Removing a nested key value pair from dictionary based on value in dataframe

Question

The question is one part help one part curiosity, so I have a dict that I'm appending to a list once all my conditions have been iterated through:

for col, row in df.iterrows():
    up_list = []

    if row['id_check'] == 'Add all':
    
        l = {'external': {'om': {'id' : row['posm']},
                                    'wd': {'id': row['wdp']},
                                    'wk': {'id': row['tw'].replace('ru: ', '')}
                                    }
                 }

         up_list.append(l)

Basically, I'm adding multiple keys and values to the dict l, and my main question is, provided one of values for 'id' == 'None' I don't want to add the entire key value pair to the dictionary.

So best case output looks like:

final_l = {'external': {'om': {'id' : '123'},
                        'wd': {'id': '456'},
                        'wk': {'id': '789'}
                                }}

BUT: provided one of those values == 'None' based on its corresponding dataframe value, I don't want to replace the 'id' with None, I don't want to have it there at all, so ideally say 'wk' == 'None' then the output dict would look like:

final_l = {'external': {'om': {'id' : '123'},
                        'wd': {'id': '456'}
                                }}

But the only thing I can get is:

final_l = {'external': {'om': {'id' : '123'},
                        'wd': {'id': '456'},
                        'wk': {'id': 'None'}
                                }}

Which is not optimal for my use case. So, How do you delete (or not even add) specific key value pairs from a dictionary based on its corresponding dataframe value? Also if there is a better way of doing this I'm very to open to this, as this "works" but by god is it not elegant.

EDIT Sample Dataframe:

   id_check   om    wd    wk
0   Add all  123  None   789
1   Add all  472   628  None
2  Add None  528   874   629

Answer 1

So I tried the provided answers, and the biggest issue I ran into was truth evaluation and speed. I coded this which "works" but I'm not too happy with it from an efficiency standpoint:

if row['id_check'] == 'Add all IDs':
        
        link_d, ex_link = {}, {}
        if row['posm'] != 'None':
            link_d['om'] = {'id': row['posm']}
        if row['pd'] != 'None':
            link_d['wd'] = {'id': row['pd']}
        if row['tw'] != 'None':
            link_d['wk'] = {'id': row['tw']}
            
        ex_link['external'] = link_d
        up_list.append(ex_link)
    
    up_d[row['id']] = up_list
    all_list.append(up_d)

Which outputs:

{'external': {'om': {'id' : '123'},
                     'wd': {'id': '456'},
                     'wk': {'id': '789'}}}

and ignores keys where the value == None:

{'external': {'om': {'id' : '123'},
                     'wd': {'id': '456'}}}

Answer 2

IIUC, you could try with to_dict, dropna, eq and to_list:

final_l=df[df['id_check'].eq('Add all')].drop('id_check',1)
                         .apply(lambda x : {'external':x.dropna().to_dict()},axis=1)
                         .to_list()

Output:

final_l
[{'external': {'om': 123.0, 'wk': '789'}},
 {'external': {'om': 472.0, 'wd': '628'}}]

Answer 3

I am editing my previous answer both based on your response that you are trying to alter the dictionary and not the dataframe and because my previous answer was incorrect.

I couldn't find a way to do what you are asking using a nice simple way - e.g. list comprehension, but was able to do it with this converter I created:

class Converter:
    
    def __init__(self):
        self.rows = []
        self.cols = []
    
    @classmethod
    def from_dict(cls, d):
        conv_df = cls()
        conv_df.cols = list(d.keys())
        conv_df.rows = list(zip(*d.values()))
        return conv_df

    def as_dict(self):
        vals = []
        
        for idx, _ in enumerate(self.cols):
            vals.append([j[idx] for j in self.rows if None not in j])
        return {k: v for k, v in zip(self.cols, vals)

Example usage:

>>> z = {'a': [1, 2, 3], 'b': ['a', 'b', 'c'], 'c': ['q', 'r', None]}
>>> conv = Converter.from_dict(z)
>>> conv.cols
['a', 'b', 'c']
>>> conv.rows
[(1, 'a', 'q'), (2, 'b', 'r'), (3, 'c', None)]
>>> "Get as dict and we expect last row not to appear in it:"
'Get as dict and we expect last row not to appear in it:'
>>> conv.as_dict()
{'a': [1, 2], 'b': ['a', 'b'], 'c': ['q', 'r']}