Reputation: 23
I have three lists and three keys for each list, and would like to convert them into a list of dictionaries.
list1 = [1, 2, 3]
list2 = ['a', 'b', 'c']
list3 = [5.0, 6.0, 7.0]
keys = ['key1', 'key2', 'key3']
my expected output is this,
[[{'key1': 1, 'key2': 'a', 'key3': 5.0}],
[{'key1': 2, 'key2': 'b', 'key3': 6.0}],
[{'key1': 3, 'key2': 'c', 'key3': 7.0}]]
What's the most pythonic way of achieving this output?
Upvotes: 2
Views: 300
Reputation: 1319
In this solution, the idea is to reduce all problem to only one dictionary with the following way: {key: array_to_assign}
from functools import reduce
from collections import OrderedDict
list_response = []
list_dicts = []
list_values = [list1, list2, list3]
for index, k in enumerate(keys):
list_dicts.append(
OrderedDict({k: list_values[index]})
)
my_dictionary = reduce(lambda d1, d2: OrderedDict(list(d1.items()) + list(d2.items())) , list_dicts)
Finally, we iterate the dictionary and assign the nested list's item to the correct key.
for key, value in my_dictionary.items():
for i in value:
list_response.append({
key: i
})
print(list_response)
Output:
[
{'key1': 1},
{'key1': 2},
{'key1': 3},
{'key2': 'a'},
{'key2': 'b'},
{'key2': 'c'},
{'key3': 5.0},
{'key3': 6.0},
{'key3': 7.0}
]
Upvotes: 0
Reputation: 68
list1 = [1, 2, 3]
list2 = ['a', 'b', 'c']
list3 = [5.0, 6.0, 7.0]
keys = ['key1', 'key2', 'key3']
lists = [list1, list2, list3]
result = []
for list in lists:
d = {}
for i, j in zip(keys, list):
d[i] = j
result.append([d])
print(result)
Upvotes: 0
Reputation: 195418
Try:
list1 = [1, 2, 3]
list2 = ["a", "b", "c"]
list3 = [5.0, 6.0, 7.0]
keys = ["key1", "key2", "key3"]
out = []
for t in zip(list1, list2, list3):
out.append([dict(zip(keys, t))])
print(out)
Prints:
[[{'key1': 1, 'key2': 'a', 'key3': 5.0}],
[{'key1': 2, 'key2': 'b', 'key3': 6.0}],
[{'key1': 3, 'key2': 'c', 'key3': 7.0}]]
Or:
out = [[dict(zip(keys, t))] for t in zip(list1, list2, list3)]
print(out)
Upvotes: 4