How could I create a dictionary with a list as key and a list as value-CodePudding

I'm trying to create a dict in the following scenario

One list to save jobtype(s)

JOB_TYPE_LIST = ["PARTTIME", "FULLTIME", "WFH"]

Corresponding list to each jobtype

PARTTIME_GROUP_LIST = ["PTGroup2", "PTGroup3"]
FULLTIME_GROUP_LIST = ["FTgroup1", "FTgroup2"]
WFH_GROUP_LIST = ["WFHgroup"]

Expected dicts

dict {
    "PTGroup2": "PARTTIME",
    "PTGroup3": "PARTTIME",
    "FTgroup1": "FULLTIME",
    "FTgroup2": "FULLTIME",
    "WFHgroup": "WFH"
}

My current way to create this dict is do it manually. However, I would like to create this dict in a more pragramming way. Otherwise, every time I add a new jobtype and the corresponding list, I need to manually modify the dict.

Thanks in advance!

CodePudding user response：

Maybe use dictionary instead of lists to store job types.

JOB_TYPES = {
    "PARTIME": ["PTGroup2", "PTGroup3"],
    "FULLTIME": ["FTgroup1", "FTgroup2"],
    "WFH": ["WFHgroup"]
}
expected_dict = {v: key for key, value in JOB_TYPES.items() for v in value}
print(expected_dict)

Output:

{'PTGroup2': 'PARTIME', 'PTGroup3': 'PARTIME', 'FTgroup1': 'FULLTIME', 'FTgroup2': 'FULLTIME', 'WFHgroup': 'WFH'}

CodePudding user response：

output = {}

JOBS_MAPPING = {
    "PARTTIME": PARTTIME_GROUP_LIST,
    "FULLTIME": FULLTIME_GROUP_LIST,
    "WFH": WFH_GROUP_LIST
}


for job_label, job_group_list in JOBS_MAPPING.items():
    print(f"Job label is {job_label}, job groups list is {job_group_list}")
    for job_group in job_group_list:
        output.update({job_group: job_label})

print(ouput)
> {'PTGroup2': 'PARTTIME', 'PTGroup3': 'PARTTIME', 'FTgroup1': 'FULLTIME', 'FTgroup2': 'FULLTIME', 'WFHgroup': 'WFH'}

CodePudding user response：

You may find this approach more flexible. Note that the json module is not required - it's just there to pretty-print the output.

Doing it this way means that you can change your list contents without having to change the code although the number of elements in the JOB_TYPE_LIST must match the number of other lists passed to the function.

import json

JOB_TYPE_LIST = ["PARTTIME", "FULLTIME", "WFH"]
PARTTIME_GROUP_LIST = ["PTGroup2", "PTGroup3"]
FULLTIME_GROUP_LIST = ["FTgroup1", "FTgroup2"]
WFH_GROUP_LIST = ["WFHgroup"]

def func(job_type_list, *args):
    dict_ = {}
    for i, jt in enumerate(job_type_list):
        for group in args[i]:
            dict_[group] = jt
    return dict_

print(json.dumps(func(JOB_TYPE_LIST, PARTTIME_GROUP_LIST, FULLTIME_GROUP_LIST, WFH_GROUP_LIST), indent=2))

Output:

{
  "PTGroup2": "PARTTIME",
  "PTGroup3": "PARTTIME",
  "FTgroup1": "FULLTIME",
  "FTgroup2": "FULLTIME",
  "WFHgroup": "WFH"
}

CodePudding user response：

JOB_TYPE_LIST = ["PARTTIME", "FULLTIME", "WFH"]
PARTTIME_GROUP_LIST = ["PTGroup2", "PTGroup3"]
FULLTIME_GROUP_LIST = ["FTgroup1", "FTgroup2"]
WFH_GROUP_LIST = ["WFHgroup"]

d = {}
i, x, j = 0, 0, 0  # to prevent 'RuntimeError: dictionary changed size during iteration'
for i in locals().items():  # loop over local symbol table
    x = i[0].split('_')[0]  # get first part of variable name
    if x in JOB_TYPE_LIST:
        for j in i[1]:
            d.update({j: x})
print(d)

Prints:

{'PTGroup2': 'PARTTIME', 'PTGroup3': 'PARTTIME', 'FTgroup1': 'FULLTIME', 'FTgroup2': 'FULLTIME', 'WFHgroup': 'WFH'}