I have the following list of dicts:
list_of_dicts = [
{"type": "X", "hour": 22},
{"type": "A", "measure": "1"},
{"type": "B", "measure": "2"},
{"type": "X", "hour": 23},
{"type": "A", "measure": "3"},
{"type": "X", "hour": 24},
{"type": "A", "measure": "4"},
{"type": "B", "measure": "5"},
{"type": "C", "measure": "6"}
]
How can I split it into a dict where keys are the 'hour' values from 'type' = 'X' dicts and values are the other dicts between two 'type' = 'X' dicts? That's what I want to obtain using this example, but the interval between two 'type' = 'X' dicts can be variable.
dict_of_dicts = {
22: [
{"type": "A", "measure": "1"},
{"type": "B", "measure": "2"},
],
23:[
{"type": "A", "measure": "3"}
],
24:[
{"type": "A", "measure": "4"},
{"type": "B", "measure": "5"},
{"type": "C", "measure": "6"},
]
}
Thanks in advance!
CodePudding user response:
This code should do the trick.
It creates new elements in a result dictionary, every time it comes across a dictionary which contains the "hour" key.
res = {}
cur_key = None
for d in list_of_dicts:
if "hour" in d:
cur_key = d["hour"]
res[cur_key] = []
elif cur_key is not None:
res[cur_key].append(d)
CodePudding user response:
Here is another approach using list comprehension and a neat little trick (borrowed from here) to make a while loop work inside it:
list_of_dicts = [
{"type": "X", "hour": 22},
{"type": "A", "measure": "1"},
{"type": "B", "measure": "2"},
{"type": "X", "hour": 23},
{"type": "A", "measure": "3"},
{"type": "X", "hour": 24},
{"type": "A", "measure": "4"},
{"type": "B", "measure": "5"},
{"type": "C", "measure": "6"}
]
def while_generator(lst):
i = 0
while i < len(lst) and lst[i]['type'] != 'X':
yield lst[i]
i = 1
dict_of_dicts = {
d['hour']: [e for e in while_generator(list_of_dicts[i 1:])]
for i, d in enumerate(list_of_dicts) if d['type'] == 'X'
}
print(dict_of_dicts)
Prints:
{
22: [
{'type': 'A', 'measure': '1'},
{'type': 'B', 'measure': '2'}
],
23: [
{'type': 'A', 'measure': '3'}
],
24: [
{'type': 'A', 'measure': '4'},
{'type': 'B', 'measure': '5'},
{'type': 'C', 'measure': '6'}
]
}