 4]>>> l_2 = [5, 6, 7, 8]>>> key_1 = 'mkt_o'>>> key_2 = 'mkt_c'>>> pd.DataFrame([l_1, l_2], index=[key_1, key_2], columns = idx)       a  b  c  dmkt_o  1  2  3  4mkt_c  5  6  7  8>>> pd.DataFrame([l_1, l_2], index=[key_1, key_2], columns = idx).T   mkt_o  mkt_ca      1      5b      2      6c      3      7d      4      8>>> pd.DataFrame([l_1, l_2], index=[key_1, key_2], columns = idx).to_dict(){'a': {'mkt_o': 1, 'mkt_c': 5}, 'b': {'mkt_o': 2, 'mkt_c': 6}, 'c': {'mkt_o': 3, 'mkt_c': 7}, 'd': {'mkt_o': 4, 'mkt_c': 8}}